I am studying properties of the kernels of different operators and the following question arose.
Given the operator $-(-\Delta)^{\alpha}u=\mathcal{F}^{-1}(|\xi|^{2\alpha}\widehat{u}(\xi))$, this operator has the kernel $K_t$ defined by $$K_t(x)=(2\pi)^{-n/2}\int_{\mathbb{R}^n}\mathrm{e}^{ix\cdot \xi}\mathrm{e}^{-t|\xi|^{2\alpha}}\,d\xi$$
This kernel satisfies that $$K_t\in L^1(\mathbb{R}^n) (\text{ for all }t>0)$$ (In fact, it is in any $L^p(\mathbb{R}^n)$ (https://arxiv.org/abs/math/0607456))
Question. If I have the following kernel $$G_t(x)=(2\pi)^{-n/2}\int_{\mathbb{R}^n}\mathbb{e}^{ix\cdot\xi}\mathrm{e}^{-t(1+|\xi|^2)^{\alpha}}\,d\xi$$
Is the same still true? I tried to test it but I can't do the scaling trick to "kill" the time variable $t$ (since the "1+" ruins everything for me.)as done in the reference. Is there any extra trick? thank you.
Below is the full answer. I know it looks rather complicated. But the main idea is just to simply follow Miao's paper, with minor adaptaions. It is normal to make mistakes in this kind of computations. So, if there is any mistake, don't hesitate to inform me, please.
Let's change the variable $\eta=t^{\frac1{2\alpha}}\xi$, then \begin{align*} G_t(x)&=(2\pi)^{-n/2}\int_{\mathbb{R}^n}e^{ix\cdot\xi}e^{-t(1+|\xi|^2)^{\alpha}}\,d\xi\\ &=(2\pi)^{-n/2}t^{-\frac n{2\alpha}}\int_{\mathbb{R}^n}e^{it^{-\frac1{2\alpha}}x\cdot\eta}e^{-\left(t^{\frac1\alpha}+|\eta|^2\right)^{\alpha}}\,d\eta\\ &=(2\pi)^{-n/2}t^{-\frac n{2\alpha}}P_{t^{\frac1\alpha}}\left(t^{-\frac1{2\alpha}}x\right), \end{align*} where $$P_A(x)=\int_{\mathbb{R}^n}e^{ix\cdot\xi}e^{-(|\xi|^2+A)^\alpha}\,d\xi=\int_{\mathbb{R}^n}e^{ix\cdot\xi}p_A(\xi)\,d\xi,\qquad x\in\mathbb R^n$$ for $A>0, \alpha>0$ with $p_A(\xi):=e^{-(|\xi|^2+A)^\alpha}$.
We start with several lemmas which will make the proof of $(1)$ more comprehensible.
Proof of Lemma 1. Let $$\varphi_A(t)=e^{-(t+A)^\alpha},\qquad t\geq0.$$ By induction we can show that for each positive integer $k$ there exists a polynomial $Q_{k-1}$ of degree $k-1$ with coefficients independent of $A$ such that the $k-$th derivative of $\varphi_A$ has the form $$\varphi_A^{(k)}(t)=e^{-(t+A)^\alpha}(t+A)^{\alpha-k}Q_{k-1}\left((t+A)^\alpha\right).$$ Hence $$\left|\varphi_A^{(k)}(t)\right|\lesssim e^{-(t+A)^\alpha}(t+A)^{\alpha-k}\left[(t+A)^{\alpha(k-1)}+1\right].$$ Now we have $p_A(\xi)=\varphi_A(|\xi|^2)$. By induction we have $$\nabla_{\xi_{j_1}\xi_{j_2}\cdots\xi_{j_k}}p_A(\xi)=\sum_{\ell=1}^{k}\varphi_A^{(\ell)}(|\xi|^2)Q_\ell^{(j_1, j_2, \cdots j_k)}(\xi_1, \cdots, \xi_n),\qquad k\in\mathbb N_{\geq 1},$$ where $Q_\ell^{(j_1, j_2, \cdots j_k)}$ are homogeneous polynomials with $\operatorname{deg}Q_\ell^{(j_1, j_2, \cdots j_k)}\leq\ell$ and with coefficients independent of $A$. Therefore, \begin{align*} \left|\nabla_\xi^kp_A(\xi)\right|&\lesssim \sum_{\ell=1}^k\left|\varphi_A^{(\ell)}(|\xi|^2)\right||\xi|^\ell\\ &\lesssim \sum_{\ell=1}^ke^{-(|\xi|^2+A)^\alpha}\left(|\xi|^2+A\right)^{\alpha-\ell}\left[\left(|\xi|^2+A\right)^{\alpha(\ell-1)}+1\right]\left(|\xi|^2+A\right)^{\frac\ell2}\\ &\lesssim e^{-(|\xi|^2+A)^\alpha}\left[\sum_{\ell=1}^k\left(|\xi|^2+A\right)^{\alpha\ell-\frac\ell2}+\sum_{\ell=1}^k\left(|\xi|^2+A\right)^{\alpha-\frac\ell2}\right]\\ &\lesssim e^{-(|\xi|^2+A)^\alpha}\Big[\left(|\xi|^2+A\right)^{\alpha-\frac12}+\left(|\xi|^2+A\right)^{\alpha k-\frac k2}\\ &\qquad\qquad+\left(|\xi|^2+A\right)^{\alpha-\frac 12}+\left(|\xi|^2+A\right)^{\alpha-\frac k2}\Big]\\ &\lesssim e^{-(|\xi|^2+A)^\alpha}\left[\left(|\xi|^2+A\right)^{\alpha k-\frac k2}+\left(|\xi|^2+A\right)^{\alpha-\frac k2}\right], \end{align*} where in the last line we note that $\alpha-\frac k2\leq \alpha-\frac12\leq \alpha k-\frac k2$.
Proof of Lemma 2. Changing the variable $s=Au$, we have $$\int_0^\varepsilon(s+A)^\beta s^\gamma\,ds=A^{\beta+\gamma+1}\int_0^{\frac\varepsilon A}(u+1)^\beta u^\gamma\,du.$$ If $\varepsilon\leq A$, then $$\int_0^{\frac\varepsilon A}(u+1)^\beta u^\gamma\,du\lesssim_\beta \int_0^{\frac\varepsilon A}u^\gamma\,du\lesssim_{\beta,\gamma}\left(\frac\varepsilon A\right)^{\gamma+1}.$$ If $\varepsilon>A$, then $$\int_0^{\frac\varepsilon A}(u+1)^\beta u^\gamma\,du\lesssim 1+\int_1^{\frac\varepsilon A}u^{\beta+\gamma}\,du\lesssim1+\left(\frac\varepsilon A\right)^{\beta+\gamma+1}\lesssim\left(\frac\varepsilon A\right)^{\beta+\gamma+1}.$$ This completes the proof of Lemma 2.
The proof of Lemma 3 is similar to the proof of Lemma 2 and we omit it.
Now we prove $(1)$. Define the invariant derivative operator $L=L(x, D)=\frac{x\cdot\nabla_\xi}{i|x|^2}$, then $L(x, D)e^{ix\cdot\xi}=e^{ix\cdot\xi}$ and the conjugate operator is $L^*=-L$. Let $\rho\in C_c^\infty(\mathbb R^n)$ be a bump function satisfying $\rho(\xi)=1$ for $|\xi|\leq 1$ and $\rho(\xi)=0$ for $|\xi|\geq 2$. Let $\delta>0$ to be determined later. We have \begin{align*} P_A(x)&=\int_{\mathbb{R}^n}e^{ix\cdot\xi}p_A(\xi)\,d\xi=\int_{\mathbb{R}^n}e^{ix\cdot\xi}(L^*p_A)(\xi)\,d\xi\\ &=\int_{\mathbb{R}^n}\rho\left(\frac\xi\delta\right)e^{ix\cdot\xi}(L^*p_A)(\xi)\,d\xi+\int_{\mathbb{R}^n}e^{ix\cdot\xi}\left(1-\rho\left(\frac\xi\delta\right)\right)(L^*p_A)(\xi)\,d\xi\\ &=: I+J. \end{align*}
For $I$, we have \begin{align*} |I|&\lesssim \frac1{|x|}\int_{|\xi|\leq2\delta}\left|\nabla_\xi p_A(\xi)\right|\,d\xi{\overset{\text{Lemma 1}}{\lesssim}}\frac1{|x|}\int_{|\xi|\leq2\delta}e^{-(|\xi|^2+A)^\alpha}\left(|\xi|^2+A\right)^{\alpha-\frac12}\,d\xi\\ &\lesssim \frac{e^{-A^\alpha}}{|x|}\int_0^{2\delta}(r^2+A)^{\alpha-\frac12}r^{n-1}\,dr\lesssim \frac{e^{-A^\alpha}}{|x|}\int_0^{4\delta^2}(s+A)^{\alpha-\frac12}s^{\frac n2-1}\,ds. \end{align*} It follows from Lemma 2 that $$|I|\lesssim \begin{cases} \frac{e^{-A^\alpha}}{|x|} \delta^n A^{\alpha-\frac12} & \text{if } 4\delta^2\leq A,\\ \frac{e^{-A^\alpha}}{|x|} \delta^{2\alpha+n-1} & \text{if } 4\delta^2> A. \end{cases}$$
For $J$, we fix a positive integer $N$ such that $N>2\alpha+n+1$, then \begin{align*} J&=\int_{\mathbb{R}^n}e^{ix\cdot\xi}\left(1-\rho\left(\frac\xi\delta\right)\right)(L^*p_A)(\xi)\,d\xi\\ &=\int_{\mathbb{R}^n}e^{ix\cdot\xi}(L^*)^{N-1}\left(\left(1-\rho\left(\frac\xi\delta\right)\right)L^*p_A\right)(\xi)\,d\xi. \end{align*} By Lemma 1, we have \begin{align*} |J|&\lesssim |x|^{-N}\int_{|\xi|\geq2\delta}\left|\nabla_\xi^Np_A(\xi)\right|\,d\xi+|x|^{-N}\sum_{k=1}^{N-1}\delta^{-k}\int_{\delta\leq|\xi|\leq2\delta}\left|\nabla_\xi^{N-k}p_A(\xi)\right|\,d\xi\\ &\lesssim |x|^{-N}\int_{|\xi|\geq2\delta}e^{-(|\xi|^2+A)^\alpha}\left[\left(|\xi|^2+A\right)^{\alpha N-\frac N2}+\left(|\xi|^2+A\right)^{\alpha-\frac N2}\right]\,d\xi\\ &\qquad+|x|^{-N}\sum_{k=1}^{N-1}\delta^{-k}\int_{\delta\leq|\xi|\leq2\delta}e^{-(|\xi|^2+A)^\alpha}\left[\left(|\xi|^2+A\right)^{\alpha (N-k)-\frac {N-k}2}+\left(|\xi|^2+A\right)^{\alpha-\frac {N-k}2}\right]\,d\xi. \end{align*} Since $$e^{-(|\xi|^2+A)^\alpha}\left(|\xi|^2+A\right)^{\alpha (N-k-1)}\lesssim e^{-\frac12(|\xi|^2+A)^\alpha}\lesssim e^{-\frac12A^\alpha},\qquad 0\leq k\leq N-1,$$ we get \begin{align*} |J|&\lesssim |x|^{-N}e^{-\frac12A^\alpha}\int_{|\xi|\geq2\delta}\left(|\xi|^2+A\right)^{\alpha-\frac N2}\,d\xi\\ &\qquad+|x|^{-N}e^{-\frac12A^\alpha}\sum_{k=1}^{N-1}\delta^{-k}\int_{\delta\leq|\xi|\leq2\delta}\left(|\xi|^2+A\right)^{\alpha-\frac {N-k}2}\,d\xi\\ &\lesssim |x|^{-N}e^{-\frac12A^\alpha}\int_{4\delta^2}^\infty (s+A)^{\alpha-\frac N2}s^{\frac n2-1}\,ds+|x|^{-N}e^{-\frac12A^\alpha}\sum_{k=1}^{N-1}\delta^{-k+n}\left(\delta^2+A\right)^{\alpha-\frac{N-k}2}, \end{align*} where in the last line we used $$\delta^2+A\leq |\xi|^2+A\leq 4\delta^2+A\leq 4(\delta^2+A),\qquad \text{for }\ \ \delta\leq|\xi|\leq2\delta.$$ It follows from Lemma 3 that $$\int_{4\delta^2}^\infty (s+A)^{\alpha-\frac N2}s^{\frac n2-1}\,ds\lesssim\begin{cases} A^{\alpha-\frac N2+\frac n2} & \text{if } 4\delta^2\leq A,\\ \delta^{2\alpha-N+n} & \text{if } 4\delta^2>A. \end{cases}$$
Now, for $|x|\geq \frac14$, we take $\delta=\sqrt{1+A}|x|^{-1}$. If $4\delta^2>A$, then $$|I|\lesssim |x|^{-1}e^{-A^\alpha}\delta^{2\alpha+n-1}\lesssim |x|^{-1}e^{-A^\alpha}(1+A)^{\frac{2\alpha+n-1}2}|x|^{-2\alpha-n+1}\lesssim|x|^{-n-2\alpha},$$ and \begin{align*} |J|&\lesssim |x|^{-N}e^{-\frac12A^\alpha}\delta^{2\alpha-N+n}+|x|^{-N}e^{-\frac12A^\alpha}\sum_{k=1}^{N-1}\delta^{-k+n}\delta^{2\alpha-N+k}\\ &\lesssim |x|^{-N}e^{-\frac12A^\alpha}\delta^{2\alpha-N+n}\lesssim |x|^{-N}e^{-\frac12A^\alpha}(1+A)^{\frac{2\alpha-N+n}2}|x|^{-2\alpha+N-n}\\ &\lesssim|x|^{-n-2\alpha}. \end{align*} Hence \begin{align*} |P_A(x)|\lesssim |x|^{-n-2\alpha},\qquad \text{for }\ \ |x|\geq\frac14, 4\delta^2>A\tag{2}. \end{align*} If $4\delta^2\leq A$, then \begin{align*} |I|&\lesssim |x|^{-1}e^{-A^\alpha}\delta^n A^{\alpha-\frac12}\lesssim \begin{cases} |x|^{-1}e^{-A^\alpha} \delta^n A^{\alpha-\frac12} & \text{if } \alpha\geq\frac12\\ |x|^{-1}e^{-A^\alpha}\delta^{\delta+2\alpha-1} & \text{if }0<\alpha<\frac12 \end{cases}\\ &\lesssim \begin{cases} |x|^{-n-1} & \text{if } \alpha\geq\frac12\\ |x|^{-n-2\alpha} & \text{if }0<\alpha<\frac12 \end{cases}\lesssim |x|^{-n-\min\{2\alpha,1\}}, \end{align*} and \begin{align*} |J|&\lesssim |x|^{-N}e^{-\frac12A^\alpha}A^{\alpha-\frac N2+\frac n2}+|x|^{-N}e^{-\frac12A^\alpha}\sum_{k=1}^{N-1}\delta^{-k+n}A^{\alpha-\frac{N-k}2}\\ &\lesssim |x|^{-N}e^{-\frac12A^\alpha}A^{\alpha-\frac N2+\frac n2}+|x|^{-N}e^{-\frac12A^\alpha}\delta^nA^{\alpha-\frac N2}\sum_{k=1}^{N-1}\left(\frac{\sqrt A}{\delta}\right)^k\\ &\lesssim |x|^{-N}e^{-\frac12A^\alpha}A^{\alpha-\frac N2+\frac n2}+|x|^{-N}e^{-\frac12A^\alpha}\delta^nA^{\alpha-\frac N2}\left(\frac{\sqrt A}{\delta}\right)^{N-1}\\ &\lesssim |x|^{-N}e^{-\frac12A^\alpha}\delta^{2\alpha-N+n}+|x|^{-N}e^{-\frac12A^\alpha}\delta^{n-N+1}A^{\alpha-\frac 12}\\ &\lesssim |x|^{-N}e^{-\frac12A^\alpha}(1+A)^{\alpha-\frac{N-n}2}|x|^{-2\alpha+N-n}\\ &\qquad\qquad+|x|^{-N}e^{-\frac12A^\alpha}(1+A)^{\frac{n-N}2+1}|x|^{-n+N-1}A^{\alpha-\frac 12}\\ &\lesssim |x|^{-n-2\alpha}+|x|^{-n-1}e^{-\frac12A^\alpha}A^{\alpha-\frac12}\lesssim |x|^{-n-\min\{2\alpha,1\}}. \end{align*} Hence \begin{align*} |P_A(x)|\lesssim |x|^{-n-\min\{2\alpha,1\}},\qquad \text{for }\ \ |x|\geq\frac14, 4\delta^2\leq A\tag{3}. \end{align*} Combining $(2)$ and $(3)$ we obtain $$|P_A(x)|\lesssim |x|^{-n-\min\{2\alpha,1\}},\qquad \text{for }\ \ |x|\geq\frac14.$$
On the other hand, $$|P_A(x)|\leq \int_{\mathbb R^n}p_A(\xi)\,d\xi\leq \int_{\mathbb R^n}e^{-|\xi|^{2\alpha}}\,d\xi\lesssim 1,\qquad x\in\mathbb R^n.$$ The proof of $(1)$ is complete now.
Remark. It seems that the bound $(1)$ is not optimal for $\alpha>\frac12$, since in Miao's paper they showed that $|P_0(x)|\lesssim (1+|x|)^{-n-2\alpha}$ for all $\alpha>0$. Indeed, in our proof, we can still get Miao's result for $A=0$, since for $A=0$ we only have the $4\delta^2>A$ case. When $A>0$, it seems that $p_A$ will decay faster than $p_0$ and we may expect that $P_A$ has an estimate not bad than $P_0$. However, I haven't found a way to show a better estimate for $P_A$ when $\alpha>\frac12$. Luckily, the bound $(1)$ is enough to give a uniform bound of $\|G_t\|_{L^1(\mathbb R^n)}$.