How to use the dense theorem to prove this exercise? Let $\frac{2n}{n+1}\leqslant p<n,\quad q=\frac{np}{2n-p},\quad u\in L^1(R^n)\cap W^{1,p}(R^n)$, 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!
 A: 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 $W^{1,q}$.
A: 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! :-)
