integration by parts in Sobolev space $W^{1,1}(\mathbb R^d)$

Question

Let $f,g\in W^{1,1}(\mathbb R^d)$ such that: $$ \int_{\mathbb R^d}|\nabla f|\,|g|\,<\infty\,,\quad \int_{\mathbb R^d}|f|\,|\nabla g| <\infty \,.$$ Can I say that: $$ \int_{\mathbb R^d} \nabla\!f\ g \,=\, -\int_{\mathbb R^d} f\;\nabla g \quad?$$ I remember this holds true if $f,g\in W^{1,2}(\mathbb R^d)$. Can this hypothesis be replaced by my weaker hypothesis?

Answer 1

This is not a full answer, since additional assumptions are required.

The answer is yes if we assume also $$ \int |f\,g| \,<\infty \;.$$

Since both $f,g$ have weak derivative, also $fg$ does and $$ \nabla(fg) \,=\, \nabla\!f\ g \,+\, f\ \nabla g \;.$$ Moreover $\nabla(fg)\in L^1(\mathbb R^d)$ thanks to the hypothesis of finite integrals. And $fg\in L^1(\mathbb R^d)$ by the additonal hypothesis.

Now, for $N\in\mathbb N\,$ let $\phi_N\in C^1_c(\mathbb R^d)$ such that $\phi_N(x)=1$ if $|x|\leq N$ and the sequences $\phi_N,\nabla\phi_N$ are uniformly bounded (notice this is possible beacause we are on $\mathbb R^d$). By definition of weak derivative we have: $$ \int\nabla(fg)\,\phi_N \,=\, - \int fg\,\nabla\phi_N $$ and by dominated convergence as $N\to\infty$ we find: $$ \int\nabla(fg) = 0 $$ which proves the thesis.