1 Introduction
Let
$\mathbb{C}^{n}$
be the complex
$n$
-space and
$dV$
be the ordinary volume measure on
$\mathbb{C}^{n}$
. If
$z=(z_{1},\ldots ,z_{n})$
and
$w=(w_{1},\ldots ,w_{n})$
are points in
$\mathbb{C}^{n}$
, we write
$$\begin{eqnarray}z\cdot \overline{w}=\mathop{\sum }_{j=1}^{n}z_{j}\overline{w}_{j},\quad |z|=(z\cdot \overline{z})^{1/2}.\end{eqnarray}$$
For any
$0<p\leqslant \infty$
we let
$L_{G}^{p}$
denote the space of Lebesgue measurable functions
$f$
on
$\mathbb{C}^{n}$
such that the function
$f(z)e^{-(1/2)|z|^{2}}$
is in
$L^{p}(\mathbb{C}^{n},dV)$
. When
$0<p<\infty$
, it is clear that
$$\begin{eqnarray}L_{G}^{p}=L^{p}(\mathbb{C}^{n},e^{-(p/2)|z|^{2}}\,dV(z)).\end{eqnarray}$$
We define
$$\begin{eqnarray}\Vert f\Vert _{p}=\biggl[\biggl(\frac{p}{2\unicode[STIX]{x1D70B}}\biggr)^{n}\int _{\mathbb{C}^{n}}|f(z)e^{-(1/2)|z|^{2}}|^{p}\,dV(z)\biggr]^{1/p}.\end{eqnarray}$$
For
$p=\infty$
the norm in
$L_{G}^{\infty }$
is defined by
$$\begin{eqnarray}\Vert f\Vert _{\infty }=\text{esssup}\{|f(z)|e^{-(1/2)|z|^{2}}:z\in \mathbb{C}^{n}\}.\end{eqnarray}$$
Let
$F^{p}$
denote the space of entire functions in
$L_{G}^{p}$
. Then
$F^{2}$
is a closed subspace of the Hilbert space
$L_{G}^{2}$
(see [Reference Zhu15]) with inner product
$$\begin{eqnarray}\langle f,g\rangle =\frac{1}{\unicode[STIX]{x1D70B}^{n}}\int _{\mathbb{C}^{n}}f(z)\overline{g(z)}e^{-|z|^{2}}\,dV(z).\end{eqnarray}$$
To give a motivation for our study of Fock–Sobolev spaces, recall that the annihilation operator
$A_{j}$
and the creation operator
$A_{j}^{\ast }$
from the quantum theory are defined by the commutation relation
$[A_{j},A_{k}^{\ast }]=\unicode[STIX]{x1D6FF}_{jk}I$
, where
$I$
is the identity operator. A natural representation of these operators is achieved on the Fock space
$F^{2}$
, namely,
$$\begin{eqnarray}A_{j}f(z)=\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}z_{j}}f(z),\qquad A_{j}^{\ast }f(z)=z_{j}f(z),\quad 1\leqslant j\leqslant n,f\in F^{2}.\end{eqnarray}$$
Both
$A_{j}$
and
$A_{j}^{\ast }$
, as defined above, are densely defined linear operators on
$F^{2}$
(unbounded though) and satisfy the commutation relation
$[A_{j},A_{k}^{\ast }]=\unicode[STIX]{x1D6FF}_{jk}I$
. Therefore, it is important to study the operator of multiplication by
$z_{j}$
and the operator of differentiation on the Fock space
$F^{2}$
.
We define the radial derivative
$\mathscr{R}$
by
$$\begin{eqnarray}\mathscr{R}:=\mathop{\sum }_{j=1}^{n}(A_{j}A_{j}^{\ast }+A_{j}^{\ast }A_{j})\end{eqnarray}$$
and the Fock–Sobolev space
$F_{\mathscr{R}}^{s,p}$
of fractional order
$s$
for which
$\mathscr{R}^{s/2}f$
is given by an
$F^{p}$
function. Then
$F_{\mathscr{R}}^{s,2}$
is a Hilbert space with inner product
$$\begin{eqnarray}\langle f,g\rangle _{F_{\mathscr{R}}^{s,2}}=\frac{1}{\unicode[STIX]{x1D70B}^{n}}\int _{\mathbb{C}^{n}}\mathscr{R}^{s/2}f(z)\overline{\mathscr{R}^{s/2}g(z)}e^{-|z|^{2}}\,dV(z)\end{eqnarray}$$
for
$f,g\in F_{\mathscr{R}}^{s,2}$
. Each point evaluation is a bounded linear functional on
$F_{\mathscr{R}}^{s,2}$
. So, to each
$z\in \mathbb{C}^{n}$
there corresponds the reproducing kernel
$K_{z}^{s}$
such that
$$\begin{eqnarray}f(z)=\langle f,K_{z}^{s}\rangle _{F_{\mathscr{R}}^{s,2}}\end{eqnarray}$$
for
$f\in F_{\mathscr{R}}^{s,2}$
. Let
$K_{s}(z,w):=K_{w}^{s}(z)$
. Defining
$\unicode[STIX]{x1D6EC}(z,w)$
by
$$\begin{eqnarray}\unicode[STIX]{x1D6EC}(z,w)=e^{(1/2)|z|^{2}+(1/2)|w|^{2}-(1/8)|z-w|^{2}},\end{eqnarray}$$
we have the following estimates of the reproducing kernel
$K_{s}(z,w)$
for
$F_{\mathscr{R}}^{s,2}$
.
Theorem 1.1. Let
$s\in \mathbb{R}$
. Then
$$\begin{eqnarray}K_{s}(z,w)=\mathscr{R}_{z}^{-s}(e^{z\cdot \overline{w}})\end{eqnarray}$$
and there are positive constants
$C=C(s)>0$
such that
$$\begin{eqnarray}|K_{s}(z,w)|\leqslant C\times \left\{\begin{array}{@{}ll@{}}(1+|z||w|)^{-s}\unicode[STIX]{x1D6EC}(z,w) & \text{if }s>0,\\ (1+|z\cdot \overline{w}|)^{-s}\unicode[STIX]{x1D6EC}(z,w) & \text{if }s\leqslant 0,\end{array}\right.\end{eqnarray}$$
for
$z,w\in \mathbb{C}^{n}$
.
It will turn out that polynomially growing/decaying weights quite naturally come into play in the study of our fractional Fock–Sobolev spaces. So, we first introduce such weighted Fock spaces. Given
$s$
real we introduce the following norm on
$F_{s}^{p}$
when
$0<p<\infty$
:
$$\begin{eqnarray}\Vert f\Vert _{F_{s}^{p}}^{p}=\unicode[STIX]{x1D714}_{n,s,p}\int _{\mathbb{C}^{n}}|(1+|z|)^{s}f(z)e^{-(1/2)|z|^{2}}|^{p}\,dV(w),\end{eqnarray}$$
where
$\unicode[STIX]{x1D714}_{n,s,p}$
is a normalizing constant so that the constant function
$1$
has norm
$1$
in
$F_{s}^{p}$
. When
$p=\infty$
, we define
$$\begin{eqnarray}\Vert f\Vert _{F_{s}^{\infty }}=\unicode[STIX]{x1D714}_{s}\sup _{z\in \mathbb{C}^{n}}[(1+|z|)^{s}|f(z)|e^{-(1/2)|z|^{2}}],\end{eqnarray}$$
where
$\unicode[STIX]{x1D714}_{s}$
is a normalizing constant so that the constant function
$1$
has norm
$1$
in
$F_{s}^{\infty }$
. Let
$L_{G,s}^{p}$
denote the space of Lebesgue measurable functions
$f$
on
$\mathbb{C}^{n}$
such that the function
$(1+|z|)^{s}f(z)$
is in
$L_{G}^{p}$
. Then
$F_{s}^{p}$
is a closed subspace of
$L_{G,s}^{p}$
.
It follows that the fractional Fock–Sobolev spaces are realized as the weighted Fock spaces that do not involve derivatives as following Theorem 1.2. So, the study on the Fock–Sobolev spaces is reduced to that on the weighted Fock spaces. It is very convenient to study function theoretic and operator theoretic properties on the weighted Fock spaces instead of the Fock–Sobolev spaces (see [Reference Cho, Choe and Koo3, Reference Choe and Yang7, Reference Mengestie9–Reference Mengestie11, Reference Wang, Cao and Xia13]).
Theorem 1.2. Suppose
$0<p\leqslant \infty$
and
$s$
is a real number. Then
$F_{\mathscr{R}}^{s,p}=F_{s}^{p}$
with equivalent norms.
Constants. In this paper we use the same letter
$C$
to denote various positive constants which may vary at each occurrence but do not depend on the essential parameters. Variables indicating the dependency of constants
$C$
will be often specified in parenthesis. For nonnegative quantities
$X$
and
$Y$
the notation
$X\lesssim Y$
or
$Y\gtrsim X$
means
$X\leqslant CY$
for some inessential constant
$C$
. Similarly, we write
$X\approx Y$
if both
$X\lesssim Y$
and
$Y\lesssim X$
hold.
2 Fractional radial derivatives
We note that
$$\begin{eqnarray}\mathscr{R}=\mathop{\sum }_{j=1}^{n}(A_{j}A_{j}^{\ast }+A_{j}^{\ast }A_{j})=2\mathop{\sum }_{j=1}^{n}z_{j}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}z_{j}}+n.\end{eqnarray}$$
It is easy to see that
$\mathscr{R}$
is unbounded, positive, self-adjoint, and invertible on
$F^{2}$
. In fact,
$\mathscr{R}^{-1}$
is a compact operator.
Example 2.1. Let
$$\begin{eqnarray}f(z)=\mathop{\sum }_{k=0}^{\infty }\frac{z_{1}^{k}}{(k+1)\sqrt{k!}}.\end{eqnarray}$$
Then
$f\in F^{2}$
, but
$\mathscr{R}f\notin F^{2}$
.
For
$f\in F^{2}$
let
$$\begin{eqnarray}f(z)=\mathop{\sum }_{\unicode[STIX]{x1D6FC}\in \mathbb{N}_{0}^{n}}c_{\unicode[STIX]{x1D6FC}}e_{\unicode[STIX]{x1D6FC}}(z)\end{eqnarray}$$
be the orthonormal decomposition of
$f$
, where
$e_{\unicode[STIX]{x1D6FC}}(z)=z^{\unicode[STIX]{x1D6FC}}/\Vert z^{\unicode[STIX]{x1D6FC}}\Vert _{2}.$
Associated with the operator
$\mathscr{R}$
is a semigroup
$\{{B_{t}\}}_{t\geqslant 0}$
defined by the expansion
$$\begin{eqnarray}B_{t}f(z)=\mathop{\sum }_{\unicode[STIX]{x1D6FC}\in \mathbb{N}_{0}^{n}}e^{-(2|\unicode[STIX]{x1D6FC}|+n)t}c_{\unicode[STIX]{x1D6FC}}e_{\unicode[STIX]{x1D6FC}}(z).\end{eqnarray}$$
We can check that
$u(z,t):=B_{t}f(z)$
is the solution of the heat-type equation:
$$\begin{eqnarray}\left\{\begin{array}{@{}ll@{}}(\unicode[STIX]{x2202}_{t}+\mathscr{R})u=0\quad & \text{on }\mathbb{C}^{n}\times (0,\infty ),\\ u(\cdot ,0)=f\quad & \text{on }\mathbb{C}^{n}.\end{array}\right.\end{eqnarray}$$
It is easy to see that
$$\begin{eqnarray}\Vert B_{t}f\Vert _{2}^{2}\leqslant e^{-2nt}\Vert f\Vert _{2}^{2}.\end{eqnarray}$$
Thus
$B_{t}$
is contractive. Moreover, we can see that
$-\mathscr{R}$
is the infinitesimal generator of
$\{{B_{t}\}}_{t\geqslant 0}$
. That is,
$$\begin{eqnarray}B_{t}=e^{-t\mathscr{R}}.\end{eqnarray}$$
See [Reference Cho, Choi and Lee5] for more properties concerning the heat semigroup as well as the spectral property of the operator
$\mathscr{R}$
.
Since
$\mathscr{R}$
has discrete spectrum
$\{2|\unicode[STIX]{x1D6FC}|+n:\unicode[STIX]{x1D6FC}\in \mathbb{N}_{0}^{n}\}$
, by using the spectral theorem, we define the fractional radial derivative
$\mathscr{R}^{s}$
for
$s\in \mathbb{R}$
as following:
Definition 2.2. Let
$s\in \mathbb{R}$
. For
$f\in F^{2}$
let
$$\begin{eqnarray}f(z)=\mathop{\sum }_{\unicode[STIX]{x1D6FC}\in \mathbb{N}_{0}^{n}}c_{\unicode[STIX]{x1D6FC}}e_{\unicode[STIX]{x1D6FC}}(z)\end{eqnarray}$$
be the orthonormal decomposition of
$f$
. By the spectral theorem,
$\mathscr{R}^{s}$
is given by
$$\begin{eqnarray}\mathscr{R}^{s}f(z)=\mathop{\sum }_{\unicode[STIX]{x1D6FC}\in \mathbb{N}_{0}^{n}}(2|\unicode[STIX]{x1D6FC}|+n)^{s}c_{\unicode[STIX]{x1D6FC}}e_{\unicode[STIX]{x1D6FC}}(z),\quad f\in {\mathcal{D}}om(\mathscr{R}^{s}).\end{eqnarray}$$
Definition 2.3. Let
$s$
be a real number. The Fock–Sobolev space
$F_{\mathscr{R}}^{s,p}$
of fractional order
$s$
is the space of all entire functions for which
$\mathscr{R}^{s/2}f$
is given by an
$F^{p}$
function. The Fock–Sobolev norm of
$f$
of fractional order
$s$
is defined accordingly,
$$\begin{eqnarray}\Vert f\Vert _{F_{\mathscr{R}}^{s,p}}=\Vert \mathscr{R}^{s/2}f\Vert _{p}.\end{eqnarray}$$
By using the semigroup, we have the integral representations for the fractional radial derivatives as following. See [Reference Bongioanni and Torrea2] for analogues in the context of other type of Sobolev spaces.
Proposition 2.4. Let
$f\in F^{2}$
and
$z\in \mathbb{C}^{n}$
. Then the following identities hold:
(i) For
$0<s<1$
we have where$$\begin{eqnarray}\mathscr{R}^{s}f(z)=\frac{1}{\unicode[STIX]{x1D6E4}(-s)}\int _{0}^{\infty }\frac{[e^{-t\mathscr{R}}f(z)-f(z)]}{t^{s}}\,\frac{dt}{t},\end{eqnarray}$$
$\unicode[STIX]{x1D6E4}(-s)$
is the gamma function to negative numbers defined by choosing$$\begin{eqnarray}\unicode[STIX]{x1D6E4}(-s)=\frac{\unicode[STIX]{x1D6E4}(-s+n)}{(-s)(-s+1)\cdots (-s+n-1)}\end{eqnarray}$$
$n$
such that
$-s+n$
is positive.(ii) For
$s>0$
we have $$\begin{eqnarray}\mathscr{R}^{-s}f(z)=\frac{1}{\unicode[STIX]{x1D6E4}(s)}\int _{0}^{\infty }t^{s}e^{-t\mathscr{R}}f(z)\,\frac{dt}{t}.\end{eqnarray}$$
Proof. We prove (i); the proof for (ii) is simpler.
In [Reference Cho, Choe and Koo4, Proposition 2.2], we calculated the size of Taylor coefficients as following:
$$\begin{eqnarray}\biggl|\frac{\unicode[STIX]{x2202}^{\unicode[STIX]{x1D6FC}}f(0)}{\unicode[STIX]{x1D6FC}!}\biggr|\lesssim e^{|\unicode[STIX]{x1D6FC}|/2}\biggl(\mathop{\prod }_{j=1}^{n}\unicode[STIX]{x1D6FC}_{j}^{-\unicode[STIX]{x1D6FC}_{j}/2}\biggr)\Vert f\Vert _{F^{2}}\end{eqnarray}$$
for a given multi-index
$\unicode[STIX]{x1D6FC}$
where
$\unicode[STIX]{x1D6FC}_{j}^{-\unicode[STIX]{x1D6FC}_{j}/2}$
is understood to be
$1$
when
$\unicode[STIX]{x1D6FC}_{j}=0$
.
For
$f\in F^{2}$
let
$$\begin{eqnarray}f(z)=\mathop{\sum }_{\unicode[STIX]{x1D6FC}\in \mathbb{N}_{0}^{n}}c_{\unicode[STIX]{x1D6FC}}e_{\unicode[STIX]{x1D6FC}}(z),\end{eqnarray}$$
be the orthonormal decomposition of
$f$
, where
$c_{\unicode[STIX]{x1D6FC}}=\unicode[STIX]{x2202}^{\unicode[STIX]{x1D6FC}}f(0)/\sqrt{\unicode[STIX]{x1D6FC}!}$
and
$e_{\unicode[STIX]{x1D6FC}}(z)=z^{\unicode[STIX]{x1D6FC}}/\sqrt{\unicode[STIX]{x1D6FC}!}$
. Note that
$$\begin{eqnarray}\int _{0}^{\infty }(e^{-t}-1)\,\frac{dt}{t^{1+s}}=\unicode[STIX]{x1D6E4}(-s).\end{eqnarray}$$
By (2.1) and (2.2), it follows that
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{\unicode[STIX]{x1D6FC}}|c_{\unicode[STIX]{x1D6FC}}||e_{\unicode[STIX]{x1D6FC}}(z)|\int _{0}^{\infty }|e^{-(2|\unicode[STIX]{x1D6FC}|+n)t}-1|\,\frac{dt}{t^{1+s}}\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{\unicode[STIX]{x1D6FC}}|c_{\unicode[STIX]{x1D6FC}}||e_{\unicode[STIX]{x1D6FC}}(z)|\,(2|\unicode[STIX]{x1D6FC}|+n)^{s}|\unicode[STIX]{x1D6E4}(-s)|\nonumber\\ \displaystyle & & \displaystyle \quad \lesssim \,\mathop{\sum }_{\unicode[STIX]{x1D6FC}}e^{|\unicode[STIX]{x1D6FC}|/2}\biggl(\mathop{\prod }_{j=1}^{n}\unicode[STIX]{x1D6FC}_{j}^{-\unicode[STIX]{x1D6FC}_{j}/2}\biggr)(2|\unicode[STIX]{x1D6FC}|+n)^{s}|z^{\unicode[STIX]{x1D6FC}}|.\nonumber\end{eqnarray}$$
We note that the power series on the right side of the inequality above is convergent for every
$z\in \mathbb{C}^{n}$
. By the dominated convergence theorem, we have
$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6E4}(-s)\mathscr{R}^{s}f(z) & = & \displaystyle \mathop{\sum }_{\unicode[STIX]{x1D6FC}}c_{\unicode[STIX]{x1D6FC}}e_{\unicode[STIX]{x1D6FC}}(z)\,(2|\unicode[STIX]{x1D6FC}|+n)^{s}\unicode[STIX]{x1D6E4}(-s)\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{\unicode[STIX]{x1D6FC}}c_{\unicode[STIX]{x1D6FC}}e_{\unicode[STIX]{x1D6FC}}(z)\int _{0}^{\infty }(e^{-(2|\unicode[STIX]{x1D6FC}|+n)t}-1)\,\frac{dt}{t^{1+s}}\nonumber\\ \displaystyle & = & \displaystyle \int _{0}^{\infty }\mathop{\sum }_{\unicode[STIX]{x1D6FC}}c_{\unicode[STIX]{x1D6FC}}e_{\unicode[STIX]{x1D6FC}}(z)(e^{-(2|\unicode[STIX]{x1D6FC}|+n)t}-1)\,\frac{dt}{t^{1+s}}\nonumber\\ \displaystyle & = & \displaystyle \int _{0}^{\infty }\frac{[e^{-t\mathscr{R}}f(z)-f(z)]}{t^{s}}\,\frac{dt}{t}.\square\nonumber\end{eqnarray}$$
Remark 2.5. We refer to [Reference Cho, Choe and Koo4] for another fractional derivatives. In [Reference Cho, Choe and Koo4], the following derivative
${\mathcal{D}}^{s}f$
is given by
$$\begin{eqnarray}{\mathcal{D}}^{s}f(z)=\mathop{\sum }_{\unicode[STIX]{x1D6FC}\in \mathbb{N}_{0}^{n}}\frac{\unicode[STIX]{x1D6E4}(n+s+|\unicode[STIX]{x1D6FC}|)}{\unicode[STIX]{x1D6E4}(n+|\unicode[STIX]{x1D6FC}|)}c_{\unicode[STIX]{x1D6FC}}e_{\unicode[STIX]{x1D6FC}}(z),\quad f\in {\mathcal{D}}om({\mathcal{D}}^{s}).\end{eqnarray}$$
We remark that our definition of
$\mathscr{R}^{s}f$
is slightly different from
${\mathcal{D}}^{s}f$
, but they are asymptotically the same in the sense that
$\unicode[STIX]{x1D6E4}(n+s+|\unicode[STIX]{x1D6FC}|)/\unicode[STIX]{x1D6E4}(n+|\unicode[STIX]{x1D6FC}|)\approx (2|\unicode[STIX]{x1D6FC}|+n)^{s}$
as
$|\unicode[STIX]{x1D6FC}|\rightarrow \infty$
by Stirling’s formula.
3 Estimates of the reproducing kernel for
$F_{\mathscr{R}}^{s,2}$
In what follows we use the conventional multi-index notation. Thus for an
$n$
-tuple
$\unicode[STIX]{x1D6FC}=(\unicode[STIX]{x1D6FC}_{1},\ldots ,\unicode[STIX]{x1D6FC}_{n})$
of nonnegative integers we write
$$\begin{eqnarray}|\unicode[STIX]{x1D6FC}|=\unicode[STIX]{x1D6FC}_{1}+\cdots +\unicode[STIX]{x1D6FC}_{n},\qquad \unicode[STIX]{x1D6FC}!=\unicode[STIX]{x1D6FC}_{1}!\cdots \unicode[STIX]{x1D6FC}_{n}!,\qquad \unicode[STIX]{x2202}^{\unicode[STIX]{x1D6FC}}=\unicode[STIX]{x2202}_{1}^{\unicode[STIX]{x1D6FC}_{1}}\cdots \unicode[STIX]{x2202}_{n}^{\unicode[STIX]{x1D6FC}_{n}},\end{eqnarray}$$
where
$\unicode[STIX]{x2202}_{j}$
denotes partial differentiation with respect to the
$j$
th component. If
$z=(z_{1},\ldots ,z_{n})$
, then
$z^{\unicode[STIX]{x1D6FC}}=z_{1}^{\unicode[STIX]{x1D6FC}_{1}}\cdots z_{n}^{\unicode[STIX]{x1D6FC}_{n}}$
.
First we get pointwise size estimates for the fractional radial derivatives of the Fock kernel as following.
Theorem 3.1. Given
$s$
real, there are positive constants
$C=C(s)>0$
such that
$$\begin{eqnarray}|\mathscr{R}_{z}^{s}(e^{z\cdot \overline{w}})|\leqslant C\times \left\{\begin{array}{@{}ll@{}}(1+|z\cdot \overline{w}|)^{s}\unicode[STIX]{x1D6EC}(z,w) & \text{if }s\geqslant 0,\\ (1+|z||w|)^{s}\unicode[STIX]{x1D6EC}(z,w) & \text{if }s<0,\end{array}\right.\end{eqnarray}$$
for
$z,w\in \mathbb{C}^{n}$
.
Proof. Since
$$\begin{eqnarray}\displaystyle |e^{z\cdot \overline{w}}| & = & \displaystyle e^{\text{Re}(z\cdot \overline{w})}=e^{(1/2)|z|^{2}+(1/2)|w|^{2}-(1/2)|z-w|^{2}}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle e^{(1/2)|z|^{2}+(1/2)|w|^{2}-(1/8)|z-w|^{2}}=\unicode[STIX]{x1D6EC}(z,w),\nonumber\end{eqnarray}$$
the cases
$s=0,1$
are trivial.
Let
$0<s<1$
. By (i) of Proposition 2.4, we have
$$\begin{eqnarray}\mathscr{R}^{s}(e^{z\cdot \overline{w}})=\frac{1}{\unicode[STIX]{x1D6E4}(-s)}\int _{0}^{\infty }(e^{-t\mathscr{R}}e^{z\cdot \overline{w}}-e^{z\cdot \overline{w}})\,\frac{dt}{t^{1+s}}.\end{eqnarray}$$
Now
$$\begin{eqnarray}\displaystyle e^{-t\mathscr{R}}(e^{z\cdot \overline{w}}) & = & \displaystyle \mathop{\sum }_{\unicode[STIX]{x1D6FC}}\frac{z^{\unicode[STIX]{x1D6FC}}\overline{w}^{\unicode[STIX]{x1D6FC}}}{\unicode[STIX]{x1D6FC}!}e^{-(2|\unicode[STIX]{x1D6FC}|+n)t}\nonumber\\ \displaystyle & = & \displaystyle \exp (e^{-2t}z\cdot \overline{w})e^{-nt}.\nonumber\end{eqnarray}$$
Thus
$$\begin{eqnarray}\mathscr{R}^{s}(e^{z\cdot \overline{w}})=\frac{1}{\unicode[STIX]{x1D6E4}(-s)}\int _{0}^{\infty }[\exp (e^{-2t}z\cdot \overline{w})e^{-nt}-e^{z\cdot \overline{w}}]\,\frac{dt}{t^{1+s}}.\end{eqnarray}$$
We write the integral on the right-hand side of (3.1) as the sum of two pieces
$I_{1}$
and
$I_{2}$
defined by
$$\begin{eqnarray}I_{1}=\int _{0}^{1}[\exp (e^{-2t}z\cdot \overline{w})e^{-nt}-e^{z\cdot \overline{w}}]\,\frac{dt}{t^{1+s}}\end{eqnarray}$$
and
$$\begin{eqnarray}I_{2}=\int _{1}^{\infty }[\exp (e^{-2t}z\cdot \overline{w})e^{-nt}-e^{z\cdot \overline{w}}]\,\frac{dt}{t^{1+s}}.\end{eqnarray}$$
Given
$z,w\in \mathbb{C}^{n}$
, put
$x=\text{Re}(z\cdot \overline{w})$
for short. Then
$$\begin{eqnarray}\displaystyle |I_{2}| & {\lesssim} & \displaystyle \int _{1}^{\infty }[\exp (e^{-2t}x)e^{-nt}+e^{x}]\,\frac{dt}{t^{1+s}}\nonumber\\ \displaystyle & {\lesssim} & \displaystyle \left\{\begin{array}{@{}ll@{}}1 & \text{if }x\leqslant 1,\\ e^{x} & \text{if }x>1.\end{array}\right.\nonumber\end{eqnarray}$$
Also,
$$\begin{eqnarray}\displaystyle |I_{1}| & {\lesssim} & \displaystyle e^{x}\int _{0}^{1}|\text{exp}(-(1-e^{-2t})z\cdot \overline{w}-nt)-1|\,\frac{dt}{t^{1+s}}\nonumber\\ \displaystyle & = & \displaystyle e^{x}\int _{0}^{1}|E_{1}(-(1-e^{-2t})z\cdot \overline{w}-nt)|\,\frac{dt}{t^{1+s}}.\nonumber\end{eqnarray}$$
Here
$E_{s}(x)$
is the truncated exponential function (see Definition A.1 in Appendix A). Note that (see (A1))
$$\begin{eqnarray}\frac{E_{1}(\unicode[STIX]{x1D706})}{\unicode[STIX]{x1D706}}=\int _{0}^{1}e^{\unicode[STIX]{x1D70C}\unicode[STIX]{x1D706}}\,d\unicode[STIX]{x1D70C}\quad \text{for }\unicode[STIX]{x1D706}\in \mathbb{C}.\end{eqnarray}$$
Then
$$\begin{eqnarray}\frac{|E_{1}(-(1-e^{-2t})z\cdot \overline{w}-nt)|}{|(1-e^{-2t})z\cdot \overline{w}+nt|}\leqslant \int _{0}^{1}e^{-\unicode[STIX]{x1D70C}\{(1-e^{-2t})x+nt\}}\,d\unicode[STIX]{x1D70C}.\end{eqnarray}$$
Hence we have
$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{0}^{1}|E_{1}(-(1-e^{-2t})z\cdot \overline{w}-nt)|\,\frac{dt}{t^{1+s}}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \,\int _{0}^{1}|(1-e^{-2t})z\cdot \overline{w}+nt|\int _{0}^{1}e^{-\unicode[STIX]{x1D70C}\{(1-e^{-2t})x+nt\}}\,d\unicode[STIX]{x1D70C}\,\frac{dt}{t^{1+s}}\nonumber\\ \displaystyle & & \displaystyle \quad =\int _{0}^{1}\biggl|\frac{1-e^{-2t}}{t}z\cdot \overline{w}+n\biggr|\int _{0}^{1}e^{-\unicode[STIX]{x1D70C}\{(1-e^{-2t})x+nt\}}\,d\unicode[STIX]{x1D70C}\,\frac{dt}{t^{s}}.\nonumber\end{eqnarray}$$
Since
$$\begin{eqnarray}|1-e^{-2t}|\approx t\quad \text{for }0<t<1,\end{eqnarray}$$
there exist
$c>0$
such that
$$\begin{eqnarray}\int _{0}^{1}|E_{1}(-(1-e^{-2t})z\cdot \overline{w}-nt)|\,\frac{dt}{t^{1+s}}\lesssim |z\cdot \overline{w}|\int _{0}^{1}\int _{0}^{1}e^{-c\unicode[STIX]{x1D70C}tx}\,d\unicode[STIX]{x1D70C}\,\frac{dt}{t^{s}}.\end{eqnarray}$$
If
$x\leqslant 1$
, then
$$\begin{eqnarray}\int _{0}^{1}\int _{0}^{1}e^{-c\unicode[STIX]{x1D70C}tx}\,d\unicode[STIX]{x1D70C}\,\frac{dt}{t^{s}}\lesssim 1.\end{eqnarray}$$
So, in case
$x\leqslant 1$
, we have
$$\begin{eqnarray}\displaystyle |\mathscr{R}^{s}(e^{z\cdot \overline{w}})| & {\lesssim} & \displaystyle 1+|z\cdot \overline{w}|\nonumber\\ \displaystyle & {\lesssim} & \displaystyle (1+|z\cdot \overline{w}|)^{s}e^{(1/2)|z||w|}\frac{(1+|z\cdot \overline{w}|)^{1-s}}{e^{(1/2)|z||w|}}\nonumber\\ \displaystyle & {\lesssim} & \displaystyle (1+|z\cdot \overline{w}|)^{s}e^{(1/2)|z||w|}\nonumber\\ \displaystyle & {\lesssim} & \displaystyle (1+|z\cdot \overline{w}|)^{s}e^{(1/2)|z|^{2}+(1/2)|w|^{2}-(1/8)|z-w|^{2}}.\end{eqnarray}$$
Here we used the following inequality
$$\begin{eqnarray}\displaystyle e^{(1/2)|z||w|} & = & \displaystyle e^{(3/4)|z||w|-(1/4)|z||w|}\leqslant e^{(3/8)|z|^{2}+(3/8)|w|^{2}+(1/4)\text{Re}(z\cdot \overline{w})}\nonumber\\ \displaystyle & = & \displaystyle e^{(1/2)|z|^{2}+(1/2)|w|^{2}-(1/8)|z-w|^{2}}.\nonumber\end{eqnarray}$$
If
$x>1$
, by Fubini’s theorem, it follows that
$$\begin{eqnarray}\displaystyle \int _{0}^{1}\int _{0}^{1}e^{-c\unicode[STIX]{x1D70C}tx}\,d\unicode[STIX]{x1D70C}\,\frac{dt}{t^{s}} & = & \displaystyle \int _{0}^{1}\int _{0}^{1}e^{-c\unicode[STIX]{x1D70C}tx}t^{-s}\,dt\,d\unicode[STIX]{x1D70C}\nonumber\\ \displaystyle & {\lesssim} & \displaystyle x^{s-1}\unicode[STIX]{x1D6E4}(1-s)\int _{0}^{1}\unicode[STIX]{x1D70C}^{s-1}\,d\unicode[STIX]{x1D70C}\nonumber\\ \displaystyle & {\lesssim} & \displaystyle x^{s-1}.\nonumber\end{eqnarray}$$
Hence, in case
$x>1$
, we have
$$\begin{eqnarray}|\mathscr{R}^{s}(e^{z\cdot \overline{w}})|\lesssim e^{x}|z\cdot \overline{w}|x^{s-1}.\end{eqnarray}$$
For the case
$x=\text{Re}(z\cdot \overline{w})>1$
, we write
$\text{Re}(z\cdot \overline{w})=|z||w|\cos \unicode[STIX]{x1D703}$
, where
$\unicode[STIX]{x1D703}$
is the angle between
$z$
and
$w$
identified as real vectors in
$\mathbb{R}^{2n}$
, and
$\unicode[STIX]{x1D6FF}=\cos ^{-1}(\frac{1}{4})$
. If
$|\unicode[STIX]{x1D703}|\leqslant \unicode[STIX]{x1D6FF}$
, then
$$\begin{eqnarray}x=\text{Re}(z\cdot \overline{w})\approx |z\cdot \overline{w}|\approx |z||w|.\end{eqnarray}$$
Hence we have
$$\begin{eqnarray}e^{x}|z\cdot \overline{w}|x^{s-1}\lesssim e^{(1/2)|z|^{2}+(1/2)|w|^{2}-(1/8)|z-w|^{2}}|z\cdot \overline{w}|^{s}.\end{eqnarray}$$
If
$\unicode[STIX]{x1D6FF}<\unicode[STIX]{x1D703}<\unicode[STIX]{x1D70B}/2$
, then
$$\begin{eqnarray}x=\text{Re}(z\cdot \overline{w})=|z||w|\cos \unicode[STIX]{x1D703}<{\textstyle \frac{1}{4}}|z||w|.\end{eqnarray}$$
Hence
$$\begin{eqnarray}\displaystyle e^{x}|z\cdot \overline{w}|x^{s-1} & {\leqslant} & \displaystyle e^{(1/2)|z||w|}|z\cdot \overline{w}|^{s}\frac{|z\cdot \overline{w}|^{1-s}}{x^{1-s}e^{(1/4)|z\cdot \overline{w}|}}\nonumber\\ \displaystyle & {\lesssim} & \displaystyle e^{(1/2)|z|^{2}+(1/2)|w|^{2}-(1/8)|z-w|^{2}}|z\cdot \overline{w}|^{s}.\end{eqnarray}$$
This, together with (3.2), yields the asserted estimate for
$0<s<1$
.
Now, assume
$s>1$
. Let
$m$
be the greatest nonnegative integer less than
$s$
. Then
$$\begin{eqnarray}\displaystyle \mathscr{R}^{s}(e^{z\cdot \overline{w}}) & = & \displaystyle \mathscr{R}^{s-m}\mathscr{R}^{m}(e^{z\cdot \overline{w}})\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{\unicode[STIX]{x1D6E4}(m-s)}\int _{0}^{\infty }[\mathscr{R}^{m}\exp \big(e^{-2t}z\cdot \overline{w}\big)e^{-nt}-\mathscr{R}^{m}e^{z\cdot \overline{w}}]\,\frac{dt}{t^{1+s-m}}.\nonumber\end{eqnarray}$$
Note that
$$\begin{eqnarray}\mathscr{R}^{m}(e^{z\cdot \overline{w}})=\mathop{\sum }_{j=0}^{m}\ell _{j}(z\cdot \overline{w})^{j}e^{z\cdot \overline{w}}\end{eqnarray}$$
and
$$\begin{eqnarray}\mathscr{R}^{m}\exp (e^{-2t}z\cdot \overline{w})=\mathop{\sum }_{j=0}^{m}\ell _{j}(e^{-2t}z\cdot \overline{w})^{j}\exp (e^{-2t}z\cdot \overline{w}),\end{eqnarray}$$
for some nonnegative integers
$\ell _{j}$
. Thus
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathscr{R}^{s}(e^{z\cdot \overline{w}})\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{1}{\unicode[STIX]{x1D6E4}(m-s)}\mathop{\sum }_{j=0}^{m}\ell _{j}(z\cdot \overline{w})^{j}\int _{0}^{\infty }[\exp (e^{-2t}z\cdot \overline{w})e^{-(2j+n)t}-e^{z\cdot \overline{w}}]\,\frac{dt}{t^{1+s-m}}.\nonumber\end{eqnarray}$$
We write the integral on the right-hand side of the above equation as the sum of two pieces
$J_{1}$
and
$J_{2}$
defined by
$$\begin{eqnarray}J_{1}=\int _{0}^{1}[\exp (e^{-2t}z\cdot \overline{w})e^{-(2j+n)t}-e^{z\cdot \overline{w}}]\,\frac{dt}{t^{1+s-m}}\end{eqnarray}$$
and
$$\begin{eqnarray}J_{2}=\int _{1}^{\infty }[\exp (e^{-2t}z\cdot \overline{w})e^{-(2j+n)t}-e^{z\cdot \overline{w}}]\,\frac{dt}{t^{1+s-m}}.\end{eqnarray}$$
Then
$$\begin{eqnarray}|J_{2}|\lesssim \int _{1}^{\infty }[\exp (e^{-2t}x)e^{-(2j+n)t}+e^{x}]\,\frac{dt}{t^{1+s-m}}\end{eqnarray}$$
and
$$\begin{eqnarray}\displaystyle |J_{1}| & {\lesssim} & \displaystyle e^{x}\int _{0}^{1}|E_{1}(-(1-e^{-2t})z\cdot \overline{w}-(2j+n)t)|\,\frac{dt}{t^{1+s-m}}\nonumber\\ \displaystyle & {\lesssim} & \displaystyle e^{x}|z\cdot \overline{w}|\int _{0}^{1}\int _{0}^{1}e^{-c\unicode[STIX]{x1D70C}tx}\,d\unicode[STIX]{x1D70C}\,\frac{dt}{t^{s-m}}.\nonumber\end{eqnarray}$$
These yield the asserted estimate for
$s>1$
.
Now for
$s>0$
, by (ii) of Proposition 2.4, we have
$$\begin{eqnarray}\displaystyle \mathscr{R}^{-s}(e^{z\cdot \overline{w}}) & = & \displaystyle \frac{1}{\unicode[STIX]{x1D6E4}(s)}\int _{0}^{\infty }e^{-t\mathscr{R}}(e^{z\cdot \overline{w}})\,\frac{dt}{t^{1-s}}\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{\unicode[STIX]{x1D6E4}(s)}\int _{0}^{\infty }\exp (e^{-2t}z\cdot \overline{w})e^{-nt}\,\frac{dt}{t^{1-s}}.\nonumber\end{eqnarray}$$
Hence
$$\begin{eqnarray}|\mathscr{R}^{-s}(e^{z\cdot \overline{w}})|\leqslant \frac{1}{\unicode[STIX]{x1D6E4}(s)}\int _{0}^{\infty }\exp (e^{-2t}x)e^{-nt}\,\frac{dt}{t^{1-s}}.\end{eqnarray}$$
If
$x=\text{Re}(z\cdot \overline{w})\leqslant 1$
, then
$$\begin{eqnarray}|\mathscr{R}^{-s}(e^{z\cdot \overline{w}})|\lesssim \frac{1}{\unicode[STIX]{x1D6E4}(s)}\int _{0}^{\infty }e^{-nt}\,\frac{dt}{t^{1-s}}=\frac{1}{n^{s}}.\end{eqnarray}$$
Now we assume that
$x=\text{Re}(z\cdot \overline{w})>1$
. Then
$$\begin{eqnarray}\displaystyle |\mathscr{R}^{-s}(e^{z\cdot \overline{w}})| & {\leqslant} & \displaystyle \frac{1}{\unicode[STIX]{x1D6E4}(s)}\mathop{\sum }_{k=0}^{\infty }\int _{0}^{\infty }\frac{(e^{-2t}x)^{k}}{k!}e^{-nt}\,\frac{dt}{t^{1-s}}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{k=0}^{\infty }\frac{x^{k}}{k!}\frac{1}{\unicode[STIX]{x1D6E4}(s)}\int _{0}^{\infty }e^{-(2k+n)t}\,\frac{dt}{t^{1-s}}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{k=0}^{\infty }\frac{x^{k}}{(2k+n)^{s}k!}.\nonumber\end{eqnarray}$$
By Stirling’s formula, it follows that
$$\begin{eqnarray}k!(2k+n)^{s}\approx \unicode[STIX]{x1D6E4}(k+1+s)\quad \text{for large }k.\end{eqnarray}$$
Hence, by Corollary A.4, we have
$$\begin{eqnarray}\displaystyle |\mathscr{R}^{-s}(e^{z\cdot \overline{w}})| & {\lesssim} & \displaystyle \mathop{\sum }_{k=0}^{\infty }\frac{x^{k+s}}{\unicode[STIX]{x1D6E4}(k+1+s)}x^{-s}\nonumber\\ \displaystyle & = & \displaystyle E_{s}(x)x^{-s}\nonumber\\ \displaystyle & {\approx} & \displaystyle e^{x}x^{-s},\quad x>1.\end{eqnarray}$$
For the case
$x=\text{Re}(z\cdot \overline{w})>1$
, we write
$\text{Re}(z\cdot \overline{w})=|z||w|\cos \unicode[STIX]{x1D703}$
, where
$\unicode[STIX]{x1D703}$
is the angle between
$z$
and
$w$
identified as real vectors in
$\mathbb{R}^{2n}$
, and
$\unicode[STIX]{x1D6FF}=\cos ^{-1}(\frac{1}{4})$
. It is easily seen from (3.5) that the required estimate holds when
$|\unicode[STIX]{x1D703}|\leqslant \unicode[STIX]{x1D6FF}$
, because
$x\approx |z||w|$
for such
$z$
and
$w$
. So, assume
$\unicode[STIX]{x1D6FF}<\unicode[STIX]{x1D703}<\unicode[STIX]{x1D70B}/2$
. Note
$x<\frac{1}{4}|z||w|$
for such
$z$
and
$w$
. We thus have by our choice of
$\unicode[STIX]{x1D6FF}$
$$\begin{eqnarray}\displaystyle \frac{e^{x}}{x^{s}} & {\leqslant} & \displaystyle \frac{e^{(1/4)|z||w|}}{x^{s}}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle e^{(1/4)|z||w|}\nonumber\\ \displaystyle & = & \displaystyle \frac{e^{(1/2)|z||w|}}{(|z||w|)^{s}}\frac{(|z||w|)^{s}}{e^{(1/4)|z||w|}}\nonumber\\ \displaystyle & {\lesssim} & \displaystyle \frac{e^{(1/2)|z||w|}}{(|z||w|)^{s}},\quad x>1.\end{eqnarray}$$
This, together with (3.5), yields the asserted estimate for
$x>1$
. This completes the proof.◻
It is the well-known formula [Reference Aronszajn1] that
$$\begin{eqnarray}K_{s}(z,w):=K_{w}^{s}(z)=\mathop{\sum }_{\unicode[STIX]{x1D6FC}}\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D6FC}}(z)\overline{\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D6FC}}(w)}\end{eqnarray}$$
where
$\{\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D6FC}}\}$
is any orthonormal basis for
$F_{\mathscr{R}}^{s,2}$
.
Lemma 3.2. Let
$s$
be real and
$\unicode[STIX]{x1D6FC}$
be a multi-index of nonnegative integers. Then
$$\begin{eqnarray}\Vert z^{\unicode[STIX]{x1D6FC}}\Vert _{F_{\mathscr{R}}^{s,2}}^{2}=(2|\unicode[STIX]{x1D6FC}|+n)^{s}\unicode[STIX]{x1D6FC}!.\end{eqnarray}$$
Proof. Since
$\mathscr{R}^{s/2}z^{\unicode[STIX]{x1D6FC}}=(2|\unicode[STIX]{x1D6FC}|+n)^{s/2}z^{\unicode[STIX]{x1D6FC}}$
, we have
$$\begin{eqnarray}\Vert z^{\unicode[STIX]{x1D6FC}}\Vert _{F_{\mathscr{R}}^{s,2}}^{2}=\Vert \mathscr{R}^{s/2}z^{\unicode[STIX]{x1D6FC}}\Vert _{2}^{2}=(2|\unicode[STIX]{x1D6FC}|+n)^{s}\Vert z^{\unicode[STIX]{x1D6FC}}\Vert _{2}^{2}=(2|\unicode[STIX]{x1D6FC}|+n)^{s}\unicode[STIX]{x1D6FC}!.\square\end{eqnarray}$$
Theorem 3.3. Let
$s\in \mathbb{R}$
. Then
$$\begin{eqnarray}K_{s}(z,w)=\mathscr{R}_{z}^{-s}(e^{z\cdot \overline{w}})\end{eqnarray}$$
and there are positive constants
$C=C(s)>0$
such that
$$\begin{eqnarray}|K_{s}(z,w)|\leqslant C\times \left\{\begin{array}{@{}ll@{}}(1+|z||w|)^{-s}\unicode[STIX]{x1D6EC}(z,w) & \text{if }s>0,\\ (1+|z\cdot \overline{w}|)^{-s}\unicode[STIX]{x1D6EC}(z,w) & \text{if }s\leqslant 0,\end{array}\right.\end{eqnarray}$$
for
$z,w\in \mathbb{C}^{n}$
.
Proof. By Lemma 3.2, we get
$$\begin{eqnarray}\displaystyle K_{s}(z,w) & = & \displaystyle \mathop{\sum }_{\unicode[STIX]{x1D6FC}\in \mathbb{N}_{0}^{n}}\frac{z^{\unicode[STIX]{x1D6FC}}\overline{w}^{\unicode[STIX]{x1D6FC}}}{\Vert z^{\unicode[STIX]{x1D6FC}}\Vert _{F_{\mathscr{R}}^{s,2}}^{2}}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{\unicode[STIX]{x1D6FC}}\frac{z^{\unicode[STIX]{x1D6FC}}\overline{w}^{\unicode[STIX]{x1D6FC}}}{(2|\unicode[STIX]{x1D6FC}|+n)^{s}\unicode[STIX]{x1D6FC}!}\nonumber\\ \displaystyle & = & \displaystyle \mathscr{R}_{z}^{-s}(e^{z\cdot \overline{w}}).\nonumber\end{eqnarray}$$
Hence the size estimates of
$K_{s}(z,w)$
follow from Theorem 3.1.◻
4 Auxiliary integral estimates
It follows that the fractional Fock–Sobolev spaces are realized as the weighted Fock spaces that do not involve derivatives. To prove the results we introduce an auxiliary integral estimate for
$\unicode[STIX]{x1D6EC}$
defined by
$$\begin{eqnarray}\unicode[STIX]{x1D6EC}(z,w)=e^{(1/2)|z|^{2}+(1/2)|w|^{2}-(1/8)|z-w|^{2}}.\end{eqnarray}$$
To handle the case
$1\leqslant p<\infty$
and for other purposes later, we introduce an integral operator induced by
$\unicode[STIX]{x1D6EC}$
. Given
$s$
real, we consider an integral operator
$L_{s}$
defined by
$$\begin{eqnarray}L_{s}\unicode[STIX]{x1D713}(z):=\int _{\mathbb{C}^{n}}\unicode[STIX]{x1D713}(w)\biggl(\frac{1+|z|}{1+|w|}\biggr)^{s}\unicode[STIX]{x1D6EC}(z,w)e^{-|w|^{2}}\,dV(w),\quad z\in \mathbb{C}^{n}\end{eqnarray}$$
for
$\unicode[STIX]{x1D713}$
which makes the above integral well-defined.
Lemma 4.1. [Reference Cho, Choe and Koo4]
Given
$s$
real, the operator
$L_{s}$
is bounded on
$L_{G}^{p}$
for any
$1\leqslant p\leqslant \infty$
.
The following Jensen-type inequality is needed to handle the case
$0<p\leqslant 1$
.
Lemma 4.2. [Reference Cho, Choe and Koo4]
Given
$0<p\leqslant 1$
,
$a>0$
and
$s$
real, there is a constant
$C=C(p,a,s)>0$
such that
$$\begin{eqnarray}\left\{\int _{\mathbb{C}^{n}}|(1+|z|)^{s}f(z)e^{-a|z|^{2}}|\,dV(z)\right\}^{p}\leqslant C\int _{\mathbb{ C}^{n}}|(1+|z|)^{s}f(z)e^{-a|z|^{2}}|^{p}\,dV(z)\end{eqnarray}$$
for
$f\in H(\mathbb{C}^{n})$
.
Lemma 4.3. [Reference Cho, Choe and Koo4]
Let
$0<p<\infty$
and
$\unicode[STIX]{x1D6FC}$
be an arbitrary real number. Then there is
$C=C(p,\unicode[STIX]{x1D6FC})>0$
such that
$$\begin{eqnarray}\int _{\mathbb{C}^{n}}(1+|w|)^{\unicode[STIX]{x1D6FC}}\unicode[STIX]{x1D6EC}(z,w)^{p}e^{-(p/2)|w|^{2}}\,dV(w)\leqslant C(1+|z|)^{\unicode[STIX]{x1D6FC}}e^{(p/2)|z|^{2}}.\end{eqnarray}$$
5 Fourier type characterization
Cho and Zhu [Reference Cho and Zhu6] studied Fock–Sobolev spaces of positive integer order. For any positive integer
$m$
and
$0<p\leqslant \infty$
we consider the space
$F^{m,p}$
consisting of entire functions
$f$
on
$\mathbb{C}^{n}$
such that
$$\begin{eqnarray}\mathop{\sum }_{|\unicode[STIX]{x1D6FC}|\leqslant m}\Vert \unicode[STIX]{x2202}^{\unicode[STIX]{x1D6FC}}f\Vert _{p}<\infty ,\end{eqnarray}$$
where
$\Vert ~\Vert _{p}$
is the norm in
$F^{p}$
. See [Reference Hall and Lewkeeratiyutkul8, Reference Radha and Thangavelu12] for other similar Sobolev spaces. Cho and Zhu [Reference Cho and Zhu6] proved a useful Fourier type characterization of the Fock–Sobolev space of integer order as following.
Theorem 5.1. [Reference Cho and Zhu6]
Suppose
$0<p\leqslant \infty$
,
$m$
is a nonnegative integer, and
$f$
is an entire function on
$\mathbb{C}^{n}$
. Then
$f\in F^{m,p}$
if and only if the function
$z^{\unicode[STIX]{x1D6FC}}f(z)$
is in
$F^{p}$
for all multi-indices
$\unicode[STIX]{x1D6FC}$
with
$|\unicode[STIX]{x1D6FC}|=m$
. Moreover,
$\Vert f\Vert _{F^{m,p}}$
is comparable to the norm of the function
$|z|^{m}f(z)$
in
$L_{G}^{p}$
.
The purpose of the current paper is to extend the notion of the Fock–Sobolev spaces to the case of fractional orders allowed to be any real number.
Theorem 5.2. Let
$s\in \mathbb{R}$
and
$0<p\leqslant \infty$
. There is a constant
$C=C(s,p)>0$
such that
$$\begin{eqnarray}\Vert f\Vert _{F_{\mathscr{R}}^{s,p}}\leqslant C\Vert f\Vert _{F_{s}^{p}}.\end{eqnarray}$$
Proof. We now consider the cases
$0<p<1$
and
$1\leqslant p\leqslant \infty$
separately.
Assume
$1\leqslant p\leqslant \infty$
. If the function
$(1+|w|)^{s}f(w)$
is in
$L_{G}^{p}$
, then
$$\begin{eqnarray}f(z)=\frac{1}{\unicode[STIX]{x1D70B}^{n}}\int _{\mathbb{C}^{n}}e^{z\cdot \overline{w}}f(w)e^{-|w|^{2}}\,dV(w),\quad z\in \mathbb{C}^{n}.\end{eqnarray}$$
Thus we obtain
$$\begin{eqnarray}\mathscr{R}^{s/2}f(z)=\frac{1}{\unicode[STIX]{x1D70B}^{n}}\int _{\mathbb{C}^{n}}\mathscr{R}^{s/2}(e^{z\cdot \overline{w}})f(w)e^{-|w|^{2}}\,dV(w).\end{eqnarray}$$
The convergence of the integrals above follows from pointwise estimates for functions in Fock spaces. Hence it follows that
$$\begin{eqnarray}\displaystyle |\mathscr{R}^{s/2}f(z)| & {\leqslant} & \displaystyle \frac{1}{\unicode[STIX]{x1D70B}^{n}}\int _{\mathbb{C}^{n}}|\mathscr{R}^{s/2}(e^{z\cdot \overline{w}})||f(w)|e^{-|w|^{2}}\,dV(w)\nonumber\\ \displaystyle & {\lesssim} & \displaystyle \int _{\mathbb{C}^{n}}|f(w)|(1+|w|)^{s}\biggl(\frac{1+|z|}{1+|w|}\biggr)^{s/2}\unicode[STIX]{x1D6EC}(z,w)e^{-|w|^{2}}\,dV(w)\nonumber\\ \displaystyle & = & \displaystyle L_{s}((1+|w|)^{s}|f|)(z).\nonumber\end{eqnarray}$$
By Lemma 4.1, we have
$$\begin{eqnarray}\Vert \mathscr{R}^{s/2}f\Vert _{p}^{p}\lesssim \Vert (1+|z|)^{s}f\Vert _{L_{G}^{p}}.\end{eqnarray}$$
Now let
$0<p<1$
. Then, by Lemma 4.2 and Theorem 3.1,
$$\begin{eqnarray}\displaystyle |\mathscr{R}^{s/2}f(z)|^{p} & {\lesssim} & \displaystyle \int _{\mathbb{C}^{n}}|\mathscr{R}^{s/2}(e^{z\cdot \overline{w}})|^{p}|f(w)|^{p}e^{-p|w|^{2}}\,dV(w)\nonumber\\ \displaystyle & {\lesssim} & \displaystyle \int _{\mathbb{C}^{n}}|f(w)|^{p}(1+|w|)^{sp/2}(1+|z|)^{sp/2}\unicode[STIX]{x1D6EC}(z,w)^{p}e^{-p|w|^{2}}\,dV(w)\nonumber\end{eqnarray}$$
or
$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{\mathbb{C}^{n}}|\mathscr{R}^{s/2}f(z)|^{p}e^{-(p/2)|z|^{2}}\,dV(z)\nonumber\\ \displaystyle & & \displaystyle \quad \lesssim \,\int _{\mathbb{C}^{n}}|f(w)|^{p}(1+|w|)^{sp/2}e^{-p|w|^{2}}\,dV(w)\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\int _{\mathbb{C}^{n}}(1+|z|)^{sp/2}\unicode[STIX]{x1D6EC}(z,w)^{p}e^{-(p/2)|z|^{2}}\,dV(z).\nonumber\end{eqnarray}$$
Now, by Lemma 4.3, it follows that
$$\begin{eqnarray}\int _{\mathbb{C}^{n}}(1+|z|)^{sp/2}\unicode[STIX]{x1D6EC}(z,w)^{p}e^{-(p/2)|z|^{2}}\,dV(z)\lesssim (1+|w|)^{sp/2}e^{(p/2)|w|^{2}}.\end{eqnarray}$$
Hence
$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{\mathbb{C}^{n}}|\mathscr{R}^{s/2}f(z)|^{p}e^{-(p/2)|z|^{2}}\,dV(z)\nonumber\\ \displaystyle & & \displaystyle \quad \lesssim \int _{\mathbb{C}^{n}}|f(w)|^{p}(1+|w|)^{sp}e^{-(p/2)|w|^{2}}\,dV(w).\square\nonumber\end{eqnarray}$$
Theorem 5.3. Suppose
$0<p\leqslant \infty$
and
$s$
is a real number. Then there is a constant
$C=C(s,p)>0$
such that
$$\begin{eqnarray}\Vert f\Vert _{F_{s}^{p}}\leqslant C\Vert f\Vert _{F_{\mathscr{R}}^{s,p}}\end{eqnarray}$$
for all
$f\in F_{\mathscr{R}}^{s,p}$
.
Proof. Let
$1\leqslant p\leqslant \infty$
. From the reproducing formula for
$\mathscr{R}^{s/2}f$
we obtain
$$\begin{eqnarray}f(z)=\mathscr{R}^{-s/2}\mathscr{R}^{s/2}f(z)=\frac{1}{\unicode[STIX]{x1D70B}^{n}}\int _{\mathbb{C}^{n}}\mathscr{R}_{w}^{s/2}f(w)\mathscr{R}_{z}^{-s/2}(e^{z\cdot \overline{w}})e^{-|w|^{2}}\,dV(w).\end{eqnarray}$$
This together with Theorem 3.1 shows that
$$\begin{eqnarray}\displaystyle (1+|z|)^{s}|f(z)| & {\lesssim} & \displaystyle (1+|z|)^{s}\int _{\mathbb{C}^{n}}|\mathscr{R}^{s/2}f(w)||\mathscr{R}^{-s/2}(e^{z\cdot \overline{w}})|e^{-|w|^{2}}\,dV(w)\nonumber\\ \displaystyle & {\lesssim} & \displaystyle \int _{\mathbb{C}^{n}}|\mathscr{R}^{s/2}f(w)|\biggl(\frac{1+|z|}{1+|w|}\biggr)^{s/2}\unicode[STIX]{x1D6EC}(z,w)e^{-|w|^{2}}\,dV(w)\nonumber\\ \displaystyle & = & \displaystyle L_{s}(|\mathscr{R}^{s/2}f|)(z).\nonumber\end{eqnarray}$$
By Lemma 4.1, we have
$$\begin{eqnarray}\Vert (1+|z|)^{s}f\Vert _{L_{G}^{p}}\lesssim \Vert \mathscr{R}^{s/2}f\Vert _{p}.\end{eqnarray}$$
When
$0<p<1$
, it follows from Lemma 4.2 and Theorem 3.1 that
$$\begin{eqnarray}\displaystyle |f(z)|^{p} & {\lesssim} & \displaystyle \biggl|\int _{\mathbb{C}^{n}}\mathscr{R}^{s/2}f(w)\mathscr{R}^{-s/2}(e^{z\cdot \overline{w}})e^{-|w|^{2}}\,dV(w)\biggr|^{p}\nonumber\\ \displaystyle & {\lesssim} & \displaystyle \int _{\mathbb{C}^{n}}|\mathscr{R}^{s/2}f(w)\mathscr{R}^{-s/2}(e^{z\cdot \overline{w}})e^{-|w|^{2}}|^{p}\,dV(w)\nonumber\\ \displaystyle & {\lesssim} & \displaystyle \int _{\mathbb{C}^{n}}|\mathscr{R}^{s/2}f(w)|^{p}\frac{e^{(p/2)|z|^{2}-(p/2)|w|^{2}-(p/8)|z-w|^{2}}}{(1+|z|)^{sp/2}(1+|w|)^{sp/2}}\,dV(w).\nonumber\end{eqnarray}$$
Fubini’s theorem shows that the integral
$$\begin{eqnarray}I=\int _{\mathbb{C}^{n}}|(1+|z|)^{s}f(z)e^{-(1/2)|z|^{2}}|^{p}\,dV(z)\end{eqnarray}$$
satisfies the following estimates:
$$\begin{eqnarray}\displaystyle I\lesssim \int _{\mathbb{C}^{n}}|\mathscr{R}^{s/2}f(w)|^{p}e^{-(p/2)|w|^{2}}\,dV(w)\int _{\mathbb{C}^{n}}\biggl(\frac{1+|z|}{1+|w|}\biggr)^{sp/2}e^{-(p/8)|z-w|^{2}}\,dV(z). & & \displaystyle \nonumber\end{eqnarray}$$
Note that
$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{\mathbb{C}^{n}}\biggl(\frac{1+|z|}{1+|w|}\biggr)^{sp/2}e^{-(p/8)|z-w|^{2}}\,dV(z)\nonumber\\ \displaystyle & & \displaystyle \quad \lesssim \,\int _{\mathbb{C}^{n}}(1+|z-w|)^{sp/2}e^{-(p/8)|z-w|^{2}}\,dV(z)\nonumber\\ \displaystyle & & \displaystyle \quad \lesssim \,1.\nonumber\end{eqnarray}$$
The proof is complete. ◻
Appendix. Truncated exponential functions
Let
$m$
be a positive integer. We consider the left truncated exponential function of integer order
$m$
,
$E_{m}(\unicode[STIX]{x1D706})$
, defined by
$$\begin{eqnarray}\displaystyle E_{m}(\unicode[STIX]{x1D706}) & = & \displaystyle e^{\unicode[STIX]{x1D706}}-1-\unicode[STIX]{x1D706}-\frac{\unicode[STIX]{x1D706}^{2}}{2!}-\cdots -\frac{\unicode[STIX]{x1D706}^{m-1}}{(m-1)!}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{k=0}^{\infty }\frac{\unicode[STIX]{x1D706}^{k+m}}{\unicode[STIX]{x1D6E4}(k+1+m)},\quad \unicode[STIX]{x1D706}\in \mathbb{C},\nonumber\end{eqnarray}$$
where
$\unicode[STIX]{x1D6E4}$
is the classical gamma function.
It is easy to check that
$$\begin{eqnarray}\frac{E_{m}(\unicode[STIX]{x1D706})}{\unicode[STIX]{x1D706}^{m}}=\frac{1}{(m-1)!}\int _{0}^{1}(1-t)^{m-1}e^{t\unicode[STIX]{x1D706}}\,dt,\end{eqnarray}$$
which immediately yields a useful inequality
$$\begin{eqnarray}|E_{m}(\unicode[STIX]{x1D706})|\leqslant \biggl(\frac{|\unicode[STIX]{x1D706}|}{\text{Re}\,\unicode[STIX]{x1D706}}\biggr)^{m}E_{m}(\text{Re}\,\unicode[STIX]{x1D706}),\quad \unicode[STIX]{x1D706}\in \mathbb{C}.\end{eqnarray}$$
Now we consider the truncated exponential function of fractional order.
Definition A.1. Let
$s\in \mathbb{R}$
. We define the generalized exponential function of fractional order
$s$
,
$E_{s}(x)$
, by
$$\begin{eqnarray}E_{s}(x)=\mathop{\sum }_{k=0}^{\infty }\frac{x^{k+s}}{\unicode[STIX]{x1D6E4}(k+1+s)},\quad x\in \mathbb{R}.\end{eqnarray}$$
We have the following integral representation of
$E_{s}(x)$
:
Proposition A.2. Let
$s>0$
. Then
$$\begin{eqnarray}E_{s}(x)=\frac{e^{x}}{\unicode[STIX]{x1D6E4}(s)}\int _{0}^{x}t^{s-1}e^{-t}\,dt,\quad x\in \mathbb{R}.\end{eqnarray}$$
Proof. Note that the following well-known property of gamma functions
$$\begin{eqnarray}\frac{\unicode[STIX]{x1D6E4}(a)\unicode[STIX]{x1D6E4}(b)}{\unicode[STIX]{x1D6E4}(a+b)}=\int _{0}^{1}t^{a-1}(1-t)^{b-1}\,dt\quad \text{when }a,b>0.\end{eqnarray}$$
Thus
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{k=0}^{\infty }\frac{x^{k+s}}{\unicode[STIX]{x1D6E4}(k+1+s)} & = & \displaystyle \mathop{\sum }_{k=0}^{\infty }\frac{x^{k+s}}{\unicode[STIX]{x1D6E4}(s)\unicode[STIX]{x1D6E4}(k+1)}\int _{0}^{1}(1-t)^{s-1}t^{k}\,dt\nonumber\\ \displaystyle & = & \displaystyle \frac{x^{s}}{\unicode[STIX]{x1D6E4}(s)}\int _{0}^{1}(1-t)^{s-1}e^{tx}\,dt\nonumber\\ \displaystyle & = & \displaystyle \frac{e^{x}}{\unicode[STIX]{x1D6E4}(s)}\int _{0}^{x}t^{s-1}e^{-t}\,dt.\square\nonumber\end{eqnarray}$$
Proposition A.3. Let
$s=m+r$
where
$m$
is a nonnegative integer and
$0\leqslant r<1$
. Then
$$\begin{eqnarray}E_{-s}(x)=\frac{e^{x}}{\unicode[STIX]{x1D6E4}(1-r)}\int _{0}^{x}t^{-r}e^{-t}\,dt+\mathop{\sum }_{k=0}^{m}\frac{x^{k-s}}{\unicode[STIX]{x1D6E4}(k+1-s)}.\end{eqnarray}$$
Proof. We have
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{k=0}^{\infty }\frac{x^{k-s}}{\unicode[STIX]{x1D6E4}(k+1-s)} & = & \displaystyle \mathop{\sum }_{k=m+1}^{\infty }\frac{x^{k-s}}{\unicode[STIX]{x1D6E4}(k+1-s)}+\mathop{\sum }_{k=0}^{m}\frac{x^{k-s}}{\unicode[STIX]{x1D6E4}(k+1-s)}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{k=0}^{\infty }\frac{x^{k+1-r}}{\unicode[STIX]{x1D6E4}(k+2-r)}+\mathop{\sum }_{k=0}^{m}\frac{x^{k-s}}{\unicode[STIX]{x1D6E4}(k+1-s)}\nonumber\\ \displaystyle & = & \displaystyle \frac{e^{x}}{\unicode[STIX]{x1D6E4}(1-r)}\int _{0}^{x}t^{-r}e^{-t}\,dt+\mathop{\sum }_{k=0}^{m}\frac{x^{k-s}}{\unicode[STIX]{x1D6E4}(k+1-s)}.\square\nonumber\end{eqnarray}$$
By Propositions A.2 and A.3, we have the following.
Corollary A.4. Let
$s\in \mathbb{R}$
. Then
$$\begin{eqnarray}\lim _{x\rightarrow \infty }\frac{E_{s}(x)}{e^{x}}=1.\end{eqnarray}$$
