# Green operator of Dirichlet problem is compact

Let $$\alpha \in (0,1]$$, $$\Omega \subset \mathbb R^n$$ open bounded and the problem $$\left\lbrace \begin{array}{r c l c l} - \Delta u&=&0&\operatorname{on}& \Omega \\ u&= & h&\operatorname{on}& \partial \Omega \end{array} \right. \tag{DP}$$ My exercise assume I know that forall $$h \in C^{0,\alpha}(\overline \Omega)$$ there is a unique $$u \in C^{2,\alpha}(\overline \Omega)$$ solution of (DP). Then let $$T : C^{0,\alpha}(\overline \Omega) \longrightarrow C^{2,\alpha}(\overline \Omega)$$ be the mapping $$h \longmapsto u$$, we admit it is linear bounded (linear easy).

Exercise: Show that $$T : C^{0,\alpha}(\overline \Omega) \longrightarrow C^{0,\alpha}(\overline \Omega)$$ is a compact operator.

Attempt: Let $$(h_j)$$ be a bounded sequence of $$C^{0,\alpha}(\overline \Omega)$$, we show $$(Th_j)$$ has a limit point in $$C^{0,\alpha}(\overline \Omega)$$. Since $$T : C^{0,\alpha}(\overline \Omega) \longrightarrow C^{2,\alpha}(\overline \Omega)$$ is bounded, $$(Th_j)$$ is bounded in $$C^{2,\alpha}(\overline \Omega)$$. So by Ascoli-Arzelà there is $$\sigma$$ extraction and $$u \in C^1(\overline \Omega)$$ such that $$Th_{\sigma(j)} \xrightarrow[j \infty]{C^1(\overline \Omega)} u.$$ It remains only to get the convergence of the $$[\cdot ]_\alpha$$ semi-norm, it is enough because then $$u \in C^{0,\alpha}(\overline \Omega)$$ and $$(Th_{\sigma(j)})$$ goes to $$u$$ in $$C^{0,\alpha}(\overline \Omega)$$-norm. If $$\Omega$$ is convex I can use the mean value inequality \begin{align} [Th_{\sigma(j)} - u ]_\alpha &= \sup_{x \neq y \in \Omega} \frac{|(Th_{(\sigma(j)}-u)(x)- (Th_{(\sigma(j)}-u)(y)|}{|x-y|^\alpha} \\ &\leq \sup_{x \neq y \in \Omega} \frac{1}{|x-y|^\alpha} \sup_{(x,y)}|(Th_{\sigma(j)}-u)'| \times |x-y| \\ &\leq \operatorname{diam}(\Omega)^{1-\alpha}||Th_{\sigma(j)}-u||_{C^1(\Omega)} \end{align} and finish the proof.$$\square$$

The problem is to control $$[Th_{\sigma(j)} - u ]_\alpha$$ when $$\Omega$$ is not convex. I suspect the problem is not consistent when $$\Omega$$ is not connected because in general solving differential equations on non connected sets boils down to solve the equation on each component and patch everything.

If $$\Omega$$ is connected (by path) I think there is a way to save my proof method, for any $$\gamma :[0,1] \longrightarrow \Omega$$ curve that verifies the Rolle's hypothesis and that links $$x$$ and $$y$$ the mean value inequality applied to $$(Th_{(\sigma(j)}-u )\circ \gamma$$ gives $$|(Th_{(\sigma(j)}-u)(x)- (Th_{(\sigma(j)}-u)(y)| \leq || \gamma ||_{C^1(0,1)} \sup_{(0,1)} |(Th_{(\sigma(j)}-u)' \circ \gamma| \leq || \gamma ||_{C^1(0,1)} ||Th_{(\sigma(j)}-u||_{C^1(\Omega)}.$$ Then denoting $$\mathcal A_x^y$$ the set of such curves we can consider the following metric on $$\Omega$$ $$d(x,y) = \inf\{ || \gamma ||_{C^1(0,1)} : \gamma \in \mathcal A_x^y \}$$ which looks like the geodesic metric and is such that $$|(Th_{(\sigma(j)}-u)(x)- (Th_{(\sigma(j)}-u)(y)| \leq d(x,y) ||Th_{(\sigma(j)}-u||_{C^1(\Omega)}$$ whence $$[Th_{\sigma(j)} - u ]_\alpha \leq ||Th_{(\sigma(j)}-u||_{C^1(\Omega)} \sup_{x \neq y \in \Omega} \frac{d(x,y)}{|x-y|^\alpha}.$$ If the last supremum is finite the I am done.

Question:

1. Is the last supremum finite? I think it should depend on the geometry of $$\Omega$$ but still, having a result for a small class of open sets would be nice

2. How could I finish the proof of $$T$$ compact?

Here is an answer to my problem, in fact we just need to prove that a $$C^1$$ function on a compact set is Lipschitz.
Claim: Let $$\Omega \subset \mathbb R^n$$ open bounded with $$C^1$$ boundary and $$f \in C^1(\overline \Omega ; \mathbb R)$$. Then $$f$$ is Lipschitz.
Proof: First note that $$f'$$ is bounded on $$\overline \Omega$$ so by the mean value inequality, any $$x \in \Omega$$ has a convex neighborhood on which $$f$$ is Lipschitz. Then assume by contradiction that $$f$$ is not Lipschitz, there is $$(x_j),(y_j)$$ two sequences of $$\Omega$$ such that $$x_j \neq y_j,\quad \frac{|f(x_j)-f(y_j)|}{|x_j-y_j|} \longrightarrow + \infty.$$ Then by compactness of $$\overline \Omega$$ we can assume without loss of generality that $$x_j \longrightarrow x \in \overline \Omega,\quad y_j \longrightarrow y \in \overline \Omega$$ and since $$f$$ is continuous on $$\overline \Omega$$ we must have $$x = y$$, otherwise the latter quotient has a finite limit. It must be that $$x \in \partial \Omega$$ because otherwise $$x \in \Omega$$ where $$f$$ is locally Lipschitz. So use the regularity of $$\partial \Omega$$ to introduce $$\phi : B(x,r) \longrightarrow B(0,1)$$ a $$C^1$$-diffeomorphism such that $$\phi(\Omega \cap B(x,r)) = B(0,1) \cap \{ x_n > 0 \},\quad \phi(\partial \Omega \cap B(x,r)) = B(0,1) \cap \{ x_n = 0 \}.$$ Then $$f \circ \phi^{-1} : B(0,1) \cap \{ x_n > 0 \} \longrightarrow \mathbb R$$ is of class $$C^1$$ on an open convex set and up to shrink $$r$$ and $$B(0,1)$$ we can assume that $$(\phi^{-1})'$$ is bounded, so that $$f \circ \phi^{-1}$$ has bounded derivative. Thus it is Lipschitz and for $$j$$ large so that $$x_j,y_j \in B(x,r)$$ we have \begin{align} \frac{|f(x_j)-f(y_j)|}{|x_j-y_j|} &= \frac{|f \circ \phi^{-1} \circ \phi (x_j) - f \circ \phi^{-1} \circ \phi (y_j)|}{|\phi(x_j) - \phi(y_j)|} \frac{|\phi(x_j)-\phi(y_j)|}{|x_j-y_j|} \\ &\leq \operatorname{Lip}(f \circ \phi^{-1})\operatorname{Lip}(\phi), \end{align} contradiction.$$\square$$