Does $a_n \in H^1(\mathbb R^n)$ and $b_n \rightharpoonup 0$ in $H^1(\mathbb R^n)$ imply $\langle a_n, b_n \rangle \to 0$?

Question

I have a question mainly in functional analysis.

Suppose that $H^1(\mathbb R^n)$ is the standard Sobolev space that we all know. My question is as follows:

Does $a_n \in H^1(\mathbb R^n)$, $|a_n| <1$, $a_nb_n \in L^1(\mathbb R^n)$, and $b_n \rightharpoonup 0$ weakly in $H^1(\mathbb R^n)$ imply $\int_{\mathbb R^n} a_n b_n dx \to 0$?

Thank you in advance.

Answer 1

No.

Let $b$ be a smooth function with support of $b$ included in the unit ball. Assume that $\|b\|_{H^1}=1$. Then define $b_n$ to be translates of $b$: $$ b_n(x) = b(x + 2n e_1). $$ Then $b_n\rightharpoonup 0$ in $H^1$, but $b_n$ does not converge stronlgy to zero in $H^1$ or $L^2$. Set $a_n:=b_n$. Then $\|a_n\|_{H^1}\le 1$, $a_nb_n\in L^1$, but $\int_{\mathbb R^n} a_nb_n = \|b\|_{L^2}^2$ converges not to zero.

Answer 2

I think it is not true. The following should work. Take $g_n:=a_n = b_n \in C_c^\infty$ with $\text{supp} g_n \in [n,n+1]$ and $0\leq g_n \leq 1$ and $g_n = 1$ on $[n+\frac 14, n+\frac 34]$.