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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 $||\cdot||_{X_0}$.

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  • $\begingroup$ Should it be also $n\ge3$? For $n\le2$ the expression may go to infinity then $x\to y$. $\endgroup$ – Andrew Nov 18 '14 at 12:04
  • $\begingroup$ Yes, I was supposing that $n\geq 3$. $\endgroup$ – foo90 Nov 18 '14 at 12:09
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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.

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