I have a fairly simple question. Suppose that $p>n$ and $u,v\in W^{1,p}_0$, how can I prove that the product is also in $W^{1,p}_0$ ? Of course, we have to employ Morrey's inequality. My idea was: $u$ and $v$ have uniformly continuous representatives. We can estimate the Hölder norm of the product in terms of the product of the Hölder norms. We can also use Hölder's inequality to ensure the product has finite $L_p$ norm. The only piece missing is that we have to show that the weak derivative exists. For this we need some kind of nontrivial product rule. And secondly, I am not really sure why Morrey's inequality should help us (this was hinted), since Morrey requires our function to be $W^{1,p}_0$ already. Anyway, thank you all :)
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MickG
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Liealgebrabach
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Or you can proceed by approximation. Take approximating sequences $u_n, v_n\in C^\infty_0$. I have the impression that with your approach you can show that the product $u_nv_n$ is Cauchy in $W^{1,p}_0$. Therefore it converges to something in $W^{1,p}$, and this something can only be the product $uv$. It should work but I am not entirely sure – Giuseppe Negro Apr 18 '16 at 21:06

[Somewhat related](http://math.stackexchange.com/q/314820/8157) – Giuseppe Negro Apr 18 '16 at 21:11

it sounds like a really promising hint. But I somehow can't figure out how to see it is Cauchy in $W^{1,p}$, since the statement is false if $p
– Liealgebrabach Apr 18 '16 at 21:25 
Okay, our emedding operator is injective, and thus, it converges in the Hoelder space, it must converge in the Sobolev space, could that be it? – Liealgebrabach Apr 18 '16 at 21:49

No, that would be too easy. If you are convergent *after* an embedding you might not be convergent before it (take for example the embedding of $C([0,1])$ into $L^2(0,1)$: not all $L^2$ converging sequences converge uniformly, even if each term of the sequence is continuous). You are right that my previous idea is incomplete and needs more work. – Giuseppe Negro Apr 18 '16 at 23:03

true, and we cannot invoke the inverse mapping theorem, since the range of the embedding is no Banach space :( – Liealgebrabach Apr 18 '16 at 23:59