# a directional derivative estimate

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 .

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 $$\operatorname {diam}\Omega$$.