Product of two functions in $H^1$

Question

Can you help me proving the following statement?

Let $u,v \in H^1(I)$ where $I$ is an open interval in $\mathbb{R}$. Prove that $u \cdot v \in H^1(I)$ and $(u\cdot v)^\prime = (u^\prime \cdot v) + (u \cdot v^\prime)$.

Answer 1

Hint: Note that the space $H^1(I)$ is the space of absolutely continuous functions with weak derivative in $L^2(I)$. Hence the product is still absolutely continuous and the Leibniz formula holds.

Hint for higher dimensions: if one of the two functions is assumed to be $C_{c}^{\infty}$, then by Mayer and Serrin (or sometimes it is taken as a definition), for every element $u \in H^{k}$ there exists a sequence of $C^{\infty}$ functions that converge to $u$. Then it’s all about using the properties for smooth functions (which hold) and then pass to the limit.