# Help explain why the proof works - Gradient Estimate Using the Maximum Principle

I found the following proposition and proof in a textbook for one of my classes, and the end of the proof seems a little "unfinished" to me. In other words, as the author left it, I'm not entirely sure how what he/she did actually gives us what we want.

Here is the Proposition:

Let $\Omega \subset \subset \mathbb{R}^{n}$, and let $u \in C^{3}(\Omega) \cap C^{1} (\overline{\Omega})$ be a solution of the non-linear PDE $$Lu=f(x,u)\quad \text{on}\,\Omega,$$ where $L = a_{ij}D_{i}{j}$ (the Laplacian) is an elliptic partial differential operator with $a_{ij}\in C^{1}(\overline{\Omega})$, and $0 < \lambda \leq (a_{ij}) \leq \Lambda$ (the ellipticity condition). The function $f$ is in $C^{1}(\overline{\Omega} \times \mathbb{R})$. Show that we must have

$$\sup_{\Omega}|Du|\leq \sup_{\partial \Omega}|Du|+C,$$

where $C$ is a constant depending only on the diameter of the domain $\Omega$, the lower bound $\lambda$, and the $C^{1}$ norms $||a_{ij}||_{C^{1}(\overline{\Omega}\times I)}$, where $I = \left[ \inf_{\overline{\Omega}}u, \sup_{\overline{\Omega}}u\right]$ is the range of the solution $u$.

The Essence of the Proof:

The author begins by calculating $L|Du|^{2}$, and then, by differentiating, etc., and using the ellipticity condition, establishes the following lower bound: $$L|Du|^{2} \geq \frac{\lambda}{2}|D^{2}u|^{2}-C_{1}|Du|^{2}-C_{2}.$$

He then goes on to establish the following lower bound for $L(|Du|^{2}+\alpha u^{2})$:

$$L(|Du|^{2}+\alpha u^{2}) \geq \frac{\lambda}{2}|D^{2}Du|^{2} + |Du|^{2} - C_{3},$$

where we have chosen $\alpha > 0$ to be large.

However, there is apparently a problem with $C_{3}$, because it came from the terms $2\alpha u f - C^{\prime}$, where $C^{\prime}$ was another constant from an intermediate step, and is dependent on the sup norm of $u$. (At least, I think that's what the problem is, but I'm not entirely sure: maybe it's just to make sure that the $C_{3}$ is not too negative?)

To control the $C_{3}$ term, we introduce yet another function, $e^{\beta x_{1}}$, for $\beta > 0$.

Then, the following lower bound is established for $L(|Du|^{2}+\alpha u^{2}+e^{\beta x_{1}}$:

$$L(|Du|^{2}+\alpha u^{2}+e^{\beta x_{1}}) \geq \frac{\lambda}{2}|D^{2}Du|^{2} + |Du|^{2} + \left\{\beta^{2}a_{11}e^{\beta x_{1}}-C_{4} \right\}.$$

If we put the domain $\Omega \subset \{ x_{1} > 0 \}$, then $e^{\beta x_{1}}\geq 1$, $\forall x \in \Omega$.

So, by choosing $\beta$ large, we may make the $\displaystyle \beta^{2}a_{11}e^{\beta x_{1}}-C_{4}$ positive.

Then, if we set $$w = |Du|^{2}+\alpha|u|^{2} + e^{\beta x_{1}}$$ for large $\alpha$, $\beta$ depending only on the diameter of $\Omega$, $diam(\Omega)$, $\lambda$, and the $C^{1}$ norms $||a_{ij}||_{C^{1}(\overline{\Omega})}$ and $||f||_{C^{1}(\overline{\Omega} \times I)}$, where $I = \left[ \inf_{\overline{\Omega}}u, \sup_{\overline{\Omega}}u\right]$ is the range of the solution $u$, then we obtain that

$$Lw \geq 0 \quad \text{in}\,\Omega.$$

And by the Maximum Principle, we have that

$$\sup_{\Omega}w \leq \sup_{\partial \Omega}w.$$

Now, my question is how does the very last statement in the proof give us that $\mathbf{\sup_{\Omega}|Du|\leq \sup_{\partial \Omega}|Du| + C}$, where $C$ is a constant fulfilling all the conditions mentioned in the statement of the proposition?

If someone could show me how this follows, I would be eternally grateful!

In the definition of $w$, I suggest to replace $|u|^2$ by $|u-x_0|^2$. This of course does not change values of second order differential operator $L$. (BTW, I suppose a derivative is missing in the definition of $L$.) I do not understand the exponential term with $\beta$, therefore let us omit that for the purpose of this note (or set $\beta=0$).
This way (with minus $x_0$ and no $\beta$) we see that for any fixed $x_0$, we have that $|Du(x_0)|^2 = w(x_0) \le \sup_\Omega w$ which is left hand side of the inequality. Then the other side of the inequality is supposed to be less than $\sup _ {\partial \Omega} |Du|^2$ which is, in the PDF file cited above, achieved by a different form of $w$ (zeroing the other terms on the boundary).
Thus for any fixed $x_0$ we estimated $|Du(x_0)|^2$. (Or we should/would.)