# Global boundedness for weak solutions in Gilbarg/Trudinger [Chapter 8.5]

I have a question concerning a step in the proof of Theorem 8.15 in Gilbarg/Trudinger "Elliptic PDEs of Second Order".

I really hope someone might be familiar with this and would be so kind as to go through the trouble of reading the proof again. This may be a bit much to ask for, but it would certainly be greatly appreciated! Here is the statement of the theorem

Theorem 8.15: Let the operator $L$ satisfy conditions (8.5), (8.6) and suppose that $f^i \in L^q(\Omega)$, $i=1, \ldots, n$, $q\in L^{q/2}(\Omega)$ for some $q>n$. Then if $u$ is a $W^{1,2}(\Omega)$ subsolution of $Lu = g + D_if^i$ in $\Omega$ satisfying $u\le 0$ on $\partial \Omega$, we have $$\sup_\Omega u \le C(\Vert u^+\Vert_2 + k)$$ where $k=\lambda^{-1}(\Vert \mathbf{f}\Vert_q + \Vert g \Vert_{q/2})$ and $C = C(n,\nu, q, |\Omega|)$.

Conditions (8.5), (8.6) are strict ellipticity (with smallest eigenvalue $\lambda$) and uniform boundedness conditions on $L$.

Now to my question: I can follow the proof up to $$\tag{8.36} \Vert H(w) \Vert_{2\hat n/(\hat n - 2)} \le C \Vert H'(w)w \Vert_{2q/(q-2)}$$ where $C = C(n, \nu, |\Omega|)$. With $H(w) = w^\beta - k^\beta$ for $k$ as in the statement of the theorem. But how can we deduce from this that $$\tag{8.37} \Vert w \Vert_{\beta\chi q^\ast} \le (C\beta)^{1/\beta}\Vert w \Vert_{\beta q^\ast}$$ for $\beta\ge 1$, $q^\ast = 2q/(q-2)$ and $\chi = \hat n(q-2)/q(\hat n - 1)$?

It is clear to me that (8.36) is equivalent to $$\Vert w^\beta - k^\beta \Vert_{\chi q^\ast}^{1/\beta} \le (C\beta)^{1/\beta}\Vert w \Vert_{\beta q^\ast}$$ What happens to the summand $-k^\beta$?

At first I thought that some scaling argument could work, but as $w = u^+ + k$, I don't seem to be able to scale $w$ withouth also scaling $k$. And then I can't make $-k^\beta$ negligible...

-
Maybe you can estimate $\int w^{\beta\chi q^*}$ separately on the sets $w\ge 2k$ and $w\le 2k$. When $w\ge 2k$, we have $w^\beta\le (1+1/(2^\beta-1))(w^\beta-k^\beta)$, which I think is okay for the subsequent proof because the extra multiplicative constant tends to 1 as $\beta\to\infty$. On the region $w\le 2k$ you can try $w^{\beta\chi q^*}\le (2k)^{\beta (\chi-1)q^*}w^{\beta q^*}$. – user31373 Jul 16 '12 at 1:37
@LeonidKovalev: Your approach worked out nicely. Thanks again! – Sam Jul 17 '12 at 23:23

Leonid Kovalev's approach works perfectly fine (with a minor adjustment). A big thank you from me again!

Everything written out explicitely (and throwing away the irrelevant information), one simply needs to prove that the assumption

$$\Vert w^\beta - k^\beta \Vert_{\chi q^\ast} \le C\beta \Vert w^\beta\Vert_{q^\ast}$$

for $w$ satisfying $w\ge k$ and $\beta \ge 1$ implies

$$\Vert w^\beta \Vert_{\chi q^\ast} \le C\beta \Vert w^\beta\Vert_{q^\ast}$$

On the one hand, if $w\le 2^{1/\beta} k$, then $$\Vert w^\beta \Vert_{\chi q^\ast}\le 2 \Vert k^\beta \Vert_{\chi q^\ast} = 2|\Omega|^{1/q^\ast\chi - 1/q^\ast}\Vert k^\beta\Vert_{q^\ast}\le 2|\Omega|^{1/q^\ast\chi - 1/q^\ast} \Vert w^\beta \Vert_{q^\ast}$$ If on the other hand $w\ge 2^{1/\beta} k$, then $$w^\beta = \frac{1}{1-k^\beta/w^\beta}(w^\beta - k^\beta) \le 2(w^\beta - k^\beta)$$ So putting things together, we obtain $\Vert w^\beta \Vert_{\chi q^\ast} \le C\beta \Vert w^\beta \Vert_{q^\ast}$ for some new constant $C$, depending on the same parameters as the old constant.

-
Yay! Thanks for posting this as an answer. – user31373 Jul 18 '12 at 0:26

I think it may also be viewed as a consequence of the triangular inequality together with the fact that $w\geq k$ by construction.

The triangular inequality gives $$\Vert w^\beta\Vert_{\chi q*} \leq \Vert w^\beta - k^\beta\Vert_{\chi q*} + k^\beta |\Omega|^{1/\chi q*}$$ and the inequality $w\geq k$ (by construction if I don't mistake) $$\kappa^\beta \leq \frac{\Vert w^\beta\Vert_{q*}}{|\Omega|^{1/q*}}$$

Therefore we get $$\Vert w^\beta\Vert_{\chi q*} \leq \Vert w^\beta-\kappa^\beta\Vert_{\chi q*} + \Vert w^\beta\Vert_{q*} \big(|\Omega|^{(1-\frac{1}{\chi})}\big)^{1/q*}$$ When $|\Omega|<+\infty$, the proof is done, at least for large $\beta$.

You may notice that if $|\Omega|=+\infty$, we can apply the last computation on some subset $\Omega'\subseteq\Omega$ of finite Lebesgue measure, and then go to the limit using Fatou's lemma (if ever $w^\beta\in L^{q*}(\Omega)$).

-