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.


  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?


1 Answer 1


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.$\square$


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