Product of functions in $H^1(B)$ where $B \subset \mathbb{R}^2$ I'm rather new to Sobolev spaces and finding myself rather deficient of intuition. So when given a problem like the below where I need to "prove or disprove", I'm finding myself stuck.
Suppose $B$ is a ball in $\mathbb{R}^2$ and $u,v$ are in the Sobolev space $H^1(B)$ (that is, $u,v \in L^2(B)$ and their weak derivatives are also in $L^2(B)$). I need to either prove $uv \in H^1(B)$ or find a counterexample where $uv \not \in H^1(B)$. 
Here are my ideas so far.
I know that $H^1(B)=W^{1,2}(B)$ is in the "edge case" for the Sobolev-type inequalities, exactly between where we get integrability ($p<n$) and where we get regularity ($p>n$). This means that $H^1(B)$ embeds into $L^q(B)$ for each $1 \leq q < \infty$. I also know that the requirement $q < \infty$ cannot be dropped. This means the easy argument for the corresponding problem on $W^{1,1}(I)$ for an interval $I$ does not work.
So here $u,v \in L^q$ for every $1 \leq q < \infty$. So $u,v \in L^4$, hence $u^2,v^2 \in L^2$, hence we can use the Cauchy-Schwarz inequality to conclude that $uv \in L^2$. 
Now the problem is with the integral of $|\nabla(uv)|^2$. As I recall the Leibniz rule goes through and we get $|u \nabla v + v \nabla u|^2$. This is at most $(|u \nabla v| + |v \nabla u|)^2 = |u \nabla v|^2 + |v \nabla u|^2 + 2|uv \nabla u \cdot \nabla v|$. And here I get stuck, not knowing where to go. Any help?
Edit: having given a little more thought, I am thinking that we can make $uv \not \in H^1$. First of all I think we can take $u=v$. Then $|\nabla(u^2)|^2=2u^2|\nabla u|^2$. Since $\nabla u \in L^2$, $|\nabla u|^2 \in L^1$. Since $u^2$ is in $L^p$ for $1 \leq p < \infty$, we need to get $|\nabla u|^2 \not \in L^p$ for all $p>1$ and we need $u \not \in L^\infty$. I think this can probably be done with polar coordinates and a radial gradient.
 A: Generally, you should expect a Sobolev space to be closed under multiplication if and only if it embeds into $L^\infty$. This has to do with the (formal) product rule: 
$$\nabla (uv) = u\nabla v+v\nabla u$$
If the factors $u,v$ are bounded, they don't worsen integrability; otherwise they may.  
A concrete example is obtained by observing that $u(x) = |\log(1/x)|^p$ is in $H^1$ when  $0<p<1$ but not in $H^1$ for $p\ge 1 $.  So, any $p\in [1/2,1)$ yields $u\in H^1$ and $u^2\notin H^1$.
A: Let me add the counterexample I gave in the comments for completeness and with further explanations:
Following Alt (Funktionalanaysis, section 8.7), we have the charaterization:
$$|x|^\rho \in H^{m,p}(B_1) \iff \rho > m -\frac{n}{p} \tag{1}$$
From this characterization it is easy to obtain counterexamples. We are interested in the case $m=1$ and $p=2$.
For simplicity we consider $n=3$. Thus $$|x|^\rho \in H^{1,2}(B_1) \iff \rho >-\frac{1}{2}\tag{2}.$$
Consequently $f(x)=|x|^{-\frac{1}{4}}\in H^{1,2}$. However, $f^2=|x|^{-\frac{1}{2}}\not\in H^{1,2}(B_1)$ due to $(1)$ and $(2)$. In particular, these functions are both not in $L^\infty(B_1)$.
With the same idea, counter examples can be constructed for any $n\geq 3$
