Let $n\geq 3$. Let $\Omega$ be an open and bounded subset of $\mathbb{R}^n$. Let define $X\_0$ as the space of functions $f:\bar\Omega\times\partial\Omega\to\mathbb{R}$ such that $f(x,\cdot)$ is continuous on $\partial\Omega$ for all $x\in\bar\Omega$. Moreover let suppose that 
\begin{equation}
||f||\_{X\_0}:=\sup\_{x\in\bar\Omega}||f(x,\cdot)||\_{C^0(\partial\Omega)}<\infty.
\end{equation}


Let $\alpha\in]0,1[$. Let $\Omega$ be of class $C^{1+\alpha}$.


I want to prove this result, but I can't succeed. If $\varphi\in C^{\frac{\alpha}{2};\alpha}([0,T]\times\partial\Omega)$ then the function $\psi:[0,T]\times\bar\Omega\times\partial\Omega\to \mathbb{R}$ defined as
\begin{equation}
\begin{split}
&\psi(t,x,y)=|x-y|^{n-2}\int\_0^t\frac{1}{[4\pi(t-\tau)]^\frac{n}{2}}\exp\Big\{-\frac{|x-y|^2}{4(t-\tau)}\Big\}\varphi(\tau,y)d\tau \;\;\mbox{ for } x\neq y\\
&\psi(t,x,x)=0\\
&\psi(0,x,y)=0
\end{split}
\end{equation}


stays in $C^{\frac{1+\alpha}{2}}([0,T],X\_0)$, where $X\_0$ is equipped with the norm $||\cdot||\_{X\_0}$.



On the Hölder regularity of an integral function
The condition $\psi(t,x,x)=0$ seems to be irrelevant here, because under the implied conditions on $\varphi$ $\lim\_{x\to y}\psi(x,y,t)=c\_n \varphi(y,t)$. If the limit values of $\psi$ is from $C^{\alpha/2}$ with respect to $t$ one cannot expect it to have uniform estimate in $C^{(1+\alpha)/2}$ up to the boundary.


