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The goal is to show that $\sup{|u|} \leq C|\Omega|^{\frac{1}{n} - \frac{1}{p}}||Du||_p$ in the case $p>n$.

There is a point in the proof that I just am not getting. First, the notation:
Let $n' = \frac{n}{n-1}, p' = \frac{p}{p-1}, \delta = \frac{n'}{p'}$ It is shown in the proof that if $\tilde{u} = \frac{\sqrt{n}|u|}{||Du||_p}$ that

$$||\tilde{u}||_{n'\delta^v} \leq \delta^{v\delta^{-v}}||\tilde{u}||_{n'\delta^{v-1}}^{1-\delta^{-v}} $$

This part I'm ok with. Then the book makes the statement "Iterating from $v=1$ ,we get for any $v$"

$$||\tilde{u}||_{\delta^v} \leq \delta^{\sum{v\delta^{-v}}}.$$

I don't see how this follows. Could you maybe give some insight?

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Could you write the indexes in the sum? –  Davide Giraudo Sep 29 '12 at 9:19
    
Write down the inequality when $\nu=1$, then $\nu=2$, then $\nu=3$. Can you plug the first case into the second, then nto the third, and so on? –  Siminore Sep 29 '12 at 11:03
    
The book doesn't give indices in the sum. Presumably it's a sum over positive integers $v$. @Siminore I'm not sure exactly what you mean? –  Euler....IS_ALIVE Sep 30 '12 at 1:04
    
Probably $$\|\tilde{u}\|_{\delta^\nu} \leq \delta^{\nu \delta^{-\nu}}\delta^{(\nu -1)\delta^{-\nu+1}} \delta^{(\nu -2)\delta^{-\nu+2}}\cdots$$ –  Siminore Sep 30 '12 at 8:21
    
@Siminore Yes, but what happened to the $n'$? The statement just has $\delta^v$ –  Euler....IS_ALIVE Oct 1 '12 at 23:33

1 Answer 1

This looks like the presentation given in Gilbarg-Trudinger. The sum is in fact $\sum_{\nu = 1}^\infty$. The lack of $n'$ is likely a typo, you should insert it in yourself (my lecture notes on Sobolev spaces uses the same proof as G-T but has the corresponding factor of $n'$; so most likely I've found that the factor is beneficial to be shown). It doesn't effect the final conclusion though:

for $\delta >1$ and any $n' > 0$, we identically have $\lim_{k\to\infty} n' \delta^k = \infty$ so the concluding step in the end does not depend on the value of $n'$.

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