Let $\Omega$ be a bounded $C^1$-domain in $\mathbb{R}^n$ satisfying the exterior sphere condition at every boundary point and $f$ be a bounded continuous function in $\Omega$ . Suppose $u\in C^2(\Omega)\cap C^1(\overline{\Omega})$ is a solution of $$\Delta u=f \ \ \text{in} \ \Omega \\ u=0 \ \ \text{on} \ \partial\Omega$$ Prove that $$\sup_{\partial\Omega}\left|\frac{\partial u}{\partial\nu}\right|\leq C\sup_{\Omega}|f|$$ where $C$ is a positive constant depending only on $n$ and $\Omega$ .

It is easy to see by maximum principle that $$\sup_{\Omega}|u|\leq\frac{R^2}{2n}\sup_{\Omega}|f|$$ if $\Omega\subseteq B_R$ , a ball of radius $R$ . Next how to relate $\displaystyle\sup_{\partial\Omega}\left|\frac{\partial u}{\partial\nu}\right|\leq\sup_{\Omega}|u|$ ? I tried to use gradient estimate but unfortunately it works only for harmonic functions .

Any help is appreciated . Regards .


1 Answer 1


Let $x_0\in \partial \Omega$ be arbitrary. The trick is to construct a function $\varphi\in C^1(\overline \Omega)$ such that $-\Delta \varphi \geqslant 1$ in $\Omega$, $\varphi(x_0)=0$, and $\varphi \geqslant 0$ on $\partial \Omega$. Given such a $\varphi$, you can obtain the required estimate easily (this is an exercise in Chapter 5 of Evans). Indeed, $$-\Delta(u - (\sup_\Omega \vert f \vert )\varphi)=-f+(\sup_\Omega \vert f \vert ) \Delta \varphi \leqslant-f-\sup_\Omega \vert f \vert\leqslant 0 \qquad \text{in } \Omega $$ and $$u -( \sup_\Omega \vert f \vert )\varphi = - (\sup_\Omega \vert f \vert) \varphi \leqslant 0, \qquad \text{on } \partial \Omega$$ so by the maximum principle $$u \leqslant \sup_\Omega \vert f \vert \varphi \qquad \text{in } \Omega .$$ Similarly, by replacing $u$ with $-u$, we obtain that $u \geqslant - (\sup_\Omega \vert f \vert) \varphi$ in $\Omega$ (note that $\varphi \geqslant 0$ in $\Omega$ by the max. principle), so we actually have that $$ \vert u\vert \leqslant (\sup_\Omega \vert f \vert) \varphi \qquad \text{in } \Omega .$$ Thus, $$\bigg \vert \frac{\partial u}{\partial \nu}(x_0)\bigg \vert = \lim_{h\to 0} \frac{\vert u(h\nu(x_0))\vert }h \leqslant (\sup_\Omega \vert f \vert) \lim_{h\to 0} \frac{ \varphi(h\nu(x_0)) }h\leqslant C\sup_\Omega \vert f \vert. \tag{$\ast$} $$ Since $x_0$ was arbitrary, taking the supremum over $x_0$ gives the required estimate.

All that is left to be done is to show that there exists such a $\varphi$. Note that the above argument didn't require that $\partial \Omega$ satisfies the exterior ball condition; however, we will need this assumption in the construction of $\varphi$. In fact, I think you need the uniform exterior ball condition (this may be what you mean), that is, there exists $\rho_\Omega>0$ such that for all $x\in \partial \Omega$, there is an exterior ball that touches $\partial \Omega$ at $x$ and has radius $\rho_\Omega$ i.e. the radius of the exterior touching ball can be chosen independent of the point.

Assuming the uniform exterior ball condition, let $x_0$ be as above and let $B_{\rho_\Omega}$ be an exterior ball that touches $\partial \Omega$ at $x_0$. After a translation, we may assume that $B_{\rho_\Omega}$ is centred at $0$. Now let $\alpha >0$ be a constant to be chosen later and define $$ \tilde \varphi(x) = e^{-\alpha \rho_\Omega^2}-e^{-\alpha \vert x \vert^2}.$$ Then $\tilde \varphi =0 $ on $\partial B_{\rho_\Omega}$, so $\tilde \varphi(x_0)=0$ and $\tilde \varphi \geqslant 0$ in $\mathbb R^n \setminus B_{\rho_\Omega} \supset\partial \Omega$. Moreover, $$-\Delta \tilde \varphi (x)=2\alpha(2\alpha \vert x \vert^2-1)e^{-\alpha \vert x \vert^2}. $$ Taking $\alpha = \frac {\sqrt 2} {\rho_\Omega}$, we have that for all $x\in\Omega$, $$-\Delta \tilde \varphi (x) \geqslant 2\alpha(2\alpha \rho_\Omega^2-1)e^{-\alpha (\operatorname {diam}\Omega +\rho_\Omega)^2}=:C_0(\rho_\Omega,\operatorname {diam}\Omega)>0. $$ Finally, defining $\varphi := \frac 1 {C_0}\tilde \varphi$ finishes the proof.

As a final remark, one can calculate $\lim_{h\to 0} \frac{ \varphi(h\nu(x_0)) }h$ explicitly since $\varphi$ is radial and one can see that the constant $C$ in ($\ast$) depends only on $\rho_\Omega$ and $\operatorname {diam}\Omega$.

  • $\begingroup$ This is a very nice solution. $\endgroup$ Commented Jan 10, 2023 at 0:24

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