Sobolev $W^{1,1}$ space

Question

I am new in Sobolev spaces and I would like to show that for every $u,v\in W^{1,1}$ holds $D(uv)\in W^{1,1}$ or $uv\in W^{1,1}$. I am a litle confused, what to do. Many thanks for any hints!

Answer 1

The property $D(uv)\in W^{1,1}$ is false in general, as it requires the existence of second derivative by product rule.

The other one is true: The prove is by density of smooth functions. Let $u,v$ be in $C^\infty(0,1)\cap W^{1,1}(0,1)$. In order to estimate $\|uv\|_{L^1}$ we need $u,v\in L^2$ to be able to apply Hoelder inequality: Using $u(x)-u(y) = \int_x^y u'(t)dt$ we have $|u(x)-u(y)| \le \|u'\|_{L^1}$. Squaring this inequality and integrating with respect to $x,y$ yields $$ \int_0^1 \int_0^1 u(x)^2 - 2u(x)u(y) + u(y)^2 \le \|u'\|_{L^1}^2, $$ evaluating the left-hand side leads to $$ 2 \|u\|_{L^2}^2 -2 \left(\int_0^1 u(x)dx\right)^2 \le \|u'\|_{L^1}^2, $$ which proves $\|u\|_{L^2} \le \|u\|_{W^{1,1}}$ for smooth $u$, which carries over by density to all $u\in W^{1,1}$. This also proves $uv\in L^1$ for all $u,v\in W^{1,1}$.

Similarly one can prove $u\in L^\infty$: $|u(x)-u(y)| \le \|u'\|_{L^1}$ implies $$ |u(x)|\le |u(y)| + \|u'\|_{L^1}, $$ integrating wrt $y$, taking sup wrt $x$ yields $\|u\|_{L^\infty} \le \|u\|_{W^{1,1}}$.

For smooth $u,v$, we have $D(uv) =uDv + vDu$, where we get by Hoelder inequality $$ \|D(uv)\|_{L^1}\le \|u\|_{L^\infty}\|Dv\|_{L^1} + \|v\|_{L^\infty}\|Du\|_{L^1} \le 2 \|u\|_{W^{1,1}}\|v\|_{W^{1,1}}. $$ This holds for smooth $u,v$. Using density, it is easy to see that it also holds for $uv,\in W^{1,1}$.

Note that the implication $u,v\in W^{1,1}$ $\Rightarrow$ $u,v\in W^{1,1}$ is only true for one-dimensional domains.