Weak and strong convergence of a sequence of operators in $L^2(\mathbb{R})$

Question

Let $M_\alpha = sin(\frac{x}{\alpha})f(x)$ a sequence of operators in $L^2(\mathbb{R})$. Prove that $M_\alpha$ does not converge strongly to $0$, but converges weakly to $0$, for $\alpha \to 0$.

Attempt at a solution To prove that it converges weakly to $0$, I thought of using Riemann-Lebesgue's Lemma. I chose a function $g(x) \in C^{\infty}_C(\mathbb{R})$ (a regular function with compact support). Then, I know that $\int sin(\frac{x}{\alpha})f(x)g(x) \leq m\int sin(\frac{x}{\alpha})f(x) \to 0$, with $m= ||g||_{\infty}$. Then, by a density argument, I prove it for $g(x) \in L^2(\mathbb{R})$. Is this correct?

My hardest doubts rely now on the strong convergence: how could I prove that $\int sin(\frac{x}{\alpha})^2 f(x)^2$ does not converge to zero?

Answer 1

Your argument is not entirely right because you act as if all functions involved were positive, which they definitely arent. And you don't need any estimate: Riemann-Lebesgue gives you directly that $$\int_{\mathbb R}\sin\tfrac x\alpha \,f(x)g(x)\,dx\to0.$$

Let $f(x)=x^2$. We have $$\int_{\mathbb R}\sin^2\tfrac x\alpha \,\frac1{x^2}\,dx=\frac1\alpha\,\int_{\mathbb R}\sin^2u\,\frac1{u^2}\,du\to\infty. $$

Answer 2

As it has been elaborated in an answer,you do not need to approximate by compactly supported continuous maps,Riesz representation theorem would do just fine.For strong convergence,it's enough to show that if $\int_{\mathbb{R}}sin^{2}(\frac{x}{a})f(x)^{2}\hspace {0.5mm}d\mu (x)$ converges to $0$ for some $f$ in $L^{2}$,then $f=0$ almost everywhere.To prove this,use the identity $sin^{2}\theta = \frac{1-cos\hspace{0.5mm}2\theta}{2}$.Take $\theta = \frac{x}{a}$,take product by $f(x)^{2}$ on both sides,integrate both sides over $\mathbb{R}$ and use Riemann-Lebesgue lemma as you take the limit.The two integrals weighted by $sin^{2}(\frac{x}{a})$ and $cos(\frac{2x}{a})$ respectively would vanish in the limit,giving you $\int_{\mathbb{R}} f^{2} d\mu \hspace{1mm}= \hspace{1mm} 0$,which implies $f$ is zero a.e.