Can we always extend a Holder-boundary continuous function to whole domain? Let $\Omega\subseteq\mathbb{R}^{n}$ be a smooth domain, and let $f\in C^{\alpha}\left(\partial\Omega\right),$
where $\alpha\in\left(0,1\right).$ Do we always have that there exists
a function $\widetilde{f}\in C^{\alpha}\left(\overline{\Omega}\right)$
so that $\left.\widetilde{f}\right|_{\partial\Omega}\equiv f?$
Note that, if $\alpha>1$ the result is true and can be found in the book by Gilbarg + Trundinger (Lemma 6.38, p 137).
 A: We will use the fact that
$$
|x_{1}-x_{2}|^{\alpha}\leq(|x_{1}-x_{3}|+|x_{2}-x_{3}|)^{\alpha}\leq
|x_{1}-x_{3}|^{\alpha}+|x_{2}-x_{3}|^{\alpha}.
$$
Let $E\subseteq\mathbb{R}^{n}$ and let $f:E\rightarrow\mathbb{R}$ be such that
$
|f(x)-f(y)|\leq L|x-y|^{\alpha}
$
 for all $x,y\in E$. Define
$$
h(x):=\inf\left\{  f(y)+L|x-y|^{\alpha}:\,y\in E\right\}  ,\quad
x\in\mathbb{R}^{n}.
$$
If $x\in E$, then taking $y=x$ we get that $h(x)\leq f(x)$. To prove that $h(x)$ is finite for every $x\in\mathbb{R}^{n}$, fix $y_{0}\in
E$. If $y\in E$ then
$$
f(y)-f(y_{0})+L|x-y|^{\alpha}\geq-L|y-y_{0}|^{\alpha}+L|x-y|^{\alpha}%
\geq-L|x-y_{0}|^{\alpha},
$$
and so
\begin{align*}
h(x)   =\inf\left\{  f(y)+L|x-y|^{\alpha}:\,y\in E\right\}  
  \geq f(y_{0})-L|x-y_{0}|^{\alpha}>-\infty.
\end{align*}
Note that if $x\in E$, then we can choose $y_{0}:=x$ in the previous
inequality to obtain $h(x)\geq f\left(  x\right)  $. Thus $h$ extends $f$. 
Next we prove that
$$
\left\vert h(x_{1})-h\left(  x_{2}\right)  \right\vert \leq L|x_{1}%
-x_{2}|^{\alpha}%
$$
for all $x_{1}$,$\,x_{2}\in\mathbb{R}^{n}$. Given $\varepsilon>0$, by the definition of $h$ there exists
$y_{1}\in E$ such that
$$
h(x_{1})\geq f(y_{1})+L|x_{1}-y_{1}|^{\alpha}-\varepsilon.
$$
Since $h\left(  x_{2}\right)  \leq f(y_{1})+L|x_{2}-y_{1}|^{\alpha}$, we get
\begin{align*}
h(x_{1})-h\left(  x_{2}\right)   &  \geq L|x_{1}-y_{1}|^{\alpha}-L|x_{2}%
-y_{1}|^{\alpha}-\varepsilon\\
&  \geq-L|x_{1}-x_{2}|^{\alpha}-\varepsilon.
\end{align*}
Letting $\varepsilon\rightarrow0$ gives $h(x_{1})-h\left(  x_{2}\right)
\geq-L|x_{1}-x_{2}|^{\alpha}$. Interchanging the roles of $x_{1}$ and $x_{2}$
proves that $h$ is Holder continuous.
