Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 \begin{equation}\tag{8.36} \Vert H(w) \Vert_{2\hat n/(\hat n - 2)} \le C \Vert H'(w)w \Vert_{2q/(q-2)} \end{equation} 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 \begin{equation}\tag{8.37} \Vert w \Vert_{\beta\chi q^\ast} \le (C\beta)^{1/\beta}\Vert w \Vert_{\beta q^\ast} \end{equation} 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...

Thanks for your help!

share|cite|improve this question
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
up vote 2 down vote accepted

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.

share|cite|improve this answer
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 $w^\beta\in L^{q*}(\Omega)$).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.