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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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    I did not write down the computation, but I guess $n'$ is irrelevant since it appears in both sides. Probably you can absorb it as a factor, somewhere.2012-10-02

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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 $n'$.