2
$\begingroup$

Suppose $H$ is a non-negative self-adjoint operator acting on $L^2(\mathbb{R}^n)$, and generates an analytic semigroup with kernel satisfying a Gaussian upper bound, i.e., if we denote $K(t,x,y)$ the kernel of the semigroup $e^{-tH}$, then $$ K(t,x,y)\leq t^{-\frac{n}{2}}e^{-\frac{|x-y|^2}{4t}} $$ By using Hadamard's three lines lemma, one can further get the pointwise estimates for the kernel $K(z,x,y)$ of the analytic semigroup $e^{-zH}$ with $\Re z>0$ as the following $$ K(re^{i\theta},x,y)\leq (r\cos\theta)^{-\frac{n}{2}}e^{-\frac{|x-y|^2\cos \theta}{2r}},~~~ r>0, -\frac{\pi}{2} <\theta<\frac{\pi}{2}, $$ I'm interested in the distribution of the value of $|K(re^{i\theta},x,y)|$ on a fixed radius $r>0$. And my question is that for each fixed $r>0$ and $|x-y|>0$, is the kernel $|K(re^{i\theta},x,y)|$ a logarithmic convex function with respect to $\theta$? i.e., Is $\ln|K(re^{i\theta},x,y)|$ a convex function for $-\frac{\pi}{2} <\theta<\frac{\pi}{2}$ with each $r>0$ fixed?

When $H=-\Delta$, this is the case, since we can compute the heat kernel with complex $t$ explicitly. However, in general, the difficulty comes when we compute the derivative with respect to $\theta$.

$\endgroup$

0

You must log in to answer this question.