# Integrability of fractional heat kernel

In Estimates of fractional heat kernel, it was stated that $$\partial_{x_j} p_t^{(n)}(x) = -\frac{x_j}{2 \pi} \, p_t^{(n+2)}(\tilde x)$$ where $$x = (x_1, \ldots, x_n) \in \mathbb R^n$$, $$\tilde x = (x_1, \ldots, x_n, 0, 0) \in \mathbb R^{n+2}$$, and $$p_t^{(n)}(x)$$ is the heat kernel for $$(-\Delta)^s$$ in $$n$$ dimensions.

Can we use this information to answer the following question?

Question: What are the values of $$p$$ and $$q$$ such that $$\nabla_x p_t^{(n)}(x) \in L^p((0,T); L^q(\mathbb R^n))$$?

And if we actually do, then we find that $$\phi(t, x) = |\nabla_x p_t^{(n)}(x)|$$ satisfies $$\phi(t, x) = t^{-(n+1)/(2s)} \phi(1, t^{-1/(2s)} x)$$ and, with $$\tilde x = (x, 0, 0) \in \mathbb R^{n+2}$$), $$\phi(1,x) = |x| p_1^{(n+2)}(\tilde x) \approx C |x| \min\{1, |x|^{-(n+2+2s)}\} = C \min\{|x|, |x|^{-(n+1+2s)}\}$$ (here we use the well-known fact that $$p_1^{(n)}(x) \approx C \min\{1, |x|^{-n-2s}\}$$). It follows that $$\|\phi(1,\cdot)\|_q < \infty \qquad \text{if and only if } q > \tfrac{n}{n+1+2s}$$ and $$\|\phi(t,\cdot)\|_q = t^{-(n+1)/2s + n/(2qs)} ,$$ and the above is in $$L^p([0,T])$$ if and only if $$p (\tfrac{n+1}{2s} - \tfrac{n}{2qs}) < 1 .$$ So the answer seems to be rather ugly: $$q > \tfrac{n}{n+1+2s} \qquad \text{and} \qquad p (\tfrac{n+1}{2s} - \tfrac{n}{2qs}) < 1 ,$$ unless I made some mistake.
• Thank you. How did you obtain $\phi(1,x) \approx C \min\{|x|, |x|^{-(n+1+2s)}\}$ and $\|\phi(t,\cdot)\|_q = t^{-(n+1)/2s + n/(2qs)}$?
• Using $p_1^{(n)}(x) \approx C \min\{1, |x|^{-n-2s}\}$. I just added some details. Mar 17, 2021 at 19:06