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If $\Omega \subseteq \mathbb{R}^N$ is an open bounded domain and $1<p<N$, then the classical Sobolev Inequality: $$\| u\|_{p^*,\Omega} \leq C\ \| \nabla u\|_{p,\Omega}$$ holds with $C=C(p,N,\Omega)>0$ and $p^*:= Np/(N-p)$ for any $u\in W_0^{1,p}(\Omega)$.

What about the case $p\geq N$? May I take the $L^\infty$-norm in the LHside?

If I remember correctly, in general I cannot get the inequality with $\| \cdot \|_\infty$, for there are counterexemples of unbounded Sobolev functions... But, what if I know "a priori" that $u\in L^\infty(\Omega) \cap W_0^{1,p}(\Omega)$?

Any reference? (Adams-Fournier? Brezis?)

Thanks in advance.

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  • $\begingroup$ I don't think that you can have an estimate like this one: $$\lVert u \rVert_\infty \le C \lVert \nabla u \rVert_p.$$ What you do have, as you already know very well, is Morrey's inequality (cfr. Brezis, Liguori ed., pag. 264): $$\frac{\lvert u(x)-u(y)\rvert}{\lvert x-y\rvert^\alpha}\le C \lVert \nabla u \rVert_p.$$ The reason why I am telling this is that in $\mathbb{R}^n$ the first inequality does not scale well while the second does. Sure, you are talking of bounded domains and so there is no scaling. But still I've got this feeling. $\endgroup$ – Giuseppe Negro Mar 26 '13 at 0:14
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For the case $p=N$ take a look here. There you will find all you want. Note that the function $f$ defined by me is a counter example for what you want, also, even if you ask $u\in L^\infty$, you dont get what you want. Take a look in the answer and you will have all the explanations you need.

When $p>N$ and you have some regularity in the boundary, then your functions are continuous, even Holder continuous. I suggest you to take a look in any good book about Sobolev Spaces. For example the book ok Leoni is a good one, there you will find all you need.

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When $\Omega\subset% %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion ^{n}$, $n\geq2$ is a bounded domain, we have by the Sobolev embedding theorems that $W_{0}^{k,p}\left( \Omega\right) \subset L^{q}\left( \Omega\right) ,~1\leq q\leq\frac{np}{n-kp},~kp<n$ and that $W_{0}^{k,\frac{n}{k}}\left( \Omega\right) \subset L^{q}\left( \Omega\right) ,~1\leq q<\infty$. However, there are counterexamples to the embedding $W_{0}^{k,\frac{n}{k}}\left( \Omega\right) \subset L^{\infty}\left( \Omega\right) $. In this case, it was proposed independently by Yudovich, Pohozaev and Trudinger that $W_{0}^{1,n}\left( \Omega\right) \subset L_{\varphi_{n}}\left( \Omega\right) $ where $L_{\varphi_{n}}\left( \Omega\right) $ is the Orlicz space associated with the Young function $\varphi_{n}(t)=\exp\left( \beta_{n} \left\vert t\right\vert ^{n/(n-1)}% \right) -1$ for some positive $\beta_{n}$. Moreover, Moser explored more in this direction and further found the largest positive real number $\beta_{n}$. In fact, in his 1971 paper [A sharp form of an inequality by N. Trudinger. Indiana Univ. Math. J. 20 (1970/71), 1077–1092.], J. Moser proved the following result: There exist sharp constant $\beta_{n}% =n\omega_{n-1}^{\frac{1}{n-1}}$, where $\omega_{n-1}$ is the area of the surface of the unit $n-$ball, such that $$ \frac{1}{\left\vert \Omega\right\vert }\int_{\Omega}\exp\left( \beta \left\vert u\right\vert ^{\frac{n}{n-1}}\right) dx\leq c_{0}% $$ for any $\beta\leq\beta_{n}$, any $u\in W_{0}^{1,n}\left( \Omega\right) $ with $\int_{\Omega}\left\vert \nabla u\right\vert ^{n}dx\leq1$. This constant $\beta_{n}$ is sharp in the sense that if $\beta>\beta_{n}$, then the above inequality can no longer hold with some $c_{0}$ independent of $u$. The same result for $W_{0}^{k,p}\left( \Omega\right)$ was proved by D. Adams in his paper [A sharp inequality of J. Moser for higher order derivatives. Ann. of Math. (2) 128 (1988), no. 2, 385–398.]

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Sobolev inequality gives you that $W_0^{m,p}$ is embedded $L^{p^*}$, and by an interpolation argument, you can embed $W_0^{m,p}$ in $L^q$ for $p\leq q\leq p^*$ for $mp<N$, in your case, $m = 1$, $p<N$.

If you want $mp=N$, $W_0^{m,p}$ is embedded in $L^q$ with $1<p\leq q <\infty$. In your case $m=1$, $p=N$, but I am not sure what happens if $p>N$.

For a proof of these facts, see page 20 of http://people.bath.ac.uk/masgrb/Sobolev/notes.pdf

I am not sure if this answers your question. I am also only a beginner in this area.

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  • $\begingroup$ @Tomás you are right, let me correct this. $\endgroup$ – Lost1 Mar 25 '13 at 23:29

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