Proof Below, we always assume that $\alpha>n$ . By (i) of Proposition 2.1, it holds

$$ \begin{align*} G_{t}(x)\gtrsim\left\{\begin{aligned} &\frac{t^{\beta}}{|x|^{n+\alpha}}\gtrsim \frac{t^{\beta}}{(|x|+t^{\beta/\alpha})^{n+\alpha}},\quad &|x|\geq t^{\beta/\alpha},\\ &\frac{1}{t^{n\beta/\alpha}}\gtrsim\frac{t^{\beta}}{(|x|+t^{\beta/\alpha})^{n+\alpha}},\quad &|x|<t^{\beta/\alpha}. \end{aligned} \right.\end{align*} $$

On the other hand, it can be deduced from (ii) of Proposition 2.1 that

$$ \begin{align*} G_{t}(x)\lesssim\left\{\begin{aligned} &\frac{t^{\beta}}{|x|^{n+\alpha}}\lesssim \frac{t^{\beta}}{(|x|+t^{\beta/\alpha})^{n+\alpha}},\quad &|x|\geq t^{\beta/\alpha},\\ &\frac{1}{t^{\beta n/\alpha}}\lesssim\frac{t^{\beta}}{(|x|+t^{\beta/\alpha})^{n+\alpha}},\quad &|x|<t^{\beta/\alpha}. \end{aligned}\right.\end{align*} $$

Finally,

$$ \begin{align*} G_{t}(x)\simeq \frac{t^{\beta}}{(|x|+t^{\beta/\alpha})^{n+\alpha}}, \end{align*} $$

which, via a direct computation, gives

$$ \begin{align*} \int_{\mathbb R^{n}}G_{t}(x)dx \simeq \int_{\mathbb R^{n}}\frac{t^{\beta}}{(|x|+t^{\beta/\alpha})^{n+\alpha}} dx\lesssim 1.\\[-42pt] \end{align*} $$

Proof Let q obey $1/r+1/q=1/p+1$ . By Young’s inequality,

$$ \begin{align*} \|R_{\alpha,\beta}(\varphi)\|_{L^{p}_{x}(\mathbb R^{n})} = \|G_{t}\ast\varphi\|_{L^{p}_{x}(\mathbb R^{n})}\lesssim\|\varphi\|_{L^{r}_{x}(\mathbb R^{n})}\|G_{t}(\cdot)\|_{L^{q}_{x}(\mathbb R^{n})}. \end{align*} $$

It follows from Corollary 2.2 that

$$ \begin{align*} \|G_{t}(\cdot)\|_{L^{q}_{x}(\mathbb R^{n})} \simeq \Bigg(\int_{\mathbb R^{n}}\frac{t^{\beta q}}{(|x|+t^{\beta/\alpha})^{q(n+\alpha)}}dx\Bigg)^{1/q}\lesssim t^{\beta n(1/q-1)/\alpha}, \end{align*} $$

which implies

$$ \begin{align*} \|R_{\alpha,\beta}(\varphi)\|_{L^{p}_{x}(\mathbb R^{n})} \lesssim t^{\beta n(1/q-1)/\alpha}\|\varphi\|_{L^{r}_{x}(\mathbb R^{n})}=t^{-\beta n(1/r-1/p)/\alpha}\|\varphi\|_{L^{r}_{x}(\mathbb R^{n})}.\\[-35pt] \end{align*} $$

Proof (i) We divide the argument into two cases.

Case 1: $p=r$ and $q=\infty $ . By Lemma 2.3, we obtain

$$ \begin{align*} \|R_{\alpha,\beta}(\varphi)\|_{L^{\infty}_{t}(I; L^{r}_{x}(\mathbb R^{n}))} \lesssim \sup_{t>0}t^{-n\beta(1/r-1/r)/\alpha}\|\varphi\|_{L^{r}_{x}(\mathbb R^{n})}\lesssim \|\varphi\|_{L^{r}_{x}(\mathbb R^{n})}. \end{align*} $$

Case 2: $p\neq r$ . Denote $F(t)(\varphi )=\|R_{\alpha ,\beta }(\varphi )\|_{L^{p}_{x}(\mathbb R^{n})}$ . Since $1/q=\beta n(1/r-1/p)/\alpha $ , a further use of Lemma 2.3 can deduce that

(2.6) $$ \begin{align} \qquad F(t)(\varphi)= \|R_{\alpha,\beta}(\varphi)\|_{L^{p}_{x}(\mathbb R^{n})}\lesssim t^{-n\beta(1/r-1/p)/\alpha}\|\varphi\|_{L^{r}_{x}(\mathbb R^{n})} =t^{-1/q}\|\varphi\|_{L^{r}_{x}(\mathbb R^{n})}. \end{align} $$

It follows from (2.6) that $F(t)$ is a weak $(r,q)$ -type operator since

$$ \begin{align*} \Big|\Big\{t:\ |F(t)(\varphi)|>\tau\Big\}\Big| \leq \Big|\Big\{t:\ t<\left(\frac{\|\varphi\|_{L^{r}_{x}(\mathbb R^{n})}}{\tau}\right)^{q}\Big\}\Big| \lesssim\left(\frac{\|\varphi\|_{L^{r}_{x}(\mathbb R^{n})}}{\tau}\right)^{q}. \end{align*} $$

On the other hand, the inequality

$$ \begin{align*} |R_{\alpha,\beta}(\varphi)(x)|\lesssim\int_{\mathbb R^{n}}\frac{t^{\beta}}{(|x-y|+t^{\beta/\alpha})^{n+\alpha}}|\varphi(y)|dy \end{align*} $$

implies that

$$ \begin{align*} \|F(t)(\varphi)\|_{L^{\infty}_{t}(I)}=\sup_{t>0}\|R_{\alpha,\beta}(\varphi)\|_{L^{p}_{x}(\mathbb R^{n})}\lesssim t^{-n\beta(1/p-1/p)/\alpha}\|\varphi\|_{L^{p}_{x}(\mathbb R^{n})}, \end{align*} $$

which means that $F(t)$ is a $(p,\infty )$ -type operator.

We can find another triplet $(q_{1}, p, r_{1})$ such that $q_{1}<q<\infty $ and $r_{1}<r<p$ satisfying

$$ \begin{align*}\begin{cases} 1/q=\theta/q_{1}+(1-\theta)/\infty,\\ 1/r=\theta/r_{1}+(1-\theta)/p,\\ 1/q_{1}=\beta n(1/r_{1}-1/p)/\alpha. \end{cases} \end{align*} $$

The Marcinkiewicz interpolation theorem implies that $F(t)$ is a strong $(r,q)$ -type operator and

$$ \begin{align*} \|R_{\alpha,\beta}(\varphi)\|_{L^{q}_{t}(I; L^{p}_{x}(\mathbb R^{n}))}\lesssim\|\varphi\|_{L^{r}_{x}(\mathbb R^{n})}. \end{align*} $$

(ii) The argument can be also divided into two cases.

Case 3: $p=r$ and $q=\infty $ . We have

$$ \begin{align*} \|R_{\alpha,\beta}(\varphi)\|_{L^{\infty}_{t}(I; L^{p}_{x}(\mathbb R^{n}))}=\sup_{t>0}\|R_{\alpha,\beta}(\varphi)\|_{L^{p}_{x}(\mathbb R^{n})}\lesssim\|\varphi\|_{L^{p}_{x}(\mathbb R^{n})}. \end{align*} $$

Case 4: $p\neq r$ . Because $1/q=\beta n(1/r-1/p)/\alpha $ , upon taking $q^\ast $ such that $1/p+1=1/r+1/q^\ast $ , we obtain

$$ \begin{align*} \|R_{\alpha,\beta}(\varphi)\|_{C_{q}(I; L^{p}_{x}(\mathbb R^{n}))} \lesssim\sup_{t>0}t^{1/q}t^{-\beta n(1/r-1/p)/\alpha}\|\varphi\|_{L^{r}_{x}(\mathbb R^{n})}\lesssim\|\varphi\|_{L^{r}_{x}(\mathbb R^{n})}. \end{align*} $$

On the other hand, for $t\in I$ , $\|R_{\alpha ,\beta }(\varphi )\|_{L^{r}_{x}(\mathbb R^{n})}\lesssim \|\varphi \|_{L^{r}_{x}(\mathbb R^{n})}.$ Consequently, $R_{\alpha ,\beta }(\varphi )\in C_{0}(I; L^{r}_{x}(\mathbb R^{n}))$ .

Proof An application of Lemma 2.3 yields

$$ \begin{align*} \left\|\int_{\mathbb R^{n}}G_{t-s}(x-y)g(s,y)dy\right\|_{L^{p}_{x}(\mathbb R^{n})}\lesssim |t-s|^{-n\beta(1/\widetilde{p}-1/p)/\alpha}\|g(s,\cdot)\|_{L_{x}^{\widetilde{p}}(\mathbb R^{n})}. \end{align*} $$

It follows from the boundedness of fractional integrals that

$$ \begin{align*} &\left\|\int^{t}_{0}\int_{\mathbb R^{n}}G_{t-s}(x-y)g(s,y)dyds\right\|_{L^{q}_{t}(I; L^{p}_{x}(\mathbb R^{n}))} \\&\quad\lesssim \left\|\int^{t}_{0}\left\|\int_{\mathbb R^{n}}G_{t-s}(x-y)g(s,y)dy\right\|_{L^{p}_{x}(\mathbb R^{n})}ds\right\|_{L^{q}_{t}(I)}\\ &\quad\lesssim \left\|\int^{t}_{0}|t-s|^{-n\beta(1/\widetilde{p}-1/p)/\alpha}\|g(s,\cdot)\|_{L_{x}^{\widetilde{p}}(\mathbb R^{n})}ds\right\|_{L^{q}_{t}(I)}\\ &\quad\lesssim \|g\|_{L^{\widetilde{q}}_{t}(I; L^{\widetilde{p}}_{x}(\mathbb R^{n}))}, \end{align*} $$

which finishes the proof.

Proof (i) For the case $p<r(b+1)$ , we have

$$ \begin{align*} \|G_{\alpha,\beta}(g)\|_{L^{\infty}_{t}(I; L^{r}_{x}(\mathbb R^{n}))}&= \sup_{t\in I}\left\|\int^{t}_{0}\int_{\mathbb R^{n}}G_{t-s}(x-y)g(s,y)dyds\right\|_{L^{r}_{x}(\mathbb R^{n})}\\ &\lesssim \sup_{t\in I}\int^{t}_{0}\left\|\int_{\mathbb R^{n}}G_{t-s}(x-y)g(s,y)dy\right\|_{L^{r}_{x}(\mathbb R^{n})}ds. \end{align*} $$

Take $q^\ast $ such that $(b+1)/p+1/q^{\ast }=1+1/r$ . Then, by Lemma 2.3,

$$ \begin{align*} \left\|\int_{\mathbb R^{n}}G_{t-s}(x-y)g(s,y)dy\right\|_{L^{r}_{x}(\mathbb R^{n})}\lesssim (t-s)^{-n\beta(1-1/q^\ast)/\alpha}\|g(s,\cdot)\|_{L^{p/(b+1)}_{x}(\mathbb R^{n})}, \end{align*} $$

which implies that

$$ \begin{align*}\|G_{\alpha,\beta}(g)\|_{L^{\infty}_{t}(I; L^{r}(\mathbb R^{n}))}\lesssim\sup_{t\in I}\int^{t}_{0}(t-s)^{-n\beta((b+1)/p-1/r)/\alpha}\|g(s,\cdot)\|_{L^{p/(b+1)}_{x}(\mathbb R^{n})}ds.\end{align*} $$

Let $\widetilde {q}$ be the conjugate of $q/(b+1)$ , i.e., $(b+1)/q+1/\widetilde {q}=1$ . Because $r>r_{0}:=b\beta n/\alpha $ , then

$$ \begin{align*} 0<1-n\beta((b+1)/p-1/r)\widetilde{q}/\alpha=\widetilde{q}(1-b\beta n/r\alpha)<1. \end{align*} $$

A direct computation, together with change of variables, gives

$$ \begin{align*} \Bigg(\int^{t}_{0}(t-s)^{-n\beta((b+1)/p-1/r)\widetilde{q}/\alpha}ds\Bigg)^{1/\widetilde{q}}\lesssim T^{1/\widetilde{q}-n\beta((b+1)/p-1/r)/\alpha}. \end{align*} $$

Then we obtain

$$ \begin{align*} &\|G_{\alpha,\beta}(g)\|_{L^{\infty}(I; L^{r})}\\&\quad\lesssim \sup_{t\in I}\Bigg(\int^{t}_{0}(t-s)^{-n\beta((b+1)/p-1/r)\widetilde{q}/\alpha}ds\Bigg)^{1/\widetilde{q}} \Bigg(\int^{t}_{0}\|g(s,\cdot)\|^{q/(b+1)}_{L^{p/(b+1)}_{x}(\mathbb R^{n})}ds\Bigg)^{(b+1)/q}\\ &\quad\lesssim T^{1/\widetilde{q}-n\beta((b+1)/p-1/r)/\alpha}\|g\|_{L^{q/(b+1)}_{t}(I; L^{p/(b+1)}_{x}(\mathbb R^{n}))}\\ &\quad\lesssim T^{1-b\beta n/r\alpha}\|g\|_{L^{q/(b+1)}_{t}(I; L^{p/(b+1)}_{x}(\mathbb R^{n}))}, \end{align*} $$

where in the last inequality we have used the fact that

$$ \begin{align*} 1/\widetilde{q}-n\beta[(b+1)/p-1/r]/\alpha=1-b\beta n/r\alpha. \end{align*} $$

When $p\geq r(b+1)$ , Lemma 2.3 gives

$$ \begin{align*} &\|G_{\alpha,\beta}(g)\|_{L^{\infty}_{t}(I; L^{r}_{x}(\mathbb R^{n}))}\lesssim\sup_{t\in I}\int^{t}_{0}\left\|\int_{\mathbb R^{n}}G_{t-s}(x-y)g(s,y)dy\right\|_{L^{r}_{x}(\mathbb R^{n})}ds\\&\quad\lesssim\sup_{t\in I}\int^{t}_{0}\big\|g(s,\cdot)\big\|_{L^{r}_{x}(\mathbb R^{n})}ds. \end{align*} $$

Notice that $\|g(s,\cdot )\|_{L^{r}_{x}(\mathbb R^{n})}=\||g(s,\cdot )|^{1/(b+1)}\|_{L^{r(b+1)}_{x}(\mathbb R^{n})}^{b+1}.$ Hence,

$$ \begin{align*} \|G_{\alpha,\beta}(g)\|_{L^{\infty}_{t}(I; L^{r}_{x}(\mathbb R^{n}))}\lesssim\sup_{t\in I}\int^{t}_{0}\Big\||g(s,\cdot)|^{1/(b+1)}\Big\|_{L^{r(b+1)}_{x}(\mathbb R^{n})}^{b+1}ds. \end{align*} $$

Take $\theta \in (0,1)$ such that $1/(rb+r)=\theta /r+(1-\theta )/p$ . Let $p_{1}=(b(1+\theta ))^{-1}$ and $q_{1}=p/(r(b+1)(1-\theta ))$ such that $1/p_{1}+1/q_{1}=1$ . Applying Hölder’s inequality on the spatial variable, we obtain

$$ \begin{align*} &\||g(s,\cdot)|^{1/(b+1)}\|_{L^{r(b+1)}_{x}(\mathbb R^{n})}^{b+1}\\&\quad\lesssim \left\{\int_{\mathbb R^{n}}|g(s,x)|^{r\theta p_{1}}dx\right\}^{1/(rp_{1})}\left\{\int_{\mathbb R^{n}}|g(s,x)|^{r(1-\theta) q_{1}}dx\right\}^{1/(rq_{1})}\\ &\quad\lesssim \left\{\int_{\mathbb R^{n}}|g(s,x)|^{r/(b+1)}dx\right\}^{\theta(b+1)/r}\left\{\int_{\mathbb R^{n}}|g(s,x)|^{p/(b+1)}dx\right\}^{(b+1)(1-\theta)/p}, \end{align*} $$

which indicates that

$$ \begin{align*} &\|G_{\alpha,\beta}(g)\|_{L^{\infty}_{t}(I; L^{r}_{x}(\mathbb R^{n}))}\\[4pt]&\quad\lesssim \sup_{t\in I}\int^{t}_{0}\big\||g(s,\cdot)|^{1/(b+1)}\big\|_{L^{r}_{x}(\mathbb R^{n})}^{\theta(b+1)}\cdot \big\||g(s,\cdot)|^{1/(b+1)}\big\|_{L^{p}_{x}(\mathbb R^{n})}^{(1-\theta)(b+1)}ds\\[4pt] &\quad\lesssim \big\||g(\cdot,\cdot)|^{1/(b+1)}\big\|^{(b+1)\theta}_{L^{\infty}_{t}(I; L^{r}_{x}(\mathbb R^{n}))}\sup_{t\in I}\int^{t}_{0}\big\||g(s,\cdot)|^{1/(b+1)}\big\|_{L^{p}_{x}(\mathbb R^{n})}^{(1-\theta)(b+1)}ds\\[4pt] &\quad\lesssim \big\||g(\cdot,\cdot)|^{1/(b+1)}\big\|^{(b+1)\theta}_{L^{\infty}_{t}(I; L^{r}_{x}(\mathbb R^{n}))}\sup_{t\in I}\left\{\left(\int^{t}_{0}1ds\right)^{1-(b+1)(1-\theta)/q}\right.\\[4pt] &\qquad \times\left.\left(\int^{t}_{0}\big\||g(s,\cdot)|^{1/(b+1)}\big\|_{L^{p}_{x}(\mathbb R^{n})}^{(1-\theta)(b+1)q/(1-\theta)(1+b)}ds\right)^{(b+1)(1-\theta)/q}\right\}\\[4pt] &\quad\lesssim T^{1-(b+1)(1-\theta)/q}\big\||g(\cdot,\cdot)|^{1/(b+1)}\big\|^{(b+1)\theta}_{L^{\infty}_{t}(I; L^{r}_{x}(\mathbb R^{n}))}\big\||g|^{1/(b+1)}\big\|^{(b+1)(1-\theta)}_{L^{q}_{t}(I; L^{p}(\mathbb R^{n}))}. \end{align*} $$

Since $1/q=\beta n(1/r-1/p)/\alpha $ and $\theta =(p-rb-r)/((p-r)(b+1))$ , then

$$ \begin{align*} \|G_{\alpha,\beta}(g)\|_{L^{\infty}_{t}(I; L^{r}_{x}(\mathbb R^{n}))}\lesssim T^{1-n\beta b/(r\alpha)}\Big\||g|^{1/(b+1)}\Big\|^{\theta(b+1)}_{L^{\infty}_{t}(I; L^{r}_{x}(\mathbb R^{n}))}\Big\||g|^{1/(b+1)}\Big\|^{(1-\theta)(b+1)}_{L^{q}_{t}(I; L^{p}_{x}(\mathbb R^{n}))}. \end{align*} $$

This completes the proof of (i).

(ii) For the case $p<r(b+1)$ , using Minkowski’s inequality and Lemma 2.3, we obtain

$$ \begin{align*} &\|G_{\alpha,\beta}(g)\|_{L^{q}_{t}(I; L^{p}_{x}(\mathbb R^{n}))} \\[4pt]&\quad\lesssim \left\{\int^{T}_{0}\left(\int^{t}_{0}\Big\|\int_{\mathbb R^{n}}G_{t-s}(x-y)g(s,y)dy\Big\|_{L^{p}_{x}(\mathbb R^{n})}ds\right)^{q}dt\right\}^{1/q}\\[4pt] &\quad\lesssim \left\{\int^{T}_{0} \left(\int^{t}_{0}(t-s)^{-n\beta((b+1)/p-1/p)/\alpha}\|g(s,\cdot)\|_{L^{p/(b+1)}_{x}(\mathbb R^{n})}ds\right)^{q}dt\right\}^{1/q}\\[4pt] &\quad= \left\|\int^{t}_{0}(t-s)^{-n\beta b/p\alpha}\|g(s,\cdot)\|_{L^{p/(b+1)}_{x}(\mathbb R^{n})}ds\right\|_{L^{q}_{t}(I)}. \end{align*} $$

Let $\chi $ be the number such that $1/q+1=(1+b)/q+1/\chi $ . An application of Young’s inequality gives

$$ \begin{align*} \|G_{\alpha,\beta}(g)\|_{L^{q}_{t}(I; L^{p}_{x}(\mathbb R^{n}))}&\lesssim \|g(\cdot,\cdot)\|_{L^{q/(b+1)}_{t}(I; L^{p/(b+1)}_{x}(\mathbb R^{n}))}\Bigg(\int^{T}_{0}t^{-n\beta b\chi/p\alpha}dt\Bigg)^{1/\chi}\\[4pt] &\lesssim T^{1-n\beta b/r\alpha}\|g(\cdot,\cdot)\|_{L^{q/(b+1)}_{t}(I; L^{p/(b+1)}_{x}(\mathbb R^{n}))}, \end{align*} $$

where in the last inequality we have used the fact that $1/q=n\beta (1/r-1/p)/\alpha $ .

For the case $p\geq r(b+1)$ , we apply Lemma 2.3 again to get

$$ \begin{align*} \|G_{\alpha,\beta}(g)\|_{L^{q}_{t}(I; L^{p}_{x}(\mathbb R^{n}))}&\lesssim \left\|\int^{t}_{0}\left\|\int_{\mathbb R^{n}}G_{t-s}(x-y)g(s,y)dy\right\|_{L^{p}_{x}(\mathbb R^{n})}ds\right\|_{L^{q}_{t}(I)}\\ &\lesssim \left\|\int^{t}_{0}(t-s)^{-n\beta(1/r-1/p)/\alpha} \left\||g(s,\cdot)|^{1/(b+1)}\right\|_{L^{p}_{x}(\mathbb R^{n})}ds\right\|_{L^{q}_{t}(I)}. \end{align*} $$

Choose $\theta \in (0,1)$ such that $1/(b+1)=\theta +(1-\theta )r/p$ . Letting $p_{2}=(b+b\theta )^{-1}$ and $q_{2}=p/(r(b+1)(1-\theta ))$ , we use Hölder’s inequality on the spatial variable to deduce

$$ \begin{align*} &\int^{t}_{0}(t-s)^{-\beta(1/r-1/p)/\alpha}\Big\||g(s,\cdot)|^{1/(b+1)}\Big\|_{L^{r(b+1)}_{x}(\mathbb R^{n})}^{b+1}ds\\ &\lesssim\int^{t}_{0}(t-s)^{-n\beta(1/r-1/p)/\alpha}\Bigg(\int_{\mathbb R^{n}}|g(s,x)|^{r\theta p_{2}}dx\Bigg)^{1/(rp_{2})}\Bigg(\int_{\mathbb R^{n}}|g(s,x)|^{r(1-\theta)q_{2}}dx\Bigg)^{1/(rq_{2})}ds\\ &\lesssim\int^{t}_{0}(t-s)^{-n\beta(1/r-1/p)/\alpha}\big\||g(s,\cdot)|^{1/(b+1)}\big\|^{\theta(b+1)}_{L^{r}_{x}(\mathbb R^{n})} \big\||g(s,\cdot)|^{1/(b+1)}\big\|^{(1-\theta)(b+1)}_{L^{p}_{x}(\mathbb R^{n})}ds\\ &\lesssim\big\||g(s,\cdot)|^{1/(b+1)}\big\|^{\theta(b+1)}_{L^{\infty}_{t}(I;L^{r}_{x}(\mathbb R^{n}))}\int^{t}_{0}(t-s)^{-n\beta(1/r-1/p)/\alpha} \big\||g(s,\cdot)|^{1/(b+1)}\big\|^{(1-\theta)(b+1)}_{L^{p}_{x}(\mathbb R^{n})}ds, \end{align*} $$

which gives

$$ \begin{align*} \|G_{\alpha,\beta}(g)\|_{L^{q}_{t}(I; L^{p}_{x}(\mathbb R^{n}))}&\lesssim \big\||g(\cdot,\cdot)|^{1/(b+1)}\big\|^{\theta(b+1)}_{L^{\infty}_{t}(I;L^{r}_{x}(\mathbb R^{n}))}\\ &\quad\times\left\|\int^{t}_{0}\!(t\kern1.3pt{-}\kern1.3pts)^{-n\beta(1/r-1/p)/\alpha} \big\||g(s,\cdot)|^{1/(b+1)}\big\|^{(1-\theta)(b+1)}_{L^{p}_{x}(\mathbb R^{n})}ds\right\|_{L^{q}_{t}(I)}\!\kern-1pt. \end{align*} $$

Suppose that $\widetilde {\chi }$ obeys $1/q+1=(1+b)(1-\theta )/q+1/\widetilde {\chi }$ . Young’s inequality on the time variable gives

$$ \begin{align*} &\|G_{\alpha,\beta}(g)\|_{L^{q}_{t}(I; L^{p}_{x}(\mathbb R^{n}))}\\&\quad\lesssim \big\||g(\cdot,\cdot)|^{1/(b+1)}\big\|^{\theta(b+1)}_{L^{\infty}_{t}(I;L^{r}_{x}(\mathbb R^{n}))}\big\||g(\cdot,\cdot)|^{1/(b+1)}\big\|^{(b+1)(1-\theta)}_{L^{q}_{t}(I; L^{p}_{x}(\mathbb R^{n}))}\\ &\qquad\times\Bigg(\int^{T}_{0}t^{-n\beta(1/r-1/p)\widetilde{\chi}/\alpha}dt\Bigg)^{1/\widetilde{\chi}}\\ &\quad\lesssim T^{1-b\beta n/(r\alpha)}\||g(\cdot,\cdot)|^{1/(b+1)}\big\|^{\theta(b+1)}_{L^{\infty}_{t}(I;L^{r}_{x}(\mathbb R^{n}))}\big\||g(\cdot,\cdot)|^{1/(b+1)}\big\|^{(b+1)(1-\theta)}_{L^{q}_{t}(I; L^{p}_{x}(\mathbb R^{n}))}, \end{align*} $$

where in the last inequality we have used the fact that $1/\widetilde {\chi }-n\beta (1/r-1/p)/\alpha =1-b\beta n/(r\alpha )$ . This completes the proof of Theorem 2.7.

Proof (i) We first consider the case $p<r(b+1)$ , and it follows from Lemma 2.3 that

$$ \begin{align*}\left\|\int_{\mathbb R^{n}}G_{t-s}(x-y)g(s,y)dy\right\|_{L^{r}_{x}(\mathbb R^{n})}\lesssim (t-s)^{-n\beta[(b+1)/p-1/r]/\alpha}\|g(s,\cdot)\|_{L^{p/(b+1)}_{x}(\mathbb R^{n})},\end{align*} $$

which implies that

$$ \begin{align*} &\big\|G_{\alpha,\beta}(g)\big\|_{L^{\infty}_{t}(I; L^{r}_{x}(\mathbb R^{n}))}\\&\quad\lesssim \sup_{t\in[0, T)}\Bigg\{\int^{t}_{0}\left\|\int_{\mathbb R^{n}}G_{t-s}(x-y)g(s,y)dy\right\|_{L^{r}_{x}(\mathbb R^{n})}ds\Bigg\}\\ &\quad\lesssim \sup_{t\in[0, T)}\Bigg\{\int^{t}_{0}(t-s)^{-n\beta[(b+1)/p-1/r]/\alpha}\|g(s,\cdot)\|_{L^{p/(b+1)}_{x}(\mathbb R^{n})}ds\Bigg\}\\ &\lesssim \|g\|_{C_{q/(b+1)}(I; L^{p/(b+1)}_{x}(\mathbb R^{n}))}\sup_{t\in[0, T)}\Bigg\{\int^{t}_{0}(t-s)^{-n\beta[(b+1)/p-1/r]/\alpha}s^{-(b+1)/q}ds\Bigg\}. \end{align*} $$

A direct computation, together with the change of variables: $u=s/t$ , gives

$$ \begin{align*} \int^{t}_{0}(t-s)^{-n\beta[(b+1)/p-1/r]/\alpha}s^{-(b+1)/q}ds\lesssim t^{1-n\beta b/(r\alpha)}, \end{align*} $$

which indicates that

$$ \begin{align*} \|G_{\alpha,\beta}(g)\|_{L^{\infty}_{t}(I; L^{r}_{x}(\mathbb R^{n}))}\lesssim T^{1-n\beta b/r\alpha}\|g\|_{C_{q/(b+1)}(I; L^{p/(b+1)}_{x}(\mathbb R^{n}))}. \end{align*} $$

Now, we consider the case $p\geq (b+1)r$ . By Lemma 2.3, we have

$$ \begin{align*} \left\|\int_{\mathbb R^{n}}G_{t-s}(x-y)g(s,y)dy\right\|_{L^{r}_{x}(\mathbb R^{n})}\lesssim \|g(s,\cdot)\|_{L^{r}_{x}(\mathbb R^{n})}. \end{align*} $$

If $\theta \in (0,1)$ , $p_{3}=\frac {1}{(b+1)\theta }$ , and $ q_{3}=\frac {p}{r(b+1)(1-\theta )}$ , then we obtain

$$ \begin{align*} \|G_{\alpha,\beta}(g)\|_{L^{\infty}_{t}(I; L^{r}_{x}(\mathbb R^{n}))}&= \sup_{t\in I}\Bigg\|\int^{t}_{0}\int_{\mathbb R^{n}}G_{t-s}(x-y)g(s,y)dyds\Bigg\|_{L^{r}_{x}(\mathbb R^{n})}\\ &\lesssim \sup_{t\in I}\int^{t}_{0}\|g(s,\cdot)\|_{L^{r}_{x}(\mathbb R^{n})}ds. \end{align*} $$

Notice that $1-(b+1)(1-\theta )/q=1-b\beta n/(r\alpha )$ . An application of Hölder’s inequality gives

$$ \begin{align*} &\|G_{\alpha,\beta}(g)\|_{L^{\infty}_{t}(I; L^{r}_{x}(\mathbb R^{n}))}\\&\lesssim \sup_{t\in I}\int^{t}_{0}\big\||g(s,\cdot)|^{1/(b+1)}\big\|^{b+1}_{L^{r(b+1)}_{x}(\mathbb R^{n})}ds\\ &\lesssim \sup_{t\in [0,T)}\left\{\int^{t}_{0}\big\||g(s,\cdot)|^{1/(b+1)}\big\|^{\theta(b+1)}_{L^{r}_{x}(\mathbb R^{n})}\big\||g(s,\cdot)|^{1/(b+1)}\big\|^{(1-\theta)(b+1)}_{L^{p}_{x}(\mathbb R^{n})}ds\right\}\\ &\lesssim \big\||g|^{1/(b+1)}\big\|^{\theta(b+1)}_{L^{\infty}_{t}(I;L^{r}_{x}(\mathbb R^{n}))}\big\||g|^{1/(b+1)}\big\|^{(1-\theta)(b+1)}_{C_{q}(I;L^{p}_{x}(\mathbb R^{n}))}\sup_{t\in [0,T)}\left\{\int^{t}_{0}s^{-(b+1)(1-\theta)/q}ds\right\}\\ &\lesssim T^{1-(b+1)(1-\theta)/q}\big\||g|^{1/(b+1)}\big\|^{\theta(b+1)}_{L^{\infty}_{t}(I;L^{r}_{x}(\mathbb R^{n}))}\big\||g|^{1/(b+1)}\big\|^{(1-\theta)(b+1)}_{C_{q}(I;L^{p}_{x}(\mathbb R^{n}))}. \end{align*} $$

We begin to prove (ii). Let $g\in C_{q/(b+1)}(I; L^{p/(b+1)}_{x}(\mathbb R^{n}))$ . For the case $p<r(b+1)$ , we have

$$ \begin{align*} \Bigg\|\int_{\mathbb R^{n}}G_{t-s}(x-y)g(s,y)dy\Bigg\|_{L^{p}_{x}(\mathbb R^{n})}\lesssim (t-s)^{-n\beta b/p\alpha}\|g(s,\cdot)\|_{L^{p/(b+1)}_{x}(\mathbb R^{n})}, \end{align*} $$

and hence

$$ \begin{align*} \Big\|G_{\alpha,\beta}(g)\Big\|_{C_{q}(I; L^{p}_{x}(\mathbb R^{n}))} &\lesssim \sup_{t\in [0,T)}t^{1/q}\int^{t}_{0}\left\|\int_{\mathbb R^{n}}G_{t-s}(x-y)g(s,y)dy\right\|_{L^{p}_{x}(\mathbb R^{n})}ds\\ &\lesssim \sup_{t\in [0,T)}t^{1/q}\int^{t}_{0}(t-s)^{-n\beta b/p\alpha}\|g(s,\cdot)\|_{L^{p/(b+1)}_{x}(\mathbb R^{n})} ds\\ &\lesssim \|g\|_{C_{q/(b+1)}(I; L^{p/(b+1)}_{x}(\mathbb R^{n}))}\sup_{t\in [0,T)}t^{1/q}\int^{t}_{0}(t-s)^{-n\beta b/p\alpha}s^{-(b+1)/q} ds\\ &\lesssim \|g\|_{C_{q/(b+1)}(I; L^{p/(b+1)}_{x}(\mathbb R^{n}))}T^{1-\beta bn/p\alpha}. \end{align*} $$

For the case $p\geq r(b+1)$ , it holds

$$ \begin{align*} \left\|\int_{\mathbb R^{n}}G_{t-s}(x-y)g(s,y)dy\right\|_{L^{p}_{x}(\mathbb R^{n})}&\lesssim (t-s)^{-n\beta(1/r-1/p)/\alpha}\|g(s,\cdot)\|_{L^{r}_{x}(\mathbb R^{n})}\\ &\lesssim (t-s)^{-n\beta(1/r-1/p)/\alpha}\big\||g(s,\cdot)|^{1/(b+1)}\big\|^{b+1}_{L^{r(b+1)}_{x}(\mathbb R^{n})}. \end{align*} $$

Taking $\theta =(p-r(b+1))/(p-r)$ , we use the Hölder inequality to deduce

$$ \begin{align*} \Big\|G_{\alpha,\beta}(g)\Big\|_{C_{q}(I; L^{p}_{x}(\mathbb R^{n}))}&\lesssim \sup_{t\in [0,T)}t^{1/q}\int^{t}_{0}(t-s)^{-\beta n(1/r-1/p)/\alpha}\big\||g(s,\cdot)|^{1/(b+1)}\big\|^{(b+1)\theta}_{L^{r}_{x}(\mathbb R^{n})}\\ &\quad\times \big\||g(s,\cdot)|^{1/(b+1)}\big\|^{(b+1)(1-\theta)}_{L^{p}_{x}(\mathbb R^{n})}ds\\ &\lesssim \big\||g|^{1/(b+1)}\big\|^{(b+1)\theta}_{L^{\infty}_{t}(I; L^{r}_{x}(\mathbb R^{n}))} \big\||g|^{1/(b+1)}\big\|^{(b+1)(1-\theta)}_{C_{q}(I;L^{p}_{x}(\mathbb R^{n}))}\\ &\quad\times\sup_{t\in [0, T)}\left\{t^{1/q}\int^{t}_{0}(t-s)^{-\beta n(1/r-1/p)/\alpha}s^{-(b+1)(1-\theta)/q}ds\right\}\\ &\lesssim T^{1-\beta bn/r\alpha}\big\||g|^{1/(b+1)}\big\|^{(b+1)\theta}_{L^{\infty}_{t}(I; L^{r}_{x}(\mathbb R^{n}))} \big\||g|^{1/(b+1)}\big\|^{(b+1)(1-\theta)}_{C_{q}(I;L^{p}_{x}(\mathbb R^{n}))}. \end{align*} $$

This completes the proof of Theorem 2.8.

Proof We first verify (2.7) for $|h|<|x|/2$ . Recall that

$$ \begin{align*} G_{t}(x)=\int^{\infty}_{0}K_{\alpha/2,s}(x)f_{E_{t}}(s)ds, \end{align*} $$

where $f_{E_{t}}(x)=t\beta ^{-1}x^{-1-1/\beta }g_{\beta }(tx^{-1/\beta })$ . An application of change of variables gives

$$ \begin{align*} G_{t}(x)=\int^{\infty}_{0}K_{\alpha/2, (t/u)^{\beta}}(x)g_{\beta}(u)du, \end{align*} $$

which, together with Proposition 2.9(i) with $\epsilon \in (0,\alpha -n)$ , implies that

(2.8) $$ \begin{align} \qquad\!\Big|G_{t}(x+h)-G_{t}(x)\Big|&\lesssim \int^{\infty}_{0}\Big|K_{\alpha/2, (t/u)^{\beta}}(x+h)-K_{\alpha/2, (t/u)^{\beta}}(x)\Big|g_{\beta}(u)du\\ &\lesssim \int^{\infty}_{0}\left(\frac{|h|}{({t}/{u})^{\beta/\alpha}}\right)^{\epsilon} \frac{(t/u)^{\beta}}{((t/u)^{\beta/\alpha}+|x|)^{n+\alpha}}g_{\beta}(u)du\nonumber\\ &\lesssim |h|^{\epsilon}\int^{\infty}_{0}\left(\frac{t}{u}\right)^{\beta-\beta\epsilon/\alpha}\frac{1}{((t/u)^{\beta/\alpha}+|x|)^{n+\alpha}}g_{\beta}(u)du\nonumber\\ &\lesssim \frac{|h|^{\epsilon}}{t^{\beta(n+\epsilon)/\alpha}}\int^{\infty}_{0}u^{\beta(n+\epsilon)/\alpha}g_{\beta}(u)du\nonumber\\ &\lesssim \left(\frac{|h|}{t^{\beta/\alpha}}\right)^{\epsilon}\!\frac{1}{t^{n\beta/\alpha}}\Bigg\{\!\kern-1pt\int^{1}_{0}u^{\beta(n+\epsilon)/\alpha}du+ \!\int^{\infty}_{1}u^{\beta(n+\epsilon)/\alpha-\beta-1}du\kern-1.5pt\Bigg\}\nonumber\\ &\lesssim \left(\frac{|h|}{t^{\beta/\alpha}}\right)^{\epsilon}\frac{1}{t^{n\beta/\alpha}},\nonumber \end{align} $$

where in the last inequality we have used the facts that $\alpha>n$ and $0<\epsilon <\alpha -n$ .

On the other hand, by Proposition 2.9,

$$ \begin{align*} \Big|K_{\alpha/2,s}(x+h)-K_{\alpha/2,s}(x) \Big|\lesssim\frac{s}{(s^{1/\alpha}+|x|)^{n+\alpha}}\left(\frac{|h|}{s^{1/\alpha}}\right)^{\epsilon} \lesssim\left(\frac{|h|}{s^{1/\alpha}}\right)^{\epsilon}\frac{s}{|x|^{n+\alpha}}, \end{align*} $$

and we have

(2.9) $$ \begin{align} \Big|G_{t}(x+h)-G_{t}(x)\Big|&\lesssim \int^{\infty}_{0}\frac{(t/u)^{\delta}}{|x|^{n+\alpha}}\left(\frac{|h|}{(t/u)^{\beta/\alpha}}\right)^{\epsilon}g_{\beta}(u)du\\ &\lesssim \frac{|h|^{\epsilon}}{|x|^{n+\alpha}}t^{\beta-\beta\epsilon/\alpha}\int^{\infty}_{0}u^{-\beta(1-\epsilon/\alpha)}g_{\beta}(u)du\nonumber\\ &\lesssim \frac{t^{\beta}}{|x|^{n+\alpha}}\left(\frac{|h|}{t^{\beta/\alpha}}\right)^{\epsilon}.\nonumber \end{align} $$

Finally, it can be deuced from (2.8) and (2.9) that

$$ \begin{align*} \Big|G_{t}(x+h)-G_{t}(x)\Big|\lesssim\min\Bigg\{\left(\frac{|h|}{t^{\beta/\alpha}}\right)^{\epsilon}\frac{1}{t^{n\beta/\alpha}},\ \frac{t^{\beta}}{|x|^{n+\alpha}}\left(\frac{|h|}{t^{\beta/\alpha}}\right)^{\epsilon}\Bigg\}; \end{align*} $$

equivalently,

$$ \begin{align*} \Big|G_{t}(x+h)-G_{t}(x)\Big|\leq C\left(\frac{|h|}{t^{\beta/\alpha}}\right)^{\epsilon}\frac{t^{\beta}}{(t^{\beta/\alpha}+|x|)^{n+\alpha}}. \end{align*} $$

Now, we assume that $|h|<t^{\beta /\alpha }$ . If $|h|<|x|/2<t^{\beta /\alpha }$ or $|h|<t^{\beta /\alpha }<|x|/2$ , it is obvious that (2.7) holds. Now, we consider the case $|x|/2<|h|<t^{\beta /\alpha }$ . We split $|G_{t}(x+h)-G_{t}(x)|\leq S_{1}+S_{2},$ where

$$ \begin{align*}\left\{\begin{aligned} S_{1}&:=\int_{u: |h|<(t/u)^{\beta/\alpha}}\Big|K_{\alpha/2, (t/u)^{\beta}}(x+h)-K_{\alpha/2, (t/u)^{\beta}}(x)\Big|g_{\beta}(u)du,\\ S_{2}&:=\int_{u: |h|\geq(t/u)^{\beta/\alpha}}\Big|K_{\alpha/2, (t/u)^{\beta}}(x+h)-K_{\alpha/2, (t/u)^{\beta}}(x)\Big|g_{\beta}(u)du. \end{aligned}\right.\end{align*} $$

For $S_{1}$ , since $|h|<(t/u)^{\beta /\alpha }$ , it follows from (ii) of Proposition 2.9 that

$$ \begin{align*} S_{1}&\lesssim \int_{u: |h|<(t/u)^{\beta/\alpha}}\left(\frac{|h|}{({t}/{u})^{\beta/\alpha}}\right)^{\epsilon} \frac{(t/u)^{\beta}}{((t/u)^{\beta/\alpha}+|x|)^{n+\alpha}}g_{\beta}(u)du\nonumber\\ &\lesssim \left(\frac{|h|}{t^{\beta/\alpha}}\right)^{\epsilon}\frac{1}{t^{n\beta/\alpha}}\Bigg\{\int^{1}_{0}u^{\beta(n+\epsilon)/\alpha}du+ \int^{\infty}_{1}u^{\beta(n+\epsilon)/\alpha-\beta-1}du\Bigg\}\nonumber\\ &\lesssim \left(\frac{|h|}{t^{\beta/\alpha}}\right)^{\epsilon}\frac{1}{t^{n\beta/\alpha}}. \end{align*} $$

We further divide $S_{2}$ as $S_{2}\leq S_{2,1}+S_{2,2}$ , where

$$ \begin{align*}\left\{\begin{aligned} S_{2,1}&:=\int_{u: |h|\geq(t/u)^{\beta/\alpha}}\Big|K_{\alpha/2, (t/u)^{\beta}}(x)\Big|g_{\beta}(u)du,\\ S_{2,2}&:=\int_{u: |h|\geq(t/u)^{\beta/\alpha}}\Big|K_{\alpha/2, (t/u)^{\beta}}(x+h)\Big|g_{\beta}(u)du. \end{aligned}\right.\end{align*} $$

Noticing that $|h|\geq (t/u)^{\beta /\alpha }$ , we deduce from (2.5) that

$$ \begin{align*} S_{2,1}&\lesssim \int_{u: |h|\geq(t/u)^{\beta/\alpha}}\left(\frac{|h|}{(t/u)^{\beta/\alpha}}\right)^{\epsilon}\frac{(t/u)^{\beta}}{(|x|+(t/u)^{\beta/\alpha})^{n+\alpha}}g_{\beta}(u)du\\ &\lesssim \left(\frac{|h|}{t^{\beta/\alpha}}\right)^{\epsilon}\frac{1}{t^{n\beta/\alpha}}\int^{\infty}_{0}u^{\beta(n+\epsilon)/\alpha}g_{\beta}(u)du\\ &\lesssim \left(\frac{|h|}{t^{\beta/\alpha}}\right)^{\epsilon}\frac{1}{t^{n\beta/\alpha}}. \end{align*} $$

The term $S_{2,2}$ can be dealt with similar to $S_{2,1}$ . Finally, due to $|x|/2<t^{\beta /\alpha }$ , it holds

$$ \begin{align*} \Big|G_{t}(x+h)-G_{t}(x)\Big| \lesssim \left(\frac{|h|}{t^{\beta/\alpha}}\right)^{\epsilon}\frac{1}{t^{n\beta/\alpha}}\lesssim \left(\frac{|h|}{t^{\beta/\alpha}}\right)^{\epsilon}\frac{t^{\beta}}{(t^{\beta/\alpha}+|x|)^{n+\alpha}}, \end{align*} $$

which completes the proof of Proposition 2.11.

Proof At first, for fixed $t_{0}>0$ , choose $(t_{0},x), ( t_{0},x_{0})\in \mathbb R^{1+n}_{+}$ . By Proposition 2.9 and Hölder’s inequality, we can get

$$ \begin{align*} &\left|R_{\alpha,\beta}(\varphi)(t_{0},x)-R_{\alpha,\beta}(\varphi)(t_{0},x_{0})\right| \\&\quad\lesssim \int_{\mathbb R^{n}}\left\{\int^{\infty}_{0}\Big|K_{\alpha/2, (t_{0}/u)^{\beta}}(x-y)-K_{\alpha/2,(t_{0}/u)^{\beta}}(x_{0}-y)\Big|g_{\beta}(u)du\right\}|\varphi(y)|dy\\ &\quad\lesssim \int^{\infty}_{0}\left\{\int_{\mathbb R^{n}}\Big|K_{\alpha/2, (t_{0}/u)^{\beta}}(x-y)-K_{\alpha/2,(t_{0}/u)^{\beta}}(x_{0}-y)\Big|\cdot|\varphi(y)|dy\right\}g_{\beta}(u)du\\ &\quad\lesssim M_{1}+M_{2}, \end{align*} $$

where

$$ \begin{align*}\left\{\begin{aligned}\! M_{1}&:=\!\int_{u: (t_{0}/u)^{\beta/\alpha}>|x-x_{0}|}\!\left\{\!\int_{\mathbb R^{n}}\Big|K_{\alpha/2, (t_{0}/u)^{\beta}}(x\kern1.3pt{-}\kern1.3pt y)\kern1.3pt{-}\kern1.3pt K_{\alpha/2,(t_{0}/u)^{\beta}}(x_{0}\kern1.3pt{-}\kern1.3pt y)\Big|\cdot |\varphi(y)|dy\!\right\}g_{\beta}(u)du,\\ \!M_{2}&:=\!\int_{u: (t_{0}/u)^{\beta/\alpha}\leq|x-x_{0}|}\!\left\{\!\int_{\mathbb R^{n}}\Big|K_{\alpha/2, (t_{0}/u)^{\beta}}(x\kern1.3pt{-}\kern1.3pt y)\kern1.3pt{-}\kern1.3pt K_{\alpha/2,(t_{0}/u)^{\beta}}(x_{0}\kern1.3pt{-}\kern1.3pt y)\Big| \cdot |\varphi(y)|dy\!\right\}g_{\beta}(u)du.\\ \end{aligned}\right.\end{align*} $$

For $M_{1}$ , since $|x-x_{0}|<(t_{0}/u)^{\beta /\alpha }$ , it follows from (ii) of Proposition 2.9 that for $\epsilon \in (0, \alpha -n/p)$ ,

$$ \begin{align*}\Big|K_{\alpha/2, (t_{0}/u)^{\beta}}(x\kern1.2pt{-}\kern1.2pt y)\kern1.2pt{-}\kern1.2pt K_{\alpha/2,(t_{0}/u)^{\beta}}(x_{0}\kern1.2pt{-}\kern1.2pt y)\Big|\lesssim \left(\frac{|x-x_{0}|}{(t_{0}/u)^{\beta/\alpha}}\right)^{\epsilon} \kern-1pt\!\frac{(t_{0}/u)^{\beta}}{((t_{0}/u)^{\beta/\alpha}+|x\kern1.2pt{-}\kern1.2pt y|)^{n+\alpha}}, \end{align*} $$

and hence

(2.10) $$ \begin{align} M_{1}&\lesssim \!\int_{u: (t_{0}/u)^{\beta/\alpha}>|x-x_{0}|}\!\left\{\kern-2pt\int_{\mathbb R^{n}}\!\left(\kern-1.2pt\frac{|x\kern1.2pt{-}\kern1.2pt x_{0}|}{(t_{0}/u)^{\beta/\alpha}}\kern-1.2pt\right)^{\epsilon} \kern-1pt\!\frac{(t_{0}/u)^{\beta}}{((t_{0}/u)^{\beta/\alpha}+|x\kern1.2pt{-}\kern1.2pt y|)^{n+\alpha}}|\varphi(y)|dy\!\right\}g_{\beta}(u)du\\ &\lesssim \int_{u: (t_{0}/u)^{\beta/\alpha}>|x-x_{0}|}\left(\frac{|x-x_{0}|}{(t_{0}/u)^{\beta/\alpha}}\right)^{\epsilon}\frac{(t_{0}/u)^{\beta}}{(t_{0}/u)^{\beta(1+n/\alpha)}} \|\varphi\|_{L_{x}^{p}}(t_{0}/u)^{\beta n/(\alpha p')}g_{\beta}(u)du\nonumber\\ &\lesssim \|\varphi\|_{L_{x}^{p}}\frac{|x-x_{0}|^{\epsilon}}{t_{0}^{\beta(\epsilon+n/p)/\alpha}}\int_{0}^{\infty}u^{\beta(\epsilon+n/p)/\alpha}g_{\beta}(u)du\nonumber\\ &\lesssim \|\varphi\|_{L_{x}^{p}}{t_{0}^{-\beta(\epsilon+n/p)/\alpha}}{|x-x_{0}|^{\epsilon}},\nonumber \end{align} $$

where in the last inequality we have applied (2.2) to estimate the integral as

$$ \begin{align*} \int_{0}^{\infty}u^{\beta(\epsilon+n/p)/\alpha}g_{\beta}(u)du\lesssim\int_{0}^{1}u^{\beta(\epsilon+n/p)/\alpha} du+\int^{\infty}_{1}u^{\beta(\epsilon+n/ p)/\alpha-\beta-1}du<\infty. \end{align*} $$

For $M_{2}$ , noticing that $|x-x_{0}|\geq (t/u)^{\beta /\alpha }$ , we can use (i) of Proposition 2.9 to get

$$ \begin{align*} &\Big|K_{\alpha/2, (t_{0}/u)^{\beta}}(x-y)-K_{\alpha/2,(t_{0}/u)^{\beta}}(x_{0}-y)\Big|\\ &\lesssim \left(\frac{|x-x_{0}|}{(t_{0}/u)^{\beta/\alpha}}\right)^{\epsilon} \left\{\frac{(t_{0}/u)^{\beta}}{((t_{0}/u)^{\beta/\alpha}+|x-y|)^{n+\alpha}}+\frac{(t_{0}/u)^{\beta}}{((t_{0}/u)^{\beta/\alpha}+|x_{0}-y|)^{n+\alpha}}\right\}. \end{align*} $$

We split $M_{2}\leq M_{2,1}+M_{2,2}$ , where

$$ \begin{align*}\left\{\begin{aligned}\! M_{2,1}&:=\!\int_{u: (t_{0}/u)^{\beta/\alpha}>|x-x_{0}|}\!\left\{\kern-1.2pt\int_{\mathbb R^{n}}\kern-2pt\left(\frac{|x\kern-1.3pt{-}\kern-1.3pt x_{0}|}{(t_{0}/u)^{\beta/\alpha}}\right)^{\epsilon} \!\frac{(t_{0}/u)^{\beta}}{((t_{0}/u)^{\beta/\alpha}+|x-y|)^{n+\alpha}}|\varphi(y)|dy\kern-1.2pt\right\}g_{\beta}(u)du,\\ \!M_{2,2}&:=\!\int_{u: (t_{0}/u)^{\beta/\alpha}>|x-x_{0}|}\!\left\{\kern-1.2pt\int_{\mathbb R^{n}}\kern-2pt\left(\frac{|x\kern-1.3pt{-}\kern-1.3pt x_{0}|}{(t_{0}/u)^{\beta/\alpha}}\right)^{\epsilon} \!\frac{(t_{0}/u)^{\beta}}{((t_{0}/u)^{\beta/\alpha}+|x-y|)^{n+\alpha}}|\varphi(y)|dy\kern-1.2pt\right\}g_{\beta}(u)du. \end{aligned}\right.\end{align*} $$

We can follow the procedure of (2.10) to deduce that

$$ \begin{align*}M_{2,1}+M_{2,2}\lesssim \|g\|_{L_{x}^{p}}{t_{0}^{-\beta(\epsilon+n/p)/\alpha}}{|x-x_{0}|^{\epsilon}}.\end{align*} $$

This means

(2.11) $$ \begin{align} \left|R_{\alpha,\beta}(\varphi)(t_{0},x)-R_{\alpha,\beta}(\varphi)(t_{0},x_{0})\right|\lesssim \|g\|_{L_{x}^{p}}{t_{0}^{-\beta(\epsilon+n/p)/\alpha}}{|x-x_{0}|^{\epsilon}}, \end{align} $$

which indicates that $R_{\alpha ,\beta }(\varphi )(t_{0}, x)$ is continuous with respect to the spatial variable x. Now, we investigate the continuity respect to the time variable t. For fixed $x\in \mathbb R^{n}$ ,

$$ \begin{align*} &R_{\alpha,\beta}(\varphi)(t,x)-R_{\alpha,\beta}(\varphi)(t_{0},x) \\&\quad= \int_{\mathbb R^{n}}\left(G_{t}(x-y)-G_{t_{0}}(x-y)\right)\varphi(y)dy\\ &\quad= \int_{\mathbb R^{n}}\left\{\int^{\infty}_{0}\left(K_{\alpha/2, (t/u)^{\beta}}(x-y)-K_{\alpha/2,(t_{0}/u)^{\beta}}(x-y)\right)g_{\beta}(u)du\right\}\varphi(y)dy\\ &\quad= \int^{\infty}_{0}\left\{\int_{\mathbb R^{n}}\left(K_{\alpha/2,(t/u)^{\beta}}(x-y)-K_{\alpha/2,(t_{0}/u)^{\beta}}(x-y)\right)\varphi(y)dy\right\}g_{\beta}(u)du\\ &\quad= \int^{\infty}_{0} \left(e^{-(t/u)^{\beta}(-\Delta)^{\alpha/2}}\varphi(x)-e^{-(t_{0}/u)^{\beta}(-\Delta)^{\alpha/2}}\varphi(x)\right)g_{\beta}(u)du, \end{align*} $$

which gives

$$ \begin{align*} &\left|R_{\alpha,\beta}(\varphi)(t,x)-R_{\alpha,\beta}(\varphi)(t_{0},x)\right| \\&\quad\lesssim \int^{\infty}_{0}\left|e^{-(t/u)^{\beta}(-\Delta)^{\alpha/2}} \varphi(x)-e^{-(t_{0}/u)^{\beta}(-\Delta)^{\alpha/2}}\varphi(x)\right|g_{\beta}(u)du. \end{align*} $$

Case 1: $p\in [1,\infty )$ . We have

$$ \begin{align*} &\left|e^{-(t/u)^{\beta}(-\Delta)^{\alpha/2}}\varphi(x)-e^{-(t_{0}/u)^{\beta}(-\Delta)^{\alpha/2}}f(x)\right|\\ &\lesssim \|\varphi\|_{L^{p}_{x}(\mathbb R^{n})}\left|\left({t}/{u}\right)^{\beta(1-1/\alpha-n/\alpha p)}-\left({t_{0}}/{u}\right)^{\beta(1-1/\alpha-n/\alpha p)}\right|\\ &\lesssim \|\varphi\|_{L^{p}_{x}(\mathbb R^{n})}u^{-\beta(1-1/\alpha-n/\alpha p)}\left|t^{\beta(1-1/\alpha-n/\alpha p)}-t_{0}^{\beta(1-1/\alpha-n/\alpha p)}\right|. \end{align*} $$

Because $\alpha>n$ and $p\geq 1$ , then $1+n/p\leq 1+n<\alpha $ . The above estimate gives

$$ \begin{align*} &\left|R_{\alpha,\beta}(\varphi)(t,x)-R_{\alpha,\beta}(\varphi)(t_{0},x)\right|\\ &\lesssim\|\varphi\|_{L^{p}_{x}(\mathbb R^{n})}\left|t^{\beta(1-1/\alpha-n/\alpha p)}-t_{0}^{\beta(1-1/\alpha-n/\alpha p)}\right|\int^{\infty}_{0}g_{\beta}(u)u^{-\beta(1-1/\alpha-n/\alpha p)}du. \end{align*} $$

It can be seen from (2.2) that if $u\in (0,1)$ , for any $N>0$ ,

$$ \begin{align*} g_{\beta}(u)&\sim K(\beta/u)^{(1-\beta/2)/(1-\beta)}\exp\{-|1-\beta|(u/\beta)^{\beta/(\beta-1)}\}\\ &\lesssim u^{-(1-\beta/2)/(1-\beta)}\frac{1}{u^{\beta N/(\beta-1)}}\\ &\lesssim u^{(N\beta-1+\beta/2)/(1-\beta)}. \end{align*} $$

Then, we split the integral

$$ \begin{align*} \int^{\infty}_{0}g_{\beta}(u)u^{-\beta(1-1/\alpha-n/\alpha p)}du\lesssim I_{1}+I_{2}, \end{align*} $$

where

$$ \begin{align*}\left\{\begin{aligned} I_{1}&:=\int^{1}_{0}u^{(N\beta-1+\beta/2)/(1-\beta)}u^{-\beta(1-1/\alpha-n/\alpha p)}du,\\ I_{2}&:=\int^{\infty}_{1}u^{-\beta-1}u^{-\beta(1-1/\alpha-n/\alpha p)}du. \end{aligned}\right.\end{align*} $$

Taking N large enough, we have $I_{1}<\infty $ . For $I_{2}$ , because $\beta \in (0,1)$ and $\alpha>n$ , then $1/\alpha +n/\alpha p-2<0$ and

$$ \begin{align*}I_{2}=\int^{\infty}_{1}u^{-\beta(2-1/\alpha-n/\alpha p)-1}du<\infty,\end{align*} $$

which indicates that

(2.12) $$ \begin{align} \left|R_{\alpha,\beta}(\varphi)(t,x)-R_{\alpha,\beta}(\varphi)(t_{0},x)\right| \lesssim\|\varphi\|_{L^{p}_{x}(\mathbb R^{n})}\left|t^{\beta(1-1/\alpha-n/\alpha p)}-t_{0}^{\beta(1-1/\alpha-n/\alpha p)}\right|. \end{align} $$

Case 2: $p=\infty $ . We can also get

(2.13) $$ \begin{align} &\Big|R_{\alpha,\beta}f(t,x)-R_{\alpha,\beta}f(t_{0},x)\Big| \\&\lesssim \|\varphi\|_{L^{\infty}_{x}(\mathbb R^{n})}\Big|t^{\beta(1-1/\alpha)}-t_{0}^{\beta(1-1/\alpha)}\Big|\left(\int^{\infty}_{0}g_{\beta}(u)u^{-(1-1/\alpha)\beta}du\right)\nonumber\\ &\lesssim \|\varphi\|_{L^{\infty}_{x}(\mathbb R^{n})}\Big|t^{\beta(1-1/\alpha)}-t_{0}^{\beta(1-1/\alpha)}\Big|.\nonumber \end{align} $$

Now, fix $(t_{0}, x_{0})\in \mathbb R^{1+n}_{+}$ . Since

$$ \begin{align*} &\Big|R_{\alpha,\beta}\varphi(t,x)-R_{\alpha,\beta}\varphi(t_{0},x_{0})\Big|\\&\quad\leq \Big|R_{\alpha,\beta}\varphi(t,x)-R_{\alpha,\beta}\varphi(t_{0},x)\Big|+\Big|R_{\alpha,\beta}f(t_{0},x)-R_{\alpha,\beta}\varphi(t_{0},x_{0})\Big|, \end{align*} $$

it follows from (2.11)–(2.13) that

$$ \begin{align*} \lim_{(t,x)\rightarrow(t_{0},x_{0})}R_{\alpha,\beta}\varphi(t,x)=R_{\alpha,\beta}\varphi(t_{0},x).\\[-42pt] \end{align*} $$

Proof Given a point $(t_{0}, x_{0})\in \mathbb R^{1+n}_{+}$ , let $x\in \mathbb R^{n}$ be sufficiently close to $x_{0}$ such that $\delta =|x-x_{0}|$ small enough. Then $|G_{\alpha ,\beta }(g)(t_{0},x)-G_{\alpha ,\beta }(t_{0}, x_{0})|\lesssim I+II,$ where

$$ \begin{align*}\left\{\begin{aligned} I&:=\int_{0}^{t_{0}}\int_{B(x_{0}, 3\delta)}\Big|G_{t_{0}-s}(x_{0}-y)-G_{t_{0}-s}(x-y)\Big||g(s,y)|dyds,\\ II&:=\int_{0}^{t_{0}}\int_{\mathbb R^{n}\setminus B(x_{0}, 3\delta)}\Big|G_{t_{0}-s}(x_{0}-y)-G_{t_{0}-s}(x-y)\Big||g(s,y)|dyds. \end{aligned}\right.\end{align*} $$

We first estimate I. Write

$$ \begin{align*} I \lesssim \!\int^{t_{0}}_{0}\!\int_{B(x_{0}, 3\delta)}\kern-1pt\Big|G_{t_{0}-s}(x_{0}-y)\Big||g(s,y)|dyds+\!\int^{t_{0}}_{0}\!\int_{B(x_{0}, 3\delta)}\Big|G_{t_{0}-s}(x-y)\Big||g(s,y)|dyds. \end{align*} $$

We further split

$$ \begin{align*}\int^{t_{0}}_{0}\int_{B(x_{0}, 3\delta)}\Big|G_{t_{0}-s}(x_{0}-y)\Big||g(s,y)|dyds\lesssim I_{1}+I_{2},\end{align*} $$

where

$$ \begin{align*}\left\{\begin{aligned} I_{1}&:=\int^{t_{0}-(2\delta)^{\alpha/\beta}}_{0}\int_{B(x_{0}, 3\delta)}\Big|G_{t_{0}-s}(x_{0}-y)\Big||g(s,y)|dyds,\\ I_{2}&:=\int^{t_{0}}_{t_{0}-(2\delta)^{\alpha/\beta}}\int_{B(x_{0}, 3\delta)}\Big|G_{t_{0}-s}(x_{0}-y)\Big||g(s,y)|dyds.\\ \end{aligned}\right.\end{align*} $$

Applying Hölder’s inequality to the variables y and s, respectively, we have

$$ \begin{align*} I_{1}&\lesssim \int^{t_{0}-(2\delta)^{\alpha/\beta}}_{0}\int_{B(x_{0}, 3\delta)}\frac{|t_{0}-s|^{\beta}}{(|t_{0}-s|^{\beta/\alpha}+|x_{0}-y|)^{n+\alpha}}|g(s,y)|dyds\\ &\lesssim \int^{t_{0}-(2\delta)^{\alpha/\beta}}_{0}\frac{1}{|t_{0}-s|^{\beta n/\alpha}}\Bigg(\int_{B(x_{0}, 3\delta)}|g(y,s)|^{p}dy\Bigg)^{1/p} \Bigg(\int_{B(x_{0}, 3\delta)}1 dy\Bigg)^{1-1/p}ds\\ &\lesssim \int^{t_{0}-(2\delta)^{\alpha/\beta}}_{0}\frac{1}{|t_{0}-s|^{\beta n/\alpha}}\|g(s,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})}\delta^{n(1-1/p)}ds\\ &\lesssim \|g(\cdot,\cdot)\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}\delta^{\alpha(1-1/q)/\beta-n/p}. \end{align*} $$

Similarly, for $I_{2}$ , we have

$$ \begin{align*} I_{2}&\lesssim \int^{t_{0}}_{t_{0}-(2\delta)^{\alpha/\beta}}\int_{B(x_{0}, 3\delta)}\frac{|t_{0}-s|^{\beta}}{(|t_{0}-s|^{\beta/\alpha}+|x_{0}-y|)^{n+\alpha}}|g(s,y)|dyds\\ &\lesssim \int^{t_{0}}_{t_{0}-(2\delta)^{\alpha/\beta}}\|g(s,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})}\Bigg(\int_{B(x_{0}, 3\delta)}\frac{|t_{0}-s|^{p'\beta}}{(|t_{0}-s|^{\beta/\alpha}+|x_{0}-y|)^{(n+\alpha)p'}}dy\Bigg)^{1/p'}ds\\ &\lesssim \int^{t_{0}}_{t_{0}-(2\delta)^{\alpha/\beta}}\|g(s,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})}\Bigg(\int_{\mathbb R^{n}}\frac{|t_{0}-s|^{p'\beta}}{(|t_{0}-s|^{\beta/\alpha}+|x_{0}-y|)^{(n+\alpha)p'}}dy\Bigg)^{1/p'}ds. \end{align*} $$

Split

$$ \begin{align*} &\int_{\mathbb R^{n}}\frac{|t_{0}-s|^{p'\beta}}{(|t_{0}-s|^{\beta/\alpha}+|x_{0}-y|)^{(n+\alpha)p'}}dy\\ &\lesssim \int_{|y-x_{0}|<|t_{0}-s|^{\beta/\alpha}}\frac{|t_{0}-s|^{p'\beta}}{(|t_{0}-s|^{\beta/\alpha}+|x_{0}-y|)^{(n+\alpha)p'}}dy\\ &\quad+\sum^{\infty}_{k=0}\int_{2^{k}|t_{0}-s|^{\beta/\alpha}\leq|y-x_{0}|<2^{k+1}|t_{0}-s|^{\beta/\alpha}} \frac{|t_{0}-s|^{p'\beta}}{(|t_{0}-s|^{\beta/\alpha}+|x_{0}-y|)^{(n+\alpha)p'}}dy\\ &\lesssim \int_{|y-x_{0}|<|t_{0}-s|^{\beta/\alpha}}\frac{|t_{0}-s|^{p'\beta}}{|t_{0}-s|^{\beta(n+\alpha)p'/\alpha}}dy\\ &\quad+\sum^{\infty}_{k=0}\int_{2^{k}|t_{0}-s|^{\beta/\alpha}\leq|y-x_{0}|<2^{k+1}|t_{0}-s|^{\beta/\alpha}} \frac{|t_{0}-s|^{p'\beta}}{(2^{k}|t_{0}-s|^{\beta/\alpha})^{(n+\alpha)p'}}dy\\ &\lesssim \frac{|t_{0}-s|^{p'\beta}}{|t_{0}-s|^{\beta(n+\alpha)p'/\alpha}}|t_{0}-s|^{n\beta/\alpha} +\sum^{\infty}_{k=0}\frac{|t_{0}-s|^{p'\beta}}{(2^{k}|t_{0}-s|)^{\beta(n+\alpha)p'/\alpha}}(2^{k+1}|t_{0}-s|^{\beta/\alpha})^{n}\\ &\lesssim \frac{1}{|t_{0}-s|^{n\beta(p'-1)/\alpha}}. \end{align*} $$

Then we obtain

$$ \begin{align*} I_{2}&\lesssim \int^{t_{0}}_{t_{0}-(2\delta)^{\alpha/\beta}}\|g(s,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})}\frac{ds}{|t_{0}-s|^{n\beta/\alpha p}}\\ &\lesssim \|g(\cdot,\cdot)\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}\Bigg(\int^{t_{0}}_{t_{0}-(2\delta)^{\alpha/\beta}}\frac{ds}{|t_{0}-s|^{n\beta q'/\alpha p}}\Bigg)^{1/q'}\\ &\lesssim \|g(\cdot,\cdot)\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}\delta^{\alpha(1-1/q)/\beta-n/p}. \end{align*} $$

Now, we estimate $II$ . It follows from (ii) of Proposition 2.9 that there exists a constant $\epsilon>0$ such that for $|x-x_{0}|<|x|/2$ ,

$$ \begin{align*} |K_{\alpha/2,s}(x)-K_{\alpha/2,s}(x_{0})|\lesssim\frac{s|x-x_{0}|^{\epsilon}}{(s^{1/\alpha}+|x|)^{n+\alpha+\epsilon}}\lesssim \frac{s|x-x_{0}|^{\epsilon}}{|x|^{n+\alpha+\epsilon}}. \end{align*} $$

We have

$$ \begin{align*} \Big|G_{t}(x)-G_{t}(x_{0})\Big|&\lesssim \int^{\infty}_{0}\frac{(t/u)^{\beta}}{|x|^{n+\alpha+\epsilon}}|x-x_{0}|^{\epsilon}g_{\beta}(u)du\\ &\lesssim \frac{t^{\beta}|x-x_{0}|^{\epsilon}}{|x|^{n+\alpha+\epsilon}}\int^{\infty}_{0}u^{-\beta}g_{\beta}(u)du\\ &\lesssim \frac{t^{\beta}|x-x_{0}|^{\epsilon}}{|x|^{n+\alpha+\epsilon}}, \end{align*} $$

which implies that

$$ \begin{align*} \Big|G_{t}(x)-G_{t}(x_{0})\Big|\lesssim\min\left\{\left(\frac{|x-x_{0}|}{t^{\beta/\alpha}}\right)^{\epsilon} \frac{1}{t^{n\beta/\alpha}},\ \frac{t^{\beta}|x-x_{0}|^{\epsilon}}{|x|^{n+\alpha+\epsilon}}\right\}. \end{align*} $$

If $|x|\leq t^{\beta /\alpha }$ , we have

$$ \begin{align*} \Big|G_{t}(x)-G_{t}(x_{0})\Big|\lesssim\frac{|x-x_{0}|^{\epsilon}t^{\beta}}{t^{\beta(n+\alpha+\epsilon)/\alpha}}\lesssim \frac{|x-x_{0}|^{\epsilon}t^{\beta}}{(t^{\beta/\alpha}+|x|)^{n+\alpha+\epsilon}}. \end{align*} $$

If $|x|> t^{\beta /\alpha }$ , it still holds

(2.14) $$ \begin{align} \Big|G_{t}(x)-G_{t}(x_{0})\Big|\lesssim\frac{|x-x_{0}|^{\epsilon}t^{\beta}}{|x|^{n+\alpha+\epsilon}}\lesssim \frac{|x-x_{0}|^{\epsilon}t^{\beta}}{(t^{\beta/\alpha}+|x|)^{n+\alpha+\epsilon}}. \end{align} $$

Applying (2.14), we can deduce that

$$ \begin{align*} II&\lesssim \int^{t_{0}}_{0}\int_{\mathbb R^{n}\setminus B(x_{0}, 3\delta)}\frac{\delta^{\epsilon}|t_{0}-s|^{\beta}}{(|t_{0}-s|^{\beta/\alpha}+|x_{0}-y|)^{n+\alpha+\epsilon}}|g(s,y)|dyds\\ &\lesssim \int^{t_{0}-(2\delta)^{\alpha/\beta}}_{0}\int_{\mathbb R^{n}\setminus B(x_{0}, 3\delta)}\frac{\delta^{\epsilon}|t_{0}-s|^{\beta}}{(|t_{0}-s|^{\beta/\alpha}+|x_{0}-y|)^{n+\alpha+\epsilon}}|g(s,y)|dyds\\ &\quad +\int^{t_{0}}_{t_{0}-(2\delta)^{\alpha/\beta}}\int_{\mathbb R^{n}\setminus B(x_{0}, 3\delta)}\frac{\delta^{\epsilon}|t_{0}-s|^{\beta}}{(|t_{0}-s|^{\beta/\alpha}+|x_{0}-y|)^{n+\alpha+\epsilon}}|g(s,y)|dyds\\ &:= II_{1}+II_{2}, \end{align*} $$

where

$$ \begin{align*} II_{1}&\lesssim \int^{t_{0}-(2\delta)^{\alpha/\beta}}_{0}\int_{\mathbb R^{n}\setminus B(x_{0}, 3\delta)}\frac{\delta^{\epsilon}|t_{0}-s|^{\beta}}{(|t_{0}-s|^{\beta/\alpha}+|x_{0}-y|)^{n+\alpha+\epsilon}}|g(s,y)|dyds\\ &\lesssim \int^{t_{0}-(2\delta)^{\alpha/\beta}}_{0}\!\delta^{\epsilon}|t_{0}-s|^{\beta}\|g(s,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})}\Bigg(\kern-2pt\int_{\mathbb R^{n}\setminus B(x_{0}, 3\delta)} \!\frac{dy}{(|t_{0}-s|^{\beta/\alpha}+|x_{0}-y|)^{p'(n+\alpha+\epsilon)}}\kern-1pt\Bigg)^{\kern-0.1pt1/p'}ds \end{align*} $$

and, similarly,

$$ \begin{align*} II_{2} \lesssim \int_{t_{0}-(2\delta)^{\alpha/\beta}}^{t_0}\delta^{\epsilon}|t_{0}-s|^{\beta}\|g(s,\cdot)\|_{L^{p}_{x} (\mathbb R^{n})}\Bigg(\int_{\mathbb R^{n}\setminus B(x_{0}, 3\delta)}\frac{dy}{|x_{0}-y|^{p'(n+\alpha+\epsilon)}}\Bigg)^{1/p'}ds. \end{align*} $$

We first estimate $II_{1}$ . The following integral can be estimated as

$$ \begin{align*} \int_{\mathbb R^{n}\setminus B(x_{0}, 3\delta)}\frac{\delta^{p'\epsilon}|t_{0}-s|^{p'\beta}dy}{(|t_{0}-s|^{\beta/\alpha}+|x_{0}-y|)^{p'(n+\alpha+\epsilon)}} &\lesssim \sum^{\infty}_{j=1}\frac{\delta^{p'\epsilon}|t_{0}-s|^{p'\beta}(3^{j}\delta)^{n}} {(|t_{0}-s|^{\beta/\alpha}+3^{j}\delta)^{p'(n+\alpha+\epsilon)}}\\ &\lesssim \sum^{\infty}_{j=1}\frac{1}{3^{j}}\int^{3^{j+1}\delta}_{3^{j}\delta} \frac{\delta^{p'\epsilon-1}|t_{0}-s|^{p'\beta}}{(|t_{0}-s|^{\beta/\alpha}+r)^{p'(n+\alpha+\epsilon)-n}}dr\\ &\lesssim \int^{\infty}_{3\delta} \frac{\delta^{p'\epsilon-1}|t_{0}-s|^{p'\beta}}{(|t_{0}-s|^{\beta/\alpha}+r)^{p'(n+\alpha+\epsilon)-n}}dr. \end{align*} $$

A direct computation gives

$$ \begin{align*} \int^{\infty}_{3\delta} \frac{\delta^{p'\epsilon-1}|t_{0}-s|^{p'\beta}}{(|t_{0}-s|^{\beta/\alpha}+r)^{p'(n+\alpha+\epsilon)-n}}dr &\lesssim \delta^{p'\epsilon-1}|t_{0}-s|^{p'\beta}\int^{\infty}_{3\delta+|t_{0}-s|^{\beta/\alpha}} \frac{1}{u^{p'(n+\alpha+\epsilon)-n}}du\\ &\lesssim \frac{\delta^{p'\epsilon-\beta}|t_{0}-s|^{p'\beta}}{(3\delta+|t_{0}-s|^{\beta/\alpha})^{p'(n+\alpha+\epsilon)-n-1}}, \end{align*} $$

which yields

$$ \begin{align*} II_{1}&\lesssim \int^{t_{0}-(2\delta)^{\alpha/\beta}}_{0}\|g(s,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})} \frac{\delta^{\epsilon-1/p'}|t_{0}-s|^{\beta}}{(\delta+|t_{0}-s|^{\beta/\alpha})^{(n+\alpha+\epsilon)-(n+1)/p'}}ds\\ &\lesssim \int^{t_{0}-(2\delta)^{\alpha/\beta}}_{0}\|g(s,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})} \frac{\delta^{\epsilon-1/p'}}{(\delta+|t_{0}-s|^{\beta/\alpha})^{n/p+\epsilon-1/p'}}ds\\ &\lesssim \delta^{\epsilon-1/p'}\|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}\Bigg(\int^{t_{0}-(2\delta)^{\alpha}}_{0}\frac{1}{(\delta+|t_{0}-s|^{\beta/\alpha})^{(n/p+\epsilon-1/p')q'}}ds\Bigg)^{1/q'}\\ &\lesssim \delta^{\alpha(1-1/q)/\beta-n/p}\|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}. \end{align*} $$

For the term $II_{2}$ , an application of Hölder’s inequality yields

$$ \begin{align*} II_{2}&\lesssim \int^{t_{0}}_{t_{0}-(2\delta)^{\alpha/\beta}}|t_{0}-s|^{\beta}\delta^{\epsilon} \|g(s,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})}\Bigg(\int_{\mathbb R^{n}\setminus B(x_{0},3\delta)}\frac{dy}{|x_{0}-y|^{p'(n+\epsilon+\alpha)}}\Bigg)^{p'}ds\\ &\lesssim \int^{t_{0}}_{t_{0}-(2\delta)^{\alpha/\beta}}\|g(s,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})}|t_{0}-s|^{\beta}\delta^{-(n+\epsilon+\alpha)+n/p'}\delta^{\epsilon}ds\\ &\lesssim \delta^{-(n+\epsilon+\alpha)+n/p'}\delta^{\epsilon}\|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}\Bigg(\int^{t_{0}}_{t_{0}-(2\delta)^{\alpha/\beta}}|t_{0}-s|^{\beta q'}ds\Bigg)^{1/q'}\\ &\lesssim \|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}\delta^{\alpha(1-1/q)/\beta-n/p}. \end{align*} $$

The estimates for I and $II$ imply that

$$ \begin{align*} \Big|G_{\alpha,\beta}(g)(t_{0},x_{0})-G_{\alpha,\beta}(g)(t_{0},x)\Big| \lesssim \|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}\delta^{\alpha(1-1/q)/\beta-n/p}. \end{align*} $$

For fixed $x\in \mathbb R^{n}$ and $t_{1}>t_{2}$ , we can see that

$$ \begin{align*} \Big|G_{\alpha,\beta}(g)(t_{1},x)-G_{\alpha,\beta}(g)(t_{2},x)\Big|\lesssim III+IV, \end{align*} $$

where

$$ \begin{align*}\left\{\begin{aligned} III&:=\Bigg|\int^{t_{2}}_{0}\int_{\mathbb R^{n}}\left(G_{t_{1}-s}(x-y)-G_{t_{2}-s}(x-y)\right)g(s,y)dsdy\Bigg|,\\ IV&:=\Bigg|\int^{t_{1}}_{t_{2}}\int_{\mathbb R^{n}}G_{t_{1}-s}(x-y)g(s,y)dyds\Bigg|. \end{aligned}\right.\end{align*} $$

It follows from Hölder’s inequality that

$$ \begin{align*} IV&\lesssim \int^{t_{1}}_{t_{2}}\int_{\mathbb R^{n}}|G_{t_{1}-s}(x-y)||g(s,y)|dyds\\ &\lesssim \int^{t_{1}}_{t_{2}}\|g(s,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})}\left\{\int_{\mathbb R^{n}}\frac{|t_{1}-s|^{\beta p'}}{(|t_{1}-s|^{\beta/\alpha}+|x-y|)^{(n+\alpha)p'}}dy\right\}^{1/p'}ds\\ &\lesssim \int^{t_{1}}_{t_{2}}\|g(s,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})}|t_{1}-s|^{-\beta n/(\alpha p)}ds\\ &\lesssim \|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}\Bigg(\int^{t_{1}}_{t_{2}}|t_{1}-s|^{-\beta nq'/(\alpha p)}ds\Bigg)^{1/q'}\\ &\lesssim \|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}|t_{1}-t_{2}|^{1-1/q-n\beta/(\alpha p)}. \end{align*} $$

To estimate $III$ , we first investigate the Lipschitz continuity of $G_{t}(\cdot )$ on the time variable t. Because

$$ \begin{align*} G_{t_{1}-s}(x-y)-G_{t_{2}-s}(x-y) = \!\int^{\infty}_{0}\!\left(K_{\alpha,\left(\frac{t_{1}-s}{u}\right)^{\beta}} (x-y)-K_{\alpha,\left(\frac{t_{2}-s}{u}\right)^{\beta}}(x-y)\kern-1pt\right) g_{\beta}(u)du, \end{align*} $$

it holds

$$ \begin{align*} III&= \Bigg|\int^{t_{2}}_{0}\int_{\mathbb R^{n}}\left\{\int^{\infty}_{0}\left(K_{\alpha/2,\left(\frac{t_{1}-s}{u}\right)^{\beta}}(x-y)-K_{\alpha/2,\left(\frac{t_{2}-s}{u}\right)^{\beta}}(x-y)\right)g_{\beta}(u)du\right\} g(s,y)dsdy\Bigg|\\ &= \Bigg|\int^{t_{2}}_{0}\int^{\infty}_{0}\left\{\int_{\mathbb R^{n}}\left(K_{\alpha/2,\left(\frac{t_{1}-s}{u}\right)^{\beta}}(x-y)-K_{\alpha/2,\left(\frac{t_{2}-s}{u}\right)^{\beta}}(x-y)\right)g(s,y)dy\right\}g_{\beta}(u)dsdu\Bigg|\\ &= \Bigg|\int^{\infty}_{0}\int^{t_{2}}_{0}\Bigg\{e^{-\left(\frac{t_{1}-s}{u}\right)^{\beta}(-\Delta)^{\alpha/2}}(\varphi)(s,x)- e^{-\left(\frac{t_{2}-s}{u}\right)^{\beta}(-\Delta)^{\alpha/2}}(\varphi)(s,x)\Bigg\}g_{\beta}(u)dsdu\Bigg|. \end{align*} $$

Noticing that

$$ \begin{align*} &\left|e^{-\left(\frac{t_{1}-s}{u}\right)^{\beta}(-\Delta)^{\alpha/2}}(\varphi)(s,x)- e^{-\left(\frac{t_{2}-s}{u}\right)^{\beta}(-\Delta)^{\alpha/2}} (\varphi)(s,x)\right|\\ &\quad= \left|\int^{\left(\frac{t_{1}-s}{u}\right)^{\beta}}_{\left(\frac{t_{2}-s}{u}\right)^{\beta}} \frac{\partial}{\partial_{r}}e^{-r(-\Delta)^{\alpha/2}}(\varphi)(s,x)dr\right|\\ &\quad\lesssim \int^{\left(\frac{t_{1}-s}{u}\right)^{\beta}}_{\left(\frac{t_{2}-s}{u}\right)^{\beta}} \left|(-\Delta)^{\alpha/2} e^{-r(-\Delta)^{\alpha/2}}(\varphi)(s,x)\right|dr\\ &\quad\lesssim \int^{\left(\frac{t_{1}-s}{u}\right)^{\beta}}_{\left(\frac{t_{2}-s}{u}\right)^{\beta}}r^{-1-n/(\alpha p)}\|g(s,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})}dr, \end{align*} $$

we can obtain

$$ \begin{align*} III&\lesssim \int^{\infty}_{0}\left\{\int^{t_{2}}_{0} \left(\int^{\left(\frac{t_{1}-s}{u}\right)^{\beta}}_{\left(\frac{t_{2}-s}{u}\right)^{\beta}}r^{-1-\frac{n}{\alpha p}}\|g(s,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})}dr\right)ds\right\}g_{\beta}(u)du\\ &\lesssim \int^{\infty}_{0}\left\{\int^{t_{2}}_{0}\left(\int^{\left(\frac{t_{1}-s}{u}\right)^{\beta}}_{\left(\frac{t_{2}-s}{u}\right)^{\beta}}r^{-1-\frac{n}{\alpha p}}dr\right)\|g(s,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})}ds\right\}g_{\beta}(u)du\\ &\lesssim \int^{\infty}_{0}\left\{\int^{t_{2}}_{0}\left(\int^{\left(\frac{t_{1}-s}{u}\right)^{\beta}- \left(\frac{t_{2}-s}{u}\right)^{\beta}}_{0} \frac{dr}{\left(r+\left(\frac{t_{2}-s}{u}\right)^{\beta}\right)^{1+\frac{n}{\alpha p}}}\right)\|g(s,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})}ds\right\}g_{\beta}(u)du. \end{align*} $$

Since for $\beta \in (0,1)$ ,

$$ \begin{align*} (t_{1}/u-s/u)^{\beta}-(t_{2}/u-s/u)^{\beta}\leq \left(t_{1}/u-t_{2}/{u}\right)^{\beta}, \end{align*} $$

applying Fubini’s theorem and Hölder’s inequality, we can get

$$ \begin{align*} III&\lesssim \int^{\infty}_{0}\Bigg\{\int^{t_{2}}_{0}\left(\int^{(\frac{t_{1}-t_{2}}{u})^{\beta}}_{0}\frac{dr}{\left(r+\left(\frac{t_{2}-s}{u}\right)^{\beta}\right)^{1+\frac{n}{\alpha p}}}\right)\|g(s,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})}ds\Bigg\}g_{\beta}(u)du\\ &\lesssim \int^{\infty}_{0}\left\{\int^{(\frac{t_{1}-t_{2}}{u})^{\beta}}_{0}\left(\int^{t_{2}}_{0}\frac{\|g(s,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})}}{\left(r+\left(\frac{t_{2}-s}{u}\right)^{\beta}\right)^{1+\frac{n}{\alpha p}}}ds\right)dr\right\}g_{\beta}(u)du\\ &\lesssim \|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}\int^{\infty}_{0}\Bigg\{\int^{(\frac{t_{1}-t_{2}}{u})^{\beta}}_{0} \Bigg(\int^{t_{2}}_{0}\frac{ds}{\left(r+\left(\frac{t_{2}-s}{u}\right)^{\beta}\right)^{(1+\frac{n}{\alpha p})q'}}\Bigg)^{1/q'}dr\Bigg\}g_{\beta}(u)du. \end{align*} $$

By change of variables, we obtain

$$ \begin{align*} \int^{t_{2}}_{0}\frac{ds}{\left(r+\left(\frac{t_{2}-s}{u}\right)^{\beta}\right)^{(1+\frac{n}{\alpha p})q'}} = u\int^{(t_{2}/u)^{\beta}}_{0}\frac{s^{1/\beta-1}_{1}ds_{1}}{(r+s_{1})^{(1+\frac{n}{\alpha p})q'}} \lesssim u r^{-(1+\frac{n}{\alpha p})q'+\frac{1}{\beta}}. \end{align*} $$

Finally, we have, upon $1/q'\beta -n/(\alpha p)>0$ ,

$$ \begin{align*} III&\lesssim \|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}\int^{\infty}_{0}\Bigg\{\int^{(t_{1}/u-t_{2}/u)^{\beta}}_{0} r^{-(1+\frac{n}{\alpha p})+\frac{1}{q'\beta}}dr\Bigg\}u^{1/q'}g_{\beta}(u)du\\ &\lesssim \|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}\int^{\infty}_{0}\left(\frac{t_{1}-t_{2}}{u}\right)^{(\frac{1}{q'}-\frac{\beta n}{\alpha p})}u^{1/q'}g_{\beta}(u)du\\ &\lesssim \|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}|t_{1}-t_{2}|^{1-1/q-n\beta/(\alpha p)}, \end{align*} $$

where in the last inequality we have used the following estimate: $\int ^{\infty }_{0}u^{\beta n/\alpha p}g_{\beta }(u)du<\infty $ . In fact, by (2.2), $g_{\beta }(u)\lesssim u^{(N\beta -1+\beta /2)/(1-\beta )}$ for $u\in (0,1)$ and N being large enough, which indicates that

$$ \begin{align*} \int^{1}_{0}u^{\beta n/\alpha p}g_{\beta}(u)du \lesssim\int^{1}_{0}u^{\beta n/\alpha p}u^{(N\beta-1+\beta/2)/(1-\beta)}du<\infty. \end{align*} $$

On the other hand, under the assumption that $\alpha>n$ and $\beta \in (0,1)$ , we can see that $\beta n/(\alpha p)-\beta -1<-1 $ and therefore

$$ \begin{align*} \int^{\infty}_{1}u^{\beta n/(\alpha p)}g_{\beta}(u)du\lesssim\int^{\infty}_{1}u^{\beta n/(\alpha p)}u^{-\beta-1}du<\infty.\\[-42pt] \end{align*} $$

Proof Let $(t,x)\in \mathbb R^{1+n}_{+}$ be fixed. Then $|G_{\alpha ,\beta }(g)(t,x)|\lesssim I+II,$ where

$$ \begin{align*}\left\{\begin{aligned} I:=\int^{r}_{0}\int_{\mathbb R^{n}}G_{t-s}(x-y)|g(s,y)|dyds,\\ II:=\int^{t}_{r}\int_{\mathbb R^{n}}G_{t-s}(x-y)|g(s,y)|dyds. \end{aligned}\right.\end{align*} $$

Below, we deal with the terms I and $II$ , separately. By Hölder’s inequality, we get

$$ \begin{align*} I&\lesssim \int^{r}_{0}\int_{\mathbb R^{n}}\frac{(t-s)^{\beta}}{(|t-s|^{\beta/\alpha}+|x-y|)^{n+\alpha}}g(s,y)dyds\\ &\lesssim \int^{r}_{0}(t-s)^{\beta}\|g(s,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})}\left\{\int_{\mathbb R^{n}}\frac{dy}{(|t-s|^{\beta/\alpha}+|x-y|)^{p(n+\alpha)/(p-1)}}\right\}^{1-1/p}ds. \end{align*} $$

It can be deduced from the integral

$$ \begin{align*} \int_{\mathbb R^{n}}\frac{dy}{(|t-s|^{\beta/\alpha}+|x-y|)^{p(n+\alpha)/(p-1)}} \lesssim\frac{|t-s|^{n\beta/\alpha}}{|t-s|^{\beta(n/\alpha+1)p/(p-1)}} \end{align*} $$

that

$$ \begin{align*} I \lesssim \int^{r}_{0}(t-s)^{\beta}\|g(s,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})}\frac{|t-s|^{n\beta (1-/p)/\alpha}}{|t-s|^{\beta(n+\alpha)/\alpha}}ds \lesssim\int^{r}_{0}\|g(s,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})}\frac{ds}{|t-s|^{n\beta/(\alpha p)}}. \end{align*} $$

By the fact that $1/q+n\beta /(\alpha p)=1$ , we apply Hölder’s inequality on s to obtain

$$ \begin{align*} I \lesssim \|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}\left(\int^{r}_{0}\frac{ds}{|t-s|^{n\beta q'/(\alpha p)}}\right)^{1-1/q} \lesssim\|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}\left(\ln\frac{t}{t-r}\right)^{1-1/q}. \end{align*} $$

Denote by $M_{\mathbb R}$ the Hardy–Littlewood maximal function on $\mathbb R$ . Then, since $1-n\beta /(p\alpha )=1/q$ , it holds

$$ \begin{align*} II&\lesssim \sum^{0}_{k=-\infty}\int^{t-2^{k-1}|t-r|}_{t-2^{k}|t-r|}\|g(s,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})}\frac{ds}{|t-s|^{n\beta/(p\alpha)}}\\ &\lesssim \sum^{0}_{k=-\infty} \frac{1}{(2^{k}|t-r|)^{n\beta/(p\alpha)}}\int^{t}_{t-2^{k}|t-r|}\|g(s,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})}ds\\ &\lesssim\sum^{0}_{k=-\infty}(2^{k}|t-r|)^{1/q}M_{\mathbb R}(\|g(\cdot,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})})(t)\\ &\lesssim|t-r|^{1/q}M_{\mathbb R}(\|g(\cdot,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})})(t). \end{align*} $$

The estimates for I and $II$ indicate that

$$ \begin{align*}|G_{\alpha,\beta}(g)(t,x)|\lesssim \|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}\left(\ln\frac{t}{t-r}\right)^{1-1/q}+|t-r|^{1/q}M_{\mathbb R}(\|g(\cdot,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})})(t).\end{align*} $$

Via choosing $r\in (0,t)$ such that

$$ \begin{align*}|t-r|^{1/q}=\min\left\{t^{1/q},\ \frac{\|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}}{M_{\mathbb R}(\|g(\cdot,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})})(t)}\right\}, \end{align*} $$

we get

$$ \begin{align*}|G_{\alpha,\beta}(g)(t,x)|\lesssim \|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}\left\{1+\left(\ln\frac{t}{t-r}\right)^{1-1/q}\right\}. \end{align*} $$

Case 1: $|t-r|^{1/q}=t^{1/q}$ . Then

$$ \begin{align*}|G_{\alpha,\beta}(g)(t,x)|\lesssim \|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}.\end{align*} $$

Case 2: $|t-r|^{1/q}=\|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}M^{-1}_{\mathbb R}(\|g(\cdot ,\cdot )\|_{L^{p}_{x}(\mathbb R^{n})})(t)$ . We have

$$ \begin{align*} |G_{\alpha,\beta}(g)(t,x)|&\lesssim \|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}\left\{1+\left(\ln\frac{t\left(M_{\mathbb R}(\|g(\cdot,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})})(t)\right)^{q}}{\|g\|^{q}_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}}\right)^{1-1/q}\right\}\\ &\lesssim \|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}\left\{1+\left(q\ln\frac{t^{1/q}M_{\mathbb R}(\|g(\cdot,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})})(t)}{\|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}}\right)^{1-1/q}\right\}. \end{align*} $$

Let $r_{0}=t_{0}^{1/2\alpha }$ . Then

$$ \begin{align*} &\iint_{B^{(\alpha)}_{r_{0}}(t_{0},x_{0})} \exp\left(\frac{G_{\alpha,\beta}(g)(t,x)}{C\|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}}\right)^{q/(q-1)}dxdt\\ &\lesssim \iint_{B^{(\alpha)}_{r_{0}}(t_{0},x_{0})} \exp\left\{\frac{\|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}\left\{1+\left(q\ln\frac{t^{1/q}M_{\mathbb R}(\|g(\cdot,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})})(t)}{\|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}}\right)^{1-1/q}\right\}}{C\|g\|_{{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}}}\right\}^{q/(q-1)}dxdt\\ &\lesssim \iint_{B^{(\alpha)}_{r_{0}}(t_{0},x_{0})}\exp\left\{1+\left(\ln\frac{t^{1/q}M_{\mathbb R}(\|g(\cdot,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})})(t)}{\|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}}\right)^{1-1/q}\right\}^{q/(q-1)}dxdt\\ &\lesssim \iint_{B^{(\alpha)}_{r_{0}}(t_{0},x_{0})}\exp \left(\ln\frac{t^{1/q}M_{\mathbb R}(\|g(\cdot,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})})(t)}{\|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}}\right)dxdt\\ &\lesssim \iint_{B^{(\alpha)}_{r_{0}}(t_{0},x_{0})}\frac{t^{1/q}M_{\mathbb R}(\|g(\cdot,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})})(t)}{\|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}}dxdt\\ &\lesssim|B^{(\alpha)}_{r_{0}}(t_{0},x_{0})|t^{1/q}_{0}\int^{2t_{0}}_{0}\frac{M_{\mathbb R}(\|g(\cdot,\cdot)\|_{L^{p}_{x}(\mathbb R^{n})})(t)}{\|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}}dt \lesssim r^{2\alpha}_{0}|B^{(\alpha)}_{r_{0}}(t_{0},x_{0})|. \end{align*} $$

This completes the proof of Theorem 2.14.

Proof Set

$$ \begin{align*} \widetilde{C^{(\alpha,\beta)}_{p,q}}(K):=\sup\left\{\|\mu\|_{1}^{p\wedge q}:\, \mu\in \mathcal{M}_{+}(K)\ \&\ \|G^{\ast}_{\alpha,\beta}\mu\|_{L^{q'}_{t}L^{p'}_{x}(\mathbb R^{1+n}_{+})}\leq1\right\}. \end{align*} $$

For any $g\geq 0$ such that $\ G_{\alpha ,\beta }(g)\geq 1_{K}$ , by (3.1) and Hölder’s inequality, we have

$$ \begin{align*} \|\mu\|_{1}&= \mu(K)=\iint_{K}1d\mu(t,x)\\ &\leq \iint_{K}G_{\alpha,\beta}(g)(t,x)d\mu(t,x)\\ &\leq \iint_{\mathbb R^{1+n}_{+}}G_{\alpha,\beta}(g)(t,x)d\mu(t,x)\\ &= \iint_{\mathbb R^{1+n}_{+}}g(t,x)G^{\ast}_{\alpha,\beta}(\mu)(t,x)dxdt\\ &\leq \|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}\|G^{\ast}_{\alpha,\beta}\mu\|_{L^{q'}_{t}L^{p'}_{x}(\mathbb R^{1+n}_{+})}, \end{align*} $$

which implies that if $\mu $ satisfies $\|G^{\ast }_{\alpha ,\beta }\mu \|_{L^{q'}_{t}L^{p'}_{x}(\mathbb R^{1+n}_{+})}\leq 1$ , then $\|\mu \|_{1}^{p\wedge q}\leq \|g\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}$ , i.e.,

$$ \begin{align*} \widetilde{C^{(\alpha,\beta)}_{p,q}}(K)\leq C^{(\alpha,\beta)}_{p,q}(K). \end{align*} $$

Moreover, let

$$ \begin{align*}\left\{\begin{aligned} &\mathcal X:=\Big\{\mu:\ \mu\in\mathcal M_{+}(K)\ \&\ \mu(K)=1\Big\},\\ &\mathcal Y:=\Big\{g:\ 0\leq g\in L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})\ \&\ \|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}\leq 1\Big\},\\ &\mathcal Z:=\Big\{g:\ 0\leq g\in L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})\ \&\ G_{\alpha,\beta}(g)\geq 1_{K}\Big\}. \end{aligned}\right.\end{align*} $$

We have

$$ \begin{align*} \min_{\mu\in\mathcal{M}_{+}(K)}\left\{(\mu(K))^{-1}\|G^{\ast}_{\alpha,\beta}\mu\|_{L^{q'}_{t}L^{p'}_{x}(\mathbb R^{1+n}_{+})}\right\} &= \min_{\mu\in\mathcal{X}} \left\{(\mu(K))^{-1}\|G^{\ast}_{\alpha,\beta} \mu\|_{L^{q'}_{t}L^{p'}_{x} (\mathbb R^{1+n}_{+})}\right\} \\ &= \min_{\mu\in\mathcal{X}}\sup_{g\in\mathcal{Y}}\Bigg\{\iint_{\mathbb R^{1+n}_{+}}g(t,x) (G^{\ast}_{\alpha,\beta}\mu)(t,x)dxdt\Bigg\}. \end{align*} $$

By [Reference Adams and Hedberg3, Theorem 2.4.1], we can deduce that

$$ \begin{align*} &\min_{\mu\in\mathcal{X}}\sup_{g\in\mathcal{Y}}\Bigg\{\iint_{\mathbb R^{1+n}_{+}}g(t,x)(G^{\ast}_{\alpha,\beta}\mu)(t,x)dxdt\Bigg\}\\ &\quad= \sup_{g\in\mathcal Y}\min_{\mu\in\mathcal{X}}\Bigg\{\iint_{\mathbb R^{1+n}_{+}} G_{\alpha,\beta}(g)(t,x)d\mu(t,x)\Bigg\}\\ &\quad= \sup_{0\leq g\in L^{q}_{t}L^{p}_{x}}\frac{1}{\|g\|_{L^{q}_{t}L^{p}_{x}}}\min_{\mu\in\mathcal{X}}\left\{\iint_{\mathbb R^{1+n}_{+}} G_{\alpha,\beta}(g)(t,x)d\mu(t,x)\right\}, \end{align*} $$

which implies

$$ \begin{align*} \min_{\mu\in\mathcal{M}_{+}(K)}\left\{\frac{1}{\mu(K)}\|G^{\ast}_{\alpha,\beta}\mu\|_{L^{q'}_{t}L^{p'}_{x}(\mathbb R^{1+n}_{+})}\right\} &\leq \sup_{0\leq g\in L^{q}_{t}L^{p}_{x}}\frac{1}{\|g\|_{L^{q}_{t}L^{p}_{x}}}\min_{(t,x)\in K}\left\{G_{\alpha,\beta}(g)(t,x)\right\}\\ &= \sup_{g\in\mathcal Z}\frac{1}{\|g\|_{L^{q}_{t}L^{p}_{x}}}=\left(C^{(\alpha,\beta)}_{p,q}(K)\right)^{-1/(p\wedge q)}. \end{align*} $$

Hence, it follows

$$ \begin{align*} C^{(\alpha,\beta)}_{p,q}(K) &= \left(\frac{1}{\min\limits_{\mu\in\mathcal{M}_{+}(K)}{\|G^{\ast}_{\alpha,\beta} \mu\|_{L^{q}_{t}L^{p}_{x}}}/{\mu(K)}}\right)^{p\wedge q}\\ &=\left(\sup_{\mu\in\mathcal{M}_{+}(K)} \frac{\|\mu\|_{1}}{\|G^{\ast}_{\alpha,\beta}\mu \|_{L^{q}_{t}L^{p}_{x}}}\right)^{p\wedge q}\lesssim\widetilde{C^{(\alpha,\beta)}_{p,q}}(K).\\[-45pt] \end{align*} $$

Proof (i) If $K=\emptyset $ , then for $g(t,x)\equiv 0$ , $1_{K}\equiv 0$ and $G_{\alpha ,\beta }(g)=1_{K}=0$ , which indicates that $C^{(\alpha ,\beta )}_{p,q}(\emptyset )=0$ . Set

$$ \begin{align*}K_{\lambda}=\Big\{(t,x)\in\mathbb R^{1+n}_{+}:\ g\geq 0\ \&\ G_{\alpha,\beta}(g)\geq \lambda\Big\}.\end{align*} $$

We can see that

$$ \begin{align*} C^{(\alpha,\beta)}_{p,q}\kern-2pt\left(\kern-1pt\Big\{(t,x)\in\mathbb R^{1+n}_{+}:\ g\geq 0 \ \& \ G_{\alpha,\beta}(g)\geq \lambda\Big\}\kern-1pt\right) :=\inf\Big\{\|g\|_{L^{q}_{t}L^{p}_{x}}^{p\wedge q}:\ G\geq 0\ \& \ G_{\alpha,\beta}(G)\geq 1_{K_{\lambda}}\Big\}. \end{align*} $$

Take $g\geq 0$ and $G_{\alpha ,\beta }(g)\geq \lambda $ on $K_{\lambda }$ such that the function $G=g/\lambda $ satisfies $G_{\alpha ,\beta }(G)\geq {(G_{\alpha ,\beta }(g))}/{\lambda }\geq 1_{K_{\lambda }},$ which means

$$ \begin{align*}C^{(\alpha,\beta)}_{p,q}(K_{\lambda})\leq \|{g}/{\lambda}\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}}\leq \lambda^{-p\wedge q} \|g\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}}.\end{align*} $$

For any $(t,x)\in \mathcal {B}(G_{\alpha ,\beta }(g), p,q)$ , $G_{\alpha ,\beta }(g)(t,x)=\infty $ . For any j, $G_{\alpha ,\beta }(g)(t,x)>2^{j}$ . Then

$$ \begin{align*}(t,x)\in\Big\{(t,x)\in\mathbb R^{1+n}_{+}:\ g\geq 0\ \&\ G_{\alpha,\beta}(g)(t,x)>2^{j}\Big\},\end{align*} $$

that is,

$$ \begin{align*}\mathcal{B}(G_{\alpha,\beta}(g), p,q)\subset \Big\{(t,x)\in\mathbb R^{1+n}_{+}:\ g\geq 0\ \&\ G_{\alpha,\beta}(g)(t,x)>2^{j}\Big\}.\end{align*} $$

We obtain

$$ \begin{align*} C^{(\alpha,\beta)}_{p,q}\left(\mathcal{B}(G_{\alpha,\beta}(g), p,q)\right)&\leq C^{(\alpha,\beta)}_{p,q}\left(\Big\{(t,x)\in\mathbb R^{1+n}_{+}:\ g\geq 0\ \&\ G_{\alpha,\beta}(g)(t,x)>2^{j}\Big\}\right)\\ &\leq 2^{-j(p\wedge q)}\|g\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}}. \end{align*} $$

Letting $j\rightarrow \infty $ gives $C^{(\alpha ,\beta )}_{p,q}(\mathcal {B}(G_{\alpha ,\beta }(g), p,q))=0.$ This shows that $C^{(\alpha ,\beta )}_{p,q}(K)=0$ if $K\subseteq \mathcal {B}(G_{\alpha ,\beta }(g), p,q)$ .

Conversely, if $C^{(\alpha ,\beta )}_{p,q}(K)=0$ , from the definition of $C^{(\alpha ,\beta )}_{p,q}(\cdot )$ , we can see that

$$ \begin{align*} \inf\left\{\|g\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}}:\ g\geq 0\ \& \ G_{\alpha,\beta}(g)\geq 1_{K}\right\}=0. \end{align*} $$

Then, for any $j\in \mathbb Z$ , there exists $g_{j}$ such that $g_{j}\geq 0$ and $G_{\alpha ,\beta }(g_{j})\geq 1$ on K satisfying $\|g_{j}\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}}<2^{-j}$ . Let $g=\sum ^{\infty }_{j=1}g_{j}$ . It is obvious that $g\geq 0$ and $G_{\alpha ,\beta }(g)=\infty $ on K, which implies that $K\subset \mathcal {B}(G_{\alpha ,\beta }, g, p,q)$ .

(ii) Let $K_{1}\subseteq K_{2}$ . Then $1_{K_{1}}(t,x)\leq 1_{K_{2}}(t,x)$ and

(3.2) $$ \begin{align} \qquad\ \Big\{\|g\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}}:\ g\geq 0\ \& \ G_{\alpha,\beta}(g)\geq 1_{K_{2}}\Big\}\subseteq \Big\{\|g\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}}:\ g\geq 0\ \& \ G_{\alpha,\beta}(g)\geq 1_{K_{1}}\Big\}.\!\!\!\!\!\!\!\!\!\!\!\! \end{align} $$

Taking the infimum on both sides of (3.2) yields $C^{(\alpha ,\beta )}_{p,q}(K_{1})\leq C^{(\alpha ,\beta )}_{p,q}(K_{2}).$

(iii) The proof of this assertion is divided into two cases.

Case 1: $p\ge q$ , i.e., $q=p\wedge q$ . Let $\{g_{j}\}$ be a sequence of functions such that $G_{\alpha ,\beta }(g_{j})\geq 1$ on $E_{j}$ . Define $g:=\sup _{j}g_{j}$ which satisfies $G_{\alpha ,\beta }(g)\geq 1$ on $\bigcup ^{\infty }_{j=1}E_{j}$ . We can get

$$ \begin{align*} \|g\|^{q}_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})} = \Bigg\{\kern-1.5pt\!\int^{\infty}_{0}\Bigg(\kern-1.5pt\!\int_{\mathbb R^{n}}|g(t,x)|^{p}dx\kern-1pt\Bigg)^{q/p}dt\Bigg\} \leq\sum^{\infty}_{j=1}\Bigg\{\!\kern-1.5pt\int^{\infty}_{0}\!\Bigg(\!\int_{\mathbb R^{n}}|g_{j}(t,x)|^{p}dx\kern-1.5pt\Bigg)^{q/p}dt\Bigg\}. \end{align*} $$

This implies that

$$ \begin{align*} C^{(\alpha,\beta)}_{p,q}\left(\bigcup^{\infty}_{j=1}E_{j}\right)\leq \sum^{\infty}_{j=1}C^{(\alpha,\beta)}_{p,q}(E_{j}). \end{align*} $$

Case 2: $p<q$ , i.e., $p\wedge q=p$ . For $\{g_{j}\}$ such that $g_{j}\geq 0$ and $G_{\alpha ,\beta }(g_{j})\geq 1$ on $E_{j}$ , let $g=\sup _{j}g_{j}$ . For $(t,x)\in \bigcup ^{\infty }_{j=1}E_{j}$ , there exists a $j_{0}$ such that $(t,x)\in E_{j_{0}}$ . Then $g(t,x)\geq g_{j_{0}}(t,x)\geq 0$ , i.e., $g\ge 0$ on $\bigcup ^{\infty }_{j=1}E_{j}$ . We get $G_{\alpha ,\beta }(g)(t,x)\geq G_{\alpha ,\beta }(g_{j_{0}})(t,x)\geq 0$ for $ (t,x)\in \bigcup ^{\infty }_{j=1}E_{j}.$ This indicates that

$$ \begin{align*} \|g\|^{p}_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})} &\leq \Bigg\{\int^{\infty}_{0}\Bigg(\sum^{\infty}_{j=1}\int_{\mathbb R^{n}}|g_{j}(t,x)|^{p}dx\Bigg)^{q/p}dt\Bigg\}^{p/q}\\ &\leq \Bigg\|\sum^{\infty}_{j=1}\int_{\mathbb R^{n}}|g_{j}(t,x)|^{p}dx\Bigg\|_{L^{q/p}_{t}}\\ &\leq \sum^{\infty}_{j=1}\Bigg\|\int_{\mathbb R^{n}}|g_{j}(t,x)|^{p}dx\Bigg\|_{L^{q/p}_{t}} =\sum^{\infty}_{j=1}\|g_{j}\|^{p}_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}, \end{align*} $$

which gives

$$ \begin{align*} C^{(\alpha,\beta)}_{p,q}\left(\bigcup^{\infty}_{j=1}E_{j}\right)\leq \sum^{\infty}_{j=1}C^{(\alpha,\beta)}_{p,q}(E_{j}). \end{align*} $$

(iv) Because $g_{x_{0}}(x,t)=g(t, x+x_{0})$ , it is easy to see that $\|g_{x_{0}}\|_{L^{q}_{t}L^{p}_{x}}=\|g\|_{L^{q}_{t}L^{p}_{x}}$ and

$$ \begin{align*}C^{(\alpha,\beta)}_{p,q}(K+(0, x_{0}))=C^{(\alpha,\beta)}_{p,q}(K).\\[-36pt]\end{align*} $$

Proof Without loss of generality, we assume that $C^{(\alpha ,\beta )}_{p,q}(E)<\infty $ . By (ii) of Theorem 3.2,

$$ \begin{align*}C^{(\alpha,\beta)}_{p,q}(E)\leq\inf\Big\{C^{(\alpha,\beta)}_{p,q}(O):\ O\supset E \ \& \ O\text{ open}\Big\}.\end{align*} $$

For $\epsilon \in (0,1)$ , there exists a measurable, nonnegative function f such that $G_{\alpha ,\beta }f\geq 1$ on E and $\|f\|_{L^{q}_{t}L^{p}_{x}}^{p\wedge q}\leq C^{(\alpha ,\beta )}_{p,q}(E)+\epsilon .$ Since $G_{\alpha ,\beta }f$ is lower semicontinuous, then the set $O_{\epsilon }:=\{(x,t)\in \mathbb R^{n+1}_{+}:\ G_{\alpha ,\beta }f(x,t)>1-\epsilon \}$ is an open set. On the other hand, $E\subset O_{\epsilon }$ . This implies that

$$ \begin{align*}C^{(\alpha,\beta)}_{p,q}(O_{\epsilon})&\leq \frac{1}{(1-\epsilon)^{p\wedge q}}\left(\int^{\infty}_{0}\left(\int_{\mathbb R^{n}}|f(x)|^{p}dx\right)^{q/p}dt\right)^{(p\wedge q)/q}\\&<\frac{1}{(1-\epsilon)^{p\wedge q}}(C^{(\alpha,\beta)}_{p,q}(E)+\epsilon).\end{align*} $$

The arbitrariness of $\epsilon $ indicates that

$$ \begin{align*}C^{(\alpha,\beta)}_{p,q}(E)\geq\inf\Big\{C^{(\alpha,\beta)}_{p,q}(O):\ O\supset E \ \& \ O\text{ open}\Big\}.\\[-37pt]\end{align*} $$

Proof Since $\{E_{j}\}^{\infty }_{j=1}$ is increasing, then

$$ \begin{align*}C^{(\alpha,\beta)}_{p,q}(\cup^{\infty}_{j=1}E_{j})\geq\lim_{j\rightarrow\infty}C^{(\alpha,\beta)}_{p,q}(E_{j}).\end{align*} $$

Conversely, without loss of generality, we assume that $\lim \limits _{j\rightarrow \infty }C^{(\alpha ,\beta )}_{p,q}(E_{j})$ is finite. For each j, let $f_{E_{j}}$ be the unique function such that $f_{E_{j}}\geq 1$ on $E_{j}$ and $\|f_{E_{j}}\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}}=C^{(\alpha ,\beta )}_{p,q}(E_{j})$ . Then, for $i<j$ , it holds that $G_{\alpha ,\beta }f_{E_{j}}\geq 1$ on $E_{i}$ and further, $G_{\alpha ,\beta }((f_{E_{i}}+f_{E_{j}})/2)\geq 1$ on $E_{i}$ , which means that

$$ \begin{align*} \left(\int^{\infty}_{0}\left(\int_{\mathbb R^{n}}((f_{E_{i}}+f_{E_{j}})(x,t)/2)^{p}dx\right)^{q/p}dt\right)^{(p \wedge q)/q}\geq C^{(\alpha,\beta)}_{p,q}(E_{i}). \end{align*} $$

Since the space $L^{q}_{t}L^{p}_{x}(\mathbb R^{n+1}_{+})$ is a uniformly convex Banach space for $1<p<\infty $ and $1<q<\infty $ (cf. [Reference Benedek and Panzone7, Theorem 1, p. 317]), by [Reference Adams and Hedberg3, Corollary 1.3.3], the sequence $\{f_{E_{j}}\}^{\infty }_{j=1}$ converges strongly to a function f satisfying

$$ \begin{align*} \|f\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}}=\lim_{j\rightarrow\infty}C^{(\alpha,\beta)}_{p,q}(E_{j}). \end{align*} $$

Similar to [Reference Adams and Hedberg3, Proposition 2.3.12], we can prove that $G_{\alpha ,\beta }f\geq 1$ on $\cup ^{\infty }_{j=1}E_{j}$ , except possibly on a countable union of sets with $C^{(\alpha ,\beta )}_{p,q}(\cdot )$ zero. Hence,

$$ \begin{align*} \lim_{j\rightarrow\infty}C^{(\alpha,\beta)}_{p,q}(E_{j})\geq\|f\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}}\geq C^{(\alpha,\beta)}_{p,q}(\cup^{\infty}_{j=1}E_{j}).\\[-40pt] \end{align*} $$

Proof We first assume that $(t_{0},x_{0})=(0,0)$ . Let $g\geq 0$ satisfying $G_{\alpha ,\beta }(g)(t,x)\geq 1_{B_{r_{0}}^{(\alpha ,\beta )}(0,0)}(t,x)$ . If $(t,x)\in B_{r_{0}}^{(\alpha ,\beta )}(0,0)$ , then $|t|<2r_{0}^{\alpha }$ and $|x|<r^{\beta }_{0}/2$ . We can see that $(t,x)\in B_{r_{0}}^{(\alpha ,\beta )}(0,0)$ if and only if $(s,y)\in B_{1}^{(\alpha ,\beta )}(0,0)$ , where $s=t/r^{\alpha }_{0}$ and $y=x/r_{0}^{\beta }$ . Then, by (2.5), we utilize the change of variables $r^{-\alpha }_{0}u=v\ \&\ r^{-\beta }_{0}z=w$ to deduce that

$$ \begin{align*} G_{\alpha,\beta}(g)(t,x) &= \int^{r^{\alpha}_{0}s}_{0}\int_{\mathbb R^{n}}G_{r^{\alpha}_{0}s-u}(r^{\beta}_{0}y-z)F(u,z)dudz\\ &= \int^{r^{\alpha}_{0}s}_{0}\int_{\mathbb R^{n}}\frac{|r^{\alpha}_{0}s-u|^{\beta}}{(|r^{\alpha}_{0}s-u|^{\beta/\alpha}+|r^{\beta}_{0}y-z|)^{n+\alpha}}g(u,z)dudz\\ &= \int^{s}_{0}\int_{\mathbb R^{n}}\frac{r^{\alpha}_{0}|s-v|^{\beta}}{(|s-v|^{\beta/\alpha}+|y-w|)^{n+\alpha}} g(r^{\alpha}_{0}v, r^{\beta}_{0}w)dvdw. \end{align*} $$

Denote $r^{\alpha }_{0}g(r^{\alpha }_{0}t, r^{\beta }_{0}x)$ by $g^{\ast }(t,x)$ . The above computation gives

$$ \begin{align*}G_{\alpha,\beta}(g)(t,x)=G_{\alpha,\beta}(g^{\ast})(s,y).\end{align*} $$

This means that $G_{\alpha ,\beta }(g)(t,x)\geq 1$ for $(t,x)\in B_{r_{0}}^{(\alpha ,\beta )}(0,0)$ if and only if $G_{\alpha ,\beta }(g^{\ast })(s,y)\geq 1$ for $(s,y)\in B_{1}^{(\alpha ,\beta )}(0,0)$ . It is easy to get

$$ \begin{align*}C^{(\alpha,\beta)}_{p,q}(B_{1}^{(\alpha,\beta)}(0,0))\leq \|g^{\ast}\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}^{p\wedge q} =\|r^{\alpha}_{0}g(r^{\alpha}_{0}\cdot, r^{\beta}_{0}\cdot)\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}^{p\wedge q}.\end{align*} $$

Below a change of variables gives

$$ \begin{align*} \|r^{\alpha}_{0}g(r^{\alpha}_{0}\cdot, r^{\beta}_{0}\cdot)\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}^{p\wedge q} &=\Bigg\{\int^{\infty}_{0}\Bigg(\int_{\mathbb R^{n}}|r^{\alpha}_{0}g(r^{\alpha}_{0}t, r^{\beta}_{0}x)|^{p}dx\Bigg)^{q/p}dt\Bigg\}^{(p\wedge q)/q} \\&= r^{(\alpha-\beta n/p-\alpha/q)(p\wedge q)}\|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}^{p\wedge q}, \end{align*} $$

which indicates that

$$ \begin{align*}C^{(\alpha,\beta)}_{p,q}(B_{1}^{(\alpha,\beta)}(0,0))\leq \|g^{\ast}\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}^{p\wedge q}\leq r^{(\alpha-\beta n/p-\alpha/q)(p\wedge q)}\|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}^{p\wedge q}.\end{align*} $$

By the arbitrariness of g, we know

$$ \begin{align*} C^{(\alpha,\beta)}_{p,q}(B_{1}^{(\alpha,\beta)}(0,0))\leq r^{(\alpha-\beta n/p-\alpha/q)(p\wedge q)}C^{(\alpha,\beta)}_{p,q}(B_{r_{0}}^{(\alpha,\beta)}(0,0)). \end{align*} $$

Changing the order of $B_{r_{0}}^{(\alpha ,\beta )}(0,0)$ and $B_{1}^{(\alpha ,\beta )}(0,0)$ , similarly, it holds

$$ \begin{align*} C^{(\alpha,\beta)}_{p,q}(B_{r_{0}}^{(\alpha,\beta)}(0,0))\leq r^{-(\alpha-\beta n/p-\alpha/q)(p\wedge q)}C^{(\alpha,\beta)}_{p,q}(B_{1}^{(\alpha,\beta)}(0,0)). \end{align*} $$

This shows that

$$ \begin{align*} C^{(\alpha,\beta)}_{p,q}(B_{r_{0}}^{(\alpha,\beta)}(0,0))\simeq r_{0}^{(p\wedge q)(n\beta/p+\alpha/q-\alpha)}. \end{align*} $$

Now, for arbitrary $(t_{0},x_{0})\in \mathbb R^{1+n}_{+}$ , let $g\geq 0$ and $G_{\alpha ,\beta }(g)\geq 1_{B^{(\alpha ,\beta )}_{r_{0}}(t_{0},x_{0})}$ . Then, for $1\leq p<\infty $ and $1<q<\infty $ , there exist $\widetilde {p}$ and $\widetilde {q}$ such that

$$ \begin{align*}\begin{cases} 1\leq p<\widetilde{p}<\infty,\\ 1<q<\widetilde{q}<\infty,\\ (1/q-1/\widetilde{q})+({\beta n}/{\alpha})(1/p-1/\widetilde{p})=1. \end{cases}\end{align*} $$

By Theorem 2.6, $\|G_{\alpha ,\beta }(g)\|_{L^{\widetilde {q}}_{t}L^{\widetilde {p}}_{x}(\mathbb R^{1+n}_{+})}\lesssim \|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}.$ It is obvious that

$$ \begin{align*} \|G_{\alpha,\beta}(g)\|_{L^{\widetilde{q}}_{t}L^{\widetilde{p}}_{x}(\mathbb R^{1+n}_{+})} &\geq \Bigg\{\int^{2r^{\alpha}_{0}+t_{0}}_{r_{0}^{\alpha}+t_{0}}\Bigg(\int_{B(x_{0}, r^{\beta}_{0}/2)}|G_{\alpha,\beta}(g)(t,x)|^{\widetilde{p}}dx\Bigg)^{\widetilde{q}/\widetilde{p}}dt\Bigg\}^{1/\widetilde{q}}\\ &\geq \Bigg\{\int^{2r^{\alpha}_{0}+t_{0}}_{r_{0}^{\alpha}+t_{0}}\Bigg(\int_{B(x_{0}, r^{\beta}_{0}/2)}1dx\Bigg)^{\widetilde{q}/\widetilde{p}}dt\Bigg\}^{1/\widetilde{q}} \gtrsim r_{0}^{n\beta/\widetilde{p}+\alpha/\widetilde{q}}, \end{align*} $$

which implies $r_{0}^{(n\beta /\widetilde {p}+\alpha /\widetilde {q})(p\wedge q)}\lesssim \|g\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}.$ Since $(1/q-1/\widetilde {q})+(\beta n/\alpha )(1/p-1/\widetilde {p})=1$ , the arbitrariness of g shows that

$$ \begin{align*} r_{0}^{(\alpha/q+n\beta/p-\alpha)(p\wedge q)}\lesssim C^{(\alpha,\beta)}_{p,q}(B^{(\alpha,\beta)}_{r_{0}}(t_{0},x_{0})). \end{align*} $$

Below, we investigate the upper bound of $C^{(\alpha ,\beta )}_{p,q}(B^{(\alpha ,\beta )}_{r_{0}}(t_{0},x_{0}))$ . Consider the ball

$$ \begin{align*} B^{(\alpha,\beta)}_{r_{0},\eta}(t_{0},x_{0})=\left\{(t,x)\in\mathbb R^{1+n}_{+}:\ |t-t_{0}|<(\eta r_{0})^{\alpha}\ \& \ |x-x_{0}|<r^{\beta}_{0}\right\}, \end{align*} $$

where $\eta $ is a sufficiently large constant. Since $(t,x)\in B^{(\alpha ,\beta )}_{r_{0},\eta }(t_{0},x_{0})$ ensures that we have, for $r_{0}> 0$ small enough,

$$ \begin{align*} G_{\alpha,\beta}1_{B^{(\alpha,\beta)}_{r_{0},\eta}(t_{0},x_{0})}(t,x)&= \int^{t}_{0}\int_{\mathbb R^{n}}G_{t-s}(x-y)1_{B^{(\alpha,\beta)}_{r_{0},\eta}(t_{0},x_{0})}(s,y)dyds\\ &= \int_{(0,t)\cap\{s: |s-t_{0}|<(\eta r_{0}^{\alpha})\}}\int_{|y-x_{0}|<r^{\beta}_{0}}G_{t-s}(x-y)dyds\\ &\geq \int_{(0,t)\cap\{s: |s-t_{0}|<(\eta r_{0}^{\alpha})\}\cap \{s: t-s>(\eta^{\alpha}-1)r_{0}^{\alpha}/2\}}\!\int_{|y-x_{0}|<r^{\beta}_{0}}G_{t-s}(x\kern1.3pt{-}\kern1.3pt y)dyds. \end{align*} $$

Because $t-s>(\frac {\eta ^{\alpha }-1}{2})r_{0}^{\alpha }$ and $|y-x_{0}|<r^{\beta }_{0}$ , then $|y-x_{0}|< (\frac {\eta ^{\alpha }-1}{2})^{-\beta /\alpha }(t-s)^{\beta /\alpha }$ and

$$ \begin{align*} G_{t-s}(x-y)\gtrsim\frac{(t-s)^{\beta}}{(|x-y|+|t-s|^{\beta/\alpha})^{n+\alpha}} \gtrsim(\frac{\eta^{\alpha}-1}{2})^{\beta(n/\alpha+1)} (t-s)^{-n\beta/\alpha}. \end{align*} $$

If $|t-t_{0}|<r^{\alpha }_{0}$ ,

$$ \begin{align*} G_{\alpha,\beta}1_{B^{(\alpha,\beta)}_{r_{0},\eta}(t_{0},x_{0})}(t,x)&\geq \int_{(0,t)\cap\{s: |s-t_{0}|<(\eta r_{0}^{\alpha})\}\cap \{s: t-s>(\eta^{\alpha}-1)r_{0}^{\alpha}/2\}}\int_{|y-x_{0}|<r^{\beta}_{0}}|t-s|^{-n\beta/\alpha}dyds\\ &= \int_{(0,t)\cap\{s: |s-t_{0}|<(\eta r_{0}^{\alpha})\}\cap \{s: t-s>(\eta^{\alpha}-1)r_{0}^{\alpha}/2\}}r_{0}^{\beta n}|t-s|^{-n\beta/\alpha}ds. \end{align*} $$

It is easy to see that if $\eta>> 1$ and

$$ \begin{align*}\begin{cases} 0<s<t,\\ t_{0}-(\eta r_{0})^{\alpha}<s<t_{0}+(\eta r_{0})^{\alpha},\\ s>t+\frac{\eta^{\alpha}-1}{2}r^{\alpha}_{0}\text{ or }s<\frac{\eta^{\alpha}-1}{2}r^{\alpha}_{0}-t, \end{cases}\end{align*} $$

then

$$ \begin{align*}\begin{cases} 0<s<t,\\ t_{0}-(\eta r_{0})^{\alpha}<s<t_{0}+(\eta r_{0})^{\alpha},\\ s<\frac{\eta^{\alpha}-1}{2}r^{\alpha}_{0}-t. \end{cases}\end{align*} $$

On the other hand, it can be deduced from $r^{\alpha }_{0}<t-t_{0}<2r^{\alpha }_{0}$ and $t_{0}\lesssim r^{\alpha }_{0}$ that $t\simeq t_{0}+r_{0}^{\alpha }\leq 2r^{\alpha }_{0}$ . If $t_{0}\lesssim r^{\alpha }_{0}$ , then $t_{0}-(\eta r_{0})^{\alpha }<0$ , which implies that $0<s<t$ and $s<\frac {\eta ^{\alpha }-1}{2}r^{\alpha }_{0}-t$ . For $\eta $ large enough, we can see that $\frac {\eta ^{\alpha }-1}{2}r^{\alpha }_{0}-t\geq \frac {\eta ^{\alpha }-1}{2}r^{\alpha }_{0}-2r_{0}^{\alpha }\geq r^{\alpha }_{0}\geq t.$ Hence, we obtain

$$ \begin{align*} G_{\alpha,\beta}\left(1_{B^{(\alpha,\beta)}_{r_{0},\eta}(t_{0},x_{0})}\right)(t,x)\gtrsim\int^{t}_{0}r_{0}^{\beta n}|t-s|^{-n\beta/\alpha}ds \gtrsim r_{0}^{\beta n}(r^{\alpha}_{0}+t_{0})^{(1-\beta n/\alpha)} \gtrsim r_{0}^{\alpha}. \end{align*} $$

Equivalently, there exists a constant c independent of $r_{0}$ such that $G_{\alpha ,\beta }\left(cr^{-\alpha }_{0}1_{B^{(\alpha ,\beta )}_{r_{0},\eta }(t_{0},x_{0})}\right)(t,x)\geq 1$ for $(t,x)\in B^{(\alpha ,\beta )}_{r_{0},\eta }(t_{0},x_{0})$ . Therefore,

$$ \begin{align*} C^{(\alpha,\beta)}_{p,q}(B^{(\alpha,\beta)}_{r_{0}}(t_{0},x_{0}))&\leq \Big\|cr^{-\alpha}_{0}1_{B^{(\alpha,\beta)}_{r_{0},\eta}(t_{0},x_{0})}\Big\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}\\ &\lesssim r_{0}^{-\alpha(p\wedge q)}\left\{\int_{|t-t_{0}|<(\eta r_{0})^{\alpha}}\left(\int_{|x-x_{0}|<r^{\beta}_{0}}1dx\right)^{q/p}dt\right\}^{(p\wedge q)/q}\\ &\lesssim r_{0}^{-\alpha(p\wedge q)}r_{0}^{\beta n(p\wedge q)/p}(\eta r_{0})^{\alpha(p\wedge q)/q}\\ &\lesssim r_{0}^{(p\wedge q)(\beta n/p+\alpha/q-\alpha)}.\\[-38pt] \end{align*} $$

Proof Without loss of generality, we assume that $\|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}<\infty $ . For fixed $g\in L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})$ and any $K=K_{t}\times K_{x}\subset \mathbb R^{1+n}_{+}$ , define

$$ \begin{align*}\Phi_{g}(K):=\frac{\|g\mid_{K}\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}}{\|g\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}},\end{align*} $$

where

$$ \begin{align*} \|g\mid_{K}\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}&:= \Bigg\{\int_{K_{t}}\Bigg(\int_{K_{x}}|g(t,x)|^{p}dx\Bigg)^{q/p}dt\Bigg\}^{1/q}\\ &= \Bigg\{\int_{0}^{\infty}\Bigg(\int_{\mathbb R^{n}}|g(t,x)|^{p}\chi_{K_{x}}(x)dx\Bigg)^{q/p}\chi_{K_{t}}(t)dt\Bigg\}^{1/q}. \end{align*} $$

For any disjoint sets A and B, it follows from the identity $\chi _{A\cup B}(t,x)=\chi _{A_{t}}(t)\chi _{A_{x}}(x)+\chi _{B_{t}}(t)\chi _{B_{x}}(x)$ that

$$ \begin{align*} &\|g\mid_{A\cup B}\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})} \\&\quad\leq \Bigg\{\int_{A_{t}}\Bigg(\int_{A_{x}}|g(t,x)|^{p}dx\Bigg)^{q/p}dt\Bigg\}^{1/q}+ \Bigg\{\int_{B_{t}}\Bigg(\int_{B_{x}}|g(t,x)|^{p}dx\Bigg)^{q/p}dt\Bigg\}^{1/q}\\ &\quad\leq \|g\mid_{A}\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}+ \|g\mid_{B}\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}, \end{align*} $$

which means $\Phi _{F}(A\cup B)\leq \Phi _{F}(A)+\Phi _{F}(B)$ . Below, we prove the reverse inequality in two cases.

Case 1: $p<q$ . For this case, $q/p>1$ . Then we obtain

$$ \begin{align*} &\|g\mid_{A\cup B}\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})} \\&\quad= \Bigg\{\int_{(A\cup B)_{t}}\Bigg(\int_{(A\cup B)_{x}}|g(t,x)|^{p}dx\Bigg)^{q/p}dt\Bigg\}^{p/q}\\ &\quad= \Bigg\{\int_{0}^{\infty}\Bigg(\int_{\mathbb R^{n}}|g(t,x)|^{p}\chi_{A\cup B}(t,x)dx\Bigg)^{q/p}dt\Bigg\}^{p/q}\\ &\quad= \Bigg\{\int_{0}^{\infty}\Bigg(\int_{\mathbb R^{n}}|g(t,x)|^{p}\Bigg(\chi_{A}(t,x)+\chi_{B}(t,x)\Bigg)dx\Bigg)^{q/p}dt\Bigg\}^{p/q}\\ &\quad= \Bigg\{\int_{0}^{\infty}\Bigg(\int_{\mathbb R^{n}}|g(t,x)|^{p}\chi_{A}(x,t)dx+\int_{\mathbb R^{n}}|g(t,x)|^{p}\chi_{B}(t,x)dx\Bigg)^{q/p}dt\Bigg\}^{p/q}, \end{align*} $$

which, together with the inequality: $(a+b)^{q/p}\geq a^{q/p}+b^{q/p}, a, b>0$ , implies that

$$ \begin{align*} &\|g\mid_{A\cup B}\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}\\ &\gtrsim \Bigg\{\!\int_{0}^{\infty}\Bigg(\!\int_{\mathbb R^{n}}|g(t,x)|^{p}\chi_{A}(t,x)dx\Bigg)^{q/p}dt +\int^{\infty}_{0}\Bigg(\!\int_{\mathbb R^{n}}|g(t,x)|^{p}\chi_{B}(t,x)dx\Bigg)^{q/p}dt\Bigg\}^{p/q}\\ &\gtrsim \Bigg\{\!\kern-1pt\int_{0}^{\infty}\Bigg(\!\kern-1pt\int_{\mathbb R^{n}}|g(t,x)|^{p}\chi_{A}(t,x)dx\Bigg)^{q/p}dt\Bigg\}^{p/q} +\Bigg\{\!\kern-1pt\int^{\infty}_{0}\!\Bigg(\!\int_{\mathbb R^{n}}|g(t,x)|^{p}\chi_{B}(t,x)dx\Bigg)^{q/p}dt\Bigg\}^{p/q}\\ &=\|g\mid_{A}\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}+\|g\mid_{B}\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}. \end{align*} $$

Case 2: $p\geq q$ . For this case, $p\wedge q=q$ . Then

$$ \begin{align*} \|g\mid_{A\cup B}\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})} &= \int_{(A\cup B)_{t}}\Bigg(\int_{(A\cup B)_{x}}|g(t,x)|^{p}dx\Bigg)^{q/p}dt\\ &= \int_{0}^{\infty}\Bigg(\int_{\mathbb R^{n}}|g(t,x)|^{p}\Bigg(\chi_{A}(t,x)+\chi_{B}(t,x)\Bigg)dx\Bigg)^{q/p}dt\\ &\gtrsim \int_{0}^{\infty}\Bigg(\int_{\mathbb R^{n}}|g(t,x)|^{p}\chi_{A}(t,x)dx\Bigg)^{q/p}dt\\ &\quad +\int_{0}^{\infty}\Bigg(\int_{\mathbb R^{n}}|g(t,x)|^{p}\chi_{B}(t,x)dx\Bigg)^{q/p}dt\\ &= \|g\mid_{A}\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}+\|F\mid_{A}\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}. \end{align*} $$

Finally, we prove that

$$ \begin{align*} \Phi_{g}(A\cup B)\simeq \Phi_{g}(A)+\Phi_{g}(B). \end{align*} $$

By [Reference Cerdá, Martín and Silvestre14, Corollary 2.3, p. 187], there exists a measure $\psi $ on $\mathbb R^{1+n}_{+}$ such that $\Phi \leq \psi $ and $\psi (\mathbb R^{1+n}_{+})\leq c(n)$ , where $c(n)$ is a constant only depending on n. For $E_{\lambda }\setminus E_{a\lambda }$ with $a\in (1,2)$ , noting that $E_{\infty }=\emptyset $ , we have

$$ \begin{align*} \int^{\infty}_{0}\Phi_{F}(E_{\lambda}\setminus E_{a\lambda})\frac{d\lambda}{\lambda}&\leq \int^{\infty}_{0}\psi(E_{\lambda}\setminus E_{a\lambda})\frac{d\lambda}{\lambda}\\ &= \int^{\infty}_{0}\Bigg(\int^{a\lambda}_{\lambda}d\psi(E_{s})\Bigg)\frac{d\lambda}{\lambda}\\ &= \int^{\infty}_{0}\Bigg(\int^{s}_{s/a}\frac{d\lambda}{\lambda}\Bigg)d\psi(E_{s})\\ &\leq \psi(\mathbb R^{1+n}_{+})\log a\\ &\leq c(n)\log a. \end{align*} $$

Thus,

$$ \begin{align*} \int^{\infty}_{0}\|g\mid_{E_{\lambda}\setminus E_{a\lambda}}\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}\frac{d\lambda}{\lambda} &= \int^{\infty}_{0}\|g\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}\Phi_{F}(E_{\lambda}\setminus E_{a\lambda})\frac{d\lambda}{\lambda}\\&\leq c(n)\log a \|g\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}. \end{align*} $$

By (1.3), $u(t,x):=G_{\alpha ,\beta }(g)(t,x)$ is a solution to the space–time fractional dissipative equation:

$$ \begin{align*}\left\{\begin{aligned} \partial^{\beta}_{t}u(t,x)+(-\Delta)^{\alpha/2}u(t,x)=f(t,x),&\quad \forall\ (t,x)\in\mathbb R^{1+n}_{+},\\ u(0,x)=0,&\quad \forall\ x\in\mathbb R^{n}. \end{aligned}\right.\end{align*} $$

Set a function $\widetilde {g}$ as

$$ \begin{align*}\widetilde{g}(t,x):=\left\{\begin{aligned} &0,\ &(t,x)\in E_{a\lambda},\\ &\frac{g(t,x)}{(a-1)\lambda},\ &(t,x)\in E_{\lambda}\setminus E_{a\lambda},\\ &0,\ &(t,x)\in \mathbb R^{1+n}_{+}\setminus E_{\lambda}. \end{aligned}\right.\end{align*} $$

Define $\widetilde {f}(t,x)=I^{1-\beta }_{t}\widetilde {g}(t,x).$ Then a direct computation indicates that the following function

$$ \begin{align*}\widetilde{u}(t,x):=\left\{\begin{aligned} &1,\ &(t,x)\in E_{a\lambda},\\ &\frac{u(t,x)-\lambda}{(a-1)\lambda},\ &(t,x)\in E_{\lambda}\setminus E_{a\lambda},\\ &0,\ &(t,x)\in \mathbb R^{1+n}_{+}\setminus E_{\lambda} \end{aligned}\right.\end{align*} $$

is a solution to the equation

$$ \begin{align*}\left\{\begin{aligned} \partial^{\beta}_{t}\widetilde{u}(t,x)+(-\Delta)^{\alpha/2}\widetilde{u}(t,x)=\widetilde{f}(t,x),&\quad \forall\ (t,x)\in\mathbb R^{1+n}_{+},\\ \widetilde{u}(0,x)=0,&\quad \forall\ x\in\mathbb R^{n}. \end{aligned}\right.\end{align*} $$

Based on the definition of $C^{(\alpha ,\beta )}_{p,q}(\cdot )$ , we obtain

$$ \begin{align*} \int^{\infty}_{0}\lambda^{p\wedge q}C^{(\alpha,\beta)}_{p,q}(E_{a\lambda})\frac{d\lambda}{\lambda}&\leq \int^{\infty}_{0}\lambda^{p\wedge q}\|\widetilde{g}\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}(E_{\lambda})}\frac{d\lambda}{\lambda}\\ &\leq \int^{\infty}_{0}\frac{\|g\mid_{E_{\lambda}\setminus E_{a\lambda}}\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}(E_{\lambda})}}{(a-1)^{p\wedge q}}\frac{d\lambda}{\lambda}\\ &\leq \frac{c(n)\log a}{(a-1)^{p\wedge q}} \|g\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}, \end{align*} $$

which, via a change of variables, implies that

$$ \begin{align*} \int^{\infty}_{0}\lambda^{p\wedge q}C^{(\alpha,\beta)}_{p,q}(E_{\lambda})\frac{d\lambda}{\lambda}\lesssim \|g\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}.\\[-42pt] \end{align*} $$

Proof Assume that (3.4) holds. Then it follows from Theorem 3.8 that

$$ \begin{align*} &\|G_{\alpha,\beta}(g)\|_{L^{(p\vee q, p\wedge q)}(\mathbb R^{1+n}_{+},\mu)} \\&\quad\lesssim \Bigg(\int^{\infty}_{0}\lambda^{p\wedge q-1}C^{(\alpha,\beta)}_{p,q}\left(\Big\{(t,x)\in\mathbb R^{1+n}_{+}:\ |G_{\alpha,\beta}(g)(t,x)|>\lambda\Big\}\right)d\lambda\Bigg)^{1/(p\wedge q)}\\ &\quad\lesssim \|g\|_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}. \end{align*} $$

Conversely, suppose that (3.3) holds for all $g\in L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})$ . Fix a compact set $K\subset \mathbb R^{1+n}_{+}$ . By the definition of $C^{(\alpha ,\beta )}_{p,q}(\cdot )$ , for any $\epsilon>0$ , there exists a function $g\in L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})$ such that

$$ \begin{align*}\left\{\begin{aligned} &G_{\alpha,\beta}(g)\geq 1_{K},\\ &\|g\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})}+\epsilon<C^{(\alpha,\beta)}_{p,q}(K). \end{aligned}\right.\end{align*} $$

Hence, $K\subset \Big \{(t,x)\in \mathbb R^{1+n}_{+}:\ |G_{\alpha ,\beta }(g)(t,x)|>1\Big \}$ . We get

$$ \begin{align*} &\Bigg(\int^{\infty}_{0}\mu\left(\Big\{(t,x)\in\mathbb R^{1+n}_{+}:\ |G_{\alpha,\beta}(g)(t,x)|>\lambda\Big\}\right)^{(p\wedge q)/(p\vee q)}d\lambda^{p\wedge q}\Bigg)^{1/(p\wedge q)}\\ &\quad\geq \Bigg(\int^{1}_{0}\mu\left(\Big\{(t,x)\in\mathbb R^{1+n}_{+}:\ |G_{\alpha,\beta}(g)(t,x)|>\lambda\Big\}\right)^{(p\wedge q)/(p\vee q)}d\lambda^{p\wedge q}\Bigg)^{1/(p\wedge q)}\\ &\quad\geq \Bigg(\int^{1}_{0}\mu\left(\Big\{(t,x)\in\mathbb R^{1+n}_{+}:\ |G_{\alpha,\beta}(g)(t,x)|>1\Big\}\right)^{(p\wedge q)/(p\vee q)}d\lambda^{p\wedge q}\Bigg)^{1/(p\wedge q)}\\ &\quad\geq \left(\mu(K)\right)^{1/(p\vee q)}, \end{align*} $$

which implies that

$$ \begin{align*} \left(\mu(K)\right)^{(p\wedge q)/(p\vee q)} \leq \|G_{\alpha,\beta}(g)\|^{p\wedge q}_{L^{(p\vee q, p\wedge q)}(\mathbb R^{1+n}_{+},\mu)} \lesssim \|g\|^{p\wedge q}_{L^{q}_{t}L^{p}_{x}(\mathbb R^{1+n}_{+})} \lesssim C^{(\alpha,\beta)}_{p,q}(K)-\epsilon. \end{align*} $$

Then (3.4) follows from the arbitrariness of $\epsilon $ . This completes the proof of Theorem 3.10.