The derivation of the product in Sobolev space

Question

Let $f \in W^{1,\infty}(0,T)$ and $g \in W^{1,1}(0,T)$ $(0<T\leq\infty)$.

Is $(fg)'= f'g+fg'$ true?

Thanks

Answer 1

Yes.

Such a statement will usually be proven by smooth approximation. In the reflexive range, so the corresponding statement for $f \in W^{1,p}(0,T)$ and $g \in W^{1,p'}(0,T)$ with $1 < p < \infty$, this is fully straight forward since smooth functions on $[0,T]$ are dense. This is however not the case for $W^{1,\infty}(0,T)$, so a bit more must be done.

One way could be like this. Extend $f$ and $g$ to functions $F \in W^{1,1}(\mathbb{R})$ and $G \in W^{1,\infty}(\mathbb{R})$ by your favorite reflection or so. Then pick a standard mollifier family $(\varphi_n)$ and set $f_n := (\varphi_n \star F)$, and $g_n$ analogously. Then (the restrictions to $(0,T)$ of) $f_n,g_n$ are smooth on $[0,T]$, with $f_n \to f$ and $g_n \to g$ in $W^{1,1}(0,T)$ and also each (by embedding) uniformly on $[0,T]$. Moreover, $$\|g_n'\|_{L^{\infty}(0,T)} \leq \|g_n'\|_{L^{\infty}(\mathbb{R})} \leq \|G'\|_{L^{\infty}(\mathbb{R})}$$ (this is a standard mollifier property), so there is a weakly-$\star$ convergent subsequence w.r.t. the pair $L^1(0,T)^\star = L^\infty(0,T)$ of $(g_n')$ (Banach-Alaoglu theorem) which we do not relabel. Since $g$ is the uniform limit of $(g_n)$, the weak-$\star$ limit of $(g_n')$ must also be $g'$. But then \begin{multline*}(fg)(t) - (fg)(0) \quad \longleftarrow \quad (f_ng_n)(t) - (f_ng_n)(0) \\ = \int_0^t \bigl[f_n'g_n + f_ng_n'\bigr] \quad \longrightarrow \quad \int_0^t \bigl[f'g + fg'\bigr]\end{multline*} for every $t \in (0,T)$. In particular, $fg \in W^{1,1}(0,T)$ and the weak derivative is $$(fg)' = f'g + fg'.$$