Prove the product of two $W_0^{1,p}$ functions gives another $W_0^{1,p}$ function if $p>n$ Fix $\Omega \subset \mathbb{R}^n$ want to show if $p>n$ then
$$u,v \in W_0^{1,p}(\Omega) \implies uv \in W_0^{1,p}(\Omega).$$
I think I have an answer but I'm not sure why the condition $p>n$ is needed. I'm going to write $||\cdot||$ instead of $||\cdot||_{W_0^{1,p}(\Omega)}$.
Since $W_0^{1,p}(\Omega)$ is defined to be the closure of $C_c^{\infty}(\Omega)$ in $W^{1,p}(\Omega)$ is suffices to show the existence of a sequence in $C_c^{\infty}(\Omega)$ which converges to $uv$ in $||\cdot||$. Let $\{u_i\}$, $\{v_i\}$ be sequences in $C_c^{\infty}(\Omega)$ st
$$||u_i-u||\to0, \,\,\,\,\,||v_i-v||\to0$$
Then clearly the sequence $\{u_iv_i\}$ is in $C_c^{\infty}(\Omega)$ and 
$$||u_iv_i-uv|| \leq ||u_i||||v_i-v||+||v||||u_i-u|| \to 0$$
noting that $||u_i||$ is bounded because it is a Cauchy sequence.
Can someone tell me why this condition is needed and if his answer is correct? Thanks 

In general, we cannot conclude that the product of two $L^p$ functions is itself $L^p$.
In fact, the best sort of integrability for products of generic functions is given by Holder's inequality:
$$ \lVert fg \rVert_{L^1} \leq \lVert f \rVert_{L^p} \lVert g \rVert_{L^q} , \frac{1}{p} + \frac{1}{q} = 1$$
On the other hand, we have Sobolev control of our functions, which gives us more information. What I would suggest is a Sobolev embedding theorem, such as Morrey's inequality. This tells us that, in a bounded domain, for $p > n$, there are $M, \alpha > 0$ such that
$$ \lVert u \rVert _{C^{0,\alpha}} \leq M\lVert u \rVert_{W_0^{1,p}} $$
where $C^{0,\alpha}$ is the space of functions which are Holder continuous of exponent $\alpha$, endowed with the standard norm.
In particular this tells us that for $p>n$, Sobolev functions are actually continuous (more precisely, they have a continuous representative), and continuous functions are, of course, bounded on compact sets. 
From here the result should follow from computation of th $L^p$ norms of of $uv$ and $D(uv) = uDv + vDu$.
