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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.

Ian
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    Do you know that the proof of the product rule states that $uv\in W^{1,1}(B)$? Did you experiment with the $x^{\alpha}$ and $\log\log(x)$ functions on $B_1(0)$? – Quickbeam2k1 Mar 07 '15 at 20:56
  • @Quickbeam2k1 I'm pretty sure that just $x^\alpha$ won't do it, because a single $x^\alpha$ will blow up for a range of $p$ from infinity down to a minimal $p$, including that minimal $p$. But we can fix that with a series. (I answered such a question, actually: http://math.stackexchange.com/questions/1039064/f-in-l1-but-f-not-in-lp-for-all-p-1/1039097#1039097) This was my idea for an example: take $u(0)=0$ and $\nabla u(x)=\left ( \frac{f(1-|x|)}{1-|x|} \right )^{1/2} \frac{x}{|x|}$, where $f$ is a function which is in $L^1[0,1]$ and not any higher $L^p$. – Ian Mar 07 '15 at 21:21
  • @Quickbeam2k1 Polar coordinate integration shows $\nabla u \in L^2$, so now I just need to be able to prove that $u \not \in L^\infty$. – Ian Mar 07 '15 at 21:23
  • In the german Funktionalanalysis book of Alt (section 8.7), you can find the following $f(x)=|x|^\rho\in H^{m,p}(B_1)\iff \rho > m- n/p$. Choosing $m=1$, $n=3$ and $p=2$ yields $\rho >-1/2$. Thus it is allowed to choose $\rho =-1/4$. However, with this $\rho$ and the $\rho$ characterization, we find $f\cdot f=f^2\not\in H^{1,2}(B_1)$. I think for $n=2$ we need $\log\log$-type functions – Quickbeam2k1 Mar 07 '15 at 21:40
  • @Quickbeam2k1 Well, $u=log(log(1+1/|x|))$ would fail to be in $L^\infty$ but would indeed be in $H^1$ (we have already done this calculation in my class). So the question more or less boils down to whether $\nabla(\log(\log(1+1/|x|)))$ fails to be in $L^p$ for any $p>2$. – Ian Mar 07 '15 at 22:01
  • @Quickbeam2k1 I made a typo in my other example, I meant $\nabla u(x) = \left ( \frac{f(1-|x|}{|x|} \right )^{1/2} \frac{x}{|x|}$, with $f \in L^1[0,1]$ and $(\forall p > 1) , f \not \in L^p[0,1]$. Then $\nabla u \in L^2$ but $\nabla u \not \in L^p$ for $p>2$. Doing it this way we find that $u$ is radial: $u(x)=u(0)+\int_0^{|x|} \left ( \frac{f(1-y)}{y} \right )^{1/2} dy$. Is it obvious that this blows up as $|x| \to 1$? If it does then we should be good, since $u$ and $\nabla u$ are definitely big "in the same place". – Ian Mar 08 '15 at 02:58
  • @Quickbeam2k1 A concrete example of $f$ is $f(x)=\sum_{n=1}^\infty 4^{-n} x^{2^{-n}-1}$. – Ian Mar 08 '15 at 03:03

2 Answers2

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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 $u^2\notin H^1$.

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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 $n\geq 3$

Quickbeam2k1
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