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!

Thanks in advance :)


1 Answer 1


I do not suppose this to be an answer, but comments are not permitted without login. May be you should specify which side of the inequality makes the trouble. Obviously the argument follows a standard one which I just met in on page 14 (Theorem 8.1) of http://arxiv.org/pdf/math/0309021 . Does looking at that short proof help?

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.)

Might this be a half an answer?


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