Why do this sequence converges weakly but not strongly?

Question

Let $f_n(x)=\sin(nx)g(x)$ a sequence in $L^2(\mathbf{R})$, where $g(x)\in L^2(\mathbf{R})$ is such that $g(x)>0$. I have to prove that $$f_n(x)\rightharpoonup 0 \ \ \ \mbox{weakly}$$ but $$f_n(x)\nrightarrow 0 \ \ \ \mbox{strongly}.$$

For the weak convergence I have to show that for every $h\in L^2(\mathbf{R})$ $$\int_{-\infty}^{+\infty}\sin(nx)g(x)h(x) dx \to 0 $$ but I don't know how can I procede.

Even for the strong convergence, i.e. for the limit $$\lim_{n\to\infty}\int_{-\infty}^{+\infty}\left[\sin(nx)g(x)\right]^2 dx$$ I ask you for a hint.

Answer 1

Hint: For the weak convergence, approximate $g \cdot h$ in the $L^1$-sense with a smooth function that has a compact support. Then use partial integration.

For the strong convergence, I currently have no idea.

Answer 2

We will use the following: if $h$ is an integrable function, then $$\tag{*}\lim_{n\to +\infty}\int_{\mathbb R}\sin\left(nx\right)h\left(x\right)\mathrm dx=0=\lim_{n\to +\infty}\int_{\mathbb R}\cos\left(nx\right)h\left(x\right)\mathrm dx.$$ This can be proved by following the lines in Dominik's answer. An alternative way is to do it when $h$ is the indicator function of a bounded open interval, then when $h$ is the indicator of an open set, a Borel set, then when $h$ is a simple function.

In order to prove that the convergence to $0$ does not hold in the strong sense, use the trigonometric identity $$\sin^2\left(nx\right)=\frac 12-\frac 12\cos\left(2nx\right)$$ and the second inequality in (*).