I am reading the chapter 14.1 in Trudinger's PDE book.
Let $\Omega$ be a bounded domain with smooth boundary and uniform exterior sphere condition. Suppose we have a quasilinear elliptic equation $Qu=0$ and $u=C$ on $\partial \Omega$ where $C$ is a constant.
If we have two barrier functions $w_1$ and $w_2$ in $\Omega$ such that
- $Qw_1<0$ and $Qw_2>0$
- $w_1=w_2=C$ on $\partial \Omega$
Then by comparison principle, we have $w_2 \leq u \leq w_1$ in $\Omega$ and $\dfrac{\partial w_2}{\partial n} \leq \dfrac{\partial u}{\partial n} \leq \dfrac{\partial w_1}{\partial n}$ on $\partial \Omega$ with $n$ being inward normal.
If we can find such barrier functions $w_i$, then the theorem (theorem 14.1 page 337) in the book stated that $|Du|\leq M$ on $\partial \Omega$ where $M$ only depends on $w_1$ and $w_2$. The reason is just because of the inequality of the directional derivative in above.
I am confused why it is true since the directional derivative may 'miss' some components in the gradient $Du$.
The "missed" components are those tangent to $\partial \Omega,$ where we already know the value of $u$.
Explicitly: at any boundary point $x \in \partial \Omega,$ we can choose an orthonormal basis $e_i$ so that $e_1 = n(x)$ and $e_2,\ldots,e_n \in T_x\partial \Omega.$ We then have $Du(e_1) = \partial u/\partial n,$ while for $i>1$ we have $Du(e_i) = 0$ because $u =C$ on $\partial \Omega.$ (Hopefully this is obvious; if not, consider a curve on $\partial \Omega$ tangent to $e_i$ and apply the chain rule.) Thus $$|Du(x)|^2 = Du(e_1)^2+\cdots+Du(e_n)^2=\left|\frac{\partial u}{\partial n}(x)\right|^2.$$