1 Introduction
The well-posedness of the classical Naiver–Stokes equations and the corresponding fractional versions has attracted many mathematicians’ attention in recent decades. In this article, we will introduce new logarithmic Q-type spaces to cover many known critical spaces for the classical Naiver–Stokes equations and the fractional versions, which can be defined on the half-space
$\mathbb {R}^{n+1}_{+}= \mathbb {R}^{n}\times (0,\infty ) , n\geq 2$
as follows:
$$ \begin{align} \left\{\begin{array}{@{}ll} \partial_{t}u +(-\Delta)^{\beta} u + u \cdot \nabla u +\nabla P=0, & \mbox{ in } \mathbb{R}^{n+1}_{+}; \\ \nabla \cdot u=0, & \mbox{ in } \mathbb{R}^{n+1}_{+};\\ u|_{t=0}= a, & \mbox{ in } \mathbb{R}^{n}, \end{array} \right. \end{align} $$
where
$(-\Delta )^{\beta }$
represents the fractional Laplacian defined by the Fourier transform in the space variable:
$$ \begin{align*}\widehat{(-\Delta)^{\beta}u}(\cdot,\xi)= |\xi|^{2\beta} \widehat{u}(\cdot,\xi). \end{align*} $$
Here, it is appropriate to point out that when
$\beta =1,$
(1.1) becomes the classical Naiver–Stokes equations. Moreover, (1.1) can be seen as a generalization of the two-dimensional quasi-geostrophic equation, which has continued to attract attention extensively.
Equations (1.1) are invariant under the following dilations:
$$ \begin{align*}a_\lambda(x)=\lambda^{2\beta-1}a(\lambda x),\ u_\lambda(x,t)=\lambda^{2\beta-1}u(\lambda x,\lambda^{2\beta}t),\ P_\lambda(x,t)=\lambda^{4\beta-2}P(\lambda x,\lambda^{2\beta}t). \end{align*} $$
A function space
$X(\mathbb {R}^{n})$
is critical for (1.1) if it is invariant under the scaling
$$ \begin{align*}a_\lambda(x)=\lambda^{2\beta-1}a(\lambda x).\end{align*} $$
When
$\beta =1$
, the well-posedness of equations (1.1) has been established in the following critical spaces but not in
$\dot {B}^{-1}_{\infty ,\infty }(\mathbb {R}^n),$
$$ \begin{align} L^n(\mathbb{R}^n)\hookrightarrow \dot{B}^{-1+\frac{n}{p}}_{p<\infty,q<\infty}(\mathbb{R}^n)\hookrightarrow BMO^{-1}(\mathbb{R}^n)\hookrightarrow\dot{B}^{-1}_{\infty,\infty}(\mathbb{R}^n). \end{align} $$
Kato [Reference Kato18] initiated the study of the Naiver–Stokes equations in critical spaces by proving that (1.1) is locally well-posed in
$L^n(\mathbb {R}^n)$
and globally well-posed with the initial data being small in
$L^n(\mathbb {R}^n).$
In 2001, Koch and Tataru [Reference Koch and Tataru19] established the well-posedness of the classical Naiver–Stokes equations in
$BMO^{-1}(\mathbb {R}^{n}),$
which is the largest critical space where the well-posedness has been established.
Recently, Q-type spaces and their equivalent characterizations have been used to study the well-posedness of the Naiver–Stokes equations. In the literature, Q-type spaces were introduced as a new space between
$W^{1,n}(\mathbb {R}^{n})$
and
$BMO(\mathbb {R}^{n})$
. In 1995, Aulaskari, Xiao, and Zhao first introduced a class of Möbius invariant analytic function spaces
$Q_{p}(\mathbb {D})$
for
$p\in (0,1)$
on the unit disc
$\mathbb {D}$
of the complex plane. The class
$Q_{p}(\mathbb {D})$
,
$p\in (0,1)$
, can be seen as subspaces and subsets of
$BMOA$
and
$UBC$
on
$\mathbb {D}$
. Since then, the Q-type spaces
$Q_{p}(\mathbb {D})$
were investigated extensively and various characterizations of
$Q_{p}(\mathbb {D})$
have been established (see [Reference Aulaskari, Stegenga and Xiao1, Reference Aulaskari, Xiao and Zhao3, Reference Xiao40, Reference Xiao42] and the references therein). As a class of analytic function spaces, the boundary of
$Q_{p}(\mathbb {D})$
is
$Q_{p}(\partial \mathbb {D})$
introduced by Nicolau and Xiao in [Reference Nicolau and Xiao28], where
$\partial \mathbb D$
denotes the boundary of
$\mathbb D$
.
Correspondingly, in the setting of Euclidean spaces, the real-variable Q-type spaces
$Q_{\alpha }(\mathbb R^{n})$
were first introduced by Essén, Janson, Peng, and Xiao [Reference Essén, Janson, Peng and Xiao13]. For
$\alpha \in (-\infty ,\infty )$
, the Q-type space
$Q_{\alpha }(\mathbb R^{n})$
is defined as the set of all measurable functions on
$\mathbb {R}^n$
that satisfy
$$ \begin{align*}\|f\|^{2}_{Q_{\alpha}}:=\sup_{I}(\ell(I))^{2\alpha-n}\int_{I}\int_{I}\frac{|f(x)-f(y)|^{2}}{|x-y|^{n+2\alpha}} dx dy<\infty.\end{align*} $$
Here and henceforth, the symbol
$\sup _{I}$
denotes the supremum taken over all cubes I with the edge length
$\ell (I)$
and the edges parallel to the coordinate axes in
$\mathbb {R}^n$
. In 2004, applying Hausdorff capacities and the tent spaces, Dafni and Xiao [Reference Dafni and Xiao11] established the Carleson measure characterization and the predual space of
$Q_{\alpha }(\mathbb R^{n})$
denoted by
$HH^{1}_{-\alpha }(\mathbb R^{n})$
. Follow-up studies have shown that the spaces
$Q_{\alpha }(\mathbb R^{n})$
are essentially Campanato–Sobolev spaces (see [Reference Wu and Xie37, Reference Xiao43]). Unlike the case of
$Q_{p}(\mathbb D)$
, the Q-type spaces of meromorphic functions defined via the Green function are not equivalent to the ones generated by the Möbius transformation (see [Reference Aulaskari, Wulan and Zhao2, Reference Wulan38]). To exhibit a more general structure behind this difference, a class of Q-type spaces related to weight functions denoted by
$Q_{K}(\mathbb D)$
was introduced (see [Reference Essén and Wulan14, Reference Essén, Wulan and Xiao15] for the details). Further, Bao, and Wulan in [Reference Bao and Wulan5] introduced the analog of
$Q_{K}(\mathbb D)$
on
$\mathbb R^{n}$
denoted by
$Q_{K}(\mathbb R^{n})$
, which generalized the Q-type space
$Q_{\alpha }(\mathbb R^{n})$
. For further information on Q-type spaces, we refer the reader to [Reference Cui, Li and Lou9, Reference Cui and Yang10, Reference Dafni and Xiao12, Reference Peng and Yang29, Reference Wang and Xiao31, Reference Xiao39, Reference Xiao44, Reference Xiao45, Reference Yang and Yuan47, Reference Yang and Yuan48, Reference Yang, Qian and Li50, Reference Yang, Qian and Li51] and the references therein.
In [Reference Xiao43], by the aid of the Carleson measure characterizations of
$Q_{\alpha }(\mathbb R^{n})$
obtained in [Reference Dafni and Xiao11], Xiao proved that the equations (1.1) with
$\beta =1$
are well-posed with data in
$Q_{\alpha }^{-1}(\mathbb {R}^{n})=\nabla \cdot (Q_{\alpha }(\mathbb {R}^{n}))^{n}$
being small, where
$\alpha =(0,1)$
. Also, it was proved in [Reference Essén, Janson, Peng and Xiao13] that
$Q_\alpha (\mathbb {R}^{n})=BMO(\mathbb {R}^{n})$
for
$\alpha \in (-{n}/{2},0).$
Hence, the result obtained in [Reference Dafni and Xiao11] generalizes Koch and Tataru’s result on the well-posedness of equations (1.1) from
$BMO^{-1}(\mathbb {R}^{n})$
to
$Q_{\alpha }^{-1}(\mathbb {R}^{n})$
. For the endpoint
$\alpha =0$
, due to lack of the corresponding semigroup characterization, the method of [Reference Xiao43] is invalid for the space
$Q_{0}(\mathbb R^{n})$
. In [Reference Xiao and Zhang46], Xiao and Zhang introduced a logarithmic Q-type space
$\mathcal {Q}_{0}$
which is defined as the set of all measurable functions f satisfying
$$ \begin{align*}\|f\|_{\mathcal{Q}_{0}}:=\sup_{I}\left(\int^{\ell(I)}_{0}\int_{I}|\nabla_{x} e^{t^{2}\Delta}f(x)|^{2}\left(\ln\left(\frac{e\ell(I)}{t}\right)\right)^{2}\frac{dxdt}{t}\right)^{\frac{1}{2}}<\infty.\end{align*} $$
Here, the symbol
$\nabla _{x}$
denotes the gradient operator for the variable
$x\in \mathbb R^{n}$
,
$$ \begin{align*} \nabla_{x}:=\left(\frac{\partial }{\partial x_1},\frac{\partial }{\partial x_2},\ldots,\frac{\partial }{\partial x_n}\right). \end{align*} $$
With
$\beta =1$
and the data in
$\mathcal {Q}^{-1}_{0}(\mathbb R^{n}):=\nabla (\mathcal {Q}_{0}(\mathbb R^{n}))^{n}$
being small, the well-posedness of equations (1.1) was obtained in [Reference Xiao and Zhang46].
The well-posedness of the fractional Naiver–Stokes equations has been established in the fractional counterparts of (1.2). Letting the spatial dimension
$n=3$
, in [Reference Lions26], Lions proved the global existence of the classical solutions of (1.1) when
$\beta \geq {5}/{4}$
. For general spatial dimension n, the existence result in [Reference Lions26] was extended to
$\beta \geq {1}/{2}+{n}/{4}$
by Wu [Reference Wu33]. Moreover, for the important case
$\beta <{1}/{2}+{n}/{4}$
, Wu [Reference Wu34, Reference Wu35] established the global existence for (1.1) in the Besov spaces
$\dot {B}^{1+{n}/{p}-2\beta ,q}_{p}(\mathbb {R}^{n})$
for
$1\leq q\leq \infty $
and for either
${1}/{2}<\beta $
and
$p=2$
or
${1}/{2}<\beta \leq 1$
and
$2<p<\infty $
and in
$\dot {B}^{r,\infty }_{2}(\mathbb {R}^{n})$
with
$r>\max \{1,1+{n}/{p}-2\beta \}$
(see also [Reference Wu36] concerning the corresponding regularity). Following the idea of [Reference Koch and Tataru19, Reference Xiao43], for
$\beta \in (\frac {1}{2},1)$
, Li and Zhai introduced a new critical Q-type spaces
$Q_{\alpha }^{\beta }(\mathbb {R}^{n})$
for
$\alpha +\beta -1\geq 0.$
With data in
$Q_{\alpha }^{\beta , -1}(\mathbb {R}^{n})=\nabla \cdot (Q_{\alpha }^{\beta }(\mathbb {R}^{n}))^{n}$
being small, in [Reference Li and Zhai23], Li and Zhai obtained the well-posedness of equations (1.1) with
${\frac {1}{2}<\beta <1.}$
Zhai in [Reference Zhai53] and Zhai-Yu in [Reference Yu and Zhai52] proved the well-posedness in
$BMO^{-(2\beta -1)}(\mathbb {R}^n)$
and
$\dot {B}^{-(2\beta -1)}_{\infty , \infty }(\mathbb {R}^n),$
correspondingly. For more progress on the well-posedness/ill-posedness of fractional Naiver–Stokes equations, we refer the reader to [Reference Bourgain and Pavlovć6–Reference Colombo, De Lellis and Massaccesi8, Reference Jiao, Huang, Li and Liu17, Reference Li, Xiao and Yang21, Reference Li and Yang22, Reference Lin and Yang25, Reference Wang and Xiao32, Reference Yang and Li49] and the references therein. Moreover, function spaces have also been applied to study fractional magneto-hydrodynamic (MHD; see [Reference Wu33]), the 2
$-D$
quasi-geostrophic equation (see [Reference Li and Zhai24]) and other equations (see [Reference Jiao, Huang, Li and Liu17]). Especially in [Reference Li and Zhai24], Li and Zhai proved that the convolution singular integral operators are bounded on
$Q^{\beta }_{\alpha }(\mathbb R^{n})$
. As an application, the well-posedness of the quasi-geostrophic equation with small data in
$Q_{\alpha }^{\beta , -1}(\mathbb {R}^{n})$
was obtained in [Reference Li and Zhai24].
In this article, we aim to introduce a new class of logarithmic Q-type spaces to cover the
$BMO^{-1}(\mathbb {R}^n), Q^{-1}_\alpha (\mathbb {R}^n), \mathcal {Q}^{-1}_{0}(\mathbb {R}^n), BMO^{-(2\beta -1)}(\mathbb {R}^n), Q_{\alpha }^{\beta ,-1}(\mathbb {R}^n)$
and the largest critical space
$\dot {B}^{-(2\beta -1)}_{\infty ,\infty }(\mathbb {R}^n).$
Definition 1.1 For any
$k\in \mathbb N$
,
$1<p<\infty $
and
$n\geq 2$
, let
$l\geq k$
. Suppose that
$\lambda \geq 0$
. The space
$Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
consists of all
$f\in L_{loc}^p(\mathbb {R}^n)$
which is the set of all locally
$L^{p}$
-integrable functions, such that
$$ \begin{align*}\|f\|_{Q_{\ln,\lambda}^{p,k,l}(\mathbb{R}^n)}^p:=\sup\limits_{I}\frac{1}{(\ell(I)) ^{l+n\lambda}}\int_I\int_I\frac{|f(x)-f(y)|^p}{|x-y|^{pn-l}}\left(\ln\left(\frac{e\sqrt{n}\ell(I)}{|x-y|}\right)\right)^kdxdy<\infty.\end{align*} $$
In Lemma 3.14, we deduce a Stegenga-type inequality (3.20). This inequality enables us to establish the extension from
$Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
to the functions on
$\mathbb R^{n+1}_{+}$
belonging to
$\mathscr H_{K_{\ln }^{(l,k)}}^{p,\lambda }(\mathbb R_+^{n+1})$
via fractional heat semigroups
$\{e^{-t(-\Delta )^{\beta }}\}_{t>0}$
(see Theorems 3.15 and 3.16). As an immediate corollary, a Carleson measure characterization of
$Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
can be deduced (see Corollary 3.18).
For a function
$u(\cdot ,\cdot )$
on
$\mathbb R^{n+1}_{+}$
, the operator
$\nabla _{(x,t)}$
is defined as
$$ \begin{align*} \nabla_{(x,t)} u(x,t)=\left(\frac{\partial u(x,t)}{\partial x_1},\ldots,\frac{\partial u(x,t)}{\partial x_n},\frac{\partial u(x,t)}{\partial t}\right). \end{align*} $$
For any cube I on
$\mathbb {R}^n$
, the Carleson box
$S(I)$
is defined by
$$ \begin{align*} S(I)=\Big\{(x,t)\in\mathbb R_+^{n+1}:\ x\in I,\ t\in(0,\ell(I)]\Big\}. \end{align*} $$
Theorems 3.5 and 3.15 indicate that the following three conditions are equivalent:
$$ \begin{align*}\left\{\begin{aligned} &\sup\limits_{I\subset\mathbb{R}^n}\frac{1}{(\ell(I)) ^{l+n\lambda}}\int_I\int_I\frac{|f(x)-f(y)|^p}{|x-y|^{pn-l}}\left(\ln\left(\frac{e\sqrt{n}\ell(I)}{|x-y|}\right)\right)^kdxdy<\infty;\\ &\sup\limits_{(y,r)\in\mathbb R_+^{n+1}}\int_{\mathbb R_+^{n+1}}K_{\ln}^{(l,k)}\left(\frac{tr^{\frac{1}{n}}}{(|x-y|^2+|r+t|^2)^{\frac{n+1}{2n}}}\right)\left(\frac{tr^{\frac{1}{n}}}{(|x-y|^2+|r+t|^2)^{\frac{n+1}{2n}}}\right)^{n\lambda}\\ &\quad\times |\nabla_{(x,t)}u(x,t)|^p\frac{dxdt}{t^{(p-1)(n-1)+n\lambda}}<\infty;\\ &\sup\limits_{I\subset\mathbb{R}^n}\int_{S(I)}|\nabla_{(x,t)}u(x,t)|^p\left(\frac{t}{\ell(I)}\right)^{n\lambda+l}\left(\ln\left(\frac{e\sqrt{n}\ell(I)}{t}\right)\right)^{k} \frac{dxdt}{t^{(p-1)(n-1)+n\lambda}}<\infty, \end{aligned}\right.\end{align*} $$
when
$p\in [n/(n-1),2n/(n-1))$
,
$0\leq \lambda <2-p+\frac {p}{n}$
and
$(p-1)n-p<l<(p-1)n.$
Here,
$u(x,t):=f\ast K_{t^{2\beta }}^\beta (x), \frac {1}{2}\leq \beta \leq 1$
, and
$K_{\ln }^{(l,k)}(\cdot )$
denotes a piecewise function on
$(0,\infty )$
which is defined in Definition 3.1. For the endpoint cases
$l=(p-1)n-p$
and
$l=(p-1)n$
, we deduce an embedding relation of
$Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
(see Theorem 3.19).
Based on the extension result obtained in Theorem 3.15, it is also worth mentioning that, when
$\frac {1}{2}\leq \beta \leq 1, k\in \mathbb N$
and
$l\in (n+2\beta -4, n+2\beta -2)$
, there holds
$$ \begin{align} &\sup\limits_{(x,r)\in\mathbb{R}^{n+1}_+}\left(r^{-l+6\beta-6}\int_0^r\int_{|y-x|<r}{|\nabla_{y} e^{-s^{2\beta}(-\Delta)^\beta}f(y)|^2 \left(\ln\left(\frac{er}{s}\right)\right)^k}\frac{dyds}{s^{n-l-3+2\beta}}\right)^{\frac{1}{2}}\\ &\approx \sup\limits_{I\subset\mathbb{R}^n}\left((\ell(I)) ^{-(l+6-6\beta)}\int_I\int_I\frac{|f(x)-f(y)|^{2}}{|x-y|^{2n-l+2\beta-2}}\left(\ln\left(\frac{e\sqrt{n}\ell(I)}{|x-y|}\right)\right)^kdxdy\right)^{\frac{1}{2}}.\nonumber \end{align} $$
In [Reference Li and Zhai23], under the assumption that
$\beta \in (\frac {1}{2}, 1]$
, the equivalent relation (1.3) was used to obtain the well-posedness of equations (1.1) with data in
$$ \begin{align*}Q^{\beta,-1}_{\alpha}(\mathbb R^{n}):=\nabla(Q_{\ln,\frac{4(1-\beta)}{n}}^{2,k,l+2-2\beta}(\mathbb{R}^n))^{n}.\end{align*} $$
To investigate the well-posedness of equations (1.1) with data in Q-type spaces when
$\beta>1$
, we introduce the following logarithmic Q spaces.
Definition 1.2 Let
$l>n+2\beta -4$
,
$\beta>\frac {1}{2}$
and
$k\in \mathbb N$
. The space
$Q^{(l,k)}_\beta (\mathbb {R}^n)$
is defined as the set of all measurable functions satisfying
$\|f\|_{Q^{(l,k)}_\beta }<\infty ,$
where
$\|f\|_{Q^{(l,k)}_\beta }$
is defined as
$$ \begin{align*} \sup\limits_{(x,r)\in\mathbb{R}^{n+1}_+}\left(r^{-l+6\beta-6}\int_0^r\int_{|y-x|<r}{|\nabla_{y} e^{-s^{2\beta}(-\Delta)^\beta}f(y)|^2 \left(\ln\left(\frac{er}{s}\right)\right)^k}\frac{dyds}{s^{n-l-3+2\beta}}\right)^{\frac{1}{2}}. \end{align*} $$
We will establish some basic properties of
$Q^{(l,k)}_\beta (\mathbb {R}^n)$
and
$Q_{l,k,\beta }^{-1}(\mathbb {R}^n)=\nabla \cdot (Q^{(l,k)}_\beta (\mathbb {R}^n))^{n}$
in Section 4.1. Assume that
$\beta \in (\frac {1}{2},\min \{2,\frac {1}{2}+\frac {n}{4}\})$
,
$n+2\beta -4<l\leq n+4\beta -4$
and
$k\in \mathbb N$
. By the quadratic estimates for integral operators obtained in Section 4.2, we obtain the well-posedness of solutions to equations (1.1) with data in
$Q_{l,k,\beta }^{-1}(\mathbb {R}^n)$
being small (see Theorem 4.12). For
$\beta =1$
,
$l=n$
and
$k=2$
,
$Q^{(n,2)}_{1}(\mathbb {R}^n)$
comes back to the space
$\mathcal {Q}_{0}$
introduced by Xiao and Zhang [Reference Xiao and Zhang46]. Moreover, for
$n-2<l\leq n$
and
$k\in \mathbb N$
, the well-posedness of equations (1.1) with
$\beta =1$
(i.e., the classical Naiver–Stokes equations) holds for all critical spaces
$Q^{(l,k)}_\beta (\mathbb {R}^n).$
For
$\beta =1$
,
$k=0$
and
$l=n-2\alpha $
,
$Q^{(n-2\alpha , 0)}_{1}(\mathbb R^{n})$
coincides with the Q-type spaces
$Q_{\alpha }(\mathbb R^{n})$
which were introduced by Essén, Janson, Peng, and Xiao [Reference Essén, Janson, Peng and Xiao13]. By [Reference Essén, Janson, Peng and Xiao13, Theorem 2.3(iii)], for
$l>n$
,
$Q^{(l, 0)}_{1}(\mathbb R^{n})=BMO(\mathbb R^{n})$
. Hence, for
$k=0, l>n, \beta =1$
, the well-posedness result for equations (1.1) with data in
$Q^{-1}_{l, 0,1}(\mathbb R^{n})$
has been proved by Koch and Tataru in [Reference Koch and Tataru19]. We also mention that for the special case
$\beta =1$
,
$l=n$
and
$k=2$
, with small data in
$Q^{-1}_{n,2,1}(\mathbb R^{n})$
, the well-posedness of equations (1.1) was obtained by Xiao and Zhang in [Reference Xiao and Zhang46]. Theorem 4.12 indicates that even for the special case
$\beta =1$
, there exist a family of logarithmic Q-type spaces
$Q^{-1}_{l,k,1}(\mathbb R^{n})$
with
$k\in \mathbb N$
and
$l\in (n-2, n]$
such that the solutions to equations (1.1) are well-posed for the data in
$Q^{-1}_{l,k,1}(\mathbb R^{n})$
being small.
In [Reference Li and Zhai23], under the assumption that
$\alpha \in (0,1)$
and
$\beta \in (\max \{\alpha ,\frac {1}{2}\},1)$
with
$\alpha +\beta -1>0$
, Li and Zhai obtained the well-posedness of equations (1.1) with data in
$Q_{\alpha }^{\beta , -1}(\mathbb {R}^{n})$
, where
$Q_{\alpha }^{\beta , -1}(\mathbb {R}^{n})=\nabla \cdot (Q_{\alpha }^{\beta }(\mathbb R^{n}))^{n}$
. Here, the Q-type space
$Q^{\beta }_{\alpha }(\mathbb R^{n})$
is defined as the set of all measurable functions f with
$$ \begin{align} \sup_{I}(\ell(I))^{4\beta-4}\int_{I}\int_{I}\frac{|f(x)-f(y)|^{2}}{|x-y|^{2n}}\left(\frac{|x-y|}{\ell(I)}\right)^{n-2(\alpha-\beta+1)}dxdy<\infty, \end{align} $$
where
$\alpha \in (0, 1)$
and
$\beta \in (\frac {1}{2}, 1)$
. It can be seen from (1.4) that for
$\alpha \in (0,1)$
and
$\beta \in (\max \{\alpha ,\frac {1}{2}\},1)$
with
$\alpha +\beta -1>0$
,
$Q^{\beta }_{\alpha }(\mathbb R^{n})= Q_{\ln , \frac {4(1-\beta )}{n}}^{2, 0, n-2(\alpha -\beta +1)}(\mathbb R^{n})$
. If
$\alpha \in (0,1)$
and
$\beta \in (\max \{\alpha ,\frac {1}{2}\},1)$
with
$\alpha +\beta -1>0$
,
$n+2\beta -4<n-2(\alpha -\beta +1)<n+4\beta -4$
. Then, Theorem 4.12 holds for
$\nabla \cdot (Q_{\ln , \frac {4(1-\beta )}{n}}^{2, 0, n-2(\alpha -\beta +1)}(\mathbb R^{n}))^{n}$
and covers [Reference Li and Zhai23, Theorem 4.14.]. Compared with the well-posedness result obtained by Li and Zhai [Reference Li and Zhai23], in Theorem 4.12, the well-posedness of (1.1) holds for
$\beta \in (\frac {1}{2},\min \{2,\frac {1}{2}+\frac {n}{4}\})$
with small data in the Q-type spaces related to logarithmic type functions, while the scope of
$\beta $
is restricted to
$(\frac {1}{2},1)$
in [Reference Li and Zhai23, Theorem 4.14]. By Proposition 4.3,
$Q^{(l,k)}_{\beta }(\mathbb R^{n})$
is a subclass of the Campanato space
$L^{2, 1+\frac {4(1-\beta )}{n}}(\mathbb R^{n})$
, which indicates that the reasonable scope of
$\beta $
should be
$(\frac {1}{2}, 1+\frac {n}{4}]$
(see Remark 4.4). On the other hand, according to Lions [Reference Lions26] and Wu [Reference Wu33] if
$\beta \geq \frac {1}{2}+\frac {n}{4}$
, there exist classical solutions globally to equations (1.1). Hence, in Theorems 4.12 and 4.13, we assume that
$\beta \in (\frac {1}{2}, \min \{2, \frac {1}{2}+\frac {n}{4}\})$
. By Theorem 3.19, if
$l\leq n+2\beta -4$
, the space
$Q^{(l,k)}_{\beta }(\mathbb R^{n})$
is trivial (see Remark 3.21). On the other hand, for
$\frac {1}{2}<\beta <1$
and
$l>n+4\beta -4$
, Theorem 4.7 implies that the space
$$ \begin{align*}Q_{l,k,\beta}^{-1}(\mathbb{R}^n):=\text{div}\Big(Q^{(l,k)}_\beta(\mathbb{R}^n),\ldots,Q^{(l,k)}_\beta(\mathbb{R}^n)\Big)\end{align*} $$
coincides with the homogeneous Besov spaces
$\dot {B}^{1-2\beta }_{\infty ,\infty }(\mathbb R^{n})$
. Hence, the well-posedness of equations (1.1) with data in
$Q^{-1}_{l,k,\beta }(\mathbb R^{n})$
with
$(\beta , l)\in (\frac {1}{2},1)\times (n+4\beta -4,\infty )$
follows from Yu and Zhai [Reference Yu and Zhai52, Theorem 3.1]. In Theorems 4.12 and 4.13, we assume that
$n+2\beta -4<l\leq n+4\beta -4$
, which is different from that in [Reference Yu and Zhai52, Theorem 3.1]. The method of Theorem 4.12 can also be used to investigate the well-posedness of fractional MHD equations (see Theorem 4.13).
Remark 1.3 It is worth emphasizing that, in Theorem 4.12, the endpoint case
$\beta = 1$
holds particular significance. Observe that the embedding
$ Q_{l,k,\beta }^{-1}(\mathbb {R}^n) \hookrightarrow \dot {B}^{1-2\beta }_{\infty ,\infty }(\mathbb {R}^n) $
follows from the scaling and translation invariance of
$ Q_{l,k,\beta }^{-1}(\mathbb {R}^n) $
. However, the converse embedding
$ \dot {B}^{1-2\beta }_{\infty ,\infty }(\mathbb {R}^n) \hookrightarrow Q_{l,k,\beta }^{-1}(\mathbb {R}^n) $
requires the restriction
$\beta < 1$
(see Theorem 4.7). This indicates that
$ Q_{l,k,1}^{-1}(\mathbb {R}^n) $
is strictly smaller than
$\dot {B}^{-1}_{\infty ,\infty }(\mathbb {R}^n),$
which aligns with the known fact that the classical Navier–Stokes equations are well-posed in
$ Q_{n,2,1}^{-1}(\mathbb {R}^n) $
(obtained by Xiao and Zhang [Reference Xiao and Zhang46]), despite being ill-posed in
$\dot {B}^{-1}_{\infty ,\infty }(\mathbb {R}^n) $
.
Remark 1.4 By Theorems 2.5 and 2.8, the following relations hold:
$$ \begin{align*}CIS^{p,\lambda}(\mathbb{R}^n)\subseteq Q^{p,k,l}_{\ln,\lambda}(\mathbb{R}^n)\subseteq L^{p, p+\lambda-1}(\mathbb{R}^n).\end{align*} $$
In Example 2.4, we provide an example such that
$CIS^{p,\lambda }(\mathbb {R}^n)$
and
$Q^{p,k,l}_{\ln ,\lambda }(\mathbb {R}^n)$
are nontrivial for
$p\in (n/(n-1), 2n/(n-1))$
and
$0\leq \lambda \leq 2+\frac {p}{n}-p$
. Meanwhile if
$p+\lambda -1>1+\frac {p}{n},$
the Campanato space
$L^{p, p+\lambda -1}(\mathbb {R}^n)$
only consists of constants. Considering
$\lambda \geq 0$
in the Definition 1.1, we restrict the scope of p to
$ [n/(n-1), 2n/(n-1))$
in the main results of Section 3.
Some notations: Throughout the rest of this article, denote by
$\mathbb N$
the set of all nonnegative integers.
$U\approx V$
indicates that there is a constant
$c>0$
such that
$c^{-1}V\leq U\leq cV$
, whose right inequality is also written as
$U\lesssim V$
. Similarly, one writes
$V\gtrsim U$
for
$V\geq cU$
. On
$\mathbb {R}^n$
, denote by
$f\ast g$
the convolution of functions f and g, i.e.,
$$ \begin{align*}f\ast g(x):=\int_{\mathbb{R}^n}f(x-y)g(y)dy=\int_{\mathbb{R}^n}f(y)g(x-y)dy.\end{align*} $$
For
$1\leq k<\infty $
,
$\mathscr C^k(\mathbb {R}^n)$
denotes the set of all functions
$f:\mathbb {R}^n\rightarrow \mathbb R$
that is k-time continuously differentiable. For
$1\leq p<\infty $
,
$L^p(\mathbb {R}^n)$
denotes the set of all Lebesgue measurable functions
$f: \mathbb {R}^n\rightarrow \mathbb R$
such that
$\int _{\mathbb {R}^n}|f(x)|^pdx<\infty $
. Denote by
$\mathscr S(\mathbb {R}^n)$
the class of Schwartz functions on
$\mathbb {R}^n$
, denote by
$\mathscr S'(\mathbb {R}^n)$
the dual of
$\mathscr S(\mathbb {R}^n)$
. Denote
$$ \begin{align*}\mathscr S_0(\mathbb{R}^n):=\bigg\{\varphi\in\mathscr{S}(\mathbb R^{n}): \int_{\mathbb{R}^n}\varphi(x)x^\gamma dx=0, |\gamma|=0,1,2,\ldots\bigg\},\end{align*} $$
where
$|\gamma |=\gamma _1+\gamma _2+\cdots +\gamma _n$
and
$x^\gamma =x_1^{\gamma _1}x_2^{\gamma _2}\cdots x_n^{\gamma _n}$
.
2 Basic properties of
$Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
As we all know, the classical Q-type space
$Q_\alpha (\mathbb {R}^n)$
is invariant under the rotation, the translation and conformal mappings. It can be deduced that
$Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
has similar properties; the proof is similar to that of [Reference Bao4, Theorem 2.2.1] and we omit the details.
Lemma 2.1 Suppose that
$k\in \mathbb N$
and
$p\in (1,\infty )$
. Let
$\lambda \geq 0$
and
$l\geq k$
.
-
(i)
$Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
is rotation invariant; -
(ii)
$Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
is translation invariant; -
(iii)
$\|f_\delta \|_{Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)}=\|f\|_{Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)}$
, where
$f_{\delta }(x):=\delta ^{(2-\lambda -p)n/p}f(\delta x)$
,
$\delta>0$
.
Proposition 2.2 Let
$k\in \mathbb N$
,
$1<p<\infty , n\geq 2$
,
$\lambda \geq 0$
and
$l\geq k$
.
$f\in Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
if and only if
$$ \begin{align} \sup\limits_{I}\frac{1}{(\ell(I))^{l+n\lambda}}\int_{|y|<\ell(I)}\int_I\frac{|f(x+y)-f(x)|^p}{|y|^{pn-l}}\left(\ln\left(\frac{e\sqrt{n}\ell(I)}{|y|}\right)\right)^kdxdy<\infty. \end{align} $$
Proof This proposition can be proved by a change of variables:
$x-y\rightarrow y$
in Definition 1.1 and a direct calculation. So we omit the details.
Motivated by [Reference Xiao41], to show that
$Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
is well defined, we need the following function spaces.
Definition 2.3 Suppose that
$p\in (1,\infty )$
and
$\lambda \geq 0$
. The space
$CIS^{p,\lambda }(\mathbb {R}^n)$
is defined to be the set of all functions
$f\in \mathscr {C}^{1}(\mathbb R^{n})$
with
$$ \begin{align*}\|f\|_{CIS^{p,\lambda}}^p:=\sup\limits_I(\ell(I))^{n(1-\lambda-p)+p}\int_I|\nabla_{x} f(x)|^pdx<\infty.\end{align*} $$
We can see that the space
$CIS^{p,\lambda }(\mathbb {R}^n)$
is nontrivial.
Example 2.4 Suppose that
$p\in [n/(n-1),2n/(n-1))$
and
$0\leq \lambda \leq 2+\frac {p}{n}-p$
. Define
$$ \begin{align*}f_0(x):=(1+|x|^{2})^{n(\lambda-2+p)/(2p)}, x\in\mathbb{R}^n.\end{align*} $$
Then,
$f_0\in CIS^{p,\lambda }(\mathbb {R}^n)$
.
Proof A direct computation gives, for
$i=1,2,\ldots , n$
,
$$ \begin{align*}\partial_{x_{i}}f_{0}(x)=n\left(\frac{\lambda}{p}-\frac{2}{p}+1\right)(1+|x|^{2})^{\frac{n(\lambda-2+p)}{2p}-1}x_{i},\end{align*} $$
which gives
$$ \begin{align*}|\nabla_{x}f_{0}(x)|^{p}\lesssim (1+|x|^{2})^{\frac{n(\lambda-2+p)}{2}-\frac{p}{2}}.\end{align*} $$
Then,
$$ \begin{align*} (\ell(I))^{n(1-\lambda-p)+p}\int_I|\nabla_{x} f_{0}(x)|^pdx \lesssim (\ell(I))^{n(1-\lambda-p)+p}\int_{I}(1+|x|^{2})^{\frac{n(\lambda-2+p)}{2}-\frac{p}{2}}dx. \end{align*} $$
Denote by
$x_{0}$
the center of I. We divide the rest of the proof into two cases.
Case 1:
$|x_{0}|\leq 2\ell (I)$
. For
$x\in I$
,
$|x|\leq |x-x_{0}|+|x_{0}|<3\ell (I)$
. Notice that if
$p>n/(n-1)$
and
$\lambda \geq 0$
, then
$n\lambda -n+np-p>0$
. We can get
$$ \begin{align} &(\ell(I))^{n(1-\lambda-p)+p}\int_I|\nabla_{x} f_{0}(x)|^pdx\\ &\quad\lesssim(\ell(I))^{n(1-\lambda-p)+p}\int^{3\ell(I)}_{0}(1+|x|^{2})^{\frac{n(\lambda-2+p)}{2}-\frac{p}{2}}|x|^{n-1}d|x|\nonumber\\ &\quad\lesssim(\ell(I))^{n(1-\lambda-p)+p}\int^{3\ell(I)}_{0}(1+|x|)^{n(\lambda-2+p)-p}|x|^{n-1}d|x|\nonumber\\ &\quad\lesssim(\ell(I))^{n-n\lambda-np+p}\int^{3\ell(I)}_{0}|x|^{n\lambda-n+np-p-1}d|x|\lesssim 1.\nonumber \end{align} $$
Case 2:
$|x_{0}|>2\ell (I)$
. For this case, if
$x\in I$
, then
$|x|\geq |x_{0}|-|x-x_{0}|>\ell (I)$
. Since
$\lambda \leq 2+\frac {p}{n}-p$
, i.e.,
$p-n(\lambda -2+p)\geq 0$
, we obtain
$$ \begin{align} &(\ell(I))^{n(1-\lambda-p)+p}\int_I|\nabla_{x} f_{0}(x)|^pdx\nonumber\\ &\quad\lesssim(\ell(I))^{n(1-\lambda-p)+p}\int_{I}\frac{dx}{(1+|x|)^{p-n(\lambda-2+p)}}\\ &\quad\lesssim(\ell(I))^{n(1-\lambda-p)+p}\frac{1}{(1+\ell(I))^{p-n(\lambda-2+p)}}\int_{I}1dx\lesssim1.\nonumber \end{align} $$
Combining with (2.2) and (2.3), we can see that
$$ \begin{align*}\sup\limits_I(\ell(I))^{n(1-\lambda-p)+p}\int_I|\nabla_{x} f_{0}(x)|^pdx<\infty,\end{align*} $$
which completes the proof.
Theorem 2.5 Suppose that
$k\in \mathbb N$
and
$p\in (1,\infty )$
. Let
$\lambda \geq 0$
and
$l\geq k$
.
$CIS^{p,\lambda }(\mathbb {R}^n)\subseteq Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
. Moreover,
$Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
is nontrivial if and only if
$l>(p-1)n-p$
.
Proof We prove this theorem by following the idea of [Reference Essén, Janson, Peng and Xiao13, Theorem 2.3]. Suppose that
$f\in CIS^{p,\lambda }(\mathbb {R}^n)$
. For a cube I, since
$$ \begin{align*}|f(x+y)-f(x)|\leq|y|\int_0^1|\nabla f(x+ry)|dr,\end{align*} $$
we can deduce that
$$ \begin{align*} &\frac{1}{(\ell(I))^{l+n\lambda}}\int_{|y|<\ell(I)}\int_I\frac{|f(x+y)-f(x)|^p}{|y|^{pn-l}}\left(\ln\left(\frac{e\sqrt{n}\ell(I)}{|y|}\right)\right)^kdxdy\\ &\leq\frac{1}{(\ell(I))^{l+n\lambda}}\int_{|y|<\ell(I)}\int_I\left(\int_0^1|\nabla f(x+ry)|dr\right)^p|y|^{p(1-n)+l}\!\left(\ln\left(\frac{e\sqrt{n}\ell(I)}{|y|}\right)\!\right)^kdxdy. \end{align*} $$
By Minkowski’s inequality, we have
$$ \begin{align*} &\frac{1}{\ell(I))^{\frac{l+n\lambda}{p}}}\left(\int_{|y|<\ell(I)}\int_I\frac{|f(x+y)-f(x)|^p}{|y|^{pn-l}}\left(\ln\left(\frac{e\sqrt{n}\ell(I)}{|y|}\right)\right)^kdxdy\right)^{\frac{1}{p}}\\ \leq&\frac{1}{\ell(I))^{\frac{l+n\lambda}{p}}}\left(\int_{|y|<\ell(I)}\int_I \!\left(\! \int_0^1|\nabla f(x+ry)|dr \!\right)^p|y|^{p(1-n)+l} \!\left(\! \ln\left(\! \frac{e\sqrt{n}\ell(I)}{|y|} \!\right)\right)^kdxdy \!\right)^{\frac{1}{p}}\\ \leq&\frac{1}{\ell(I))^{\frac{l+n\lambda}{p}}}\left(\int_{3\sqrt{n}I}|\nabla f(w)|^pdw\int_{|y|<\ell(I)}|y|^{p(1-n)+l}\left(\ln\left(\frac{e\sqrt{n}\ell(I)}{|y|}\right)\right)^kdy\right)^{\frac{1}{p}}\\ \lesssim&\left(\int_0^{\sqrt{n}}t^{l-(p-1)n+p-1}\left(\ln\left(\frac{e\sqrt{n}}{t}\right)\right)^kdt\right)^{\frac{1}{p}}\|f\|_{CIS^{p,\lambda}}. \end{align*} $$
Let
$u=1/t$
,
$v=e\sqrt {n}u$
and
$w=\ln v$
. We can deduce from
$l>(p-1)n-p$
that
$$ \begin{align*} &\int_0^{\sqrt{n}}t^{l-(p-1)(n-1)}\left(\ln\left(\frac{e\sqrt{n}}{t}\right)\right)^kdt \approx\int_1^\infty e^{w((p-1)n-p-l)}w^kdw\\ &\approx -w^ke^{w((p-1)n-p-l)}\bigg|_1^\infty+k\int_1^\infty e^{w((p-1)n-p-l)}w^{k-1}dw. \end{align*} $$
Since
$$ \begin{align*} \int_1^\infty e^{w((p-1)n-p-l)}dw\approx e^{(p-1)n-p-l}, \end{align*} $$
it holds
$$ \begin{align*} \int_0^{\sqrt{n}}t^{l-(p-1)(n-1)}\left(\ln\left(\frac{e\sqrt{n}}{t}\right)\right)^kdt<\infty. \end{align*} $$
Therefore,
$CIS^{p,\lambda }(\mathbb {R}^n)\subseteq Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
and
$Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
is nontrivial.
Conversely, suppose that
$Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
is nontrivial. If
$l\leq (p-1)n-p$
, take
$f\in Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)\cap \mathscr C^1(\mathbb {R}^n)$
. Without loss of generality, assume that f is real-valued and not an identical constant. Then, there exists a point satisfying
$\nabla f(x)\neq 0$
, where
$x=(x_1,x_2,\ldots , x_n)$
. According to the Householder reflector [Reference Trefethen and Bau30, p. 71], there is an orthogonal matrix
$A=(a_{ij}), i,j=1,2,\ldots ,n$
, satisfying
$$ \begin{align*} \left\{ \begin{aligned} &\sum\limits_{i=1}^n\frac{\partial f}{\partial x_i}(x)a_{i1}=|\nabla f(x)|; \\ &\sum\limits_{i=1}^n\frac{\partial f}{\partial x_i}(x)a_{ij}=0, j=2,3,\ldots,n, \end{aligned} \right. \end{align*} $$
which implies that
$$ \begin{align*}\nabla f(x)A=(|\nabla f(x)|,0,\ldots,0).\end{align*} $$
Let
$A^T$
be the transpose matrix of A and
$\varphi (x)=f(xA^T)$
. We can see that det
$(A^T)\neq 0$
. There exists a point
$y=(y_1,y_2,\ldots ,y_n)$
such that
$yA^T=x$
and
$$ \begin{align*} \left\{ \begin{aligned} &x_j=\sum\limits_{i=1}^ny_ia_{ji};\\ &\frac{\partial \varphi(x)}{\partial y_1}=\sum\limits_{j=1}^n\frac{\partial f(xA^T)}{\partial x_j}a_{j1}=\sum\limits_{j=1}^n\frac{\partial f(x)}{\partial x_j}a_{j1}=|\nabla f(x)|;\\ &\frac{\partial \varphi(x)}{\partial y_i}=\sum\limits_{j=1}^n\frac{\partial f(xA^T)}{\partial x_j}a_{ji}=0,\ i\geq2. \end{aligned} \right. \end{align*} $$
Hence,
$\nabla \varphi (x)=(|\nabla f(x)|,0,\ldots ,0)$
. Notice that
$\varphi \in \mathscr C^1(\mathbb {R}^n)$
, for any
$\varepsilon>0$
, there exists a small cube I centered at
$x_0$
and a constant
$\delta>0$
. If
$|y-x|<\delta $
, it holds
$$ \begin{align*}\bigg|\frac{\partial\varphi(y)}{\partial y_i}-\frac{\partial\varphi(x)}{\partial y_i}\bigg|<\varepsilon.\end{align*} $$
Letting
$\varepsilon =|\nabla f(x)|/3$
, we can get
$\partial \varphi (y)/\partial y_1>2\delta $
and
$\partial \varphi (y)/\partial y_i>\delta ,\ j\geq 2$
. Define
$$ \begin{align*}M=\Big\{x=(x_1,x_2,\ldots,x_n)\in\mathbb{R}^n:\ |x_2|+|x_3|+\cdots+|x_n|<|x_1|<\ell(I)/4\Big\}.\end{align*} $$
If
$x,y\in I$
and
$x-y\in M$
, it can be deduced from the mean value theorem that
$|\varphi (x)-\varphi (y)|>\delta |x_1-y_1|.$
Therefore, by the fact that
$|z|\approx z_1, z\in M$
, we have
$$ \begin{align*} &\frac{1}{(\ell(I))^{(n\lambda+l)}}\int_I\int_I\frac{|\varphi(x)-\varphi(y)|^p}{|x-y|^{pn-l}}\left(\ln\left(\frac{e\sqrt{n}\ell(I)}{|x-y|}\right)\right)^kdxdy\\ &\quad \geq\frac{1}{(\ell(I))^{(n\lambda+l)}}\int_{I/2}\int_{I-x}\frac{|\varphi(x+z)-\varphi(x)|^p}{|z|^{pn-l}}\left(\ln\left(\frac{e\sqrt{n}\ell(I)}{|z|}\right)\right)^kdxdz\\ &\quad \geq\frac{1}{(\ell(I))^{(n\lambda+l)}}\int_{I/2}dx\int_M\frac{\delta^p}{|z_1|^{pn-p-l}}\left(\ln\left(\frac{e\sqrt{n}\ell(I)}{|z_1|}\right)\right)^kdz\\ &\quad \approx(\ell(I))^{n-(n\lambda+l)}\int_0^{\ell(I)/4}\frac{1}{|z_1|^{pn-p-l}}\left(\ln\left(\frac{e\sqrt{n}\ell(I)}{|z_1|}\right)\right)^kd{z_1}\\ &\qquad \times\int_{\{(z_2,\ldots,z_n):|z_2|+\cdots+|z_n|<z_1\}}d{z_2}\cdots d{z_n}\\ &\quad \approx(\ell(I))^{n(2-p-\lambda)+p}\int_0^{\sqrt{n}}t^{l-(p-1)(n-1)}\left(\ln\left(\frac{e\sqrt{n}}{t}\right)\right)^kdt=\infty, \end{align*} $$
which implies that
$\varphi \notin Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
and it is a contraction. For
$f\in Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
and
$\varphi \in L^1(\mathbb {R}^n)$
, notice that
$Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
is conformal invariant. By Minkowski’s inequality, we have
$f\ast \varphi \in Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
and
$$ \begin{align*}\|f\ast\varphi\|_{Q_{\ln}^{p,\lambda}(\mathbb{R}^n)}\leq\|f\|_{Q_{\ln}^{p,\lambda}(\mathbb{R}^n)}\int_{\mathbb{R}^n}|\varphi(y)|dy.\end{align*} $$
In addition, if
$\varphi $
is a compactly supported smooth function, then
$f\ast \varphi \in Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)\cap \mathscr C^1(\mathbb {R}^n)$
is a constant almost everywhere. By [Reference Bao and Wulan5, p. 3], there is a consequence
$\{\varphi _n\}$
with
$\varphi _n\geq 0$
,
$\int \varphi _n=1$
and
$\text {supp}\ \varphi _n$
shrinking to the empty set such that
$f\ast \varphi \rightarrow f$
almost everywhere. So f is constant almost everywhere. This completes the proof of Theorem 2.5.
Let f be a locally integrable function on
$\mathbb R^{n}$
. For any cube
$I\subset \mathbb R^{n}$
, denote by
$f_{I}$
the mean of f on I, i.e.,
$$ \begin{align*}f_{I}=\frac{1}{|I|}\int_{I}f(x)dx.\end{align*} $$
Definition 2.6 Let
$p\in (1,\infty )$
and
$\gamma \in (0,1+\frac {p}{n})$
. The Campanato space
$L^{p,\gamma }(\mathbb {R}^{n})$
is defined as the set of all locally integrable functions f such that
$$ \begin{align*} \|f\|_{L^{p,\gamma}(\mathbb{R}^{n})}^{p}:=\sup_{I}\frac{1}{|I|^{\gamma}}\int_{I}|f(x)-f_{I}|^{p}dx<\infty. \end{align*} $$
Remark 2.7 It follows from a direct computation that
$f\in L^{p,\gamma }(\mathbb R^{n})$
,
$p\in (1,\infty )\ \&\ \gamma \in (0,1+\frac {p}{n})$
, if and only if
$$ \begin{align*}\sup_{I}\frac{1}{|I|^{\gamma+1}}\int_{I}\int_{I}|f(x)-f(y)|^{p}dxdy<\infty.\end{align*} $$
Theorem 2.8 Suppose that
$k\in \mathbb N$
and
$p\in (1,\infty )$
. Let
$\lambda \geq 0$
and
$l\geq k$
. Then,
-
(i)
$Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)\subseteq L^{p,p+\lambda -1}(\mathbb {R}^n)$
; -
(ii) if
$l>(p-1)n$
,
$Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)=L^{p,p+\lambda -1}(\mathbb {R}^n)$
.
Proof (i) Suppose that
$f\in Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
. For any cube I and
$x,y\in I$
, if a positive constant
$C_1$
is small enough, there exists another constant
$C_2$
such that
$$ \begin{align*}C_2\lesssim(\ell(I))^{-n}\Big|\Big\{z\in I:\ \min\{|x-z|,|y-z|\}>C_1\ell(I)\Big\}\Big|.\end{align*} $$
Define
$J(x,z)=\min \bigg \{|x-z|^{l}\left (\ln \left (\frac {e\sqrt {n}\ell (I)}{|x-z|}\right )\right )^k,|y-z|^{l}\left (\ln \left (\frac {e\sqrt {n}\ell (I)}{|y-z|}\right )\right )^k\bigg \}.$
By the fact that
$t^l\left (\ln (\frac {e\sqrt {n}}{t})\right )^k$
is nondecreasing, we have
$$ \begin{align*} (\ell(I))^{-(l+n)}\int_IJ(x,z)dz &\geq (\ell(I))^{-(l+n)}\int_{\{z\in I:\text{min}(|x-z|,|y-z|)>C_1\ell(I)\}}J(x,z)dz\\ &\gtrsim C_2C_1^{l}\left(\ln\left(\frac{e\sqrt{n}}{C_1}\right)\right)^k. \end{align*} $$
Therefore,
$$ \begin{align*} &C_2C_1^{l}\left(\ln\left(\frac{e\sqrt{n}}{C_1}\right)\right)^k(\ell(I))^{n(1-\lambda-p)}\int_I|f(x)-f_I|^pdx\\ &\leq\sup\limits_I\frac{1}{(\ell(I))^{l+n(1+\lambda+p)}}\int_I\int_I\int_I|f(x)-f(y)|^pJ(x,z)dxdydz\\ &\lesssim\sup\limits_I\frac{1}{(\ell(I))^{l+n(1+\lambda+p)}}\int_I\int_I\int_I|f(x)-f(z)|^p|x-z|^{l}\left(\ln\left(\frac{e\sqrt{n}\ell(I)}{|x-z|}\right)\right)^kdxdydz\\ &\quad +\sup\limits_I\frac{1}{(\ell(I))^{l+n(1+\lambda+p)}}\int_I\int_I\int_I|f(y)-f(z)|^p|y-z|^{l}\left(\ln\left(\frac{e\sqrt{n}\ell(I)}{|y-z|}\right)\right)^kdxdydz\\ &\lesssim\sup\limits_I\frac{1}{(\ell(I))^{l+n\lambda}}\int_I\int_I\frac{|f(x)-f(z)|^p}{|x-z|^{pn-l}}\left(\ln\left(\frac{e\sqrt{n}\ell(I)}{|x-z|}\right)\right)^kdxdz, \end{align*} $$
which implies that
$Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)\subseteq L^{p,p+\lambda -1}(\mathbb {R}^n)$
.
(ii) Suppose that
$f\in L^{p,p+\lambda -1}(\mathbb {R}^n)$
. For a cube I and any
$y\in \mathbb {R}^n$
such that
$|y|<\sqrt {n}\ell (I)$
, we can deduce that
$$ \begin{align*} &\frac{1}{(\ell(I))^{l+n\lambda}}\int_I\int_I\frac{|f(x)-f(y)|^p}{|x-y|^{pn-l}}\left(\ln\left(\frac{e\sqrt{n}\ell(I)}{|x-y|}\right)\right)^kdxdy\\ &\quad \leq \frac{1}{(\ell(I))^{l+n\lambda}}\int_{|y|<\ell(I)}\int_I\frac{|f(x+y)-f(y)|^p}{|y|^{pn-l}}\left(\ln\left(\frac{e\sqrt{n}\ell(I)}{|y|}\right)\right)^kdxdy\\ &\lesssim \frac{1}{(\ell(I))^{l+n\lambda}}\bigg\{\int_I\bigg(|f(x+y)-f_{3\sqrt{n}I}|^p+|f(x)-f_{3\sqrt{n}I}|^p\bigg)dx\bigg\}\\ &\qquad\times\bigg\{\int_{|y|<\ell(I)}|y|^{l-pn}\left(\ln\left(\frac{e\sqrt{n}\ell(I)}{|y|}\right)\right)^kdy\bigg\}\\ &\quad\lesssim \|f\|_{L^{p,p+\gamma-1}(\mathbb{R}^n)}^p\int_0^{\sqrt{n}}t^{l-(p-1)n-1}\left(\ln\left(\frac{e\sqrt{n}}{t}\right)\right)^kdt. \end{align*} $$
Since
$l>(p-1)n$
, it holds
$$ \begin{align*} \int_0^{\sqrt{n}}t^{l-(p-1)n-1}\left(\ln\left(\frac{e\sqrt{n}}{t}\right)\right)^kdt<\infty. \end{align*} $$
Therefore,
$Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n) = L^{p,p+\lambda -1}(\mathbb {R}^n)$
.
Corollary 2.9 For
$p\in (2n/(n-1),\infty )$
,
$Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
contains constant functions only.
Remark 2.10 Theorems 2.5 and 2.8 indicate the following inclusion relations: for fixed
$k, l$
with
$l\geq k$
, if
$\lambda \geq 0$
,
$$ \begin{align*}CIS^{p,\lambda}(\mathbb{R}^n)\subseteq Q^{p,k,l}_{\ln,\lambda}(\mathbb{R}^n)\subseteq L^{p, p+\lambda-1}(\mathbb{R}^n),\end{align*} $$
which means that for
$\lambda> 2+\frac {p}{n}-p$
,
$Q^{p,k,l}_{\ln ,\lambda }(\mathbb {R}^n)$
is trivial, i.e., only consists of constant functions.
Theorem 2.11 Suppose that
$k\in \mathbb N$
and
$p\in (1,2n/(n-1))$
. Let
$l\geq k$
,
$0<\lambda <2+\frac {p}{n}-p$
and
$(p-1)n-p<l<(p-1)n$
. Define
$$ \begin{align*}M^{(l,k)}_{\ln}(t):=\min\Bigg\{s^l\left(\ln\left(\frac{e\sqrt{n}}{s}\right)\right)^k,\ t^l\left(\ln\left(\frac{e\sqrt{n}}{t}\right)\right)^k\Bigg\}\end{align*} $$
with
$s^l\left (\ln \left (\frac {e\sqrt {n}}{s}\right )\right )^k>0$
for some
$s>0$
. Then,
$f\in Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
if and only if
$$ \begin{align} \sup_{I}\frac{1}{(\ell(I))^{n\lambda}}\int_I\int_I\frac{|f(x)-f(y)|^p}{|x-y|^{pn}}M^{(l,k)}_{\ln}\left(\frac{|x-y|}{\ell(I)}\right)dxdy<\infty. \end{align} $$
Proof If
$f\in Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
, by the definition of
$M^{(l,k)}_{\ln }(\cdot )$
, we can deduce that
$$ \begin{align*}M^{(l,k)}_{\ln}(t)\leq t^l\left(\ln\left(\frac{e\sqrt{n}}{t}\right)\right)^k\end{align*} $$
and
$$ \begin{align*} \frac{1}{(\ell(I))^{n\lambda}}\int_I\int_I\frac{|f(x)-f(y)|^p}{|x-y|^{pn}}M^{(l,k)}_{\ln}\left(\frac{|x-y|}{\ell(I)}\right)dxdy\leq \|f\|_{Q_{\ln,\lambda}^{p,k,l}(\mathbb{R}^n)}^p. \end{align*} $$
Conversely, if (2.4) holds, for any cube I, split
$$ \begin{align*} \frac{1}{(\ell(I))^{l+n\lambda}}\int_I\int_I\frac{|f(x)-f(y)|^p}{|x-y|^{pn-l}}\left(\ln\left(\frac{e\sqrt{n}}{|x-y|}\right)\right)^kdxdy=A_1+A_2, \end{align*} $$
where
$$ \begin{align*} &A_1:=\frac{1}{(\ell(I))^{l+n\lambda}}\int_I\left(\int_{I_1(s)}\frac{|f(x)-f(y)|^p}{|x-y|^{pn-l}}\left(\ln\left(\frac{e\sqrt{n}\ell(I)}{|x-y|}\right)\right)^kdx\right)dy;\\ &A_2:=\frac{1}{(\ell(I))^{l+n\lambda}}\int_I\left(\int_{I_2(s)}\frac{|f(x)-f(y)|^p}{|x-y|^{pn-l}}\left(\ln\left(\frac{e\sqrt{n}\ell(I)}{|x-y|}\right)\right)^kdx\right)dy, \end{align*} $$
$I_1(s)=\{x\in I:|x-y|\leq s\ell (I)\}, I_2(s)=\{x\in I:s\ell (I)\leq |x-y|\leq \sqrt {n}\ell (I)\}.$
For
$A_1$
, it is easy to see that
$$ \begin{align*}A_1\leq \frac{1}{(\ell(I))^{n\lambda}}\int_I\int_I\frac{|f(x)-f(y)|^p}{|x-y|^{pn}}M^{(l,k)}_{\ln}\left(\frac{|x-y|}{\ell(I)}\right)dxdy.\end{align*} $$
For
$A_2$
, we know
$$ \begin{align*}\|f\|_{L^{p,p+\lambda-1}(\mathbb{R}^n)}^p\leq\frac{1}{(\ell(I))^{n\lambda}}\int_I\int_I\frac{|f(x)-f(y)|^p}{|x-y|^{pn}}M^{(l,k)}_{\ln}\left(\frac{|x-y|}{\ell(I)}\right)dxdy.\end{align*} $$
Since
$t^l\left (\ln \left (\frac {e\sqrt {n}}{t}\right )\right )^k$
is nondecreasing on
$(0,\sqrt {n})$
for
$l\geq k$
,
$$ \begin{align*} A_2&\leq \frac{1}{(\ell(I))^{l+n\lambda}}\int_I|f(x+y)-f(x)|^pdx \int_{I_2(s)}|y|^{l-pn}\left(\ln\left(\frac{e\sqrt{n}\ell(I)}{|y|}\right)\right)^kdy\\ &\lesssim\frac{1}{(\ell(I))^{l+n\lambda}}\left\{\int_I\bigg(|f(x+y)-f_{3\sqrt{n}I}|^p+|f(x)-f_{3\sqrt{n}I}|^p\bigg)dx\right\}\\ &\quad\times\left\{\int_{\{s\ell(I)\leq|y|<\sqrt{n}\ell(I)\}}|y|^{l-pn}\left(\ln\left(\frac{e\sqrt{n}\ell(I)}{|y|}\right)\right)^kdy\right\}\\ &\lesssim(\ell(I))^{(p-1)n}\|f\|_{L^{p,p+\lambda-1}(\mathbb{R}^n)}^p\int_{s\ell(I)}^{\sqrt{n}\ell(I)}t^{(1-p)n-1}dt\\ &\lesssim \frac{1}{(\ell(I))^{n\lambda}}\int_I\int_I\frac{|f(x)-f(y)|^p}{|x-y|^{pn}}M^{(l,k)}_{\ln}\left(\frac{|x-y|}{\ell(I)}\right)dxdy. \end{align*} $$
which implies that
$f\in Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
.
3 Extension via fractional heat semigroups
To characterize
$Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
via fractional heat semigroups, we introduce the following Q-type spaces related to piece-wise functions.
Definition 3.1 For any
$k\in \mathbb N$
,
$1<p<2n/(n-1)$
and
$n\geq 2$
, let
$l\geq k$
and
$$ \begin{align} K_{\ln}^{(l,k)}(t):=\left\{ \begin{aligned} &t^l\left(\ln\left(\frac{e\sqrt{n}}{t}\right)\right)^{k},&\ \ \ 0<t\leq \sqrt{n};\\ &t^l,&\ \ \ t>\sqrt{n}. \end{aligned} \right. \end{align} $$
Suppose that
$\lambda \geq 0$
. The space
$Q_{K_{\ln }^{(l,k)}}^{p,\lambda }(\mathbb {R}^n)$
consists of all measurable functions of
$f\in L_{loc}^p(\mathbb {R}^n)$
such that
$$ \begin{align*}\|f\|_{Q_{K_{\ln}^{(l,k)}}^{p,\lambda}(\mathbb{R}^n)}^p:=\sup\limits_{I\subset\mathbb{R}^n}\frac{1}{(\ell(I))^{n\lambda}}\int_I\int_I\frac{|f(x)-f(y)|^p}{|x-y|^{pn}} K_{\ln}^{(l,k)}\left(\frac{|x-y|}{\ell(I)}\right)dxdy<\infty.\end{align*} $$
Remark 3.2 The function
$K_{\ln }^{(l,k)}(\cdot )$
defined in (3.1) is nondecreasing. Also
$K_{\ln }^{(l,k)}(2t)\approx K_{\ln }^{(l,k)}(t)$
on
$(0,\infty )$
. In fact, when
$0<t\leq \sqrt {n}$
and
$2t>\sqrt {n}$
, it is obvious that
$K_{\ln }^{(l,k)}(2t)\geq K_{\ln }^{(l,k)}(t)$
. On the other hand,
$$ \begin{align*}K_{\ln}^{(l,k)}(t)=t^l\left(\ln\left(\frac{e\sqrt{n}}{t}\right)\right)^{k}\approx (2t)^l\left(\ln\left(\frac{e\sqrt{n}}{t}\right)\right)^{k}\geq (2t)^l=K_{\ln}^{(l,k)}(2t),\end{align*} $$
which indicates that
$K_{\ln }^{(l,k)}(2t)\approx K_{\ln }^{(l,k)}(t)$
. When
$0<t\leq 2t\leq \sqrt {n}$
, we have
$K_{\ln }^{(l,k)}(2t)\geq K_{\ln }^{(l,k)}(t)$
. By the inequality:
$$ \begin{align*} K_{\ln}^{(l,k)}(t)&=t^l\left(\ln\left(\frac{e\sqrt{n}}{t}\right)\right)^{k}\approx (2t)^l\left(\ln\left(\frac{e\sqrt{n}}{t}\right)\right)^{k}\\ &\geq (2t)^l\left(\ln\left(\frac{e\sqrt{n}}{2t}\right)\right)^{k}=K_{\ln}^{(l,k)}(2t), \end{align*} $$
we get
$K_{\ln }^{(l,k)}(2t)\approx K_{\ln }^{(l,k)}(t).$
For the case
$2t>t>\sqrt {n}$
, it is obvious that
$t^l\approx (2t)^l$
. We can also get
$K_{\ln }^{(l,k)}(2t)\approx K_{\ln }^{(l,k)}(t).$
To further investigate
$Q_{K_{\ln }^{(l,k)}}^{p,\lambda }(\mathbb {R}^n)$
, following the idea of [Reference Bao and Wulan5, Reference Essén, Wulan and Xiao15], we define the following auxiliary function:
$$ \begin{align} \Phi^{(l,k)}_{\ln}(s):=\sup\limits_{0<t\leq1}\frac{K_{\ln}^{(l,k)}(st)}{K_{\ln}^{(l,k)}(t)},\quad 0<s<\infty. \end{align} $$
Next, we will verify that
$$ \begin{align} \int_0^1\frac{\Phi^{(l,k)}_{\ln}(s)}{s^{(p-1)(n-1)}}ds<\infty \end{align} $$
and
$$ \begin{align} \int_1^\infty\frac{\Phi^{(l,k)}_{\ln}(s)}{s^{(p-1)n+1}}ds<\infty \end{align} $$
hold for
$K_{\ln }^{(l,k)}(\cdot )$
.
Proposition 3.3 Suppose that
$k\in \mathbb N$
and
$p\in (1,2n/(n-1))$
. Let
$l\geq k$
,
$n\geq 2$
and
$(p-1)n-p<l<(p-1)n.$
With
$K_{\ln }^{(l,k)}(\cdot )$
and
$\Phi ^{(l,k)}_{\ln }(\cdot )$
defined above, there hold (3.3) and (3.4).
Proof Case 1: If k is an even number, let
$k=2m, m\in \mathbb N$
. Let
$0<t\leq 1$
. First, we will investigate the case of
$0<s<\sqrt {n}.$
When
$0<s<\sqrt {n},$
we get
$0<st<\sqrt {n}$
and
$$ \begin{align*}\frac{K_{\ln}^{(l,k)}{(st)}}{K_{\ln}^{(l,k)}{(t)}}=s^l\left(1+\frac{\ln s}{\ln t-\frac{1}{2}\ln n-1}\right)^{2m}:=s^lh(t).\end{align*} $$
A direct computation gives us
$$ \begin{align*} \left\{ \begin{aligned} &h'(t)=-2m\frac{\ln s(\ln t+\ln s-\frac{1}{2}\ln n-1)^{2m-1}}{t(\ln t-\frac{n}{2}\ln n-1)^{2m+1}};\\ &\left(\ln t+\ln s-\frac{1}{2}\ln n-1\right)^{2m-1}<0;\\ &t\left(\ln t-\frac{1}{2}\ln n-1\right)^{2m+1}<0. \end{aligned} \right. \end{align*} $$
When
$0<s<1$
, then
$\ln s<0$
, which implies that
$h'(t)>0$
,
$h(\cdot )$
is increasing and
$$ \begin{align*}h(t)\leq h(1)=\left(1-\frac{\ln s}{1+\frac{1}{2}\ln n}\right)^{2m}.\end{align*} $$
When
$1\leq s<\sqrt {n}$
, it holds that
$\ln s>0$
. Then,
$h'(t)<0$
which implies that
$h(\cdot )$
is decreasing and
$$ \begin{align*}h(t)\leq h(0)=1.\end{align*} $$
Next, we investigate the case of
$s\geq \sqrt {n}$
. When
$\sqrt {n}/s<t\leq 1$
, it follows that
$$ \begin{align*}\frac{K_{\ln}^{(l,k)}{(st)}}{K_{\ln}^{(l,k)}{(t)}}=\frac{s^l}{(1+\frac{1}{2}\ln n-\ln t)^{2m}}:=s^lg(t).\end{align*} $$
It is easy to get
$g'(t)=\frac {2m}{t(1+\frac {1}{2}\ln n-\ln t)^{2m+1}}$
and
$\left (1+\frac {1}{2}\ln n-\ln t\right )^{2m+1}>0,$
which implies that
$g'(t)>0$
and
$g(\cdot )$
is increasing. Then, we get
$$ \begin{align*}g(t)\leq g(1)=\left(1+\frac{1}{2}\ln n\right)^{-2m}.\end{align*} $$
When
$0<t\leq \sqrt {n}/s$
, it follows that
$$ \begin{align*}\frac{K_{\ln}^{(l,k)}{(st)}}{K_{\ln}^{(l,k)}{(t)}}=s^l\left(1+\frac{\ln s}{\ln t-\frac{1}{2}\ln n-1}\right)^{2m}:=s^lh(t).\end{align*} $$
According to the case of
$0<s<\sqrt {n}$
, we can get
$$ \begin{align*} \left\{ \begin{aligned} &\left(\ln t+\ln s-\frac{1}{2}\ln n-1\right)^{2m-1}<0;\\ &t\left(\ln t-\frac{1}{2}\ln n-1\right)^{2m+1}<0;\\ &h'(t)=-2m\frac{\ln s(\ln t+\ln s-\frac{1}{2}\ln n-1)^{2m-1}}{t(\ln t-\frac{1}{2}\ln n-1)^{2m+1}}<0, \end{aligned} \right. \end{align*} $$
which indicates that
$h(t)$
is decreasing and
$$ \begin{align*}h(t)\leq h(0)=1.\end{align*} $$
Due to (3.2), it holds
$$ \begin{align*} \Phi^{(l,k)}_{\ln}(s)=\left\{ \begin{aligned} &s^l\left(1-\frac{\ln s}{1+\frac{1}{2}\ln n}\right)^{2m},&\ 0<s<1;\\ &s^l,&\ s\geq1. \end{aligned} \right. \end{align*} $$
By the fact that
$(p-1)n-p<l<(p-1)n$
, we can deduce that
$$ \begin{align*}\int_1^\infty\frac{\Phi^{(l,k)}_{\ln}(s)}{s^{(p-1)n+1}}ds=\int_1^\infty s^{l-(p-1)n-1}ds<\infty.\end{align*} $$
Notice that
$$ \begin{align*} \int_0^1\frac{\Phi^{(l,k)}_{\ln}(s)}{s^{(p-1)(n-1)}}ds &=\int_0^1s^{l-(p-1)(n-1)}\left(1-\frac{\ln s}{\frac{1}{2}\ln n+1}\right)^{2m}ds\\ & \lesssim \int_0^1s^{l-(p-1)(n-1)}ds+\int_0^1s^{l-(p-1)(n-1)}(\ln s)ds\\ &\quad +\int_0^1s^{l-(p-1)(n-1)}(\ln s)^2ds+\cdots+\int_0^1s^{l-(p-1)(n-1)}(\ln s)^{2m}ds. \end{align*} $$
Since
$(p-1)n-p<l<(p-1)n$
, it can be deduced that
$$ \begin{align} \left\{ \begin{aligned} &\int_0^1s^{l-(p-1)(n-1)}ds\approx1;\\ &\lim\limits_{s\rightarrow0}(\ln s)s^{l-(p-1)n+p}=0;\\ &\lim\limits_{s\rightarrow0}(\ln s)^ms^{l-(p-1)n+p}=0, \end{aligned} \right. \end{align} $$
which gives
$$ \begin{align} \left\{ \begin{aligned} \int_0^1s^{l-(p-1)(n-1)}\ln sds&\approx(\ln s)s^{l-(p-1)n+p}\bigg|_0^1-\int_0^1s^{l-(p-1)(n-1)}ds<\infty;\\ \int_0^1s^{l-(p-1)(n-1)}(\ln s)^mds&\approx(\ln s)^ms^{l-(p-1)n+p}\bigg|_0^1\\ &\quad -m\int_0^1s^{l-(p-1)(n-1)}(\ln s)^{m-1}ds<\infty. \end{aligned} \right. \end{align} $$
Hence, (3.3) holds.
Case 2: If k is odd, let
$k=2m+1, m\in \mathbb N$
. At first, we investigate the case of
$0<s\leq 1$
. By the definition of
$\Phi ^{(l,k)}_{\ln }(\cdot )$
, it follows that
$0<t\leq 1$
and
$0<st\leq 1$
. Then,
$$ \begin{align*}\frac{K_{\ln}^{(l,k)}{(st)}}{K_{\ln}^{(l,k)}{(t)}}=s^l\left(1+\frac{\ln s}{\ln t-\frac{1}{2}\ln n-1}\right)^{2m+1}:=s^lu(t).\end{align*} $$
A direct computation gives us
$$ \begin{align*}u'(t)=-(2m-1)\frac{\ln s(\ln s+\ln t-\frac{1}{2}\ln n-1)^{2m}}{t(\ln t-\frac{1}{2}\ln n-1)^{2m+2}}.\end{align*} $$
Since
$\ln s<0$
,
$(\ln s+\ln t-\frac {1}{2}\ln n-1)^{2m}>0$
and
$t(\ln t-\frac {1}{2}\ln n-1)^{2m+2}>0$
, we can see that
$u'(t)>0$
and
$u(\cdot )$
is increasing, i.e.,
$$ \begin{align*}u(t)\leq u(1)=\left(1-\frac{\ln s}{1+\frac{1}{2}\ln n}\right)^{2m+1}.\end{align*} $$
Next, we investigate the case of
$s>1$
. When
$0<t\leq \sqrt {n}/s$
, we can deduce that
$$ \begin{align*}\frac{K_{\ln}^{(l,k)}{(st)}}{K_{\ln}^{(l,k)}{(t)}}=s^l\left(1+\frac{\ln s}{\ln t-\frac{1}{2}\ln n-1}\right)^{2m+1}:=s^lu(t).\end{align*} $$
Then, we can get
$$ \begin{align*} \left\{ \begin{aligned} &\ln s>0;\\ &\left(\ln t+\ln s-\frac{1}{2}\ln n-1\right)^{2m}>0;\\ &t\left(\ln t-\frac{1}{2}\ln n-1\right)^{2m+2}>0;\\ &u'(t)=-(2m-1)\frac{\ln s(\ln t+\ln s-\frac{1}{2}\ln n-1)^{2m}}{t(\ln t-\frac{1}{2}\ln n-1)^{2m+2}}<0, \end{aligned} \right. \end{align*} $$
which implies that
$u(\cdot )$
is decreasing and
$u(t)\leq u(0)=1.$
When
$\sqrt {n}/s<t\leq 1$
, we can deduce that
$$ \begin{align*}\frac{K_{\ln}^{(l,k)}{(st)}}{K_{\ln}^{(l,k)}{(t)}}=\frac{s^l}{(1+\frac{1}{2}\ln n-\ln t)^{2m+1}}:=s^lv(t).\end{align*} $$
It is easy to get
$v'(t)=\frac {2m+1}{t(1+\frac {1}{2}\ln n-\ln t)^{2m+2}}>0$
and
$v(\cdot )$
is increasing, then
$$ \begin{align*}v(t)\leq v(1)=\left(1+\frac{1}{2}\ln n\right)^{-2m-1}.\end{align*} $$
Due to (3.2), it holds
$$ \begin{align*} \Phi^{(l,k)}_{\ln}(s)=\left\{ \begin{aligned} &s^l\left(1-\frac{\ln s}{1+\frac{1}{2}\ln n}\right)^{2m+1},&\ 0<s<1;\\ &s^l,&\ s\geq1. \end{aligned} \right. \end{align*} $$
By the fact that
$(p-1)n-p<l<(p-1)n$
, we can deduce that
$$ \begin{align*}\int_1^\infty\frac{\Phi^{(l,k)}_{\ln}(s)}{s^{(p-1)n+1}}ds=\int_1^\infty s^{l-(p-1)n-1}ds<\infty.\end{align*} $$
Notice that
$$ \begin{align*} &\int_0^1\frac{\Phi^{(l,k)}_{\ln}(s)}{s^{(p-1)(n-1)}}ds =\int_0^1s^{l-(p-1)(n-1)}\left(1-\frac{\ln s}{\frac{1}{2}\ln n+1}\right)^{2m+1}ds\\ &\quad \lesssim \int_0^1s^{l-(p-1)(n-1)}ds+\int_0^1s^{l-(p-1)(n-1)}(\ln s)ds\\ &\qquad+\int_0^1s^{l-(p-1)(n-1)}(\ln s)^2ds+\cdots+\int_0^1s^{l-(p-1)(n-1)}(\ln s)^{2m+1}ds. \end{align*} $$
According to (3.5) and (3.6), we can deduce that (3.3) holds. Then, (3.4) follows. This completes the proof of Proposition 3.3.
Definition 3.4 Suppose that
$k\in \mathbb N$
and
$p\in (1,2n/(n-1))$
. Let
$0\leq \lambda <2+\frac {p}{n}-p$
and
$l\geq k$
. Denote
$$ \begin{align*}g_{y,r}(x,t):=\frac{tr^{\frac{1}{n}}}{(|x-y|^2+|r+t|^2)^{\frac{n+1}{2}n}}.\end{align*} $$
The space
$\mathscr H_{K_{\ln }^{(l,k)}}^{p,\lambda }(\mathbb R_+^{n+1})$
consists of all measurable functions
$u(\cdot ,\cdot )$
such that
$$ \begin{align*}\sup\limits_{(y,r)\in\mathbb R_+^{n+1}}\int_{\mathbb R_+^{n+1}} (g_{y,r}(x,t))^{n\lambda}K_{\ln}^{(l,k)}\left(g_{y,r}(x,t)\right)|\nabla_{(x,t)}u(x,t)|^p\frac{dxdt}{t^{(p-1)(n-1)+n\lambda}}<\infty. \end{align*} $$
For a cube I, denote by
$S(I)$
the Carleson box centered at
$(y,r)$
, where
$r=\text {dist}\{(y,r),\mathbb {R}^n\}=\ell (I)/2$
.
Theorem 3.5 Suppose that
$k\in \mathbb N$
and
$p\in [n/(n-1),2n/(n-1))$
. Let
$$ \begin{align*}0\leq\lambda<2+\frac{p}{n}-p,\quad l\geq k\quad \text{and}\quad (p-1)(n-1)-1<l<(p-1)n.\end{align*} $$
Then,
$u\in \mathscr H_{K_{\ln }^{(l,k)}}^{p,\lambda }(\mathbb R_+^{n+1})$
if and only if
$$ \begin{align} \sup\limits_I\int_{S(I)}|\nabla_{(x,t)}u(x,t)|^p\left(\frac{t}{\ell(I)}\right)^{n\lambda+l} \left(\ln\left(\frac{e\sqrt{n}\ell(I)}{t}\right)\right)^k\frac{dxdt}{t^{(p-1)(n-1)+n\lambda}}<\infty. \end{align} $$
Proof We can see that (3.7) is equivalent to
$$ \begin{align*} \sup\limits_I\int_{S(I)}|\nabla_{(x,t)}u(x,t)|^p\left(\frac{t}{\ell(I)}\right)^{n\lambda}K_{\ln}^{(l,k)} \left(\frac{t}{\ell(I)}\right)\frac{dxdt}{t^{(p-1)(n-1)+n\lambda}}<\infty. \end{align*} $$
First, assume that
$u(\cdot ,\cdot )\in \mathscr H_{K_{\ln }^{(l,k)}}^{p,\lambda }(\mathbb R_+^{n+1})$
. If
$(x,t)\in S(I)$
, then
$|r+t|\leq \frac {3}{2}\ell (I)$
and
$|x-y|\leq \sqrt {n}\ell (I)$
. Furthermore, we have
$(|x-y|^2+|r+t|^2)^{\frac {1}{2}}\lesssim \ell (I)$
and
$$ \begin{align*} \frac{t}{\ell(I)}\lesssim \frac{tr^{\frac{1}{n}}}{(|x-y|^2+|r+t|^2)^{(n+1)/(2n)}}. \end{align*} $$
Therefore, it can be deduced that
$$ \begin{align*} &\sup\limits_I\int_{S(I)}|\nabla_{(x,t)}u(x,t)|^p\left(\frac{t}{\ell(I)}\right)^{n\lambda}K_{\ln}^{(l,k)}\left(\frac{t}{\ell(I)}\right)\frac{dxdt}{t^{(p-1)(n-1)+n\lambda}}\\ &\quad \lesssim\sup\limits_{(y,r)\in\mathbb R_+^{n+1}}\int_{\mathbb R_+^{n+1}}K_{\ln}^{(l,k)}\left(g_{y,r}(x,t)\right) \frac{\left(g_{y,r}(x,t)\right)^{n\lambda}|\nabla_{(x,t)}u(x,t)|^p}{t^{(p-1)(n-1)+n\lambda}}dxdt<\infty. \end{align*} $$
Conversely, assume that
$u(\cdot ,\cdot )$
satisfies (3.7). Let
$(y,r)\in \mathbb R^{n+1}_{+}$
and I be a cube parallel to the coordinate centered at y with edge length r. For any integer
$j>0$
, denote by
$I_j$
the cube centred at y and with edge length
$2^j\ell (I)$
. The Carleson box corresponding to
$I_{j}$
is written as
$S(I_j)$
. When
$(x,t)\in S(I)$
,
$(|x-y|^2+|r+t|^2)^{\frac {1}{2}}\geq r$
. When
$(x,t)\in S(I_{j+1})\setminus S(I)$
,
$(|x-y|^2+|r+t|^2)^{\frac {1}{2}}\geq 2^jr$
. Write
$$ \begin{align*} \int_{\mathbb R_+^{n+1}}K_{\ln}^{(l,k)}\left(g_{y,r}(x,t)\right)\left(g_{y,r}(x,t)\right)^{n\lambda}\frac{|\nabla_{(x,t)}u(x,t)|^p}{t^{(p-1)(n-1)+n\lambda}}dxdt\lesssim B_0+\sum\limits_{j=1}^\infty B_j, \end{align*} $$
where
$$ \begin{align*} \left\{ \begin{aligned} &B_0:=\int_{S(2I)}\left(\frac{t}{2r}\right)^{n\lambda}K_{\ln}^{(l,k)}\left(\frac{t}{2r}\right)|\nabla_{(x,t)}u(x,t)|^p\frac{dxdt}{t^{(p-1)(n-1)+n\lambda}};\\ &B_j:=\int_{S(2^{j+1}I)}\left(\frac{t}{2^{j+1}2^{j/n}r}\right)^{n\lambda}K_{\ln}^{(l,k)}\left(\frac{t}{2^{j+1}2^{j/n}r}\right)\frac{|\nabla_{(x,t)}u(x,t)|^p}{t^{(p-1)(n-1)+n\lambda}}dxdt. \end{aligned} \right. \end{align*} $$
It follows from (3.2) that
$$ \begin{align*}B_j\leq\Phi^{(l,k)}_{\ln}(2^{-j/n})2^{-j\lambda}\int_{S(2^{j+1}I)}\left(\frac{t}{2^{j+1}r}\right)^{n\lambda}K_{\ln}^{(l,k)}\left(\frac{t}{2^{j+1}r}\right)\frac{|\nabla_{(x,t)}u(x,t)|^p}{t^{(p-1)(n-1)+n\lambda}}dxdt.\end{align*} $$
Since
$u(\cdot ,\cdot )$
satisfies (3.7), there holds
$$ \begin{align} \sup\limits_I\int_{S(2^jI)}|\nabla_{(x,t)}u(x,t)|^p\left(\frac{t}{2^{j}r}\right)^{n\lambda}K_{\ln}^{(l,k)}\left(\frac{t}{2^{j}r}\right)\frac{dxdt}{t^{(p-1)(n-1)+n\lambda}}\lesssim1, \end{align} $$
which indicates that
$$ \begin{align*} &\sup\limits_{(y,r)\in\mathbb R_+^{n+1}}\int_{\mathbb R_+^{n+1}}\left(\frac{t}{\ell(I)}\right)^{n\lambda}K_{\ln}^{(l,k)}\left(g_{y,r}(x,t)\right)|\nabla_{(x,t)}u(x,t)|^p\frac{dxdt}{t^{(p-1)(n-1)+n\lambda}}\\ &\quad \lesssim1+\sum\limits_{j=1}^\infty2^{-j\lambda}\Phi^{(l,k)}_{\ln}(2^{-j/n})\int_{2^{-j/n-1}}^{2^{-j/n}}2^{j/n}ds. \end{align*} $$
Notice that
$K_{\ln }^{(l,k)}(t)\approx K_{\ln }^{(l,k)}(2t)$
. We can deduce from (3.2) that
$\Phi ^{(l,k)}_{\ln }(2^{-j/n})\approx \Phi ^{(l,k)}_{\ln }(2^{-j/n-1})$
. By the fact that
$2^{-j/n-1}<s<2^{-j/n}$
, we have
$2^{-j\lambda }\lesssim s^{n\lambda }$
and
$2^{j/n}<s^{-1}$
. Hence, by (3.3) and
$p\in [n/(n-1),\infty )$
, we have
$n\lambda -1+(p-1)(n-1)\geq 0$
and
$$ \begin{align*} &\sup\limits_{(y,r)\in\mathbb R_+^{n+1}}\int_{\mathbb R_+^{n+1}}\left(\frac{t}{\ell(I)}\right)^{n\lambda}K_{\ln}^{(l,k)}\left(g_{y,r}(x,t)\right) |\nabla_{(x,t)}u(x,t)|^p\frac{dxdt}{t^{(p-1)(n-1)+n\lambda}}\\ &\lesssim1+\sum\limits_{j=1}^\infty\int_{2^{-j/n-1}}^{2^{-j/n}}\Phi^{(l,k)}_{\ln}(s)\frac{ds}{s^{1-n\lambda}}\\ &\lesssim1+\int_0^1\Phi^{(l,k)}_{\ln}(s)\frac{ds}{s^{(p-1)(n-1)}}<\infty. \end{align*} $$
The proof of Theorem 3.5 is finished.
Theorem 3.6 Suppose that
$k\in \mathbb N$
and
$p\in (n/(n-1),2n/(n-1))$
. Let
$0\leq \lambda <2+\frac {p}{n}-p$
and
$l\geq k$
. Define
$$ \begin{align*} \mathcal{K}^{(l,k)}_{\ln}(t):=\left\{ \begin{aligned} &K_{\ln}^{(l,k)}(t),&\ 0<t<1;\\ &K_{\ln}^{(l,k)}(1)t^{(p-1)(n-1)},&\ t\geq1. \end{aligned} \right. \end{align*} $$
Then,
$f\in Q_{K_{\ln }^{(l,k)}}^{p,\lambda }(\mathbb {R}^n)$
if and only if
$$ \begin{align} \sup_{I}\frac{1}{(\ell(I))^{n\lambda}}\int_I\int_I\frac{|f(x)-f(y)|^p}{|x-y|^{pn}}\mathcal{K}^{(l,k)}_{\ln}\left(\frac{|x-y|}{\ell(I)}\right)dxdy<\infty. \end{align} $$
Proof At first, let (3.9) hold. For any cube I,
$$ \begin{align*} \frac{1}{(\ell(I))^{n\lambda}}\int_I\int_I\frac{|f(x)-f(y)|^p}{|x-y|^{pn}}K_{\ln}^{(l,k)}\left(\frac{|x-y|}{\ell(I)}\right)dxdy\leq U_1+U_2, \end{align*} $$
where
$$ \begin{align*}U_1:=\frac{1}{(\ell(I))^{n\lambda}}\iint_{\{y\in I:|x-y|<\ell(I)\}}\frac{|f(x)-f(y)|^p}{|x-y|^{pn}}K_{\ln}^{(l,k)}\left(\frac{|x-y|}{\ell(I)}\right)dxdy,\end{align*} $$
$$ \begin{align*}U_2:=\frac{1}{(\ell(I))^{n\lambda}}\iint_{I(x)}\frac{|f(x)-f(y)|^p}{|x-y|^{pn}}K_{\ln}^{(l,k)}\left(\frac{|x-y|}{\ell(I)}\right)dxdy\end{align*} $$
with
$I(x):=\{y\in I:\ell (I)\leq |x-y|<\sqrt {n}\ell (I)\}.$
We can estimate
$U_1$
as
$$ \begin{align*}U_1\leq\frac{1}{(\ell(I))^{n\lambda}}\int_I\int_I\frac{|f(x)-f(y)|^p}{|x-y|^{pn}}\mathcal{K}^{(l,k)}_{\ln}\left(\frac{|x-y|}{\ell(I)}\right)dxdy.\end{align*} $$
It follows from the fact
$\ell (I)\leq |x-y|<\sqrt {n}\ell (I)$
that
$1\leq |x-y|/{\ell (I)}<\sqrt {n}$
and
$$ \begin{align*} U_2&\leq\frac{1}{(\ell(I))^{n\lambda}}\iint_{I(x)}\frac{|f(x)-f(y)|^p}{|x-y|^{pn}}K_{\ln}^{(l,k)}(\sqrt{n})dxdy\\ &\lesssim\frac{1}{(\ell(I))^{n\lambda}}\int_I\int_I\frac{|f(x)-f(y)|^p}{|x-y|^{pn}}K_{\ln}^{(l,k)}(1)\left(\frac{|x-y|}{\ell(I)}\right)^{(p-1)(n-1)}dxdy. \end{align*} $$
Hence,
$$ \begin{align*} &\frac{1}{(\ell(I))^{n\lambda}}\int_I\int_I\frac{|f(x)-f(y)|^p}{|x-y|^{pn}}K_{\ln}^{(l,k)}\left(\frac{|x-y|}{\ell(I)}\right)dxdy\\ &\lesssim\frac{1}{(\ell(I))^{n\lambda}}\int_I\int_I\frac{|f(x)-f(y)|^p}{|x-y|^{pn}}\mathcal{K}^{(l,k)}_{\ln}\left(\frac{|x-y|}{\ell(I)}\right)dxdy, \end{align*} $$
which means
$f\in Q_{K_{\ln }^{(l,k)}}^{p,\lambda }(\mathbb {R}^n)$
.
Now, assume that
$f\in Q_{K_{\ln }^{(l,k)}}^{p,\lambda }(\mathbb {R}^n)$
. Then,
$$ \begin{align*} \frac{1}{(\ell(I))^{n\lambda}}\int_I\int_I\frac{|f(x)-f(y)|^p}{|x-y|^{pn}}\mathcal{K}^{(l,k)}_{\ln}\left(\frac{|x-y|}{\ell(I)}\right)dxdy\leq V_1+V_2, \end{align*} $$
where
$$ \begin{align*}V_1:=\frac{1}{(\ell(I))^{n\lambda}}\iint_{\{y\in I:|x-y|<\ell(I)\}}\frac{|f(x)-f(y)|^p}{|x-y|^{pn}}\mathcal{K}^{(l,k)}_{\ln}\left(\frac{|x-y|}{\ell(I)}\right)dxdy\end{align*} $$
and
$$ \begin{align*}V_2:=\frac{1}{(\ell(I))^{n\lambda}}\iint_{I(x)}\frac{|f(x)-f(y)|^p}{|x-y|^{pn}}\mathcal{K}^{(l,k)}_{\ln}\left(\frac{|x-y|}{\ell(I)}\right)dxdy.\end{align*} $$
Since
$\mathcal K(t)=K_{\ln }^{(l,k)}(t), 0<t<1$
and
$|x-y|<\ell (I).$
Then, for
$V_1$
, it holds
$$ \begin{align*}V_1\leq(\ell(I))^{-\lambda}\int_I\int_I\frac{|f(x)-f(y)|^p}{|x-y|^{pn}}K_{\ln}^{(l,k)}\left(\frac{|x-y|}{\ell(I)}\right)dxdy.\end{align*} $$
For
$V_2$
,
$$ \begin{align*} V_2&=\frac{1}{(\ell(I))^{n\lambda}}\iint_{I(x)}\frac{|f(x)-f(y)|^p}{|x-y|^{pn}}K_{\ln}^{(l,k)}(1)\left(\frac{|x-y|}{\ell(I)}\right)^{(p-1)(n-1)}dxdy\\ &\lesssim\frac{1}{(\ell(I))^{n\lambda}}\int_I\int_I\frac{|f(x)-f(y)|^p}{|x-y|^{pn}}K_{\ln}^{(l,k)}\left(\frac{|x-y|}{\ell(I)}\right)dxdy. \end{align*} $$
Hence,
$$ \begin{align*} &\frac{1}{(\ell(I))^{n\lambda}}\int_I\int_I\frac{|f(x)-f(y)|^p}{|x-y|^{pn}}\mathcal K\left(\frac{|x-y|}{\ell(I)}\right)dxdy\\ &\lesssim\frac{1}{(\ell(I))^{n\lambda}}\int_I\int_I\frac{|f(x)-f(y)|^p}{|x-y|^{pn}}K_{\ln}^{(l,k)}\left(\frac{|x-y|}{\ell(I)}\right)dxdy, \end{align*} $$
which implies (3.9). This completes the proof of Theorem 3.6.
Lemma 3.7 Suppose that
$k\in \mathbb N$
and
$p\in (1,2n/(n-1))$
. Let
$l\geq k$
and
$$ \begin{align*}(p-1)n-p<l<(p-1)n.\end{align*} $$
There holds
$$ \begin{align} \sup\limits_{0<s<\infty}\left(\int_0^sK_{\ln}^{(l,k)}(t)t^{(p-1)(1-n)}dt\right)^{\frac{1}{p}}\left(\int_s^\infty \left(K_{\ln}^{(l,k)}(t)t^{n(1-p)-1+2p}\right)^{\frac{1}{1-p}}dt\right)^{\frac{p-1}{p}}<\infty. \end{align} $$
Proof Define
$$ \begin{align*} K^*(t):=\left\{ \begin{aligned} &t^{p(n-1)-n}\int_0^tK_{\ln}^{(l,k)}(s)s^{(p-1)(1-n)}ds,&\ 0<t<1;\\ &t^{(p-1)(n-1)}\int_0^1K_{\ln}^{(l,k)}(s)s^{(p-1)(1-n)}ds,&\ t\geq1. \end{aligned} \right. \end{align*} $$
According to Theorem 3.6, without loss of generality,
$$ \begin{align*}K_{\ln}^{(l,k)}(t)=K_{\ln}^{(l,k)}(1)t^{(p-1)(n-1)}, t\geq1.\end{align*} $$
By a direct calculation, we know that
$K^*(\cdot )$
is increasing. When
$0< t<1$
, it follows from (3.2) that
$$ \begin{align*} \left\{\begin{aligned} K^*(t)&=\int_0^1K_{\ln}^{(l,k)}(st)s^{(p-1)(1-n)}ds\\ & \quad \leq K_{\ln}^{(l,k)}(t)\int_0^1\Phi^{(l,k)}_{\ln}(s)s^{(p-1)(1-n)}ds\approx K_{\ln}^{(l,k)}(t);\\ K^*(t)&\geq t^{p(n-1)-n}K_{\ln}^{(l,k)}\left(\frac{t}{3}\right)\int_{t/3}^ts^{(p-1)(1-n)}ds\approx K_{\ln}^{(l,k)}\left(\frac{t}{3}\right). \end{aligned} \right. \end{align*} $$
Moreover, when
$t\geq 1$
, we can see from the definition of
$\Phi ^{(l,k)}_{\ln }(\cdot )$
that there exists a constant
$C>0$
such that
$$ \begin{align*}\int_0^1K_{\ln}^{(l,k)}(s)s^{(p-1)(1-n)}ds=CK_{\ln}^{(l,k)}(1).\end{align*} $$
Hence,
$$ \begin{align*} K^*{(t)}\approx K_{\ln}^{(l,k)}(1)t^{(p-1)(n-1)}=K_{\ln}^{(l,k)}(t), \end{align*} $$
which implies that
$K^*(t)\approx K_{\ln }^{(l,k)}(t)$
on
$(0,\infty )$
. For some sufficiently small
$c>0$
,
$$ \begin{align*} (K^*(t)/t^{c+p(n-1)-n})'=\left\{ \begin{aligned} &t^{-c-(p-1)(n-1)}(K_{\ln}^{(l,k)}(t)-cK^*(t))>0,&\ 0<t<1;\\ &(K_{\ln}^{(l,k)}(1)t^{-c+1})'>0,&\ t\geq1. \end{aligned} \right. \end{align*} $$
Therefore,
$K^*(t)/t^{c+p(n-1)-n}$
is increasing on
$(0,\infty )$
. For
$0<s<\infty $
,
$$ \begin{align*} &\left(\int_0^sK^*(t)t^{(p-1)(1-n)}dt\right)^{\frac{1}{p}}\left(\int_s^\infty \left(K^*(t)t^{n(1-p)-1+2p}\right)^{\frac{1}{1-p}}dt\right)^{1-\frac{1}{p}}\\ &\leq\left(\int_0^st^{c-1}dt\right)^{\frac{1}{p}}\left(\int_s^\infty t^{-1-c/(p-1)}dt\right)^{1-\frac{1}{p}}<\infty. \end{align*} $$
This completes the proof of Lemma 3.7.
Theorem 3.8 Suppose that
$k\in \mathbb N$
and
$p\in (1,2n/(n-1))$
. Let
$l\geq k$
and
$(p-1)n-p<l<(p-1)n$
. There exists
$u\in (0,\infty ]$
such that
$$ \begin{align} \sup\limits_{0<s<u}\left(\int_s^uK_{\ln}^{(l,k)}(t)t^{(1-p)n-1}dt\right)^{\frac{1}{p}}\left(\int_0^s(K_{\ln}^{(l,k)}(t))^{\frac{1}{1-p}}t^{n-1}dt\right)^{1-\frac{1}{p}}<\infty. \end{align} $$
Proof According to (3.4) and the definition of
$\Phi ^{(l,k)}_{\ln }(\cdot )$
, it follows that
$$ \begin{align*}\frac{1}{K_{\ln}^{(l,k)}(\frac{1}{s})}\leq\frac{\Phi^{(l,k)}_{\ln}(s)}{K_{\ln}^{(l,k)}(1)}, \quad \int_0^1\frac{s^{(p-1)n-1}}{K_{\ln}^{(l,k)}(s)}ds=\int_1^\infty\frac{1}{s^{(p-1)n+1}K_{\ln}^{(l,k)}(\frac{1}{s})}ds<\infty.\end{align*} $$
Hence,
$$ \begin{align*} \frac{t^{(p-1)n}}{K_{\ln}^{(l,k)}(t)} &\lesssim \frac{1}{K_{\ln}^{(l,k)}(t)}\int_0^ts^{(p-1)n-1}ds\\ &\lesssim \int_0^1\frac{s^{(p-1)n-1}}{K_{\ln}^{(l,k)}(s)}ds<\infty, \end{align*} $$
which implies that
$\lim \limits _{t\rightarrow 0}\inf g(t)>0$
, where
$$ \begin{align*}g(t):=\frac{K_{\ln}^{(l,k)}(t)}{t^{(p-1)n}}=t^{l-(p-1)n}\left(\ln\left(\frac{e\sqrt{n}}{t}\right)\right)^k.\end{align*} $$
Since
$l<(p-1)n$
, we can deduce that
$g(\cdot )$
is decreasing and
$$ \begin{align*} \lim\limits_{t\rightarrow0}\inf g(t)=g(1)=\left(\ln\left(e\sqrt{n}\right)\right)^k=K_{\ln}^{(l,k)}(1),\ \ \ 0<t<1. \end{align*} $$
Define
$$ \begin{align*}K^\#(t):=t^{(p-1)}\int_t^\infty\frac{K_{\ln}^{(l,k)}(s)}{s^{(p-1)n+1}}ds, \quad0<t<\infty.\end{align*} $$
By a direct calculation, we can get
$(K^\#(t))'\geq 0$
and
$K_{\ln }^{(l,k)}(t)\lesssim K^\#(t),0<t<\infty $
. If
$0<t<1$
, notice that
$$ \begin{align*}\int_t^1K_{\ln}^{(l,k)}(s)s^{(1-p)n-1}ds\leq K_{\ln}^{(l,k)}(t)t^{(1-p)n}\int_1^\infty\Phi^{(l,k)}_{\ln}(s)s^{(1-p)n-1}ds;\end{align*} $$
$$ \begin{align*}\int_1^\infty K_{\ln}^{(l,k)}(s)s^{(1-p)n-1}ds\leq K_{\ln}^{(l,k)}(1)\int_1^\infty\Phi^{(l,k)}_{\ln}(s)s^{(1-p)n-1}ds\lesssim K_{\ln}^{(l,k)}(t)t^{(1-p)n}.\end{align*} $$
We can deduce that
$$ \begin{align*}K^\#(t)\lesssim K_{\ln}^{(l,k)}(t)\left(\int_1^\infty\Phi^{(l,k)}_{\ln}(s)s^{(1-p)n-1}ds+1\right)\end{align*} $$
and
$K^\#(t)\approx K_{\ln }^{(l,k)}(t)$
on
$(0,1)$
. If
$1\leq t<\infty $
, without loss of generality, assume that
$K_{\ln }^{(l,k)}(t)=K_{\ln }^{(l,k)}(1)t^{(p-1)(n-1)}$
. It holds
$$ \begin{align*}K^\#(t)=K_{\ln}^{(l,k)}(1)t^{(p-1)n}\int_t^\infty s^{-p}ds\approx K_{\ln}^{(l,k)}(t).\end{align*} $$
Hence,
$K^\#(t)\approx K_{\ln }^{(l,k)}(t),0<t<\infty $
. Suppose that
$c>0$
is sufficiently small. We can deduce that
$(K^\#(t)t^{(1-p)n+c})'<0$
, which indicates that
$K^\#(t)t^{(1-p)n+c}$
is decreasing. Moreover, for some
$0<u\leq \infty $
, we can get
$$ \begin{align*} &\left(\int_s^u K^\#(t)t^{(1-p)n-1}dt\right)^{\frac{1}{p}}\left(\int_0^s(K^\#(t))^{\frac{1}{1-p}}t^{n-1}dt\right)^{1-\frac{1}{p}}\\ &\quad \leq\left(\int_s^\infty t^{-1-c}dt\right)^{\frac{1}{p}}\left(\int_0^st^{c/(p-1)-1}dt\right)^{1-\frac{1}{p}}<\infty. \end{align*} $$
It follows that
$$ \begin{align} \sup\limits_{0<s<u}\left(\int_s^u K_{\ln}^{(l,k)}(t)t^{(1-p)n-1}dt\right)^{\frac{1}{p}}\left(\int_0^s(K_{\ln}^{(l,k)}(t))^{\frac{1}{1-p}}t^{n-1}dt\right)^{1-\frac{1}{p}}<\infty. \end{align} $$
Lemma 3.9 Suppose that
$k\in \mathbb N$
and
$p\in (1,2n/(n-1))$
. Let
$l\geq k$
. The following two statements are true.
-
(i) There holds:
(3.13)
$$ \begin{align} \sup\limits_{0<s<\infty}\left(\int_s^\infty K_{\ln}^{(l,k)}(t)t^{(1-p)n-1}dt\right)^{\frac{1}{p}}\left(\int_0^s(K_{\ln}^{(l,k)}(t))^{\frac{1}{1-p}}t^{n-1}dt\right)^{1-\frac{1}{p}}<\infty. \end{align} $$
-
(ii) There exists another nonnegative function
$\mathscr {K}^{(l,k)}_{\ln }(\cdot )$
such that
-
(a)
$\mathscr {K}^{(l,k)}_{\ln }\approx K_{\ln }^{(l,k)}$
on
$[0,\infty );$
-
(b) for some small enough
$c>0, \mathscr {K}^{(l,k)}_{\ln }(t)t^{(1-p)n+c}$
is decreasing; -
(c)
$\mathscr {K}^{(l,k)}_{\ln }(t)t^{(1-p)n}$
is also decreasing.
Proof (i) Since
$K_{\ln }^{(l,k)}$
satisfies (3.11), if
$u=\infty $
, it is easy to get (3.13) holds. We divide the proof into two cases if
$0<u<\infty $
.
Case 1:
$0<s<u$
. Since (3.11) holds for
$K_{\ln }^{(l,k)}$
,
$$ \begin{align*} &\left(\int_s^\infty K_{\ln}^{(l,k)}(t)t^{(1-p)n-1}dt\right)^{\frac{1}{p}}\left(\int_0^s(K_{\ln}^{(l,k)}(t))^{\frac{1}{1-p}}t^{n-1}dt\right)^{1-\frac{1}{p}}\\ &\lesssim 1+\left(\int_u^\infty K_{\ln}^{(l,k)}(t)t^{(1-p)n-1}dt\right)^{\frac{1}{p}}\left(\int_0^u(K_{\ln}^{(l,k)}(t))^{\frac{1}{1-p}}t^{n-1}dt\right)^{1-\frac{1}{p}}. \end{align*} $$
Notice that
$K_{\ln }^{(l,k)}(t)=K_{\ln }^{(l,k)}(1)t^{(p-1)(n-1)}, t\geq 1$
. It follows that
$$ \begin{align*} & \int_u^\infty K_{\ln}^{(l,k)}(t)t^{(1-p)n-1}dt \\ &\quad =\int_u^1 K_{\ln}^{(l,k)}(t)t^{(1-p)n-1}dt+\int_1^\infty K_{\ln}^{(l,k)}(t)t^{(1-p)n-1}dt<\infty. \end{align*} $$
Let
$s=u/2$
in (3.11). We can deduce that
$$ \begin{align*}\left(\int_0^u(K_{\ln}^{(l,k)}(t))^{\frac{1}{1-p}}t^{n-1}dt\right)^{1-\frac{1}{p}}<\infty.\end{align*} $$
Therefore,
$$ \begin{align} \sup\limits_{0<s<u}\left(\int_s^\infty K_{\ln}^{(l,k)}(t)t^{(1-p)n-1}dt\right)^{\frac{1}{p}}\left(\int_0^s(K_{\ln}^{(l,k)}(t))^{\frac{1}{1-p}}t^{n-1}dt\right)^{1-\frac{1}{p}}<\infty. \end{align} $$
Case 2:
$u\leq s<\infty $
. When
$s<1$
,
$$ \begin{align*} &\left(\int_s^\infty K_{\ln}^{(l,k)}(t)t^{(1-p)n-1}dt\right)^{\frac{1}{p}}\left(\int_0^s(K_{\ln}^{(l,k)}(t))^{\frac{1}{1-p}}t^{n-1}dt\right)^{1-\frac{1}{p}}\\ &\quad \lesssim\left(\int_u^\infty K_{\ln}^{(l,k)}(t)t^{(1-p)n-1}dt\right)^{\frac{1}{p}}\left(\int_0^u(K_{\ln}^{(l,k)}(t))^{\frac{1}{1-p}}t^{n-1}dt\right)^{1-\frac{1}{p}}\\ &\qquad +\left(\int_u^1 K_{\ln}^{(l,k)}(t)t^{(1-p)n-1}dt+\int_1^\infty K_{\ln}^{(l,k)}(t)t^{(1-p)n-1}dt\right)^{\frac{1}{p}}\\ &\qquad \times\left(\int_u^1(K_{\ln}^{(l,k)}(t))^{\frac{1}{1-p}}t^{n-1}dt\right)^{1-\frac{1}{p}}<\infty. \end{align*} $$
When
$s\geq 1$
, due to (3.11), we can see that
$$ \begin{align*}\left(\int_0^1(K_{\ln}^{(l,k)}(t))^{\frac{1}{1-p}}t^{n-1}dt\right)^{1-\frac{1}{p}}<\infty\end{align*} $$
and
$$ \begin{align*} &\lim\limits_{s\rightarrow\infty}\left(\int_s^\infty K_{\ln}^{(l,k)}(t)t^{(1-p)n-1}dt\right)^{\frac{1}{p}}\left(\int_0^s(K_{\ln}^{(l,k)}(t))^{\frac{1}{1-p}}t^{n-1}dt\right)^{1-\frac{1}{p}}\\ &\lesssim\lim\limits_{s\rightarrow\infty}\left(\int_s^\infty K_{\ln}^{(l,k)}(t)t^{(1-p)n-1}dt\right)^{\frac{1}{p}}\left(\int_0^1(K_{\ln}^{(l,k)}(t))^{\frac{1}{1-p}}t^{n-1}dt\right)^{1-\frac{1}{p}}\\ &\quad+\lim\limits_{s\rightarrow\infty}\left(\int_s^\infty K_{\ln}^{(l,k)}(t)t^{(1-p)n-1}dt\right)^{\frac{1}{p}}\left(\int_1^s(K_{\ln}^{(l,k)}(t))^{\frac{1}{1-p}}t^{n-1}dt\right)^{1-\frac{1}{p}}<\infty, \end{align*} $$
which indicates that
$$ \begin{align*}\left(\int_s^\infty K_{\ln}^{(l,k)}(t)t^{(1-p)n-1}dt\right)^{\frac{1}{p}}\left(\int_0^s(K_{\ln}^{(l,k)}(t))^{\frac{1}{1-p}}t^{n-1}dt\right)^{1-\frac{1}{p}}<\infty.\end{align*} $$
Therefore,
$$ \begin{align} \sup\limits_{u\leq s<\infty}\left(\int_s^\infty K_{\ln}^{(l,k)}(t)t^{(1-p)n-1}dt\right)^{\frac{1}{p}}\left(\int_0^s(K_{\ln}^{(l,k)}(t))^{\frac{1}{1-p}}t^{n-1}dt\right)^{1-\frac{1}{p}}<\infty. \end{align} $$
Combining (3.14) with (3.15), it holds
$$ \begin{align*}\sup\limits_{0<s<\infty}\left(\int_s^\infty K_{\ln}^{(l,k)}(t)t^{(1-p)n-1}dt\right)^{\frac{1}{p}}\left(\int_0^s(K_{\ln}^{(l,k)}(t))^{\frac{1}{1-p}}t^{n-1}dt\right)^{1-\frac{1}{p}}<\infty.\end{align*} $$
(ii) Define
$$ \begin{align*} \mathscr{K}^{(l,k)}_{\ln}(t):=\left\{ \begin{aligned} &t^{(p-1)n}\int_t^\infty K_{\ln}^{(l,k)}(s)s^{(1-p)n-1}ds,&\ 0<t<\infty;\\ &0,&\ t=0. \end{aligned} \right. \end{align*} $$
By a direct calculation,
$\mathscr K^{(l,k)}_{\ln }(t)\approx K_{\ln }^{(l,k)}(1)t^{(p-1)(n-1)},$
when
$t\geq 1$
. Since
$K_{\ln }^{(l,k)}(\cdot )$
is increasing, we can deduce
$$ \begin{align*} \left\{ \begin{aligned} \mathscr K^{(l,k)}_{\ln}(t)&\geq K_{\ln}^{(l,k)}(t)t^{(p-1)n}\int_t^\infty s^{(1-p)n-1}ds\approx K_{\ln}^{(l,k)}(t);\\ \mathscr K^{(l,k)}_{\ln}(t)&\lesssim t^{(p-1)n}\left(\int_0^t(K_{\ln}^{(l,k)}(s))^{\frac{1}{1-p}}s^{n-1}ds\right)^{1-p}\\ &\lesssim t^{(p-1)n}\left(\int_0^t(K_{\ln}^{(l,k)}(t))^{\frac{1}{1-p}}s^{n-1}ds\right)^{1-p}\approx K_{\ln}^{(l,k)}(t), \end{aligned} \right. \end{align*} $$
which implies that
$\mathscr K^{(l,k)}_{\ln }(\cdot )\approx K_{\ln }^{(l,k)}(\cdot ), 0\leq t<\infty $
. Suppose that
$c>0$
is sufficiently small. We have
$$ \begin{align*} (\mathscr K^{(l,k)}_{\ln}(t)t^{(1-p)n+c})'=t^{c+(1-p)n-1}(c\mathscr K^{(l,k)}_{\ln}(t)-K^{(l,k)}_{\ln}(t))<0. \end{align*} $$
Hence,
$\mathscr K^{(l,k)}_{\ln }(t)t^{(1-p)n+c}$
and
$\mathscr K^{(l,k)}_{\ln }(t)t^{(1-p)n}$
are decreasing. The proof of Lemma 3.9 is completed.
We need the following Hardy-type inequalities to investigate the characterizations of
$Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
.
Lemma 3.10 [Reference Kufner and Persson20, p. 4, equations (0.10) and (0.12)]
Suppose that
$f\geq 0$
is a measurable function,
$0<m\leq \infty $
and
$\frac {1}{p}+1/q=1$
such that
$1<p,q<\infty $
. Let
$\sigma $
and
$\tau $
be nonnegative functions on
$(0,h)$
. Then,
-
(i)
holds for f if and only if
$$ \begin{align*}\int_0^h\left(\int_0^sf(t)dt\right)^p\sigma(s)ds\lesssim\int_0^h(f(s))^p\tau(s)ds\end{align*} $$
$$ \begin{align*}\sup\limits_{0<s<h}\left(\int_s^h\sigma(t)dt\right)^{\frac{1}{p}}\left(\int_0^s(\tau(t))^{1-q}dt\right)^{\frac{1}{q}}<\infty.\end{align*} $$
-
(ii)
holds for f if and only if
$$ \begin{align*}\int_0^h\left(\int_s^hf(t)dt\right)^p\sigma(s)ds\lesssim\int_0^h(f(s))^p\tau(s)ds\end{align*} $$
$$ \begin{align*}\sup\limits_{0<s<h}\left(\int_0^s\sigma(t)dt\right)^{\frac{1}{p}}\left(\int_s^h(\tau(t))^{1-q}dt\right)^{\frac{1}{q}}<\infty.\end{align*} $$
Suppose that f is a measurable function on
$\mathbb R^{n}$
satisfying
$$ \begin{align} \int_{\mathbb{R}^n}\frac{|f(x)|}{1+|x|^{n+2\beta}}dx<\infty. \end{align} $$
Define the semigroup operator with a dimensional constant c as
$$ \begin{align*}e^{-t^{2\beta}(-\Delta)^{\beta}}\phi(x)=c\int_{\mathbb{R}^n}\widehat{\phi}(\zeta)e^{-t^{2\beta}|\zeta|^{2\beta}+i\zeta\cdot x}d\zeta,\ \ \phi\in \mathscr S (\mathbb{R}^n),\end{align*} $$
where
$\Delta $
denotes the Laplace operator, a second-order differential operator in an n-dimensional Euclidean space. Denote by
$K_{t^{2\beta }}^\beta (\cdot )$
the kernel of the operator
$e^{-t^{2\beta }(-\Delta )^{\beta }}$
, i.e.,
$$ \begin{align*}K_{t^{2\beta}}^\beta(x):=c\int_{\mathbb{R}^n}e^{-t^{2\beta}|\zeta|^{2\beta}+i\zeta\cdot x}d\zeta,\ \ t>0,\end{align*} $$
and denote by
$u(\cdot ,\cdot )$
the convolution of f and
$K_{t^{2\beta }}^\beta $
, i.e.,
$$ \begin{align} u(x,t):=f\ast K_{t^{2\beta}}^\beta(x)=\int_{\mathbb{R}^n}f(x-y)K_{t^{2\beta}}^\beta(y)dy. \end{align} $$
If
$\phi \in \mathscr S(\mathbb {R}^n)$
, then
$e^{-t^{2\beta }(-\Delta )^{\beta }}\phi (x):=K_{t^{2\beta }}^\beta \ast \phi (x)$
. In [Reference Miao, Yuan and Zhang27, Remark 2.1], by the Fourier transforms, Miao et al. obtained the following regularity estimate of fractional heat kernels
$K_{t^{2\beta }}^\beta (\cdot )$
:
$$ \begin{align} \left\{ \begin{aligned} &|K_{t^{2\beta}}^\beta(x)|\lesssim\frac{t^{2\beta}}{(t+|x|)^{n+2\beta}};\\ &|\nabla_{x}K_{t^{2\beta}}^\beta(x)|\lesssim\frac{1}{t^{n+1}+|x|^{n+1}},\quad \frac{1}{2}\leq\beta<\infty;\\ &\int_{\mathbb{R}^n}\nabla_{(x,t)}K_{t^{2\beta}}^\beta(x)dx=0. \end{aligned} \right. \end{align} $$
In fact, the estimate for
$|\nabla _{x}K_{t^{2\beta }}^\beta (x)|$
in (3.18) can be improved.
Lemma 3.11 Let
$\beta \in (0,1)$
,
$m\in \mathbb {Z}_{+}$
. There exists a positive constant C such that for
$t>0$
,
$$ \begin{align*}\big|t^{m}\partial_{t}^{m}K^{\beta}_{t}(x)\big|\leq \frac{Ct}{(t^{\frac{1}{2\beta}}+|x|)^{n+2\beta}}.\end{align*} $$
Proof Since
$\{e^{-t(-\Delta )^{\beta }}\}_{t>0}$
,
$\beta \in (0,1)$
, is an analytic semigroup, the estimate for
$|t^{m}\partial _{t}^{m}K^{\beta }_{t}(\cdot )|$
can be deduced by that of
$K^{\beta }_{t}(\cdot )$
and the Cauchy integral formula. We omit the details.
Lemma 3.12 Let
$\beta \in (0,1)$
. There exists a constant
$C>0$
such that for all
$x,y\in \mathbb {R}^{n}$
and
$t>0$
,
$$ \begin{align*}\big|t^{\frac{1}{2\beta}}\nabla_{x}K^{\beta}_{t}(x)\big|\leq \frac{Ct}{(t^{\frac{1}{2\beta}}+|x|)^{n+2\beta}}. \end{align*} $$
Proof Denote by
$K_{t}(\cdot )$
the heat kernel of Laplace operator
$(-\Delta )$
. The subordinate formula gives
$$ \begin{align*}\nabla_{x}K^{\beta}_{t}(x)=\int_{0}^{\infty}\eta_{t}^{\beta}(s)\nabla_{x}K_{s}(x)ds, \end{align*} $$
where
$\eta _{t}^{\beta }(\cdot )$
satisfies (cf. [Reference Grigor’yan16])
$$ \begin{align} \left\{\begin{aligned} &\eta_{t}^{\beta}(s)=\frac{1}{t^{\frac{1}{\beta}}}\eta_{1}^{\beta}(s/t^{\frac{1}{\beta}});\\ &\eta_{t}^{\beta}(s)\lesssim \frac{t}{s^{1+\beta}}\quad \forall s,t>0;\\ &\int_{0}^{\infty}s^{-\gamma}\eta_{1}^{\beta}(s)ds<\infty,\quad \gamma>0;\\ &\eta_{t}^{\beta}(s)\simeq\frac{t}{s^{1+\beta}},\quad \forall s\geq t^{\frac{1}{\beta}}>0. \end{aligned}\right. \end{align} $$
A direct computation gives
$$ \begin{align*}|\nabla_{x}K_{t}(x)|\lesssim \frac{1}{t^{\frac{n+1}{2}}}e^{-c|x|^2/t}. \end{align*} $$
By (3.19) and the change of variables:
$s=t^{\frac {1}{\beta }}u$
and
$r^2={|x|^2}/{(t^{\frac {1}{\beta }}u)},$
we obtain
$$ \begin{align*} |\nabla_{x}K^{\beta}_{t}(x)|&\leq \int_{0}^{\infty}\frac{t}{s^{1+\beta}}\frac{1}{s^{\frac{n+1}{2}}}e^{-\frac{c|x|^2}{s}}ds\\ &=\int_{0}^{\infty}\frac{t}{(t^{\frac{1}{\beta}}u)^{1+\beta+\frac{n+1}{2}}}e^{-\frac{c|x|^2}{(t^{\frac{1}{\beta}}u)}}t^{\frac{1}{\beta}}du\\ &\leq\int_{0}^{\infty}\frac{1}{t^{\frac{n+1}{2\beta}}}\frac{1}{u^{1+\beta+\frac{n+1}{2}}}e^{-\frac{c|x|^2}{(t^{\frac{1}{\beta}}u)}}du\\ &\lesssim\int_{0}^{\infty}\frac{1}{t^{\frac{n+1}{2\beta}}}\bigg(\frac{t^{\frac{1}{\beta}}r^2}{|x|^2}\bigg)^{1+\beta+\frac{n+1}{2}}e^{-cr^2} \frac{|x|^2}{t^{\frac{1}{\beta}}r^3}dr\\ &\lesssim\frac{t}{|x|^{2\beta+n+1}}\int_{0}^{\infty}\bigg(\frac{t^{\frac{1}{\beta}}r^2}{|x|^2}\bigg)^{1+\beta+\frac{n+1}{2}}e^{-cr^2} \frac{|x|^2}{t^{\frac{1}{\beta}}r^3}dr\\ &\lesssim\frac{t}{|x|^{2\beta+n+1}}\Bigg\{\int_{0}^{1}r^{2\beta+n+1}dr+\int_{1}^{\infty}r^{2\beta+n+1}e^{-cr^2}dr\Bigg\}\\ &\lesssim\frac{t}{|x|^{2\beta+n+1}}. \end{align*} $$
On the other hand, we can get
$$ \begin{align*} |\nabla_{x}K^{\beta}_{t}(x)|&\leq \int_{0}^{\infty}\frac{1}{t^{\frac{1}{\beta}}}\eta_{1}^{\beta}\bigg(\frac{s}{t^{\frac{1}{\beta}}}\bigg)\frac{1}{s^{\frac{n+1}{2}}}ds\\ &\leq\int_{0}^{\infty}\eta_{1}^{\beta}(u)\frac{1}{(ut^{\frac{1}{\beta}})^{\frac{n+1}{2}}}du\\ &\lesssim\frac{1}{t^{\frac{n+1}{2\beta}}}. \end{align*} $$
Thus, it is easy to see that
$$ \begin{align*}|\nabla_{x}K^{\beta}_{t}(x)|\lesssim \min\Bigg\{\frac{t}{|x|^{2\beta+n+1}},\frac{1}{t^{\frac{n+1}{2\beta}}}\Bigg\}. \end{align*} $$
Case 1:
$0\leq t^{\frac {1}{2\beta }}\leq |x|$
. For this case,
${t}/{|x|^{2\beta +n+1}}\leq {1}/{t^{\frac {n+1}{2\beta }}}$
. Moreover, it leads to
$$ \begin{align*} |\nabla_{x}K^{\beta}_{t}(x)|\lesssim \frac{t}{|x|^{2\beta+n+1}}\lesssim\frac{t}{|x|^{2\beta+n}}\frac{1}{t^{\frac{1}{2\beta}}} \lesssim\frac{t}{(t^{\frac{1}{2\beta}}+|x|)^{2\beta+n}}\frac{1}{t^{\frac{1}{2\beta}}}. \end{align*} $$
Case 2:
$t^{\frac {1}{2\beta }}> |x|$
. It holds
$$ \begin{align*} |\nabla_{x}K^{\beta}_{t}(x)|\lesssim \frac{1}{t^{\frac{n+1}{2\beta}}}=\frac{t}{(t^{\frac{1}{2\beta}})^{n+2\beta}}\frac{1}{t^{\frac{1}{2\beta}}} \lesssim\frac{t}{(t^{\frac{1}{2\beta}}+|x|)^{2\beta+n}}\frac{1}{t^{\frac{1}{2\beta}}}. \end{align*} $$
Remark 3.13 Notice that
$$ \begin{align*}\partial_{t}(K^{\beta}_{t^{2\beta}}(x))\approx t^{2\beta-1}(\partial_{s}K^{\beta}_{s}(x))\mid_{s=t^{2\beta}}.\end{align*} $$
It can be deduced from Lemmas 3.11 and 3.12 that
$$ \begin{align*}\big|\nabla_{(x,t)}K^{\beta}_{t^{2\beta}}(x)\big|\lesssim \frac{t^{2\beta-1}}{(t+|x|)^{n+2\beta}},\end{align*} $$
which implies
$$ \begin{align*}|\nabla_{(x,t)}K_{t^{2\beta}}^\beta(x)|\lesssim\frac{1}{t^{n+1}+|x|^{n+1}},\ \ \frac{1}{2}\leq\beta<\infty.\end{align*} $$
The semigroup characterization of
$Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
is based on the following Stegenga-type inequality.
Lemma 3.14 For any
$k\in \mathbb N$
and
$p\in (1,\infty )$
, let
$\lambda \geq 0$
,
$l\geq k$
and
$(p-1)n-p<l<(p-1)n$
. If
$f\in L_{loc}^1(\mathbb {R}^n)$
and satisfies (3.16) and
$u(x,t)=f\ast K_{t^{2\beta }}^\beta (x)$
with
$\beta \geq \frac {1}{2},$
then there holds
$$ \begin{align} &\frac{1}{(\ell(I))^{n\lambda}}\int_{S(I)}|\nabla_{(x,t)} u(x,t)|^pK_{\ln}^{(l,k)}\left(\frac{t}{\ell(I)}\right)\frac{dxdt}{t^{(p-1)(n-1)}}\\ &\lesssim\frac{1}{(\ell(J))^{n\lambda}}\int_{|y|<\ell(J)}\int_J\frac{|f(x+y)-f(x)|^p}{|y|^{pn}}K_{\ln}^{(l,k)}\left(\frac{|y|}{\ell(J)}\right)dxdy\nonumber\\ &\quad + (\ell(J))^{n(1-\lambda-p)}\int_J|f(x)-f_J|^pdx\nonumber\\ &\quad+(\ell(J))^{n(2-\lambda-p)+p}\left(\int_{\mathbb{R}^n\setminus\frac{2J}{3}}\frac{|f(x)-f_J|}{|x-x_0|^{n+1}}dx\right)^p,\nonumber \end{align} $$
where I and J are cubes centered at
$x_0$
with
$\ell (J)=3\ell (I)$
.
Proof We prove (3.20) following the idea of [Reference Essén, Janson, Peng and Xiao13, Lemma 3.1]. Without loss of generality, assume that
$x_0=0$
. Let
$\vartheta $
be a function such that
$$ \begin{align*} \left\{ \begin{aligned} &0\leq\vartheta\leq1;\\ &\vartheta=1\ \text{on}\ (2/3)J;\\ &\text{supp}\ \vartheta\subseteq(1/4)J;\\ &|\vartheta(x)-\vartheta(y)|\lesssim\ell(J)^{-1}|x-y|,\ \ x,y\in\mathbb{R}^n. \end{aligned} \right. \end{align*} $$
Split
$u(x,t){\kern-1pt}={\kern-1pt}u_1(x,t){\kern-1pt}+{\kern-1pt}u_2(x,t){\kern-1pt}+{\kern-1pt}u_3(x,t)$
, where
$u_1(x,t){\kern-1pt}={\kern-1pt}f_J\ast K_{t^{2\beta }}^\beta (x,t), u_2(x,t)= ((f-f_J)\vartheta )\ast K_{t^{2\beta }}^\beta (x,t), u_3(x,t)=((f-f_J)(1-\vartheta ))\ast K_{t^{2\beta }}^\beta (x,t)$
. Since
$f_J$
is a constant and
$\int _{\mathbb {R}^n}\nabla _{(x,t)}K_{t^{2\beta }}^\beta (x)dx=0$
, we have
$\nabla _{(x,t)} u(x,t)=\nabla _{(x,t)} u_2(x,t)+\nabla _{(x,t)} u_3(x,t)$
. Together with (3.17), we get
$$ \begin{align*} &\int_{S(I)}|\nabla_{(x,t)}u(x,t)|^pK_{\ln}^{(l,k)}\left(\frac{t}{\ell(I)}\right)\frac{dxdt}{t^{(p-1)(n-1)}}\\ &=\int_{S(I)}\bigg|\int_{\mathbb{R}^n}\left(f(x-y)\nabla_{(x,t)}K_{t^{2\beta}}^\beta(y)-f(x)\nabla_{(x,t)}K_{t^{2\beta}}^\beta(y)\right)dy\bigg|^p\frac{K_{\ln}^{(l,k)}\left(\frac{t}{\ell(I)}\right)}{t^{(p-1)(n-1)}}dxdt. \end{align*} $$
It follows from (3.18) that
$|\nabla _{(x,t)}K_{t^{2\beta }}^\beta (x)|\lesssim t^{-(n+1)}$
,
$|\nabla _{(x,t)}K_{t^{2\beta }}^\beta (x)|\lesssim |x|^{-(n+1)}$
and
$$ \begin{align*} \|\nabla_{(x,t)}u(x,t)\|_{L^p(\mathbb{R}^n)}&\lesssim t^{-(n+1)}\int_{|y|\leq t}\|f(\cdot+y)-f(\cdot)\|_{L^p(\mathbb{R}^n)}dy\\ &+\int_{|y|>t}\|f(\cdot+y)-f(\cdot)\|_{L^p(\mathbb{R}^n)}|y|^{-(n+1)}dy. \end{align*} $$
Write
$y=s\zeta \in \mathbb {R}^n$
, where
$s=|y|$
and
$|\zeta |=1$
. Define
$$ \begin{align*}\Gamma(s):=\int_{|\zeta|=1}\|f(\cdot+s\zeta)-f(\cdot)\|_{L^p(\mathbb{R}^n)}d{\zeta}.\end{align*} $$
This gives
$$ \begin{align*} \|\nabla_{(x,t)}u(x,t)\|_{L^p(\mathbb{R}^n)}\lesssim t^{-(n+1)}\int_0^t\Gamma(s)s^{n-1}ds+\int_t^\infty\Gamma(s)s^{-2}ds. \end{align*} $$
In Lemma 3.10, let
$\sigma _1(t):=K_{\ln }^{(l,k)}(t)t^{(1-2p)n-1}$
and
$\tau _1(t):=K_{\ln }^{(l,k)}(t)t^{(1-2p)n-1+p}. $
Since
$$ \begin{align*} &\left(\int_s^1 \sigma_1(t)dt\right)^{\frac{1}{p}}\left(\int_0^s\left(\tau_1(t)\right)^{\frac{1}{1-p}}dt\right)^{1-\frac{1}{p}}\\ &=\left(\int_s^1 K_{\ln}^{(l,k)}(t)t^{(1-2p)n-1}dt\right)^{\frac{1}{p}}\left(\int_0^s\left(K_{\ln}^{(l,k)}(t)t^{(1-2p)n-1+p}\right)^{\frac{1}{1-p}}dt\right)^{1-\frac{1}{p}}\\ &\lesssim \sup\limits_{0<s<1}\left(\int_s^1 K_{\ln}^{(l,k)}(t)t^{(1-p)n-1}dt\right)^{\frac{1}{p}}\left(\int_0^s(K_{\ln}^{(l,k)}(t))^{\frac{1}{1-p}}t^{n-1}dt\right)^{1-\frac{1}{p}}<\infty, \end{align*} $$
we have
$$ \begin{align} \sup\limits_{0<s<1}\left(\int_s^1 \sigma_1(t)\right)^{\frac{1}{p}}\left(\int_0^s\left(\tau_1(t)\right)^{\frac{1}{1-p}}dt\right)^{1-\frac{1}{p}}<\infty. \end{align} $$
Let
$\sigma _2(t):=K_{\ln }^{(l,k)}(t)t^{(p-1)(1-n)}$
and
$\tau _2(t):=K_{\ln }^{(l,k)}(t)t^{n(1-p)-1+2p}.$
By Lemma 3.7, we have
$$ \begin{align} \sup\limits_{0<s<\infty}\left(\int_0^s\sigma_2(t)dt\right)^{\frac{1}{p}}\left(\int_s^\infty \left(\tau_2(t)\right)^{\frac{1}{1-p}}dt\right)^{1-\frac{1}{p}}<\infty. \end{align} $$
By (3.22) and (3.21), it can be deduced from Lemma 3.10 that
$$ \begin{align*} &\frac{1}{(\ell(I))^{n\lambda}}\int_0^{\ell(I)}\|\nabla_{(x,t)}u(\cdot,t)\|_{L^p(\mathbb{R}^n)}^pK_{\ln}^{(l,k)}\left(\frac{t}{\ell(I)}\right)\frac{dt}{t^{(p-1)(n-1)}}\\ &\lesssim\frac{1}{(\ell(I))^{n\lambda}}\int_0^{\ell(I)}\left\{\left(\int_0^t\Gamma(s)s^{n-1}ds\right)^{p}\frac{1}{t^{(2p-1)n+1}}\right.\\ &\left.+\left(\int_t^\infty\Gamma(s)s^{-2}ds\right)^{p}\frac{1}{t^{(p-1)(n-1)}}\right\}K_{\ln}^{(l,k)}\left(\frac{t}{\ell(I)}\right)dt\\ &\approx (\ell(I))^{(1-p-\lambda)n}\int_{0}^{1}\left\{\left(\int_0^t\Gamma(s\ell(I))s^{n-1}ds\right)^p\frac{1}{t^{(2p-1)n+1}}\right.\\ &\left.+\int_0^1\left(\int_t^\infty\Gamma(s\ell(I))s^{-2}ds\right)^{p}\frac{1}{t^{(p-1)(n-1)}}\right\}K_{\ln}^{(l,k)}(t)dt\\ &\lesssim (\ell(I))^{(1-p-\lambda)n}\int_0^\infty\Gamma^p(t\ell(I))K_{\ln}^{(l,k)}(t)\frac{dt}{t^{(p-1)n+1}}\\ &\approx\frac{1}{(\ell(I))^{n\lambda}}\int_0^\infty\Gamma^p(t)K_{\ln}^{(l,k)}\left(\frac{t}{\ell(I)}\right)\frac{dt}{t^{(p-1)n+1}}\\ &\lesssim\frac{1}{(\ell(I))^{n\lambda}}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\frac{|f(x+y)-f(x)|^p}{|y|^{pn}}K_{\ln}^{(l,k)}\left(\frac{|y|}{\ell(I)}\right)dxdy. \end{align*} $$
Next, we deal with
$$ \begin{align*}\frac{1}{(\ell(I))^{n\lambda}}\int_{S(I)}|\nabla_{(x,t)}u_2(x,t)|^pK_{\ln}^{(l,k)}\left(\frac{t}{\ell(I)}\right)\frac{dxdt}{t^{(p-1)(n-1)}}.\end{align*} $$
Similarly to the above estimate for
$ u,$
we can get
$$ \begin{align*} &\frac{1}{(\ell(I))^{n\lambda}}\int_{S(I)}|\nabla_{(x,t)}u_2(x,t)|^pK_{\ln}^{(l,k)}\left(\frac{t}{\ell(I)}\right)\frac{dxdt}{t^{(p-1)(n-1)}}\\ &\lesssim\frac{1}{(\ell(I))^{n\lambda}}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\frac{|f_2(x+y)-f_2(x)|^p}{|y|^{pn}}K_{\ln}^{(l,k)}\left(\frac{|y|}{\ell(I)}\right)dxdy\\ &\lesssim M_1+M_2+M_3, \end{align*} $$
where
$f_2:=(f-f_J)\vartheta $
and
$$ \begin{align*} \left\{ \begin{aligned} &M_1:=\frac{1}{(\ell(J))^{n\lambda}}\int_J\int_J\frac{|f_2(x)-f_2(y)|^p}{|x-y|^{pn}}K_{\ln}^{(l,k)}\left(\frac{|y|}{\ell(J)}\right)dxdy;\\ &M_2:=\frac{1}{(\ell(J))^{n\lambda}}\int_{y\in(3/4)J}\int_{x\notin J}\frac{|f_2(x)-f_2(y)|^p}{|x-y|^{pn}}K_{\ln}^{(l,k)}\left(\frac{|y|}{\ell(J)}\right)dxdy;\\ &M_3:=\frac{1}{(\ell(J))^{n\lambda}}\int_{y\notin J}\int_{x\in(3/4)J}\frac{|f_2(x)-f_2(y)|^p}{|x-y|^{pn}}K_{\ln}^{(l,k)}\left(\frac{|y|}{\ell(J)}\right)dxdy. \end{aligned} \right. \end{align*} $$
Since
$0\leq \vartheta \leq 1$
and
$|\vartheta (x)-\vartheta (y)|\lesssim |x-y|/\ell (J), x,y\in \mathbb {R}^n,$
there holds
$$ \begin{align*}|f_2(x)-f_2(y)|\lesssim |f(x)-f(y)|+\ell(J)^{-1}|x-y||f(y)-f_J|.\end{align*} $$
Thus,
$$ \begin{align*} M_1&\lesssim \frac{1}{(\ell(J))^{n\lambda}}\int_{|y|<\ell(J)}\int_{J}\frac{|f(x+y)-f(x)|^p}{|y|^{pn}}K_{\ln}^{(l,k)}\left(\frac{|y|}{\ell(J)}\right)dxdy\\ &\quad+(\ell(J))^{-(p+n\lambda)}\int_{|y|<\ell(J)}\int_{J}\frac{|f(x+y)-f_J|^p}{|y|^{p(n-1)}}K_{\ln}^{(l,k)}\left(\frac{|y|}{\ell(J)}\right)dxdy\\ &\lesssim\frac{1}{(\ell(J))^{n\lambda}}\int_{|y|<\ell(J)}\int_{J}\frac{|f(x+y)-f(x)|^p}{|y|^{pn}}K_{\ln}^{(l,k)}\left(\frac{|y|}{\ell(J)}\right)dxdy\\ &\quad+(\ell(J))^{-(p+n\lambda)}\int_{|y|<\ell(J)}\int_{J}\frac{|f(x)-f_J|^p}{|y|^{p(n-1)}}K_{\ln}^{(l,k)}\left(\frac{|y|}{\ell(J)}\right)dxdy\\ &\lesssim\frac{1}{(\ell(J))^{n\lambda}}\int_{|y|<\ell(J)}\int_{J}\frac{|f(x+y)-f(x)|^p}{|y|^{pn}}K_{\ln}^{(l,k)}\left(\frac{|y|}{\ell(J)}\right)dxdy\\ &\quad+(\ell(J))^{n(1-\lambda-p)}\int_0^{\sqrt{n}}\frac{K_{\ln}^{(l,k)}(t)}{t^{(p-1)(n-1)}}dt\int_J|f(x)-f_J|^{p}dx. \end{align*} $$
Since
$l>(p-1)(n-1)-1$
, we can get
$$ \begin{align*} \int_0^{\sqrt{n}}\frac{K_{\ln}^{(l,k)}(t)}{t^{(p-1)(n-1)}}dt=\int_0^{\sqrt{n}}t^{l-(p-1)(n-1)}\left(\ln\frac{e\sqrt{n}}{t}\right)^kdt<\infty \end{align*} $$
and
$$ \begin{align*} M_1&\lesssim\frac{1}{(\ell(J))^{n\lambda}}\int_{|y|<\ell(J)}\int_{J}\frac{|f(x+y)-f(x)|^p}{|y|^{pn}}K_{\ln}^{(l,k)}\left(\frac{|y|}{\ell(J)}\right)dxdy\\ &+(\ell(J))^{n(1-\lambda-p)}\int_J|f(x)-f_J|^{p}dx. \end{align*} $$
For
$M_2$
, since
$x\notin J$
,
$y\in (3/4)J$
and
$l<(p-1)n$
, we can get
$$ \begin{align*}|x-y|>(1/8)\ell(J)\quad \text{and} \quad |f_2(x)-f_2(y)|\lesssim|f(y)-f_J|,\end{align*} $$
which implies that
$M_2$
can be controlled by
$$ \begin{align*} &\frac{1}{(\ell(J))^{n\lambda}}\Bigg(\int_{y\in(3/4)J}|f(y)-f_J|^pdy\Bigg) \Bigg(\int_{|z|>(1/8)\ell(J)}K_{\ln}^{(l,k)}\left(\frac{|z|}{\ell(J)}\right)|z|^{-pn}dz\Bigg)\\ &\lesssim(\ell(J))^{n(1-p-\lambda)}\Bigg(\int_{1/8}^{\infty}K_{\ln}^{(l,k)}(t)\frac{dt}{t^{(p-1)n+1}}\Bigg)\Bigg(\int_J|f(y)-f_J|^pdy\Bigg)\\ &=(\ell(J))^{n(1-p-\lambda)}\left\{\int_{1/8}^{\sqrt{n}}t^{l-(p-1)n-1}\left(\ln\left(\frac{e\sqrt{n}}{t}\right)\right)^kdt\right.\\ &\left.+\int_{\sqrt{n}}^\infty t^{l-(p-1)n-1}dt\right\}\int_J|f(y)-f_J|^pdy\\ &\lesssim(\ell(J))^{n(1-p-\lambda)}\int_J|f(y)-f_J|^pdy. \end{align*} $$
Similarly to
$M_{2}$
, for
$M_{3}$
, we obtain
$$ \begin{align*}M_3\lesssim(\ell(J))^{n(1-p-\lambda)}\int_J|f(y)-f_J|^pdy.\end{align*} $$
If
$(x,t)\in S(I)$
and
$y\in \mathbb {R}^n\setminus (2/3)J$
, then
$$ \begin{align*}|\nabla_{(x,t)}K_{t^{2\beta}}^\beta(x)|\lesssim\frac{1}{t^{n+1}+|x|^{n+1}}\lesssim|x|^{-(n+1)}\end{align*} $$
and
$$ \begin{align*} &\frac{1}{(\ell(I))^{n\lambda}}\int_{S(I)}|\nabla_{(x,t)}u_3(x,t)|^pK_{\ln}^{(l,k)}\left(\frac{t}{\ell(I)}\right)\frac{dxdt}{t^{(p-1)(n-1)}}\\ &\lesssim\frac{1}{(\ell(I))^{n\lambda}}\int_{S(I)}\left(\int_{\mathbb{R}^n}|\nabla_{(x,t)} K_{t^{2\beta}}^\beta(x-y)||f(y)-f_J|dy\right)^pK_{\ln}^{(l,k)}\left(\frac{t}{\ell(I)}\right)\frac{dxdt}{t^{(p-1)(n-1)}}\\ &\lesssim\frac{1}{(\ell(I))^{n\lambda}}\int_{S(I)}\left(\int_{\mathbb{R}^n\setminus(2/3)J}\frac{|f(y)-f_J|}{|y|^{n+1}}dy\right)^pK_{\ln}^{(l,k)}\left(\frac{t}{\ell(I)}\right)\frac{dxdt}{t^{(p-1)(n-1)}}\\ &\lesssim(\ell(I))^{n(2-\lambda-p)+p}\left(\int_0^1\frac{K_{\ln}^{(l,k)}(t)}{t^{(p-1)(n-1)}}dt\right)\left(\int_{\mathbb{R}^n\setminus(2/3)J}\frac{|f(y)-f_J|}{|y|^{n+1}}dy\right)^p\\ &\lesssim(\ell(I))^{n(2-\lambda-p)+p}\left(\int_{\mathbb{R}^n\setminus(2/3)J}\frac{|f(y)-f_J|}{|y|^{n+1}}dy\right)^p, \end{align*} $$
which completes the proof of Lemma 3.14.
Now, by the idea of [Reference Essén, Janson, Peng and Xiao13, Theorem 3.2], we give the characterization of
$Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
via fractional heat semigroups
$\{e^{-t(-\Delta )^{\beta }}\}_{t>0}$
.
Theorem 3.15 For any
$k\in \mathbb N$
and
$p\in [n/(n-1),2n/(n-1))$
, suppose that
$$ \begin{align*}0\leq\lambda<2-p+\frac{p}{n},\quad l\geq k\quad \text{and}\quad (p-1)n-p<l<(p-1)n.\end{align*} $$
Let
$f\in L_{loc}^p(\mathbb {R}^n)$
satisfying (3.16). Denote
$u(x,t)=f\ast K_{t^{2\beta }}^\beta (x)$
with
$\beta \geq \frac {1}{2}.$
There holds
$$ \begin{align*}f\in Q_{\ln,\lambda}^{p,k,l}(\mathbb{R}^n)\Longleftrightarrow u(\cdot,\cdot)\in \mathscr H_{K_{\ln}^{(l,k)}}^{p,\lambda}(\mathbb R_+^{n+1}).\end{align*} $$
Proof First, let
$f\in Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
. Let I and J be two cubes centered at the
$x_0$
such that
$\ell (J)=3\ell (I)$
. Without loss of generality, assume that
$x_0=0$
. Applying H
$\ddot {\text {o}}$
lder’s inequality, we can get
$$ \begin{align*} |f_{2^{m+1}J}-f_{2^mJ}|^p\lesssim2^{m+1}J|^{\lambda+p-2}\|f\|_{L^{p,p+\lambda-1}(\mathbb{R}^n)}^p,\ \ \ m\in\mathbb N, \end{align*} $$
and
$$ \begin{align*} \left(\int_{2^{k+1}J}|f_{2^{k+1}J}-f_J|dx\right)^p&\leq|2^{k+1}J|^{p-1}\int_{2^{k+1}J}|f_{2^{k+1}J}-f_J|^pdx\\ &\lesssim |2^{k+1}J|^{\lambda+2p-2}\|f\|_{L^{p,p+\lambda-1}(\mathbb{R}^n)}^p. \end{align*} $$
Notice that
$$ \begin{align*} \left(\int_{2^{k+1}J}|f(x)-f_{2^{k+1}J}|dx\right)^p &\leq|2^{k+1}J|^{p-1}\int_{2^{k+1}J}|f(x)-f_{2^{k+1}J}|^pdx\\ &\approx|2^{k+1}J|^{2p-2+\lambda} \|f\|_{L^{p,p+\lambda-1}(\mathbb{R}^n)}^p. \end{align*} $$
A direct calculation gives
$$ \begin{align*} \left(\int_{\mathbb{R}^n\setminus(2/3)J}\frac{|f(x)-f_J|}{|x|^{n+1}}dx\right)^p &\lesssim\left(\sum\limits_{k=0}^\infty\int_{2^k\ell(J)\leq|x|\leq2^{k+1}\ell(J)}\frac{|f(x)-f_J|}{|x|^{n+1}}dx\right)^p\\ &\lesssim\left(\sum\limits_{k=0}^\infty(2^k\ell(J))^{-(n+1)}\int_{2^{k+1}\ell(J)}|f(x)-f_{2^{k+1}J}|dx\right)^p\\ &+\left(\sum\limits_{k=0}^\infty(2^k\ell(J))^{-(n+1)}\int_{2^{k+1}\ell(J)}|f_{2^{k+1}J}-f_J|dx\right)^p\\ &\lesssim\ell(J)^{n(p-2+\lambda)-p}\|f\|_{L^{p,p+\lambda-1}(\mathbb{R}^n)}^p. \end{align*} $$
Theorem 2.8 implies
$f\in L^{p,p+\lambda -1}(\mathbb {R}^n)$
and
$\|f\|_{L^{p,p+\lambda -1}(\mathbb {R}^n)}\lesssim \|f\|_{Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)}$
. Lemma 3.14 and Theorem 3.5 indicate that
$$ \begin{align*} &\frac{1}{(\ell(I))^{n\lambda}}\sup\limits_I\int_{S(I)}|\nabla_{(x,t)}u(x,t)|^pK_{\ln}^{(l,k)}\left(\frac{t}{\ell(I)}\right)\frac{dxdt}{t^{(p-1)(n-1)}}\\ &\lesssim\|f\|_{Q_{\ln,\lambda}^{p,k,l}(\mathbb{R}^n)}^p+\|f\|_{L^{p,p+\lambda-1}(\mathbb{R}^n)}^p\\ &\lesssim\|f\|_{Q_{\ln,\lambda}^{p,k,l}(\mathbb{R}^n)}^p. \end{align*} $$
Conversely, if
$u(\cdot ,\cdot )\in \mathscr H_{K_{\ln }^{(l,k)}}^{p,\lambda }(\mathbb R_+^{n+1})$
, we will prove that
$$ \begin{align} \frac{1}{(\ell(I))^{n\lambda}}\int_{|y|<\ell(I)}\int_I\frac{|f(x+y)-f(x)|^p}{|y|^{pn}}K_{\ln}^{(l,k)}\left(\frac{|y|}{\ell(I)}\right)dxdy<\infty. \end{align} $$
It is easy to see that
$|f(x+y)-f(x)|$
can be controlled by
$$ \begin{align*} &|f(x+y)-u(x+y,|y|)|+|u(x+y,|y|)-u(x,|y|)|+|u(x,|y|)-f(x)|\\ &:=U_1+U_2+U_3. \end{align*} $$
For
$U_3$
, notice that
$U_3=|u(x,|y|)-u(x,0)|\leq \int _0^{|y|}|\nabla _ru(x,r)|dr.$
Similarly to Lemma 3.14, we can apply Lemma 3.10 and Minkowski’s inequality to deduce that
$$ \begin{align*} \left(\int_I|U_3|^pdx\right)^{\frac{1}{p}}\leq\left(\int_I\left(\int_0^{|y|}|\nabla_ru(x,r)|dr\right)^pdx\right)^{\frac{1}{p}}\leq\int_0^{|y|}\left(\int_I|\nabla_ru(x,r)|^pdx\right)^{\frac{1}{p}}dr \end{align*} $$
and
$$ \begin{align*} &\frac{1}{(\ell(I))^{n\lambda}}\int_{|y|<\ell(I)}\left(\int_I|U_3|^pdx\right)|y|^{-pn}K_{\ln}^{(l,k)}\left(\frac{|y|}{\ell(I)}\right)dy\\ &\lesssim\frac{1}{(\ell(I))^{n\lambda}}\int_0^{\sqrt{n}\ell(I)}\left(\int_0^r\left(\int_I|\nabla_ru(x,r)|^pdx\right)^{\frac{1}{p}}dr\right)^pK_{\ln}^{(l,k)}\left(\frac{t}{\ell(I)}\right)t^{(1-p)n-1}dt\\ &\lesssim\frac{1}{(\ell(I))^{n\lambda}}\int_{S(I)}|\nabla_{(x,t)}u(x,t)|^pK_{\ln}^{(l,k)}\left(\frac{t}{\ell(I)}\right)\frac{dxdt}{t^{(p-1)(n-1)}}<\infty. \end{align*} $$
According to the fact,
$$ \begin{align*}\int_I|U_1|^pdx=\int_{I+y}|U_3|^pdx\leq\int_{3I}|U_3|^pdx\end{align*} $$
for any y with
$|y|<\ell (I),$
we can get
$$ \begin{align*}\frac{1}{(\ell(I))^{n\lambda}}\int_{|y|<\ell(I)}\left(\int_I|U_1|^pdx\right)|y|^{-pn}K_{\ln}^{(l,k)}\left(\frac{|y|}{\ell(I)}\right)dy<\infty.\end{align*} $$
For
$U_2$
, it is easy to see that
$$ \begin{align*}U_2\lesssim\int_0^{|y|}\left|\nabla_ru\left(x+r\frac{y}{|y|},|y|\right)\right|dr.\end{align*} $$
It follows from Minkowski’s inequality that
$$ \begin{align*} \left(\int_I|U_2|^pdx\right)^{\frac{1}{p}}&\lesssim\left(\int_I\bigg|\int_0^{|y|}\bigg|\nabla_ru\left(x+r\frac{y}{|y|},|y|\right)\bigg|dr\bigg|^pdx\right)^{\frac{1}{p}}\\ &\lesssim|y|\left(\int_{3I}|\nabla_ru(x,|y|)|^pdx\right)^{\frac{1}{p}}. \end{align*} $$
Therefore, it yields
$$ \begin{align*} &\frac{1}{(\ell(I))^{n\lambda}}\int_{|y|<\ell(I)}\left(\int_I|U_2|^pdx\right)|y|^{-pn}K_{\ln}^{(l,k)}\left(\frac{|y|}{\ell(I)}\right)dy\\ &\lesssim\frac{1}{(\ell(I))^{n\lambda}}\sup\limits_I\int_{S(I)}|\nabla_{(x,t)}u(x,t)|^pK_{\ln}^{(l,k)}\left(\frac{t}{\ell(I)}\right)\frac{dxdt}{t^{(p-1)(n-1)}}<\infty. \end{align*} $$
Hence, the inequality (3.23) holds, and this completes the proof of Theorem 3.15.
Following the procedure of Theorem 3.15, we can deduce the following result.
Theorem 3.16 For any
$k\in \mathbb N$
and
$p\in [n/(n-1),2n/(n-1))$
, suppose that
$$ \begin{align*}0\leq\lambda<2-p+\frac{p}{n}, \quad l\geq k\quad \text{and} \quad(p-1)n-p<l<(p-1)n.\end{align*} $$
Let
$f\in L_{loc}^p(\mathbb {R}^n)$
satisfying (3.16). Denote
$u(x,t)=f\ast K_{t^{2\beta }}^\beta (x), (x,t)\in \mathbb R_+^{n+1}.$
The following statements are equivalent:
$$ \begin{align*} \left\{ \begin{aligned} &\sup\limits_{I\subset\mathbb{R}^n}\frac{1}{(\ell(I)) ^{l+n\lambda}}\int_I\int_I\frac{|f(x)-f(y)|^p}{|x-y|^{pn-l}}\left(\ln\left(\frac{e\sqrt{n}\ell(I)}{|x-y|}\right)\right)^kdxdy<\infty;\\ &\sup\limits_{(y,r)\in\mathbb R_+^{n+1}}\int_{\mathbb R_+^{n+1}}K_{\ln}^{(l,k)}\left(\frac{tr^{\frac{1}{n}}}{(|x-y|^2+|r+t|^2)^{\frac{n+1}{2n}}}\right)\left(\frac{tr^{\frac{1}{n}}}{(|x-y|^2+|r+t|^2)^{\frac{n+1}{2n}}}\right)^{n\lambda}\\ &\quad\times|\nabla_{x}u(x,t)|^p\frac{dxdt}{t^{(p-1)(n-1)+n\lambda}}<\infty;\\ &\sup\limits_{I\subset\mathbb{R}^n}\int_{S(I)}|\nabla_{x}u(x,t)|^p\left(\frac{t}{\ell(I)}\right)^{n\lambda+l}\left(\ln\left(\frac{e\sqrt{n}\ell(I)}{t}\right)\right)^k\frac{dxdt}{t^{(p-1)(n-1)+n\lambda}}<\infty, \end{aligned}\right. \end{align*} $$
By the aid of Theorems 3.15 and 3.16, we can characterize
$Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
via the following Carleson-type measures related to logarithmic functions.
Definition 3.17 Let
$(k,l,\lambda )\in \mathbb {N}\times (0,\infty ) \times (0,\infty )$
. A positive Borel measure
$\mu (\cdot ,\cdot )$
on
$\mathbb R^{n+1}_{+}$
is said to be a
$(k,l,\lambda )$
-logarithmic Carleson measure if
$$ \begin{align*}\sup_{I}\frac{1}{|I|^{\lambda}}\int_{S(I)}\left(\frac{t}{\ell(I)}\right)^{l}\left(\ln\left(\frac{e\sqrt{n}\ell(I)}{t}\right)\right)^{k}d\mu(x,t)<\infty.\end{align*} $$
The following Carleson measure characterizations can be deduced from Theorems 3.15 and 3.16 immediately.
Corollary 3.18 For any
$k\in \mathbb N$
and
$p\in [n/(n-1),2n/(n-1))$
, suppose that
$$ \begin{align*}0\leq\lambda<2-p+\frac{p}{n}, \quad l\geq k\quad \text{and} \quad(p-1)n-p<l<(p-1)n.\end{align*} $$
-
(i)
$f\in Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
if and only if is a
$$ \begin{align*}d\mu(x,t):=|\nabla_{(x,t)}e^{-t^{2\beta}(-\Delta)^{\beta}}f(x)|^{p}t^{-(p-1)(n-1)}dxdt\end{align*} $$
$(k,l,\lambda )$
-logarithmic Carleson measure.
-
(ii)
$f\in Q_{\ln ,\lambda }^{p,k,l}(\mathbb {R}^n)$
if and only if is a
$$ \begin{align*}d\nu(x,t):=|\nabla_{x}e^{-t^{2\beta}(-\Delta)^{\beta}}f(x)|^{p}t^{-(p-1)(n-1)}dxdt\end{align*} $$
$(k,l,\lambda )$
-logarithmic Carleson measure.
Notice that for
$k\geq 0$
, the term
$$ \begin{align*}\left(\ln\left(\frac{e\sqrt{n}\ell(I)}{|x-y|}\right)\right)^k\geq 1,\end{align*} $$
which, together with Theorem 3.16, for
$0\leq \lambda <2-p+\frac {p}{n}$
,
$l\geq k$
and
$(p-1)n-p<l<(p-1)n$
, indicates the following embedding relation:
$$ \begin{align} &\sup\limits_{I\subset\mathbb{R}^n}\frac{1}{\ell(I)^{n\lambda}}\int_I\int_I\frac{|f(x)-f(y)|^p}{|x-y|^{pn}}\left(\frac{|x-y|}{\ell(I)}\right)^{l}dxdy\\ &\lesssim \sup\limits_{I\subset\mathbb{R}^n}\int_{S(I)}|\nabla_{x}e^{-t^{2\beta}(-\Delta)^{\beta}}f(x)|^{p}\left(\frac{t}{\ell(I)}\right)^{n\lambda+l}\left(\ln\left(\frac{e\sqrt{n}\ell(I)}{t}\right)\right)^k \frac{dxdt}{t^{(p-1)(n-1)+n\lambda}}.\nonumber \end{align} $$
Below we will prove (3.24) still holds for the endpoint cases
$l= (p-1)n-p$
and
$l=(p-1)n$
.
Theorem 3.19 Suppose that f is a measurable function such that (3.16) and
$k\in \mathbb N$
and
$k\geq 2$
. Let
$\beta \geq \frac {1}{2}$
,
$\lambda \geq 0$
and
$$ \begin{align*} \left\{ \begin{aligned} &1<p<k+1,&\ l\leq(p-1)n-p;\\ &1<p<k,&\ l=(p-1)n. \end{aligned} \right. \end{align*} $$
If
$$ \begin{align} \sup\limits_{I}\left(\frac{1}{(\ell(I))^{n\lambda}}\int_{S(I)}|\nabla_x e^{-t^{2\beta}(-\Delta)^\beta}f(x)|^p\frac{\left(\frac{t}{\ell(I)}\right)^{l}\left(\ln\left(\frac{e\ell(I)}{t}\right)\right)^k}{t^{(p-1)(n-1)}}\right)^{\frac{1}{p}}dxdt<\infty, \end{align} $$
then
$$ \begin{align} \sup\limits_{I}\frac{1}{(\ell(I))^{n\lambda}}\int_{|y|<\ell(I)}\int_I\frac{|f(x+y)-f(x)|^p}{|y|^{pn}}\left(\frac{|y|}{\ell(I)}\right)^{l}dxdy<\infty. \end{align} $$
Proof Assume that
$\sigma (x,t)= e^{-t^{2\beta }(-\Delta )^\beta }f(x)$
and
$\sigma (x,0)=f(x)$
. Write
$$ \begin{align*}|f(x+y)-f(x)|\leq D_1+D_2,\end{align*} $$
where
$$ \begin{align*}D_1:=\bigg|\sigma\left(x+\frac{y}{2},\frac{|y|}{2}\right)-\sigma(x,0)\bigg| \quad \text{and}\quad D_2:=\bigg|\sigma\left(x+\frac{y}{2},\frac{|y|}{2}\right)-\sigma(x+y,0)\bigg|.\end{align*} $$
According to the symmetry of
$D_i, i=1,2,$
we only need to verify
$$ \begin{align*} E_1:&=\frac{1}{(\ell(I))^{n\lambda}}\int_{|y|<\ell(I)}\int_I\frac{|D_1|^p}{|y|^{pn}}\left(\frac{|y|}{\ell(I)}\right)^{l}dxdy\\ &=\frac{1}{(\ell(I))^{n\lambda+l}}\int_{|y|<\ell(I)}\left(\int_I|D_1|^pdx\right)\frac{dy}{|y|^{pn-l}}\\ &\lesssim\frac{1}{(\ell(I))^{n\lambda}}\int_{S(I)}|\nabla_x e^{-t^{2\beta}(-\Delta)^\beta}f(x)|^p\left(\frac{t}{\ell(I)}\right)^{l}\left(\ln\left(\frac{e\ell(I)}{t}\right)\right)^k\frac{dxdt}{t^{(p-1)(n-1)}}. \end{align*} $$
Letting
$z=x+\frac {ty}{2}, s=\frac {t|y|}{2}$
, we can deduce that
$$ \begin{align*} D_1&=\left|\int_0^1\frac{d}{dt}\sigma\left(x+\frac{ty}{2},\frac{t|y|}{2}\right)dt\right|\\ &\leq\int_0^1\bigg |\left(\left.\nabla_{z,s}\sigma(z,s)\right |_{(z,s)=\left(x+\frac{ty}{2},\frac{t|y|}{2}\right)}\right)\cdot \left(\frac{y}{2},\frac{|y|}{2}\right)\bigg |dt. \end{align*} $$
Applying Minkowski’s inequality and the change of variables, we obtain
$$ \begin{align*} \left(\int_I|D_1|^pdx\right)^{\frac{1}{p}} &\leq\left(\int_I\left(\int_0^1\bigg|\nabla_{z,s}\sigma(z,s)\cdot\left(\frac{y}{2}, \frac{|y|}{2}\right)\bigg|dt\right)^pdx\right)^{\frac{1}{p}}\\ &\leq\left(\int_I\left(\int_0^1|\partial_{z}\sigma(z,s)|+|\partial_{s}\sigma(z,s)|\frac{|y|}{2}dt\right)^pdx\right)^{\frac{1}{p}}\\ &\leq\int_0^1\left(\int_I\left(|\partial_{z}\sigma(z,s)|+|\partial_{s}\sigma(z,s)|\right)^pdx\right)^{\frac{1}{p}}\frac{|y|}{2}dt\\ &\lesssim\int_0^{|y|/2}\left(\int_I\bigg|\nabla_x\sigma\left(x+s\frac{y}{|y|},s\right)\bigg|^pdx\right)^{\frac{1}{p}}ds. \end{align*} $$
Therefore, we can see that
$$ \begin{align*} E_1&\lesssim\frac{1}{(\ell(I))^{(n\lambda+l)}}\int_{|y|<\ell(I)}\left(\int_0^{|y|/2}\left(\int_{4I}|\nabla_x\sigma(x,t)|^pdx\right)^{\frac{1}{p}}dt\right)^p|y|^{l-pn}dy\\ &\lesssim\frac{1}{(\ell(I))^{(n\lambda+l)}}\int_0^{\ell(I)}\left(\int_0^s\parallel\nabla_x\sigma(\cdot,t)\parallel_{L^p(4I)}dt\right)^ps^{l-pn+n-1}ds\\ &\approx(\ell(I))^{(-\lambda+1-p)n+p}\int_0^1\left(\int_0^s\parallel\nabla_x\sigma(\cdot,t\ell(I))\parallel_{L^p(4I)}dt\right)^ps^{l-pn+n-1}ds. \end{align*} $$
When
$l=(p-1)n$
, since
$1<p<k$
, it follows that
$\lim \limits _{u\rightarrow \infty }(\ln s)u^{p-k-1}=0$
$$ \begin{align*} \lim\limits_{s\rightarrow0}(\ln s)(1-\ln s)^{p-k-1}=\lim\limits_{u\rightarrow -\infty}u(1-u)^{p-k-1}=\lim\limits_{w\rightarrow\infty}(1-w)w^{p-k-1}=0, \end{align*} $$
and
$$ \begin{align*} &\lim_{0<s<1}\left(\int_s^1\frac{1}{t}dt\right)\left(\int_0^s\left(\left(\frac{t}{e}\right)^{p-1}\left(\ln\left(\frac{e}{t}\right)\right)^k\right)^{\frac{1}{1-p}}dt\right)^{p-1}\\ &\approx-(\ln s)\left(-u^{1+k/(1-p)}\bigg|_{1-\ln s}^\infty\right)^{p-1}<\infty. \end{align*} $$
When
$l\leq (p-1)n-p$
, since
$1<p<k+1$
, it follows that
$$ \begin{align*}\lim\limits_{u\rightarrow\infty}(1-s^{(p-1)n-l})u^{p-k-1}=0,\quad \lim\limits_{s\rightarrow0}(1-s^{(p-1)n-l})(1-\ln s)^{p-k-1}=0,\end{align*} $$
and
$$ \begin{align*} &\sup\limits_{0<s<1}\bigg(\int_s^1t^{l-pn+n-1}dt\bigg)\left(\int_0^s \left(\left(\frac{t}{e}\right)^{l-(p-1)(n-1)}\left(\ln\left(\frac{e}{t}\right)\right)^k\right)^{\frac{1}{1-p}}dt\right)^{p-1}\\ &\lesssim (s^{l-(p-1)n}-1)s^{(p-1)n-l}\bigg(-u^{k/(1-p)+1}\bigg|_{1-\ln s}^\infty\bigg)^{p-1}<\infty. \end{align*} $$
Therefore, for
$l=(p-1)n$
and
$l\leq (p-1)n-p$
,
$$ \begin{align*} \sup\limits_{0<s<1}\bigg(\int_s^1t^{l-pn+n-1}dt\bigg)\left(\int_0^s\left(\left(\frac{t}{e}\right)^{l-(p-1)(n-1)}\left(\ln\left(\frac{e}{t}\right)\right)^k\right)^{\frac{1}{1-p}}dt\right)^{p-1}<\infty. \end{align*} $$
By Lemma 3.10, we can get
$$ \begin{align*} E_1&\lesssim(\ell(I))^{(-\lambda+1-p)n+p}\int_0^1\left(\int_0^s\parallel\nabla_x\sigma(\cdot,t\ell(I))\parallel_{L^p(4I)}dt\right)^ps^{l-pn+n-1}ds\\ &\lesssim(\ell(I))^{(-\lambda+1-p)n+p}\int_0^1\parallel\nabla_x\sigma(\cdot,t\ell(I))\parallel_{L^p(4I)}^pt^{l-(p-1)(n-1)}\left(\ln\left(\frac{e}{t}\right)\right)^kdt\\ &\approx\frac{1}{(\ell(I))^{n\lambda}}\int_0^{\ell(I)}\parallel\nabla_x\sigma(\cdot,t)\parallel_{L^p(4I)}^p\left(\frac{t}{\ell(I)}\right)^{l}\left(\ln\left(\frac{e\ell(I)}{t}\right)\right)^{k}\frac{dt}{t^{(p-1)(n-1)}}\\ &\lesssim \frac{1}{(\ell(I))^{n\lambda}}\int_{S(I)}|\nabla_x e^{-t^{2\beta}(-\Delta)^\beta}f(x)|^p\left(\frac{t}{\ell(I)}\right)^{l}\left(\ln\left(\frac{e\ell(I)}{t}\right)\right)^k\frac{dxdt}{t^{(p-1)(n-1)}}, \end{align*} $$
which completes the proof of Theorem 3.19.
From Theorem 3.19, we can deduce the following result.
Corollary 3.20 Let
$\beta \geq \frac {1}{2}, l\leq (p-1)n-p$
,
$\lambda \geq 0$
and
$k\in \mathbb N$
with
$1<p<k+1$
. Suppose that f is a measurable function such that (3.16) holds. If f satisfies (3.25), then f is a constant almost everywhere.
Proof Similarly to the proof of Theorem 2.5, without loss of generality, we assume that f satisfies (3.25) and
$f\in \mathscr C^1(\mathbb {R}^n)$
. By Theorem 3.19, f satisfies (3.26). If f is real-valued and not an identical constant, there exists a point
$x_0$
such that
$\nabla f(x_0)\neq 0$
. According to Definition 1.1 and Lemma 2.1, we can deduce from (3.26) that f is rotation invariant. Thus, we can assume that
$\nabla f(x_0)$
is directed along the positive axis of
$x_1$
. There exists
$\theta>0$
and a small cube I about
$x_0$
such that
$$ \begin{align*} \left\{ \begin{aligned} &\partial f/\partial x_1>2\theta;\\ &|\partial f/\partial x_k|<\theta, k\geq2. \end{aligned} \right. \end{align*} $$
Define
$V=\{x: |x_2|+\cdots +|x_n|<x_1<\ell (I)/2\}$
. If
$x,y\in I$
with
$x-y\in V$
, we have
$f(x)-f(y)>\theta (x_1-y_1)$
and
$$ \begin{align*} &\frac{1}{(\ell(I))^{n\lambda}}\int_{|y|<\ell(I)}\int_I\frac{|f(x+y)-f(x)|^p}{|y|^{pn}}\left(\frac{|y|}{\ell(I)}\right)^{l}dxdy\\ &\geq \frac{1}{(\ell(I))^{l+n\lambda}}\int_{I/2}dx\int_{y\in V}\frac{\theta^py_1^p}{|y|^{pn-l}}dy\\ &\approx \theta^p(\ell(I))^{n-(l+n\lambda)}\int_0^{\ell(I)/2}z_1^{l-(p-1)(n-1)}dz_1=\infty, \end{align*} $$
which is a contradiction, and f is a constant almost everywhere.
Remark 3.21 Corollary 3.20 will be used in Section 4 to study the critical logarithmic Q-type spaces
$Q^{(l,k)}_\beta (\mathbb {R}^n)$
(see Definition 1.2) for the well-posedness of equations (1.1). Specially, for
$p=2$
and
$k\geq 2$
, we can deduce from Theorem 3.19 that
$$ \begin{align*} &\sup\limits_{I}(\ell(I))^{4\beta-4}\int_{|y|<\ell(I)}\int_I\frac{|f(x+y)-f(x)|^{2}}{|y|^{2n}}\left(\frac{|y|}{\ell(I)}\right)^{l+2-2\beta}dxdy\\ &\lesssim \sup\limits_{I}(\ell(I))^{4\beta-4}\int_{S(I)}|\nabla_x e^{-t^{2\beta}(-\Delta)^\beta}f(x)|^{2}\left(\frac{t}{\ell(I)}\right)^{l+2-2\beta} \left(\ln\left(\frac{e\ell(I)}{t}\right)\right)^k\frac{dxdt}{t^{n-1}}. \end{align*} $$
Based on Corollary 3.20, for
$l+2-2\beta \leq n-2$
, any function f satisfying
$$ \begin{align*}\sup\limits_{I}(\ell(I))^{4\beta-4}\int_{S(I)}|\nabla_x e^{-t^{2\beta}(-\Delta)^\beta}f(x)|^{2}\left(\frac{t}{\ell(I)}\right)^{l+2-2\beta} \left(\ln\left(\frac{e\ell(I)}{t}\right)\right)^k\frac{dxdt}{t^{n-1}}<\infty,\end{align*} $$
is a constant function. Hence, in the definition of
$Q^{(l,k)}_\beta (\mathbb {R}^n),$
the index l should satisfy
$l>n+2\beta -4$
.
4 Well-posedness of fractional Naiver–Stokes/MHD equations
In this section, with data in the logarithmic Q spaces, we study the well-posedness of the fractional Naiver–Stokes equations (1.1) and the fractional MHD equations with
$\beta>\frac {1}{2}$
. As a matter of convenience, we first state some basic properties of homogeneous Besov spaces. Let
$\mathscr D(\mathbb {R}^n)$
be the space of polynomials. The dual of
$\mathscr {S}_{0}(\mathbb R^{n})$
is
$$ \begin{align*}\mathscr S_0'(\mathbb{R}^n)=\mathscr S'(\mathbb{R}^n)/\mathscr D(\mathbb{R}^n)=\mathscr S'(\mathbb{R}^n)/\mathscr S^\bot(\mathbb{R}^n).\end{align*} $$
Suppose that
$\varphi $
is a function on
$\mathbb {R}^{n}$
satisfying
$\text { supp }\widehat {\varphi }\subset \{\xi \in \mathbb {R}^{n}: |\xi |\leq 1\}$
and
$\widehat {\varphi }(\xi )=1$
for
$\{\xi \in \mathbb {R}^{n}:\ |\xi |\leq {1}/{2}\}$
. Let
$$ \begin{align*}\varphi_{j}(x)=2^{n(j+1)}\varphi(2^{j+1}x)-2^{nj}\varphi(2^{j}x)\ \ \forall\ j\in \mathbb{Z}\end{align*} $$
be the Littlewood–Paley functions. For
$1\leq p, q\leq \infty $
and
$s\in \mathbb {R}^n$
, the homogeneous Besov space
$\dot {B}_{p,q}^s(\mathbb {R}^n)$
is defined as the set of all
$f\in \mathscr S_0'(\mathbb {R}^n)$
such that
$$ \begin{align*} \|f\|_{\dot{B}_{p,q}^s(\mathbb{R}^n)}=\left\{ \begin{aligned} \left(\sum\limits_{j=-\infty}^\infty(2^{js}\|\varphi_{j}\ast f\|_{L^p(\mathbb{R}^n)})^q\right)^{1/q}<\infty,&\ \ \ \ q<\infty;\\ \sup\limits_{-\infty<j<\infty}2^{js}\|\varphi_j\ast f\|_{L^p(\mathbb{R}^n)}<\infty,&\ \ \ \ q=\infty. \end{aligned} \right. \end{align*} $$
It is well known that for
$1\leq p<\infty $
,
$0<q<\infty $
and
$-\infty <s<\infty $
,
$(\dot {B}_{p,q}^s(\mathbb {R}^n))'=\dot {B}_{p',q'}^{-s}(\mathbb {R}^n)$
, where
$\frac {1}{p}+\frac {1}{p}'=1$
and
$1/q+1/q'=1$
. The following fractional semigroup characterization of
$\dot {B}_{p,q}^s(\mathbb {R}^n)$
is due to Miao et al. [Reference Miao, Yuan and Zhang27].
Proposition 4.1 [Reference Miao, Yuan and Zhang27, Proposition 2.1] Let
$1\leq p,q\leq \infty $
,
$s<0$
and assume that
$n\in \mathbb N$
and
$0<\beta <\infty $
. Then,
$f\in \dot {B}_{p,q}^s(\mathbb {R}^n)$
if and only if
$$ \begin{align*}\left\{\begin{aligned} &\Bigg(\int^{\infty}_{0}(t^{-s/(2\beta)}\|e^{-t(-\Delta)^{\beta}}f\|_{L^p})^{q}\frac{dt}{t}\Bigg)^{1/q},&\ 1\leq q<\infty;\\ &\sup_{t>0}t^{-s/(2\beta)}\|e^{-t(-\Delta)^{\beta}}f\|_{L^p},&\ q=\infty. \end{aligned}\right.\end{align*} $$
4.1 Critical Q-type spaces
By the change of variable, we can get the following equivalent norm of
$\|\cdot \|_{Q^{(l,k)}_\beta (\mathbb {R}^n)}.$
Proposition 4.2 Let
$l>n+2\beta -4$
,
$\beta>\frac {1}{2}$
and
$k\in \mathbb N$
.
$f\in Q^{(l,k)}_\beta (\mathbb {R}^n)$
if and only if
$$ \begin{align*} \sup\limits_{(x,r)\in\mathbb{R}^{n+1}_+}\left(r^{-l+6\beta-6}\int_0^{r^{2\beta}}\int_{|y-x|<r}|\nabla_{y} e^{-s(-\Delta)^\beta}f(y)|^2{\left(\ln\left(\frac{er}{s^{\frac{1}{2\beta}}}\right)\right)^k}\frac{dyds}{s^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\right)^{\frac{1}{2}}<\infty. \end{align*} $$
Proposition 4.3 Let f be a measurable function satisfying (3.16),
$k\in \mathbb N$
with
$$ \begin{align*} \left\{ \begin{aligned} &k\geq2,\quad \beta=1;\\ &k\geq1,\quad \beta\neq1, \end{aligned} \right. \end{align*} $$
and
$$ \begin{align*} \left\{ \begin{aligned} &n+2\beta-4<l\leq n+4\beta-4,\quad \frac{1}{2}<\beta<1;\\ &n+2\beta-4<l\leq n+2\beta-2,\quad \beta\geq1. \end{aligned} \right. \end{align*} $$
If
$f\in Q^{(l,k)}_\beta (\mathbb {R}^n)$
, then
$$ \begin{align*} \sup\limits_I(\ell(I))^{-n+2\beta-2}\int_{|y|<\ell(I)}\int_I|f(x+y)-f(x)|^2\frac{dxdy}{|y|^{n-2\beta+2}}<\infty. \end{align*} $$
Proof Let
$f\in Q^{(l,k)}_\beta (\mathbb {R}^n)$
and
$F(x,t):=e^{-t^{2\beta }(-\Delta )^\beta }f(x)$
. By the triangle inequality, we get
$$ \begin{align*} &|f(x+y)-f(x)|\\ &\leq|F(x+y,|y|)-F(x, |y|)|+|F(x, |y|)-f(x)|+|f(x+y)-F(x+y,|y|)|\\ &:=A_1+A_2+A_3. \end{align*} $$
Now, we prove
$$ \begin{align*} B_i=(\ell(I))^{-n+2\beta-2}\int_{|y|<\ell(I)}\left(\int_I|A_i|^2dx\right)\frac{dy}{|y|^{n+2-2\beta}}<\infty. \end{align*} $$
It is easy to see that
$$ \begin{align*} A_1&=\Big|\int_0^{|y|}\partial_tF(x+te_y, |y|)dt\Big|\leq\int_0^{|y|}\Big|\partial_tF(x+te_y, |y|)\Big|dt\\ &\leq\int_0^{|y|}\Big|\nabla_{(x,t)}F(x+te_y, |y|)\Big|dt. \end{align*} $$
On the other hand, for
$x\in I$
and
$z=x+te_{y}$
,
$$ \begin{align*} |z-x_0|\leq|x-x_0|+t\leq|x-x_0|+|y|\leq3\ell(I), \end{align*} $$
which gives
$$ \begin{align*} \left(\int_I|A_1|^2dx\right)^{\frac{1}{2}}&\leq\left(\int_I\left(\int_0^{|y|}|\nabla_{(x,t)}F(x+te_y, |y|)|dt\right)^2dx\right)^{\frac{1}{2}}\\ &\leq\int_0^{|y|}\left(\int_I|\nabla_{(x,t)}F(x+te_y, |y|)|^2dx\right)^{\frac{1}{2}}dt\\ &\leq\int_0^{|y|}\left(\int_{3I}|\nabla_{(z,t)}F(z, |y|)|^2dz\right)^{\frac{1}{2}}dt\\ &\leq|y|\left(\int_{3I}|\nabla_{(z,t)}F(z, |y|)|^2dz\right)^{\frac{1}{2}}. \end{align*} $$
For
$0<s<\ell (I)$
,
$\ln \left ({e\ell (I)}/{s}\right )=\ln e+\ln \left ({\ell (I)}/{s}\right )\geq 1$
. We have
$$ \begin{align*} B_1&\lesssim(\ell(I))^{-n+2\beta-2}\int_I\left(\int_{3I}|\nabla_{(x,t)}F(x, |y|)|^2dx\right)|y|^{-n+2\beta}dy\\ &\lesssim(\ell(I))^{-n+2\beta-2}\int_0^{\ell(I)}\left(\int_{3I}|\nabla_{(x,t)}F(x, |y|)|^2dx\right)t^{2\beta-1}dt\\ &\lesssim(\ell(I))^{-n+2\beta-2}\int_0^{\ell(I)}\left(\int_{3I}|\nabla_{(x,t)}F(x, |y|)|^2dx\right)\left(\ln\left(\frac{e\ell(I)}{t}\right)\right)^kt^{2\beta-1}dt\\ &\lesssim(\ell(I))^{-l+6\beta-6}\int_0^{\ell(I)}\left(\int_{3I}|\nabla_{(x,t)}F(x, |y|)|^2dx\right)\left(\ln\left(\frac{e\ell(I)}{t}\right)\right)^k\frac{dt}{t^{n-l-3+2\beta}}\\ &\lesssim\|f\|_{Q^{(l,k)}_\beta}^2. \end{align*} $$
Now we turn to estimate
$B_2$
. It holds
$$ \begin{align*} \left(\int_I|A_2|^2dx\right)^{\frac{1}{2}}&=\left(\int_I|F(x, |y|)-F(x,0)|^2dx\right)^{\frac{1}{2}}\\ &=\left(\int_I\bigg|\int_0^{|y|}\partial_tF(x,t)dt\bigg|^2dx\right)^{\frac{1}{2}}\\ &\leq\int_0^{|y|}\left(\int_I|\nabla_{(x,t)}F(x,t)|^2dx\right)^{\frac{1}{2}}dt, \end{align*} $$
which implies
$$ \begin{align*} B_2&\lesssim(\ell(I))^{-n+2\beta-2}\int_{|y|<\ell(I)}\left(\int_0^{|y|}\left(\int_I|\nabla_{(x,t)}F(x,t)|^2dx\right)^{\frac{1}{2}}dt\right)^2\frac{dy}{|y|^{n+2-2\beta}}\\ &\lesssim(\ell(I))^{-n+2\beta-2}\int_0^{\ell(I)}\left(\int_0^{s}\|\nabla_{(x,t)}F(\cdot, t)\|_{L^2(I)}dt\right)^2\frac{ds}{s^{3-2\beta}}. \end{align*} $$
By the change of variables:
$s=u\ell (I)$
and
$t=\ell (I)v$
, we can obtain
$$ \begin{align*} B_2&\lesssim(\ell(I))^{-n+4\beta-4}\int_0^1\left(\int_0^{u\ell(I)}\|\nabla_{(x,t)}F(\cdot, t)\|_{L^2(I)}dt\right)^2\frac{du}{u^{3-2\beta}}\\ &\lesssim(\ell(I))^{4\beta-2-n}\int_0^1\left(\int_0^u\|\nabla_{(x,v\ell(I))}F(\cdot, v\ell(I))\|_{L^2(I)}dv\right)^2\frac{du}{u^{3-2\beta}}. \end{align*} $$
Let
$$ \begin{align*}\mu(t)=t^{2\beta-3} \quad\text{and}\quad \nu(t)=\left(\frac{t}{e}\right)^{l-n+3-2\beta}\left(\ln\left(\frac{e}{t}\right)\right)^k. \end{align*} $$
Letting
$e/t=u$
and
$u=e^w$
, we get
$$ \begin{align*} \int_s^1\mu(t)dt=\left\{ \begin{aligned} &-\ln s,& \ \beta=1;\\ &\frac{1}{2\beta-2}(1-s^{2\beta-2}),& \ \beta\neq1. \end{aligned} \right. \end{align*} $$
On the other hand,
$$ \begin{align*}\int_0^s\nu^{-1}(t)dt=e\int_{\ln(e/s)}^\infty e^{w(l-n+2-2\beta)}w^{-k}dw\lesssim\frac{e(e/s)^{l-n+2-2\beta}}{(\ln(e/s))^{k-1}}.\end{align*} $$
Then, when
$\beta =1$
and
$k\geq 2$
, we get
$$ \begin{align*}\left(\int_s^1\mu(t)dt\right)\left(\int_0^s\nu^{-1}(t)dt\right)\lesssim\frac{-e^{l-n+1}s^{n-l}\ln s}{(1-\ln s)^{k-1}}.\end{align*} $$
It is easy to see that
$$ \begin{align*} \left\{ \begin{aligned} &\lim\limits_{s\rightarrow0^+}\frac{-s^{n-l}\ln s}{(1-\ln s)^{k-1}}=\lim\limits_{s\rightarrow0^+}\frac{s^{n-l}}{(1-\ln s)^{k-2}}-\lim\limits_{s\rightarrow0^+}\frac{s^{n-l}}{(1-\ln s)^{k-1}}=0;\\ &\lim\limits_{s\rightarrow1^-}\frac{-s^{n-l}\ln s}{(1-\ln s)^{k-1}}=0. \end{aligned} \right. \end{align*} $$
When
$\beta \neq 1$
and
$k\geq 1$
,
$$ \begin{align*}\left(\int_s^1\mu(t)dt\right)\left(\int_0^s\nu^{-1}(t)dt\right)\lesssim\frac{e^{l-n+3-2\beta}s^{n-l+2\beta-2}(1-s^{2\beta-2})}{(2\beta-2)(1-\ln s)^{k-1}}.\end{align*} $$
A direct computation gives
$$ \begin{align*} \left\{ \begin{aligned} \lim\limits_{s\rightarrow0^+}\frac{s^{n-l+2\beta-2}(1-s^{2\beta-2})}{(2\beta-2)(1-\ln s)^{k-1}}&=\lim\limits_{s\rightarrow0^+}\frac{s^{n-l+2\beta-2}}{(2\beta-2)(1-\ln s)^{k-1}}\\ &\quad-\lim\limits_{s\rightarrow0^+}\frac{s^{n-l+4\beta-4}}{(2\beta-2)(1-\ln s)^{k-1}}=0;\\ \lim\limits_{s\rightarrow1^-}\frac{s^{n-l+2\beta-2}(1-s^{2\beta-2})}{(2\beta-2)(1-\ln s)^{k-1}}&=0, \end{aligned} \right. \end{align*} $$
which yields
$$ \begin{align*} \sup\limits_{s\in(0,1)}\left(\int_s^1\mu(t)dt\right)\left(\int_0^s\nu^{-1}(t)dt\right)<\infty. \end{align*} $$
Applying Lemma 3.10 and the change of variable
$t\ell (I)=s$
, we obtain
$$ \begin{align*} B_2&\lesssim(\ell(I))^{4\beta-2-n}\int_0^1\|\nabla_{(x,t\ell(I))}F(\cdot,t\ell(I))\|_{L^2(I)}^2\left(\frac{t}{e}\right)^{l-n+3-2\beta}\left(\ln\left(\frac{e}{t}\right)\right)^kdt\\ &\lesssim(\ell(I))^{4\beta-3-n}\int_0^{\ell(I)}\|\nabla_{(x,s)}F(\cdot,s)\|_{L^2(I)}^2\left(\frac{s}{e\ell(I)}\right)^{l-n+3-2\beta}\left(\ln\left(\frac{e\ell(I)}{s}\right)\right)^kds\\ &\lesssim(\ell(I))^{-l+6\beta-6}\int_0^{\ell(I)}\|\nabla_{(x,s)}F(\cdot,s)\|_{L^2(I)}^2\left(\ln\left(\frac{e\ell(I)}{s}\right)\right)^k\frac{ds}{s^{n-l-3+2\beta}}\\ &\lesssim\|f\|_{Q^{(l,k)}_\beta(\mathbb{R}^n)}^2. \end{align*} $$
Now we estimate
$B_3$
. Since
$|y|<\ell (I)$
, then
$x\in I$
implies that
$x+y\in I+y$
. Hence,
$$ \begin{align*} \int_I|A_3|^2dx=\int_I|f(x+y)-F(x+y,|y|)|^2dx=\int_{I+y}|A_2|^2dx\lesssim\int_{3I}|A_2|^2dx \end{align*} $$
and
$$ \begin{align*} B_3=(\ell(I))^{-n+2\beta-2}\int_{|y|<\ell(I)}\left(\int_I|A_3|^2dx\right)\frac{dy}{|y|^{n+2-2\beta}}\lesssim\|f\|_{Q^{(l,k)}_\beta(\mathbb{R}^n)}^2. \end{align*} $$
This completes the proof of Proposition 4.3.
Remark 4.4 Proposition 4.3 enables us to identify the scope of
$\beta $
such that the space
$Q^{(l,k)}_\beta (\mathbb {R}^n)$
is nontrivial. If
$f\in Q^{(l,k)}_\beta (\mathbb {R}^n)$
, it can be deduced from Proposition 4.3 that f satisfies
$$ \begin{align*}\sup\limits_I(\ell(I))^{-n+2\beta-2}\int_{|y|<\ell(I)}\int_I|f(x+y)-f(x)|^2\frac{dxdy}{|y|^{n-2\beta+2}}<\infty.\end{align*} $$
Similarly to [Reference Li and Zhai23, Theorem 3.2], we can prove that
$$ \begin{align*}\sup_{I}(\ell(I))^{4\beta-4-2n}\int_{I}\int_{I}|f(x)-f(y)|^{2}dxdy<\infty,\end{align*} $$
i.e.,
$f\in L^{2, 1+\frac {4(1-\beta )}{n}}(\mathbb R^{n})$
. This means that if f is not a constant a.e., then
$0\leq 1+\frac {4(1-\beta )}{n}\leq 1+\frac {n}{4}$
, equivalently,
$\frac {1}{2}\leq \beta < 1+\frac {n}{4}$
.
Definition 4.5 Let
$l>n+2\beta -4$
,
$\beta>\frac {1}{2}$
and
$k\in \mathbb N$
. The symbol
$Q_{l,k,\beta }^{-1}(\mathbb {R}^n)$
denotes the divergence space of
$Q^{(l,k)}_\beta (\mathbb {R}^n)$
, i.e.,
$$ \begin{align*} Q_{l,k,\beta}^{-1}(\mathbb{R}^n)=\text{div}\bigg(Q^{(l,k)}_\beta(\mathbb{R}^n),\ldots,Q^{(l,k)}_\beta(\mathbb{R}^n)\bigg). \end{align*} $$
In other words,
$f\in Q_{l,k,\beta }^{-1}(\mathbb {R}^n)$
if and only if there are
$f_1,\ldots ,f_n\in Q^{(l,k)}_\beta (\mathbb {R}^n)$
such that
$$ \begin{align*} f(x)=\text{div}(f_1(x),\ldots,f_n(x))=\sum\limits_{j=1}^n\partial_{x_j}f_j(x_1,\ldots,x_n) \end{align*} $$
with
$\|f\|_{Q_{l,k,\beta }^{-1}(\mathbb {R}^n)}=\sum \limits _{j=1}^n\|f_j\|_{Q^{(l,k)}_\beta (\mathbb {R}^n)}$
.
Theorem 4.6 Suppose that
$k\in \mathbb N$
,
$n+2\beta -4<l\leq n+4\beta -4$
,
$\beta>\frac {1}{2}$
. Let f be a tempered distribution on
$\mathbb {R}^n$
. Then,
$f\in Q_{l,k,\beta }^{-1}(\mathbb {R}^n)$
if and only if
$$ \begin{align} &\|f\|_{Q_{l,k,\beta}^{-1},*}:=\\ &\sup\limits_{(x,r)\in\mathbb R_+^{n+1}}\left(r^{-l+6\beta-6}\int_0^{r^{2\beta}}\int_{|y-x|<r}|e^{-t(-\Delta)^\beta}f(y)|^2{\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k}\frac{dydt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\right)^{\frac{1}{2}}<\infty.\nonumber \end{align} $$
Proof Suppose that
$f\in Q_{l,k,\beta }^{-1}(\mathbb {R}^n)$
. It follows that there exist
$f_1,\ldots ,f_n\in Q^{(l,k)}_\beta (\mathbb {R}^n)$
such that
$f(x)=\sum \limits _{j=1}^n\partial _{x_j}f_j(x)$
. Hence,
$$ \begin{align*} &\|f\|_{Q_{l,k,\beta}^{-1},*}\\ &=\sup\limits_{(x,r)\in\mathbb R_+^{n+1}}\left(r^{-l+6\beta-6}\int_0^{r^{2\beta}}\int_{|y-x|<r}|e^{-t(-\Delta)^\beta}\sum\limits_{j=1}^n\partial_{x_j}f_j(y)|^2\frac{\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}dydt\right)^{\frac{1}{2}}\\ &\lesssim\sum\limits_{j=1}^n\sup\limits_{(x,r)\in\mathbb R_+^{n+1}}\left(r^{-l+6\beta-6}\int_0^{r^{2\beta}}\int_{|y-x|<r}|e^{-t(-\Delta)^\beta}\partial_{x_j}f_j(y)|^2\frac{\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}dydt\right)^{\frac{1}{2}}\\ &\lesssim\sum\limits_{j=1}^n\sup\limits_{(x,r)\in\mathbb R_+^{n+1}}\left(r^{-l+6\beta-6}\int_0^{r}\int_{|y-x|<r}|\nabla_{x} e^{-t^{2\beta}(-\Delta)^\beta}f_j(y)|^2\frac{\left(\ln\left(\frac{er}{t}\right)\right)^k}{t^{n-l-3+2\beta}}dydt\right)^{\frac{1}{2}}\\ &=\|f\|_{Q_{l,k,\beta}^{-1}(\mathbb{R}^n)}. \end{align*} $$
Conversely, suppose that (4.1) holds. Set
$f_{j,k}(x):=\partial _j\partial _k(-\Delta )^{-1}f(x)$
. If
$\|f\|_{Q_{l,k,\beta }^{-1},*}<\infty $
,
$f_k=-\partial _k(-\Delta )^{-1}f\in Q^{(l,k)}_\beta (\mathbb {R}^n)$
. In fact, we can see that
$$ \begin{align*} \left(\widehat{\sum\limits_{k=1}^n\partial_kf_k}\right)(\zeta)=-\sum\limits_{k=1}^ni\zeta_k\widehat{f_k}(\zeta)=\sum\limits_{k=1}^ni\zeta_k\cdot i\zeta_k|\zeta|^{-2}\widehat{f}(\zeta)=\widehat{f}(\zeta), \end{align*} $$
which implies that
$f=\sum \limits _{k=1}^n\partial _kf_k\in Q_{l,k,\beta }^{-1}(\mathbb {R}^n)$
.
If
$\|f_{j,k}\|_{Q_{l,k,\beta }^{-1},*}<\infty $
with
$f_{j,k}=\partial _j\partial _k(-\Delta )^{-1}f$
, we have
$$ \begin{align*} &\sup\limits_{(x,r)\in\mathbb R_+^{n+1}}\left(r^{-l+6\beta-6}\int_0^{r^{2\beta}}\int_{|y-x|<r}|e^{-t(-\Delta)^\beta}f_{j,k}(y)|^2\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k\frac{dydt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\right)^{\frac{1}{2}}\\ &=\sup\limits_{(x,r)\in\mathbb R_+^{n+1}}\left(r^{-l+6\beta-6}\int_0^{r}\int_{|y-x|<r}|e^{-t^{2\beta}(-\Delta)^\beta}f_{j,k}(y)|^2\left(\ln\left(\frac{er}{t}\right)\right)^k\frac{dydt}{t^{n-l-3+2\beta}}\right)^{\frac{1}{2}}\\ &=\sup\limits_{(x,r)\in\mathbb R_+^{n+1}}\left(r^{-l+6\beta-6}\int_0^{r}\int_{|y-x|<r}|\partial_je^{-t^{2\beta}(-\Delta)^\beta}(\partial_k(-\Delta)^{-1}f)(y)|^2\frac{\left(\ln\left(\frac{er}{t}\right)\right)^k}{t^{n-l-3+2\beta}}dydt\right)^{\frac{1}{2}} \end{align*} $$
which implies
$$ \begin{align*} &\sup\limits_{(x,r)\in\mathbb R_+^{n+1}}\left(r^{-l+6\beta-6}\int_0^{r}\int_{|y-x|<r}|\partial_je^{-t^{2\beta}(-\Delta)^\beta}(\partial_k(-\Delta)^{-1}f)(y)|^2\frac{\left(\ln\left(\frac{er}{t}\right)\right)^k}{t^{n-l-3+2\beta}}dydt\right)^{\frac{1}{2}} \\ &\quad <\infty, \end{align*} $$
that is,
$f_k\in Q^{(l,k)}_\beta (\mathbb {R}^n)$
.
Now, we prove
$\|f_{j,k}\|_{Q_{l,k,\beta }^{-1},*}<\infty $
. Take a function
$\phi \in C_0^\infty (\mathbb {R}^n)$
such that
$$ \begin{align*} \left\{ \begin{aligned} &\text{ supp }\phi\subset B(0,1)=\{x\in\mathbb{R}^n:|x|<1\};\\ &\int_{\mathbb{R}^n}\phi(x)dx=1;\\ &\phi_r(x)=r^{-n}\phi(x/r). \end{aligned} \right. \end{align*} $$
Then,
$$ \begin{align*} e^{-t(-\Delta)^\beta}f_{j,k}(x)=\partial_j\partial_k(-\Delta)^{-1}e^{-t(-\Delta)^\beta}f(x)=f_r(x,t)+g_r(x,t), \end{align*} $$
where
$g_r(x,t)=\phi _r*\partial _j\partial _k(-\Delta )^{-1}e^{-t(-\Delta )^\beta }f(x)$
. Since
$(\dot {B}_{1,1}^{2\beta -1})^*=\dot {B}_{\infty ,\infty }^{1-2\beta },$
we can get
$$ \begin{align*} \|g_r(\cdot,t)\|_{L^\infty(\mathbb{R}^n)}&\leq\|\phi\|_{\dot{B}_{1,1}^{2\beta-1}}\|\partial_j\partial_k(-\Delta)^{-1}e^{-t(-\Delta)^\beta}f\|_{\dot{B}_{\infty,\infty}^{1-2\beta}}\\ &\lesssim r^{1-2\beta}\|f\|_{\dot{B}_{\infty,\infty}^{1-2\beta}}\lesssim r^{1-2\beta}\|f\|_{Q_{l,k,\beta}^{-1},*}. \end{align*} $$
The above estimate yields
$$ \begin{align*} &r^{-l+6\beta-6}\int_0^{r^{2\beta}}\int_{|y-x|<r}|g_r(y,t)|^2\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k\frac{dydt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\\ &\lesssim\|f\|_{Q_{l,k,\beta}^{-1},*}^2r^{n-l+2\beta-4}\int_0^{r^{2\beta}}t^{l/{(2\beta)}-n/{(2\beta)}+\frac{2}{\beta}-2}\left(\ln\left(\frac{er^2}{t^{\frac{1}{\beta}}}\right)\right)^kdt. \end{align*} $$
Since
$l>n+2\beta -4$
, by the changes of variables
$u=t^{\frac {1}{\beta }}$
,
$er^2/u=v$
and
$v=e^w$
, we can get
$$ \begin{align*} &r^{n-l+2\beta-4}\int_0^{r^{2\beta}}t^{l/{(2\beta)}-n/{(2\beta)}+\frac{2}{\beta}-2}\left(\ln\left(\frac{er^2}{t^{\frac{1}{\beta}}}\right)\right)^kdt\\ &\approx\int_1^\infty e^{(-l/2+n/2+\beta-2)w}w^kdw<\infty \end{align*} $$
and
$$ \begin{align*} &r^{-l+6\beta-6}\int_0^{r^{2\beta}}\int_{|y-x|<r}|g_r(y,t)|^2\left(\ln\left(\frac{er}{t^{\frac{1}{2}\beta}}\right)\right)^k\frac{dydt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\lesssim\|f\|_{Q_{l,k,\beta}^{-1},*}^2. \end{align*} $$
Now we estimate
$f_r$
. Take
$\varphi \in C_0^\infty (\mathbb {R}^n)$
with
$$ \begin{align*} \left\{ \begin{aligned} &\varphi(y)=1,\quad \forall\ y\in \{x\in\mathbb{R}^n:\ |x|<10\};\\ &\varphi_{r,x}(y)=\varphi((y-x)/r);\\ &f_r(y)=F_{r,x}(y,t)+G_{r,x}(y,t), \end{aligned} \right. \end{align*} $$
where
$$ \begin{align*} G_{r,x}(y,t):=\partial_j\partial_k(-\Delta)^{-1}\varphi_{r,x}e^{-t(-\Delta)^\beta}f(y)-\phi_r*\partial_j\partial_k(-\Delta)^{-1}\varphi_{r,x}e^{-t(-\Delta)^\beta}f(y). \end{align*} $$
So
$$ \begin{align*} &r^{-l+6\beta-6}\int_0^{r^{2\beta}}\int_{|y-x|<r}|f_r(y,t)|^2\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k\frac{dydt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\\ &\lesssim r^{-l+6\beta-6}\int_0^{r^{2\beta}}\int_{|y-x|<r}|F_{r,x}(y,t)|^2\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k\frac{dydt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\\ &\quad+r^{-l+6\beta-6}\int_0^{r^{2\beta}}\int_{|y-x|<r}|G_{r,x}(y,t)|^2\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k\frac{dydt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}:=M_1+M_2. \end{align*} $$
We first estimate
$M_2$
. Write
$$ \begin{align*} M_2&\lesssim r^{-l+6\beta-6}\int_0^{r^{2\beta}}\bigg\|\partial_j\partial_k(-\Delta)^{-1}\varphi_{r,x}e^{-t(-\Delta)^\beta}f\bigg\|_{L^2(\mathbb{R}^n)}^2\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k\frac{dt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\\ &\ \ \ \ +r^{-l+6\beta-6}\int_0^{r^{2\beta}}\bigg\|\phi_r*\partial_j\partial_k(-\Delta)^{-1}\varphi_{r,x}e^{-t(-\Delta)^\beta}f\bigg\|_{L^2(\mathbb{R}^n)}^2\frac{\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}dt\\ &:=M_{2,1}+M_{2,2}. \end{align*} $$
Let
$R_j,$
for
$j=1,2,\ldots ,n,$
denote Riesz transforms. By the
$L^2$
- boundedness of
$R_j$
, we can get
$$ \begin{align*} M_{2,1}&\lesssim r^{-l+6\beta-6}\int_0^{r^{2\beta}}\bigg\|R_jR_k(-\Delta)^{-1}\varphi_{r,x}e^{-t(-\Delta)^\beta}f\bigg\|_{L^2(\mathbb{R}^n)}^2\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k\frac{dt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\\ &\lesssim r^{-l+6\beta-6}\int_0^{r^{2\beta}}\bigg\|\varphi_{r,x}e^{-t(-\Delta)^\beta}f\bigg\|_{L^2(\mathbb{R}^n)}^2\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k\frac{dt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}. \end{align*} $$
For
$M_{2,2}$
, notice that
$$ \begin{align*}|\widehat{\phi_r}(\zeta)|=\bigg|\int_{\mathbb{R}^n}r^{-n}\phi(x/r)e^{-ix\zeta}dx\bigg|\leq|\widehat{\phi}(rx)|\leq C.\end{align*} $$
It follows from the Plancherel formula that
$$ \begin{align*} M_{2,2}&\lesssim r^{-l+6\beta-6}\int_0^{r^{2\beta}}\bigg\|\widehat{\phi_r}(\zeta_j\zeta_k|\zeta|^{-2}(\varphi_{r,x}\widehat{e^{-t(-\Delta)^\beta}f}))\bigg\|_{L^2}^2\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k\frac{dydt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\\ &\lesssim r^{-l+6\beta-6}\int_0^{r^{2\beta}}\bigg\|\varphi_{r,x}e^{-t(-\Delta)^\beta}f\bigg\|_{L^2}^2\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k\frac{dydt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}, \end{align*} $$
thereby
$$ \begin{align*} M_2&\lesssim r^{-l+6\beta-6}\int_0^{r^{2\beta}}\bigg\|\varphi_{r,x}e^{-t(-\Delta)^\beta}f\bigg\|_{L^2}^2\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k\frac{dydt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\lesssim\|f\|_{Q_{l,k,\beta}^{-1},*}^2. \end{align*} $$
Now, we estimate
$M_1$
. Since
$$ \begin{align*} &\int_{|y-x|<r}|F_{r,x}(y,t)|^2dy\lesssim r^{n+1}\int_{|w-x|\geq10r}\bigg|e^{-t(-\Delta)^\beta}f(w)\bigg|^2\frac{dw}{|x-w|^{n+1}}, \end{align*} $$
we can see that
$$ \begin{align*} M_1&\lesssim r^{-l+6\beta-6}\int_0^{r^{2\beta}}\left(\int_{B(x,r)}|F_{r,x}(y,t)|^2dy\right)\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k\frac{dt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\\ &\lesssim r^{-l+6\beta-6}\int_0^{r^{2\beta}}\left(r^{n+1}\int_{|w-x|\geq10r}\bigg|e^{-t(-\Delta)^\beta}f(w)\bigg|^2\frac{dw}{|x-w|^{n+1}}\right)\frac{\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}dt\\ &\lesssim r^{n-l+6\beta-5}\int_{|w-x|\geq10r}\left(\int_0^{r^{2\beta}}\bigg|e^{-t(-\Delta)^\beta}f(w)\bigg|^2\frac{\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}dt\right)\frac{dw}{|x-w|^{n+1}}\\ &=\sum\limits_{k=1}^\infty M_{1,k}, \end{align*} $$
where
$M_{1,k}$
is defined as
$$ \begin{align*} r^{n-l+6\beta-5}\int_{10^kr\leq|w-x|<10^{k+1}r}\int_0^{r^{2\beta}}\bigg|e^{-t(-\Delta)^\beta}f(w)\bigg|^2 \ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)^k\frac{dtdw}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}|x-w|^{n+1}}. \end{align*} $$
Notice that
$\left (\ln \left (\frac {er}{t^{\frac {1}{2\beta }}}\right )\right )^k=\left (\ln \left (\frac {e10^{k+1}r}{10^{k+1}t^{\frac {1}{2\beta }}}\right )\right )^k\lesssim \left (\ln \left (\frac {e10^{k+1}r}{t^{\frac {1}{2\beta }}}\right )\right )^k, $
we obtain
$$ \begin{align*} M_{1,k} &\lesssim\frac{r^{n-l+6\beta-5}}{(10^kr)^{n+1}}\int_{|x-w|<10^{k+1}r}\int_0^{(10^{k+1}r)^{2\beta}}\bigg|e^{-t(-\Delta)^\beta}f(w)\bigg|^2\frac{\left(\ln\left(\frac{er}{t^{(\frac{1}{2}\beta)}}\right)\right)^k}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}dtdw\\ &\lesssim10^{k(l-n-6\beta+5)}(10^{k+1}r)^{-l+6\beta-6}\int_0^{(10^{k+1}r)^{2\beta}}\int_{|x-w|<10^{k+1}r}\bigg|e^{-t(-\Delta)^\beta}f(w)\bigg|^2\\ &\quad\times\frac{\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}dtdw. \end{align*} $$
Since
$\beta>\frac {1}{2}$
and
$l\leq n+4\beta -4$
,
$M_1\lesssim \sum \limits _{k=1}^\infty 10^{k(l-n-6\beta +5)}\|f\|_{Q_{l,k,\beta }^{-1},*}^2. $
This completes the proof of Theorem 4.6.
Theorem 4.7 Let
$k\in \mathbb N$
,
$l>n+4\beta -4$
and
$\beta \in (\frac {1}{2},1)$
.
$$ \begin{align*}Q^{-1}_{l,k,\beta}(\mathbb R^{n})=\dot{B}^{1-2\beta}_{\infty,\infty}(\mathbb R^{n}).\end{align*} $$
Proof It is easy to see that
$Q^{-1}_{l,k,\beta }(\mathbb R^{n})$
is invariant under the dilation:
$f_{\delta }(x)=\delta ^{2\beta -1}f(\delta x)$
. By [Reference Li and Zhai23, Proposition 4.6],
$Q^{-1}_{l,k,\beta }(\mathbb R^{n})\subseteq \dot {B}^{1-2\beta }_{\infty ,\infty }(\mathbb R^{n})$
. Now, we assume that
$f\in \dot {B}^{1-2\beta }_{\infty ,\infty }(\mathbb R^{n})$
. Then, for any
$(x,t)\in \mathbb R^{n+1}_{+}$
, it follows from Proposition 4.1 that
$$ \begin{align*}|e^{-t(-\Delta)^{\beta}}f(x)|\leq t^{(1-2\beta)/(2\beta)}\|f\|_{\dot{B}^{1-2\beta}_{\infty,\infty}},\end{align*} $$
which, together with Theorem 4.6, gives
$$ \begin{align*} \|f\|_{Q^{-1}_{l,k,\beta}}&\lesssim\sup\limits_{(x,r)\in\mathbb R_+^{n+1}}\left(r^{-l+6\beta-6}\int_0^{r^{2\beta}}\int_{|y-x|<r}|e^{-t(-\Delta)^\beta}f(y)|^{2} \frac{\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}dydt\right)^{\frac{1}{2}}\\ &\lesssim\|f\|_{\dot{B}^{1-2\beta}_{\infty,\infty}}\sup\limits_{(x,r)\in\mathbb R_+^{n+1}}\left(r^{-l+6\beta-6}\int_0^{r^{2\beta}}\int_{|y-x|<r}t^{\frac{1}{\beta}-2} \frac{\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}dydt\right)^{\frac{1}{2}}\\ &\lesssim\|f\|_{\dot{B}^{1-2\beta}_{\infty,\infty}}\sup\limits_{(x,r)\in\mathbb R_+^{n+1}}\left(r^{n-l+6\beta-6}\int_{0}^{r^{2\beta}}t^{\frac{1}{\beta}-2} \left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^{k}\frac{dt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\right)^{\frac{1}{2}}. \end{align*} $$
Letting
$s=t^{\frac {1}{2\beta }}$
and
$er/s=v$
, we can get
$$ \begin{align*} \|f\|_{Q^{-1}_{l,k,\beta}}&\lesssim\|f\|_{\dot{B}^{1-2\beta}_{\infty,\infty}}\sup\limits_{(x,r)\in\mathbb R_+^{n+1}}\left(r^{n-l+6\beta-6}\int_{0}^{r}s^{2-4\beta} \left(\ln\left(\frac{er}{s}\right)\right)^{k}\frac{s^{2\beta-1}ds}{s^{n-l+4\beta-4}}\right)^{\frac{1}{2}}\\ &\lesssim\|f\|_{\dot{B}^{1-2\beta}_{\infty,\infty}}\sup\limits_{(x,r)\in\mathbb R_+^{n+1}}\left(r^{n-l+6\beta-6}\left(\frac{1}{r}\right)^{n-l+6\beta-6}\int^{\infty}_{e}(\ln v)^{k}v^{n-l+6\beta-7}dv\right)^{\frac{1}{2}}. \end{align*} $$
Since
$\beta \in (\frac {1}{2},1)$
and
$l>n+4\beta -4$
, the integral
$\int ^{\infty }_{e}(\ln v)^{k}v^{n-l+6\beta -7}dv<\infty ,$
which implies that
$\|f\|_{Q^{-1}_{l,k,\beta }}\lesssim \|f\|_{\dot {B}^{1-2\beta }_{\infty ,\infty }}$
, and
$\dot {B}^{1-2\beta }_{\infty ,\infty }(\mathbb R^{n})\subseteq Q^{-1}_{l,k,\beta }(\mathbb R^{n})$
. This completes the proof of Theorem 4.7.
4.2 Quadratic estimates for an integral operator
Lemma 4.8 Given
$n\in \mathbb N_+$
and
$k\in \mathbb N$
. Let
$0<l\leq n+4\beta -4$
,
$\beta>\frac {1}{2}$
and
$$ \begin{align*} A(x,t):=\int_0^te^{-(t-s)(-\Delta)^\beta}(-\Delta)^\beta f(x,s)ds. \end{align*} $$
Then,
$$ \begin{align*} & \int_0^\infty\|A(\cdot,t)\|_{L^2(\mathbb{R}^n)}^2{\left(\ln\left(\frac{e}{t^{\frac{1}{2\beta}}}\right)\right)^k}\frac{dt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}} \\ &\quad \lesssim\int_0^\infty\|f(\cdot, t)\|_{L^2(\mathbb{R}^n)}^2{\left(\ln\left(\frac{e}{t^{\frac{1}{2\beta}}}\right)\right)^k}\frac{dt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}. \end{align*} $$
Proof By Plancherel’s formula and Cauchy–Schwarz’s inequality, we obtain
Noting that
$$ \begin{align*} \int_0^t|\zeta|^{2\beta}e^{-(t-s)|\zeta|^{2\beta}}ds=e^{-t|\zeta|^{2\beta}}\int_0^t|\zeta|^{2\beta}e^{s|\zeta|^{2\beta}}ds =e^{-t|\zeta|^{2\beta}}(e^{t|\zeta|^{2\beta}}-1)\lesssim1, \end{align*} $$
for
$0<s<t$
, we can deduce from
$0<l\leq n+4\beta -4$
that
$$ \begin{align*}\left(\ln\left(\frac{e}{t^{\frac{1}{2\beta}}}\right)\right)^k\frac{1}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\lesssim\left(\ln\left(\frac{e}{s^{\frac{1}{2\beta}}}\right)\right)^k\frac{1}{s^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}.\end{align*} $$
Therefore,
$$ \begin{align*} I_A&\lesssim\int_{\mathbb{R}^n}\left(\int_0^\infty\left(\int_0^t\frac{|\zeta|^{2\beta}}{e^{(t-s)|\zeta|^{2\beta}}}|\widehat{f}(\zeta,s)|^2ds\right)\left(\ln\left(\frac{e}{t^{\frac{1}{2\beta}}}\right)\right)^k\frac{dt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\right)d{\zeta}\\ &\lesssim\int_{\mathbb{R}^n}\int_0^\infty|\widehat{f}(\zeta,s)|^2e^{s|\zeta|^{2\beta}}\left(\int_s^\infty|\zeta|^{2\beta}e^{-t|\zeta|^{2\beta}}dt\right)\left(\ln\left(\frac{e}{s^{\frac{1}{2\beta}}}\right)\right)^k\frac{dsd{\zeta}}{s^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\\ &\lesssim\int_{\mathbb{R}^n}\int_0^\infty|\widehat{f}(\zeta,s)|^2\left(\ln\left(\frac{e}{s^{\frac{1}{2\beta}}}\right)\right)^k\frac{dsd{\zeta}}{s^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\\ &\lesssim\int_0^\infty\|f(\cdot,s)\|_{L^2(\mathbb{R}^n)}^2\left(\ln\left(\frac{e}{s^{\frac{1}{2\beta}}}\right)\right)^k\frac{ds}{s^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}. \end{align*} $$
This completes the proof of Lemma 4.8.
We need the following estimate established by Wang and Xiao [Reference Wang and Xiao32].
Proposition 4.9 [Reference Wang and Xiao32, Lemma 2.1]
For
$\beta>\frac {1}{2}$
, if
$s\in (0,1)$
and
$K_{\beta , s}(\cdot )$
is the kernel of
$(-\Delta )^{1-\beta }(e^{-(-\Delta )^\beta }-e^{-s(-\Delta )^\beta }),$
then
$$ \begin{align*}K_{\beta, s}(x)\lesssim(1+|x|)^{-n-1}+s^{-(2+n-2\beta)/{(2\beta)}}(1+s^{-\frac{1}{2\beta}}|x|)^{-n-1}. \end{align*} $$
Lemma 4.10 For
$k\in \mathbb N$
and
$\beta>\frac {1}{2}$
. Suppose that
$0<l\leq n+4\beta -4$
and
$N(x,t)$
defined on
$(0,1)\times \mathbb {R}^n$
. Let
$A(\beta ,N)$
be the quantity:
$$ \begin{align*} A(\beta,N):=\sup\limits_{x\in\mathbb{R}^n,r\in(0,1)}r^{-l+6\beta-6}\int_0^{r^{2\beta}}\int_{|y-x|<r}|N(x,t)|\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k\frac{dxdt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}. \end{align*} $$
Then,
$$ \begin{align*} &\int_0^1\left\|(-\Delta)^{\frac{1}{2}}e^{-t(-\Delta)^\beta/2}\int_0^tN(\cdot,s)ds\right\|_{L^2(\mathbb{R}^n)}^2\left(\ln\left(\frac{e}{t^{\frac{1}{2\beta}}}\right)\right)^k\frac{dt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\\ &\lesssim A(\beta,N)\int_0^1\int_{\mathbb{R}^n}|N(x,s)|\left(\ln\left(\frac{e}{s^{\frac{1}{2\beta}}}\right)\right)^k\frac{dxds}{s^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}. \end{align*} $$
Proof Let
$L(s):=\frac {\left (\ln \left (\frac {e}{s^{\frac {1}{2\beta }}}\right )\right )^k}{s^{\frac {n-l}{2\beta }+2-\frac {2}{\beta }}}, I=\int _0^1\left \|(-\Delta )^{\frac {1}{2}}e^{-t(-\Delta )^\beta /2}\int _0^tN(\cdot ,s)ds\right \|_{L^2(\mathbb {R}^n)}^2L(t)dt.$
We can see that,
$$ \begin{align*} I&=\int_0^1\bigg\langle\int_0^t(-\Delta)^{\frac{1}{2}}e^{-t(-\Delta)^\beta/2}N(\cdot,s)ds,\ \int_0^t(-\Delta)^{\frac{1}{2}}e^{-t(-\Delta)^\beta/2}N(\cdot,h)dh\bigg\rangle L(t)dt\\ &=2Re\left(\iint_{0<h<s<1}\bigg\langle N(\cdot,s),\ \int_s^1(-\Delta)e^{-t(-\Delta)^\beta}N(\cdot,h)L(t)dt\bigg\rangle dsdh\right)\\ &\lesssim\iint_{0<h<s<1}\bigg\langle|N(\cdot,s)|,\ (-\Delta)^{1-\beta}(e^{-s(-\Delta)^\beta}-e^{-(-\Delta)^\beta})|N(\cdot,h)|\bigg\rangle L(t)dhds\\ &\lesssim\int_0^1\bigg\langle|N(\cdot,s)|,\ (-\Delta)^{1-\beta}\int_0^s(e^{-s(-\Delta)^\beta}-e^{-(-\Delta)^\beta})|N(\cdot,h)|dh\bigg\rangle L(s)ds. \end{align*} $$
By Proposition 4.9, we have
$$ \begin{align*} &\int_0^s(-\Delta)^{1-\beta}(e^{-s(-\Delta)^\beta}-e^{-(-\Delta)^\beta})|N(\cdot,h)|dh\\ &\lesssim\sup\limits_{s\in[0,1]}\int_0^s\int_{\mathbb{R}^n}s^{-\frac{n+2-2\beta}{2\beta}}\frac{|N(y,h)|}{\left(1+\frac{|x-y|}{s^{\frac{1}{2\beta}}}\right)^{n+1}}dydh\\ &\lesssim\sup\limits_{s\in[0,1]}s^{-\frac{n+2-2\beta}{2\beta}}\int_0^s\sum\limits_{l\in\mathbb Z^n}\int_{s^{-\frac{1}{2\beta}}(x-y)\in l+[0,1]^n}\frac{|N(y,h)|}{\left(1+\frac{|x-y|}{s^{\frac{1}{2\beta}}}\right)^{n+1}}dydh\\ &\lesssim\sup\limits_{r\in(0,1),x\in\mathbb{R}^n}r^{-(n+2-2\beta)}r^{-(l+2-2\beta)}r^{l+2-2\beta}\int_0^{r^{2\beta}}\int_{|y-x|<r}|N(y,h)|h^{\frac{n-l}{(2\beta)}-\frac{2}{\beta}+2}\\ &\quad\times\left(\ln\left(\frac{er}{h^{\frac{1}{2\beta}}}\right)\right)^{-k}\left(\ln\left(\frac{er}{h^{\frac{1}{2\beta}}}\right)\right)^k\frac{dydh}{h^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}. \end{align*} $$
Notice that
$0<l\leq n+4\beta -4$
. A direct computation gives for
$0<h<r^{2\beta }$
,
$$ \begin{align*}h^{-\frac{l}{2\beta}+\frac{n}{2\beta}-\frac{2}{\beta}+2}r^{l+2-2\beta}\left(\ln\left(\frac{er}{h^{\frac{1}{2\beta}}}\right)\right)^{-k} \lesssim r^{n+2-2\beta}r^{4\beta-4}\left(\ln\left(\frac{er}{h^{\frac{1}{2\beta}}}\right)\right)^{-k}\leq r^{n+2-2\beta}r^{4\beta-4},\end{align*} $$
thereby
$$ \begin{align*} I&\lesssim\left(\int_0^1\|N(\cdot,s)\|_{L^1(\mathbb{R}^n)}L(s)ds\right)\bigg\|(-\Delta)^{1-\beta}\int_0^s(e^{-s(-\Delta)^\beta}-e^{-(-\Delta)^\beta})|N(\cdot,h)|dh\bigg\|_{L^\infty}\\ &\lesssim\sup\limits_{r\in(0,1),x\in\mathbb{R}^n}r^{-(n+2-2\beta)}r^{n+2-2\beta}r^{-(l+2-2\beta)}r^{4\beta-4}\left(\int_0^1\|N(\cdot,s)\|_{L^1(\mathbb{R}^n)}L(s)ds\right)\\ &\quad\times\int_0^{r^{2\beta}}\int_{|y-x|<r}|N(y,h)|\left(\ln\left(\frac{er}{h^{\frac{1}{2\beta}}}\right)\right)^k\frac{dydh}{h^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\\ &\lesssim A(\beta,N)\int_0^1\int_{\mathbb{R}^n}|N(x,s)|\left(\ln\left(\frac{e}{s^{\frac{1}{2\beta}}}\right)\right)^k\frac{dxds}{s^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}. \end{align*} $$
This completes the proof of Lemma 4.10.
4.3 Well-posedness of fractional Naiver–Stokes equations
In this section, we prove that the equations (1.1) is well-posed with data
$f\in Q^{-1}_{l,k,\beta }(\mathbb R^{n})$
being small. By Theorem 4.7,
$Q^{-1}_{l,k,\beta }(\mathbb R^{n})=\dot {B}^{1-2\beta }_{\infty ,\infty }(\mathbb R^{n})$
for
$\beta \in (\frac {1}{2},1)$
,
$k\in \mathbb N$
and
$l>n+4\beta -4$
. In [Reference Yu and Zhai52], Yu and Zhai proved that, with data
$f\in \dot {B}^{1-2\beta }_{\infty ,\infty }(\mathbb R^{n})$
being small, the well-posedness of equations (1.1) holds (see [Reference Yu and Zhai52, Theorem 3.1]). Moreover, for
$l\leq n+2\beta -4$
, Remark 3.21 indicates that if
$f\in Q^{(l,k)}_{\beta }(\mathbb R^{n})$
then f is a constant. Equivalently, the space
$Q_{l,k,\beta }^{-1}(\mathbb {R}^n)=\{0\}$
. Hence, in Theorem 4.12, the scope of l is restricted to
$(n+2\beta -4, n+4\beta -4]$
. It is worth mentioning that the scope of l in Theorem 4.12 is different from that of [Reference Yu and Zhai52, Theorem 3.1].
Definition 4.11 Let
$l>n+2\beta -4$
,
$\beta>\frac {1}{2}$
and
$k\in \mathbb N$
. A function
$v(\cdot ,\cdot )$
on
$\mathbb R_+^{n+1}$
belongs to the space
$X_{T}^{l,k,\beta }$
provided
$\|v\|_{X_{T}^{l,k,\beta }}<\infty $
, where
$$ \begin{align*} \|v\|_{X_{T}^{l,k,\beta}}&:=\sup\limits_{t\in(0,T)}t^{1-\frac{1}{2\beta}}\|v(\cdot,t)\|_{L^\infty(\mathbb{R}^n)}\\ &+\sup\limits_{x\in\mathbb{R}^n, r^{2\beta}\in(0,T)}\left(r^{-l+6\beta-6}\int_0^{r^{2\beta}}\int_{|y-x|<r}|v(y,t)|^2\frac{\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}dydt\right)^{\frac{1}{2}}. \end{align*} $$
Theorem 4.12 Suppose that
$n\geq 2$
and
$k\in \mathbb N$
. Let
$\frac {1}{2}<\beta <\min \{2, \frac {1}{2}+\frac {n}{4}\}$
and
$n+2\beta -4<l\leq n+4\beta -4$
. There is
$\epsilon _0>0$
such that if the incompressible initial data a satisfies
$$ \begin{align*} \|a=(a_1,\ldots,a_n)\|_{(Q_{l,k,\beta}^{-1}(\mathbb{R}^n))^n}:=\sum\limits_{j=1}^n\|a_j\|_{Q_{l,k,\beta}^{-1}(\mathbb{R}^n)}\leq\epsilon_0, \end{align*} $$
then equations (1.1) have a unique mild solution
$$ \begin{align*} u=(u_1,\ldots,u_n)=e^{-t(-\Delta)^\beta}a-\int_0^te^{-(t-s)(-\Delta)^\beta} {P}\nabla\cdot(u\otimes u)ds \end{align*} $$
with
$\|u\|_{(X_T^{l,k,\beta })^n}:=\sum \limits _{j=1}^n\|u_{j}\|_{X_T^{l,k,\beta }}<\infty ,$
where
$$ \begin{align*} \left\{ \begin{aligned} &e^{-t(-\Delta)^\beta}a=(e^{-t(-\Delta)^\beta}a_1,\ldots,e^{-t(-\Delta)^\beta}a_n);\\ &{P}=\{{P}_{j,k}\}_{j,k=1,\ldots,n}=\{\delta_{j,k}+R_jR_k\}_{j,k=1,\ldots,n};\\ &\delta_{j,k}=\text{the Kronecker symbol};\\ &R_j=\partial_j(-\Delta)^{-\frac{1}{2}}=\partial_{x_j}(-\Delta)^{-\frac{1}{2}}=\text{the Riesz transform}. \end{aligned} \right. \end{align*} $$
Proof First, by Proposition 4.1,
$$ \begin{align*} t^{1-\frac{1}{2\beta}}\|e^{-t(-\Delta)^\beta}a_j\|_{L^\infty(\mathbb{R}^n)}=t^{(2\beta-1)/{(2\beta)}}\|e^{-t(-\Delta)^\beta}a_j\|_{L^\infty(\mathbb{R}^n)}\lesssim\|a_j\|_{Q_{l,k,\beta}^{-1}(\mathbb{R}^n)}. \end{align*} $$
By Picard’s contraction principle, it is sufficient to verify the bilinear operator
$$ \begin{align*} B(u,v)=\int_0^te^{-(t-s)(-\Delta)^\beta}P\nabla\cdot(u\otimes v)ds \end{align*} $$
is bounded from
$(X_{T}^{l,k,\beta })^n\times (X_{T}^{l,k,\beta })^n$
to
$(X_{T}^{l,k,\beta })^n$
.
Part I:
$L^\infty $
- bounded. The aim is to prove
$$ \begin{align*} \|B(u,v)\|_{L^\infty(\mathbb{R}^n)}\lesssim t^{\frac{1}{2\beta}-1}\|u\|_{(X_{T}^{l,k,\beta})^n}\|v\|_{(X_{T}^{l,k,\beta})^n} \quad \forall\ t\in(0,T). \end{align*} $$
If
$\frac {t}{2}\leq s\leq t$
, then
$$ \begin{align*} \bigg\|e^{-(t-s)(-\Delta)^\beta}\mathbb{P}\nabla\cdot(u\otimes v)\bigg\|_{L^\infty(\mathbb{R}^n)}&\lesssim(t-s)^{-\frac{1}{2\beta}}\|u\|_{L^\infty(\mathbb{R}^n)}\|v\|_{L^\infty(\mathbb{R}^n)}\\ &\lesssim(t-s)^{-\frac{1}{2\beta}}s^{\frac{1}{\beta}-2}\|u\|_{(X_{T}^{l,k,\beta})^n}\|v\|_{(X_{T}^{l,k,\beta})^n}. \end{align*} $$
If
$0<s<\frac {t}{2}$
,
$$ \begin{align*} \bigg|e^{-(t-s)(-\Delta)^\beta}\mathbb{P}\nabla\cdot(u\otimes v)\bigg|&\lesssim\int_{\mathbb{R}^n}\frac{|u(y,s)||v(y,s)|}{((t-s)^{\frac{1}{2\beta}}+|x-y|)^{n+1}}dy\\ &\lesssim\int_{\mathbb{R}^n}\frac{|u(y,s)||v(y,s)|}{(t^{\frac{1}{2\beta}}+|x-y|)^{n+1}}dy\\ &\lesssim\sum\limits_{k\in\mathbb Z^n}\int_{x-y\in t^{\frac{1}{2\beta}}(k+[0,1]^n)}\frac{|u(y,s)||v(y,s)|}{(t^{\frac{1}{2\beta}}(1+|k|))^{n+1}}dy. \end{align*} $$
This gives
$$ \begin{align*} |B(u,v)|&\lesssim\int_0^{\frac{t}{2}}|e^{-(t-s)(-\Delta)^\beta}\mathbb{P}\nabla\cdot(u\otimes v)|ds+\int_{\frac{t}{2}}^t|e^{-(t-s)(-\Delta)^\beta}\mathbb{P}\nabla\cdot(u\otimes v)|ds\\ &\lesssim\sum\limits_{k\in\mathbb Z^n}(t^{\frac{1}{2\beta}}\frac{1}{(1+|k|))^{n+1}}\int_0^{\frac{t}{2}}\int_{x-y\in t^{\frac{1}{2\beta}}(k+[0,1]^n)}|u(y,s)||v(y,s)|dyds\\ &\ \ \ \ +\int_{\frac{t}{2}}^t(t-s)^{-\frac{1}{2\beta}}s^{\frac{1}{\beta}-2}ds\|u\|_{(X_{T}^{l,k,\beta})^n}\|v\|_{(X_{T}^{l,k,\beta})^n}\\ &:=I_3+I_4. \end{align*} $$
Obviously,
$$ \begin{align*} I_4\lesssim t^{\frac{1}{\beta}-2}t^{1-\frac{1}{2\beta}}\|u\|_{(X_{T}^{l,k,\beta})^n}\|v\|_{(X_{T}^{l,k,\beta})^n}\lesssim t^{\frac{1}{2\beta}-1}\|u\|_{(X_{T}^{l,k,\beta})^n}\|v\|_{(X_{T}^{l,k,\beta})^n}. \end{align*} $$
For
$I_3$
, by H
$\ddot {\text {o}}$
lder’s inequality, we have
$$ \begin{align*} I_3&\lesssim\sum\limits_{k\in\mathbb Z^n}(t^{\frac{1}{2\beta}}(1+|k|))^{-(n+1)}\left(\int_0^{\frac{t}{2}}\int_{|x-y|<t^{\frac{1}{2\beta}}}|u(y,s)|^2dyds\right)^{\frac{1}{2}}\\ &\quad\times\left(\int_0^{\frac{t}{2}}\int_{|x-y|<t^{\frac{1}{2\beta}}}|v(y,s)|^2dyds\right)^{\frac{1}{2}}\\ &\lesssim\sum\limits_{k\in\mathbb Z^n}(t^{\frac{1}{2\beta}}(1+|k|))^{-(n+1)}I_{3,1}\times I_{3,2}. \end{align*} $$
A direct computation gives
$$ \begin{align*} I_{3,1}&\lesssim\left(s^{\frac{n-l}{2\beta}}\int_0^{\frac{t}{2}}\int_{|x-y|<t^{\frac{1}{2\beta}}}|u(y,s)|^2s^{2-\frac{2}{\beta}}\left(\ln\left(\frac{et^{\frac{1}{2\beta}}}{s^{\frac{1}{2\beta}}}\right)\right)^k\frac{ds}{s^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\right)^{\frac{1}{2}}\\ &\lesssim t^{\frac{n}{4\beta}+\frac{1}{2\beta}-\frac{1}{2}}\|u\|_{(X_{T}^{l,k,\beta})^n}. \end{align*} $$
Similarly,
$I_{3,2}\lesssim t^{\frac {n}{4\beta }+\frac {1}{2\beta }-\frac {1}{2}}\|v\|_{(X_{T}^{l,k,\beta })^n}$
, thereby
$$ \begin{align*} I_3&\lesssim\sum\limits_{k\in\mathbb Z^n}\frac{1}{(1+|k|)^{n+1}}t^{-(n+1)/{(2\beta)}}t^{n/{(2\beta)}+\frac{1}{\beta}-1}\|u\|_{(X_{T}^{l,k,\beta})^n}\|v\|_{(X_{T}^{l,k,\beta})^n}\\ &\lesssim t^{\frac{1}{2\beta}-1}\|u\|_{(X_{T}^{l,k,\beta})^n}\|v\|_{(X_{T}^{l,k,\beta})^n}. \end{align*} $$
Part II:
$L^2$
-bounded. We want to establish that
$$ \begin{align*} r^{-l+6\beta-6}\int_0^{r^{2\beta}}\int_{|x-y|<r}|B(u,v)|^2\frac{\left(\ln\left(\frac{er}{s^{\frac{1}{2\beta}}}\right)\right)^k}{s^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}dyds\lesssim \|u\|^2_{(X_{T}^{l,k,\beta})^n}\|v\|^2_{(X_{T}^{l,k,\beta})^n}. \end{align*} $$
Define 
the characteristic function on the ball
$$ \begin{align*}\{y\in\mathbb{R}^n:|y-x|<10r\}.\end{align*} $$
We divide
$B(u,v)$
into three parts:
and 
For
$B_1$
, we get
$e^{-t(-\Delta )^\beta }\mathbb {P}\nabla \cdot f(x)=\int _{\mathbb {R}^n}\nabla K_{j,k,t}^\beta (x-y)f(y)dy.$
Since
$$ \begin{align*}|\nabla K_{j,k,t}^\beta(x-y)|\lesssim t^{-(n+1)/{(2\beta)}}(1+t^{-\frac{1}{2\beta}}|x-y|)^{-(n+1)} \lesssim(t^{\frac{1}{2\beta}}+|x-y|)^{-(n+1)},\end{align*} $$
we have
$$ \begin{align*} |B_1(u,v)|\lesssim\int_0^s\int_{|z-x|\geq10r}\frac{|u(z,h)||v(z,h)|}{((s-h)^{\frac{1}{2\beta}}+|z-y|)^{n+1}}dzdh. \end{align*} $$
Notice that
$|z-x|\geq 10r$
,
$0<s<r^{2\beta }$
and
$|y-x|<r$
, then
$|x-z|\lesssim |y-z|$
. This, together with H
$\ddot {\text {o}}$
lder’s inequality, gives
$$ \begin{align*} |B_1(u,v)|\lesssim\int_0^{r^{2\beta}}\int_{|z-x|\geq10r}\frac{|u(z,h)||v(z,h)|}{|x-z|^{n+1}}dzdh\lesssim I_1\times I_2, \end{align*} $$
where
$$ \begin{align*} \left\{ \begin{aligned} &I_1:=\left(\int_0^{r^{2\beta}}\int_{|z-x|\geq10r}\frac{|u(z,h)|^2}{|x-z|^{n+1}}dzdh\right)^{\frac{1}{2}};\\ &I_2:=\left(\int_0^{r^{2\beta}}\int_{|z-x|\geq10r}\frac{|v(z,h)|^2}{|x-z|^{n+1}}dzdh\right)^{\frac{1}{2}}. \end{aligned} \right. \end{align*} $$
We can see that
$$ \begin{align*} I^2_1&\lesssim\sum\limits_{j=3}^\infty\frac{1}{(2^jr)^{n+1}}\int_0^{r^{2\beta}}\int_{2^jr<|z-x|<2^{j+1}r}|u(z,h)|^2dzdh\\ &\lesssim\sum\limits_{j=3}^\infty\frac{1}{(2^jr)^{n+1}}\int_0^{r^{2\beta}}\int_{2^jr<|z-x|<2^{j+1}r}|u(z,h)|^2\left(\ln\left(\frac{e2^{j+1}r}{h^{\frac{1}{2\beta}}}\right)\right)^{-k}\\ &\quad\times\left(\ln\left(\frac{e2^{j+1}r}{h^{\frac{1}{2\beta}}}\right)\right)^kh^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}\frac{dh}{h^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}. \end{align*} $$
The function
$g(h):=\left (\ln \left (\frac {e2^{j+1}r}{h^{\frac {1}{2\beta }}}\right )\right )^{-k}$
is increasing for
$0<h<r^{2\beta }$
, that is,
$$ \begin{align*} g(h)\lesssim\left(\ln\left(\frac{e2^{j+1}r}{r}\right)\right)^{-k}=(1+(j+1)\ln2)^{-k}. \end{align*} $$
The above estimate implies
$$ \begin{align*} I_1&\lesssim\left(\sum\limits_{j=3}^\infty\frac{1}{(2^jr)^{n+1}}\int_0^{r^{2\beta}}\int_{|z-x|<2^{j+1}r}|u(z,h)|^2(1+(j+2)\ln2)^{-k}\right.\\ &\quad \times\left.\left(\ln\left(\frac{e2^{j+1}r}{h^{\frac{1}{2\beta}}}\right)\right)^kh^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}\frac{dh}{h^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\right)^{\frac{1}{2}}\\ &\lesssim\left(\sum\limits_{j=3}^n\frac{(r^{2\beta})^{\frac{n-l}{(2\beta)}+2-\frac{2}{\beta}}}{(2^{j+1}r)^{n+1}(1+(j+1)\ln2)^k}\int_0^{r^{2\beta}}\int_{|z-x|<2^{j+1}r}|u(z,h)|^2\right.\\ &\quad \times\left.\left(\ln\left(\frac{e2^{j+1}r}{h^{\frac{1}{2\beta}}}\right)\right)^k\frac{dh}{h^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\right)^{\frac{1}{2}}\\ &\lesssim r^{\frac{1}{2}-\beta}\|u\|_{(X_T^{l,k,\beta})^n}. \end{align*} $$
Similarly, we obtain
$I_2\lesssim r^{\frac {1}{2}-\beta }\|v\|_{(X_T^{l,k,\beta })^n}$
, which gives
$|B_1(u,v)|\lesssim r^{1-2\beta }\|u\|_{(X_T^{l,k,\beta })^n}\|v\|_{(X_T^{l,k,\beta })^n}.$
Hence,
$$ \begin{align*} &r^{-(l+2-2\beta)}\int_0^{r^{2\beta}}\int_{|x-y|<r}|B_1(u,v)|^2\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k\frac{dydt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\\ &\lesssim\int_0^{r^{2\beta}}\int_{|x-y|<r}r^{-l-2\beta}\|u\|^2_{(X_T^{l,k,\beta})^n}\|v\|^2_{(X_T^{l,k,\beta})^n} \left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k\frac{dydt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\\ &\lesssim\|u\|^2_{(X_T^{l,k,\beta})^n}\|v\|^2_{(X_T^{l,k,\beta})^n}\int_0^{r^{2\beta}}r^{n-l-2\beta} \left(1+\ln\left(\frac{r}{t^{\frac{1}{2\beta}}}\right)\right)^k\frac{dt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}. \end{align*} $$
Since
$l>n+2\beta -4$
, we have
$\int _0^\infty e^{w(-l+n-4+2\beta )}dw\approx 1$
and
$$ \begin{align*} \int_0^\infty e^{w(-l+n-4+2\beta)}(1+w)^kdw\approx1+k\int_0^\infty e^{w(-l+n-4+2\beta)}(1+w)^{k-1}dw. \end{align*} $$
By the change of variables:
$t^{\frac {1}{2\beta }}=u$
,
$v=r/u$
and
$w=\ln v$
, respectively, it holds
$$ \begin{align*} &r^{n-l-2\beta}\int_0^{r^{2\beta}}t^{l/{(2\beta)}-n/{(2\beta)}+\frac{2}{\beta}-2}\left(1+\ln\left(\frac{r}{t^{\frac{1}{2\beta}}}\right)\right)^kdt\\ &\lesssim r^{4-4\beta}\int_0^\infty e^{w(-l+n-4+2\beta)}(1+w)^kdw\lesssim r^{4-4\beta}. \end{align*} $$
Hence,
$$ \begin{align*} r^{-l+6\beta-6}\int_0^{r^{2\beta}}\int_{|x-y|<r}|B_1(u,v)|^2\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k\frac{dydt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}} \lesssim\|u\|^2_{(X_T^{l,k,\beta})^n}\|v\|^2_{(X_T^{l,k,\beta})^n}. \end{align*} $$
Now, we deal with the term
$B_2(u,v)$
. Recall that
$$ \begin{align*} \mathbb{P}=\{\mathbb{P}_{j,k}\}_{j,k=1,\ldots,n}=\{\delta_{j,k}+R_jR_k\}_{j,k=1,\ldots,n}. \end{align*} $$
Obviously, the operator
$(-\Delta )^{\frac {1}{2}}\mathbb {P}\nabla $
is bounded on
$L^2(\mathbb {R}^n)$
. By Lemma 4.8, we obtain
The convolution operator
$t^{\frac {1}{2\beta }-1}(-\Delta )^{\frac {1}{2}-\beta }(I-e^{-t(-\Delta )^\beta })$
is represented as
$$ \begin{align*}\left(t(-\Delta\right)^\beta)^{\frac{1}{2\beta}-1}(I-e^{-t(-\Delta)^\beta})(f)(x)=\int_{\mathbb{R}^n}t^{\frac{1}{2\beta}-1}|\zeta|^{1-2\beta}(1-e^{-t|\zeta|^{2\beta}})\widehat{f}(\zeta)e^{2\pi ix\zeta}d{\zeta}.\end{align*} $$
When
$\beta>\frac {1}{2}$
,
$\lim \limits _{|\zeta |\rightarrow 0}|\zeta |^{1-2\beta }(1-e^{-|\zeta |^{2\beta }})=\lim \limits _{|\zeta |\rightarrow \infty }|\zeta |^{1-2\beta }(1-e^{-|\zeta |^{2\beta }})=0, $
i.e.,
$\sup \limits _{|\zeta |\in (0,\infty )}|\zeta |^{1-2\beta }(1-e^{-|\zeta |^{2\beta }})<\infty .$
So
$t^{\frac {1}{2\beta }-1}(-\Delta )^{\frac {1}{2}-\beta }(I-e^{-t(-\Delta )^\beta }), \beta>\frac {1}{2},$
are bounded on
$L^2(\mathbb {R}^n)$
. Let 
. The Cauchy–Schwarz’s inequality implies
$$ \begin{align*} &r^{-l+6\beta-6}\int_0^{r^{2\beta}}\int_{|x-y|<r}|B_2(u,v)|^2\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k\frac{dydt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\\ &\lesssim r^{-l+6\beta-6}\int_0^{r^{2\beta}}t^{2-\frac{1}{\beta}}\|M(\cdot,t)\|_{L^2(\mathbb{R}^n)}\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k\frac{dt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\\ &\lesssim r^{-l+6\beta-6}\int_0^{r^{2\beta}}t^{2-\frac{1}{\beta}}\int_{|y-x|<r}|u(y,t)|^2|v(y,t)|^2\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k\frac{dydt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\\ &\lesssim \left(\sup\limits_{t\in(0,T)}t^{1-\frac{1}{2\beta}}\|u(\cdot,t)\|_{L^\infty(\mathbb{R}^n)}\right)\left(\sup\limits_{t\in(0,T)}t^{1-\frac{1}{2\beta}}\|v(\cdot,t)\|_{L^\infty(\mathbb{R}^n)}\right)\\ & \quad \times\left(r^{-l+6\beta-6}\int_0^{r^{2\beta}}\int_{|x-y|<r}|u(y,t)|^2\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k\frac{dydt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\right)^{\frac{1}{2}}\\ &\quad \times\left(r^{-l+6\beta-6}\int_0^{r^{2\beta}}\int_{|x-y|<r}|v(y,t)|^2\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k\frac{dydt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\right)^{\frac{1}{2}}\\ &\lesssim\|u\|^2_{(X_T^{l,k,\beta})^n}\|v\|^2_{(X_T^{l,k,\beta})^n}. \end{align*} $$
Now, for
$B_3(u,v),$
by the change of variables
$t=r^{2\beta }\tau $
and
$s=r^{2\beta }\theta $
, we can get
$$ \begin{align*} I&:=r^{-l+6\beta-6}\int_0^{r^{2\beta}}\|B_3(u,v)\|_{L^2(\mathbb{R}^n)}^2\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k\frac{dt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\\ &\lesssim r^{-l+6\beta-6}\int_0^{r^{2\beta}}\bigg\|(-\Delta)^{\frac{1}{2}}e^{-t(-\Delta)^\beta}\left(\int_0^tM(\cdot,s)ds\right)\bigg\|_{L^2(\mathbb{R}^n)}^2\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k\frac{dt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\\ &\lesssim r^{8\beta-2-n}\int_0^1\bigg\|(-\Delta)^{\frac{1}{2}}e^{-r^{2\beta}\tau(-\Delta)^\beta}\left(\int_0^\tau M(r\cdot, r^{2\beta}\theta)d{\theta}\right)\bigg\|_{L^2(\mathbb{R}^n)}^2\frac{\left(\ln\left(\frac{e}{\tau^{\frac{1}{2\beta}}}\right)\right)^k}{\tau^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}d{\tau}. \end{align*} $$
By the Fourier transform,
$$ \begin{align*} e^{-r^{2\beta}\tau(-\Delta)^\beta}f(x)=\int_{\mathbb{R}^n}\left(e^{-r^{2\beta}\tau|\zeta|^\beta}\widehat{f}(\zeta)\right)e^{-ix\zeta}d{\zeta}=\int_{\mathbb{R}^n}e^{-\tau|\eta|^{2\beta}}\widehat{f}\left(\frac{\eta}{r}\right)\frac{e^{-\frac{ix}{r}\cdot r\zeta}}{r^{n}}d{\eta}. \end{align*} $$
On the other hand,
$\widehat {f(r\cdot )}(\eta )=r^{-n}\widehat {f}(\frac {\eta }{r})$
. Denote
$$ \begin{align*}F(x/r):=\left(e^{-\tau(-\Delta)^\beta}f(r\cdot)\right)(x/r)=e^{-r^{2\beta}\tau(-\Delta)^\beta}f(x).\end{align*} $$
Then, we can get
$$ \begin{align*} \|(-\Delta)^{\frac{1}{2}}e^{-r^{2\beta}\tau(-\Delta)^\beta}\|_{L^2(\mathbb{R}^n)}^2&=\|(-\Delta)^{\frac{1}{2}}F(x/r)\|_{L^2(\mathbb{R}^n)}^2=\int_{\mathbb{R}^n}|\zeta|^2\bigg|\widehat{F\left(\frac{\cdot}{r}\right)}(\zeta)\bigg|^2d{\zeta}\\ &=\int_{\mathbb{R}^n}|\zeta|^2r^{2n}|\widehat{F}(r\zeta)|^2d{\zeta}=r^{n-2}\int_{\mathbb{R}^n}|\eta|^2|\widehat{F}(\eta)|^2d{\eta}\\ &=r^{n-2}\|(-\Delta)^{\frac{1}{2}}F(\cdot)\|_{L^2(\mathbb{R}^n)}^2. \end{align*} $$
Therefore, by Lemma 4.10, we get
$$ \begin{align*} I&\lesssim r^{8\beta-4}\int_0^1\bigg\|(-\Delta)^{\frac{1}{2}}e^{-\tau(-\Delta)^\beta}\left(\int_0^\tau M(r\cdot, r^{2\beta}\theta)d{\theta}\right)\bigg\|_{L^2(\mathbb{R}^n)}^2\frac{\left(\ln\left(\frac{e}{\tau^{\frac{1}{2\beta}}}\right)\right)^k}{\tau^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}d{\tau}\\ &\lesssim r^{8\beta-4}A(\beta,M(ry,r^{2\beta}s))\left(\int_0^1\int_{\mathbb{R}^n}|M(rx,r^{2\beta}s)|\left(\ln\left(\frac{e}{s^{\frac{1}{2\beta}}}\right)\right)^k\frac{dxds}{s^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\right), \end{align*} $$
where
$A(\beta ,M(r\cdot , r^\beta \cdot ))$
is defined as
$$ \begin{align*} \sup\limits_{(x,\rho)\in\mathbb{R}^n\times(0,1)}\rho^{-l+6\beta-6}\int_0^{\rho^{2\beta}}\int_{|y-x|<\rho}|M(ry, r^{2\beta}s)|\frac{\left(\ln\left(\frac{e\rho}{s^{\frac{1}{2\beta}}}\right)\right)^k}{s^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}dxds. \end{align*} $$
By the change of variables
$r^{2\beta }s=t$
and
$ry=x$
, we can obtain
$$ \begin{align*} &\int_0^1\int_{\mathbb{R}^n}|M(rx, r^{2\beta}s)|\left(\ln\left(\frac{e}{s^{\frac{1}{2\beta}}}\right)\right)^k\frac{dxds}{s^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\\ &=r^{-l+2\beta-4}\int_0^{r^{2\beta}}\int_{|y-x|<r}|M(y,t)|\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k\frac{dydt}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}\\ &\lesssim r^{2-4\beta}\|u\|_{(X_T^{l,k,\beta})^n}\|v\|_{(X_T^{l,k,\beta})^n}. \end{align*} $$
On the other hand, by the change of variables
$r^{2\beta }s=t$
and
$ry=z$
, we can deduce that
$A(\beta ,M(r\cdot , r^\beta \cdot ))$
can be controlled by
$$ \begin{align*} &\sup\limits_{x\in\mathbb{R}^n, \rho\in(0,1)}r^{-(l+4-2\beta)}\rho^{-l+6\beta-6}\int_0^{(r\rho)^{2\beta}}\int_{|z-rx|<r\rho}|M(z,t)|\frac{\left(\ln\left(\frac{er\rho}{t^{\frac{1}{2\beta}}}\right)\right)^k}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}dzdt\\ &\lesssim\sup\limits_{x\in\mathbb{R}^n, \rho\in(0,1)}r^{2-4\beta}(r\rho)^{-l+6\beta-6}\int_0^{(r\rho)^{2\beta}}\int_{|z-rx|<r\rho}|M(z,t)|\frac{\left(\ln\left(\frac{er\rho}{t^{\frac{1}{2\beta}}}\right)\right)^k}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}dzdt\\ &\lesssim r^{2-4\beta}\|u\|_{(X_T^{l,k,\beta})^n}\|v\|_{(X_T^{l,k,\beta})^n}, \end{align*} $$
which gives
$$ \begin{align*} r^{-l+6\beta-6}\int_0^{r^{2\beta}}\int_{|x-y|<r}|B_3(u,v)|^2\frac{\left(\ln\left(\frac{er}{t^{\frac{1}{2\beta}}}\right)\right)^k}{t^{\frac{n-l}{2\beta}+2-\frac{2}{\beta}}}dydt \lesssim\|u\|^2_{(X_T^{l,k,\beta})^n}\|v\|^2_{(X_T^{l,k,\beta})^n}. \end{align*} $$
4.4 Well-posedness of fractional MHD equations
The method of Theorem 4.12 can also be used to investigate the well-posedness of fractional MHD equations in
$\mathbb {R}^{n+1}_{+}$
,
$n\geq 2$
:
$$ \begin{align} \left\{\begin{aligned} &\partial_{t}u+(-\Delta)^{\beta}u+u\cdot \nabla u+\nabla P-b\cdot\nabla b=0;\\ &\partial_{t}b+(-\Delta)^{\beta}b+u\cdot \nabla b-b\cdot\nabla u=0;\\ &u|_{t=0}=u_{0},\ b|_{t=0}=b_{0}, \end{aligned} \right. \end{align} $$
with
$\nabla \cdot u=\nabla \cdot b=0$
and
$\beta \in (\frac {1}{2},\min \{2,\frac {1}{2}+\frac {n}{4}\}).$
Following the procedure of Theorem 4.12, we can obtain the following result.
Theorem 4.13 Let
$n\geq 2$
,
$\frac {1}{2}<\beta <\min \{2, \frac {1}{2}+\frac {n}{4}\}$
,
$n+2\beta -4<l\leq n+4\beta -4$
. The fractional MHD equation (4.2) has a unique small global mild solution in
$(X_T^{l,k,\beta })^n$
for an initial data
$(u_0,b_0)$
with
$\nabla \cdot u_0=\nabla \cdot b_0=0$
and small
$\|(u_0,b_0)\|_{(Q_{l,k,\beta }^{-1}(\mathbb {R}^n))^n}$
.
Proof The solution
$(u,b)$
to equation (4.2) can be written as
$$ \begin{align*}u(t,x)=e^{-t(-\Delta)^{\beta}}u_{0}(x)-B(u,u)+B(b,b):=F_{1}(u,b), \end{align*} $$
$$ \begin{align*}b(t,x)=e^{-(-\Delta)^{\beta}}b_{0}(x)-B(u,b)+B(b,u):=F_{2}(u,b), \end{align*} $$
with
$ B(u,v)=\int _{0}^{t}e^{-(t-s)(-\Delta )^{\beta }}P\nabla \cdot (u\otimes v)(s)ds. $
We rewrite the solution
$(u,b)$
as
$$ \begin{align*}\left(\begin{array}{@{}ccc@{}}u\\b\end{array}\right)=\left(\begin{array}{@{}ccc@{}}F_{1}(u,b)\\F_{2}(u,b)\end{array}\right):=F(u,b). \end{align*} $$
Since
$B(u,v)$
is bounded from
$(X_T^{l,k,\beta })^n(\mathbb {R}^n)\times (X_T^{l,k,\beta })^n(\mathbb {R}^n)$
to
$(X_T^{l,k,\beta })^n(\mathbb {R}^n),$
we have
$$ \begin{align*} \|F_{1}(u,b)\|_{(X_T^{l,k,\beta})^n}\leq C_1\|u_{0} \|_{(Q_{l,k,\beta}^{-1}(\mathbb{R}^n))^n}+C_2\|(u,b)\|^2_{(X_{T}^{l,k,\beta})^n}, \end{align*} $$
$$ \begin{align*} &\|F_{1}(u,b)-F_{1}(u',b')\|_{(X_{T}^{l,k,\beta})^n}\\ &\leq \|B(u,u)-B(u',u')\|_{(X_{T}^{l,k,\beta})^n}+\|B(b,b)-B(b',b')\|_{(X_{T}^{l,k,\beta})^n}\\ &\leq C\Big(\|u\|_{(X_{T}^{l,k,\beta})^n}+\|u'\|_{(X_{T}^{l,k,\beta})^n}\Big)\|u-u'\|_{(X_{T}^{l,k,\beta})^n}\\ &+C\Big(\|b\|_{(X_{T}^{l,k,\beta})^n}+\|b'\|_{(X_{T}^{l,k,\beta})^n}\Big)\|b-b'\|_{(X_{T}^{l,k,\beta})^n}\\ &\leq C\|(u,b)-(u',b')\|_{(X_{T}^{l,k,\beta})^n}\left(\|(u,b)\|_{(X_{T}^{l,k,\beta})^n}+\|(u',b')\|_{(X_{T}^{l,k,\beta})^n}\right). \end{align*} $$
Similarly, we get
$\|F_{2}(u,b)\|_{(X_{T}^{l,k,\beta })^n}\leq C_1\|b_{0} |_{(Q_{l,k,\beta }^{-1}(\mathbb {R}^n))^n}+C_2\|(u,b)\|^2_{(X_{T}^{l,k,\beta })^n} $
and
$$ \begin{align*} &\|F_{2}(u,b)(t)-F_{2}(u',b')\|_{(X_{T}^{l,k,\beta})^n}\\ &\lesssim\|(u,b)-(u',b')\|_{(X_{T}^{l,k,\beta})^n}\left(\|(u,b)\|_{(X_{T}^{l,k,\beta})^n}+\|(u',b')\|_{(X_{T}^{l,k,\beta})^n}\right). \end{align*} $$
Thus, we have
$$ \begin{align*}\|F(u,b)\|_{(X_{T}^{l,k,\beta})^n}\leq C\|(u_0,b_{0} )|_{(Q_{l,k,\beta}^{-1}(\mathbb{R}^n))^n}+C\|u,b\|^2_{(X_{T}^{l,k,\beta})^n}, \end{align*} $$
$$ \begin{align*} &\|F(u,b)(t)-F(u',b')\|_{(X_{T}^{l,k,\beta})^n}\\ &\leq C\|(u,b)-(u',b')\|_{(X_{T}^{l,k,\beta})^n}\left(\|(u,b)\|_{(X_{T}^{l,k,\beta})^n}+\|(u',b')\|_{(X_{T}^{l,k,\beta})^n}\right). \end{align*} $$
Let
$R\leq 2C\|(u_0,b_{0} )|_{(Q_{l,k,\beta }^{-1}(\mathbb {R}^n))^n}$
be small enough. So F is a contraction mapping on
$$ \begin{align*}E=\Big\{(u,b)\in(X_{T}^{l,k,\beta})^n:\ \|(u,b)\|_{(X_{T}^{l,k,\beta})^n}\leq R\Big\}\end{align*} $$
into itself. Thus, a fixed point
$(u,b)$
of the operator F exists.
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