convergence of $g_n(x)=f(x)\sin(nx)$ in $L^2(\mathbb R)$ and $L^1(\mathbb R)$

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Let $f_1\in\mathbb L^1(\mathbb R)$ and $f_2\in L^2(\mathbb R)$. I am interested in the convergence (weak and strong) of $g_1(x)=g_1^{(n)}=f_1(x)\sin(nx)$ in $L^1(\mathbb R)$ and $g_2(x)=g_2^{(n)}=f_2(x)\sin(nx)$ in $L^2(\mathbb R)$.

I think we have $g_1\rightarrow 0$ in $L^1(\mathbb R)$ by the Riemann-Lebesgue lemma and therefore also weak convergence.

For the second one, $g_2$, I can't use the lemma. Any hints?

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The Riemann-Lebesgue lemma gives the weak convergence of the sequences (to $0$), but not the strong convergence. Indeed, neither sequence converges strongly, unless $f_k = 0$ almost everywhere.

Let's generally show the weak convergence to $0$ of $g_n \colon x \mapsto f(x)\sin (nx)$ for $f\in L^p(\mathbb{R})$, where $1 \leqslant p < \infty$ (and the weak$^\ast$ convergence in the case $p = \infty$). Let $q$ the conjugate exponent to $p$, and $\varphi \in L^q(\mathbb{R})$. Then $\psi \colon x \mapsto f(x)\varphi(x)$ belongs to $L^1(\mathbb{R})$, and thus by the Riemann-Lebesgue lemma,

$$\int_{\mathbb{R}} g_n(x)\varphi(x)\,dx = \int_{\mathbb{R}} \psi(x)\sin (nx)\,dx \to 0.$$

It is easy to see that the sequence doesn't converge strongly - by the weak convergence, it could only converge to $0$ - if there is a nondegenerate interval $I$ and a $\varepsilon > 0$ such that $\lvert f(x)\rvert \geqslant \varepsilon$ for all $x\in I$. By shrinking the interval if necessary, we can assume that $I = \bigl[\frac{p}{q}\pi, \frac{p+1}{q}\bigr]$ for some $q \in \mathbb{N}\setminus \{0\}$ and $p \in \mathbb{Z}$. Clearly,

$$\lVert g_n\rVert_{L^p(\mathbb{R})}^p \geqslant \int_I \lvert g_n(x)\rvert^p\,dx \geqslant \varepsilon^p \int_I \lvert \sin (nx)\rvert^p\,dx.$$

For $n = mq$, we have

$$\int_{\frac{p}{q}\pi}^{\frac{p+1}{q}\pi} \lvert \sin (mqx)\rvert^p\,dx = \frac{1}{mq} \int_{pm\pi}^{(p+1)m\pi}\lvert \sin y\rvert^p\,dy = \frac{1}{q} \int_{0}^{\pi} \lvert \sin y\rvert^p\,dy,$$

which is a positive constant. The general case can be obtained by approximating $f$ by functions for which such an $I$ and $\varepsilon$ exist, for example continuous functions.