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Let $\frac{2n}{n+1}\leqslant p, then prove $u^2\in W^{1,q}(R^n).$

I hope someone can show me how to prove it by dense theorem approximation in the theory of Sobolev spaces! Thanks!

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    Could you give the statement of dense theorem you are thinking?2012-11-02
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    To Davide Giraudo: Dense Theorem I'm thinking is as follows: $C_0^\infty(R^n)$ is dense in the $W^{1,p}(R^n),1\leqslant p<+\infty$2012-11-02
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    Maybe the result is linked to Sobolev embeddings.2012-11-02
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    To Davide Giraudo:Yes! In fact, I can proof $Du^2\in L^q(R^n)$ by Sobolev embedding theorem, but failed in $u^2\in W^{1,q}(R^n)$ by dense theorem.2012-11-02

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Let $u \in L^1(\mathbb{R}^n) \cap W^{1,p}(\mathbb{R}^n)$. Denote by $p^*$ the Sobolev conjugate of $p$, that is $$ p^* = \frac{np}{n-p}. $$

From the Sobolev embedding theorem, we know that $u \in L^{p^*}(\mathbb{R}^n)$. Note that $1 \leq 2q \leq p^{*}$. Since $u \in L^1(\mathbb{R}^n)$ and $u \in L^{p^*}(\mathbb{R}^n)$, using Holder's inequality we can show that $u$ belongs to all intermediate spaces, and in particular $u \in L^{2q}(\mathbb{R}^n)$, or $u^2 \in L^q(\mathbb{R}^n)$. In more details, if we choose $0 \leq \theta \leq 1$ such that $$ \frac{1}{2q} = \frac{1 - \theta}{1} + \frac{\theta}{p^*} $$ then from Holder's inequality we obtain $$ ||u||_{2q} \leq ||u||_1^{1 - \theta} ||u||_{p^*}^{\theta} < \infty. $$

To estimate the gradient of $u^2$, note that $\frac{p^*}{q}$ and $\frac{p}{q}$ are conjugate exponents and use Holder's inequality again to obtain $$ ||\nabla (u^2)||_q = \left( \int |\nabla (u^2)|^q \right)^{\frac{1}{q}} = 2 \left( \int |u|^q |\nabla u|^q \right)^{\frac{1}{q}} \leq 2 ||u||_{p^*} ||\nabla u||_p < \infty. $$

Note that to justify the argument above rigoursly, you need to say why $u^2$ is weakly differentiable with a weak gradient given by $2u \nabla u$. You can quote an approximate product rule theorem or use an approximation argument. That is, first assume that $u$ is a smooth compactly supported function and write the estimates that allow you to control the $L^q$ norm of $u$ and $\nabla u$ in terms of the $L^p$ and $L^1$ norms of $u$ and $\nabla u$. Then approximate a general function in $W^{1,p}$ and $L^1$ using smooth functions and show using the estimates that the limit lies in $W^{1,q}$.

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    :In my knowledge, the Sobolev embedding is valid for exponent $1\leqslant p<\infty$. However, you can see in my assumption that $\frac{2n}{n+1}\leqslant p. So if $p=\frac{2n}{n+1}$, then $\frac{p}{2}=\frac{n}{n+1}<1$. For exponent$<1$, is there certain version of Sobolev embedding? By the way, can you find a way that use the dense theorem to proof the result? Thank you very much!2012-11-03
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    Hmm, you are absolutely correct. The proof above works only when $p \geq 2$. I'll try to think how to handle $\frac{2n}{n+1} \leq p < 2$. Since I didn't use $u \in L^1(\mathbb{R}^n)$, maybe one needs to use it now.2012-11-03
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    To levap: I'm glad to talk about the problem with you! I will also consider it. If I get some idea, I will post the answer.:-)2012-11-03
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    I've fixed the argument so it works for all $p$. It turns out to be just an interpolation argument.2012-11-03
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    To levap: Yep,this time you are totally correct.Thanks for your help!2012-11-03
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    To levap: Yor answer is according to the interpolation inequlities in $L^p$. Now,I'm thinking about another method by density theorem approximation.I wish I can succeed.2012-11-03
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Firstly, I want to say that the answer due to levap is nice!

Now I give another way by using the density theorem approximation to complete the proof of $u^2\in L^q(R^n)$ as follows:

Note that $u^2\in L^q$ is equivalent to $u\in L^{2q}$, so we only need to show the later.

We can get a approximation sequence $\{u_n\}\subset C_0^\infty(R^n)$ such that

$$ u_n\rightarrow u\quad in \quad L^1\\ u_n\rightarrow u\quad in \quad L^{p*} $$

are both valid by density theorem. Then

$$ ||u_n-u_m||_{L^{2q}}\leqslant||u_n-u_m||_{L^1}^{1-\alpha}||u_n-u_m||_{L^{p^*}}^\alpha $$

by interpolation inequalities in $L^p$ spaces.

So $\{u_n\}$ also a Cauchy sequence in $L^{2q}$, then there exist $v\in L^{2q}$ such that

$$ u_n\rightarrow v\quad in \quad L^{2q} $$

Note the fact that the convergence in $L^p$ spaces imply a subsequence converges almost everywhere. We have

$$ v=u\quad a.e.\quad in \quad R^n $$

So $u\in L^{2q}(R^n)$.

PS:In fact, we can directly use the interpolation inequality just like what levap did. My work is still valuable,I use the density theorem after all. I expect anybody can improve my proof by using density theorem approximation! :-)

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    This is quite a stilted use of the density theorem, especially that you don't even use the density theorem in **Sobolev** spaces, only in $L^p$... What is more interesting is to use the density theorem in $W^{1,p}$ (and other spaces) to justify that $u^2$ has **weak derivatives**, and that they coincide with what they are suppose to be - $2u \nabla u$.2012-11-04
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    Hmm, my proof is really stilted.Can you make your idea rigorous.Please show me,I really want to know how to skillfully use the density theorem in **Sobolev** spaces.Thank you! Of course, I will also continue to think about it.2012-11-05