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).


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.

  • $\begingroup$ Excellent. Thanks. $\endgroup$ – Binjiu Dec 6 '16 at 7:17

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