TITLE: Is the heat kernel for the hyperbolic plane uniformly continuous in $t\in(0,\infty)$?
QUESTION [5 upvotes]: let $\mathbb{H}$ be the hyperbolic plane and let $k(t,x,y)$ be the associated heat kernel.
I am wondering, if for any fixed $y\in M$ and $\epsilon >0$ the function $u_t(x):=k(t,x,y)$ is continuous in $S:=\lbrace x\in M : d(x,y)>\epsilon \rbrace$ uniformly in $t\in(0,\infty)$?
For instance in Euclidean space $\mathbb{R}^n$ the heat kernel is given by $k(t,x,y)=\frac{1}{(4\pi t)^{n/2}}e^{-\frac{\Vert x-y \Vert^2}{4t}}$. Hence for $\mathbb{R}^n$ the claim is true because the function $u_t(x)$ has bounded derivative on S, i.e. $\sup_{t>0} \vert  \nabla u_t(x) \vert\leq C$.
There is a formula for hyperbolic plane as well, given by
$$k(t,x,y)=\frac{\sqrt{2}}{(4\pi t)^{3/2}}e^{-\frac{t}{4}}\int\limits_{d(x,y)}^{\infty}\frac{s e^{-\frac{s^2}{4t}}}{\sqrt{\cosh{s}-\cosh{d(x,y)}}}ds$$
but I do not know how to estimate the derivative for this function.
Q: How can I prove/disprove that for any $x^{*}\in\mathbb{H}^2, x^{*}\neq y, \forall \epsilon>0$ $\exists\delta>0,$ s.t. $\vert u_t(x)-u_t(x^{*}) \vert\leq \epsilon$ for all $x\in \mathbb{H^2}$ with $d(x,x^{*})<\delta$? If the above statement is wrong, What happens if we restrict the domain to $t\in (0,T)$ for $T>0$ arbitrary and fixed?
Any help will be very appreciated!
Best wishes

REPLY [4 votes]: It follows from the interior  Schauder estimates for parabolic equations. Namely, put for simplicity's sake $y=0$, $k(t,x)=k(t,x,0)$ and for negative values of $t$ continue $k$ by zero: $k(x,t)=0$, $t<0.$ For $r>0$ denote $B_r(x_0)=\{x\in \mathbb{H}^2|d(x,x_0)<r\}$ a ball on the plane and $B_r=B_r(0)$. Then $k(x,t)$ is bounded on a cylindrical domain  $H_r=\mathbb R\times (\mathbb{H}^2\backslash B_r)$. Now fixing $\alpha\in(0,1)$ and applying the interior Schauder estimate in the cylinder $G_r(x_0,t_0)=(t_0-r^2,t_0)\times B_r(x_0)$, $d(x_0,0)\ge 2r$, we have
$$
\|k\|_{C^{2,\alpha}(G_{r/2}(x_0,t_0))}\le C_r\sup_{G_{r}(x_0,t_0)}|k|\le C_r\sup_{H_r}|k|.
$$
It follows that 
$$
\sup_{H_{2r}}|\nabla_x k|\le C_r\sup_{H_r}|k|.
$$