Product of test function and function in first order Sobolev space is also in first order Sobolev space

Question

Suppose $1 \leq p \leq \infty$ , $\phi \in C^{\infty}_c (\Omega) $ and $u \in W^{1,p}(\Omega)$ ,where $Ω$ is an open subset of $\mathbb{R}^n$. Prove that $\phi u \in W^{1,p} (Ω)$ and $D_{x_i}(\phi u) = \phi D_{x_i} u + u D_{x_i} \phi $ , where $D_{x_i}$ denotes the weak i-th partial derivative.

I am trying to understand why I need to show the product rule is true for $D_{x_i}(\phi u)$ when the question already states that $\phi$ is infinitely differentiable and $u$ has weak derivatives. Am I asked to show the product rule is valid for the function $\phi u$ or am I missing something? Thanks for any help.

Edit : Follow up question on how does the weak product rule imply $\phi u \in W^{1,p} (\Omega)$ ?

Answer 1

You don't necessarily know that the "weak" product rule holds. So, you need to show that if $\psi\in C_c^\infty(\Omega)$ and $D_{x_i}u=v,$ then $$\int\limits_{\Omega}\phi uD_{x_i}\psi\, dx=-\int\limits_{\Omega}(\phi v+uD_{x_i}\varphi)\psi\, dx.$$ I used here that the weak derivative of something smooth is the same as its strong derivative.

Answer 2

Still, perhaps one should not shy away from seeing how easy the product rule is to prove: with test functions $\varphi,f$ and distribution $u$, $$ (\varphi u)'(f) \;=\; - (\varphi u)(f') \;=\; -u(\varphi f') \;=\; -u((\varphi f)'-\varphi'f) \;=\; u'(\varphi f)+u(\varphi' f) $$ $$ \;=\; (\varphi u')(f) + (\varphi' u)(f) \;=\; (\varphi u'+\varphi'u)(f) $$ Quite innocent! :)

And, of course, the same pattern holds for partial derivatives in several variables.