Prove that $T_a$ does not convergent in operator norm.

Question

Let $H:=L^2 (\mathbb{R})$ and for $a>0$,$\,$$f\in H$, $T_a f(x):=\frac{1}{2a}\int _{x-a}^{x+a} f(y)dy$. Then prove that $T_a \not \to id$ as $a\to 0$ in operator norm.

My idea: I proved $T_a$ strongly convergent to $id$ by using Schwartz's inequality and Fubini's theorem. If $T_a$ are compact operator and $T_a$ convergent $id$, $id$ is compact operator. It is a contradiction to dimension of $H$. So, I tried to prove $T_a$ is compact operator, but I can't.

Answer 1

For each $a>0$, notice that $T_a$ is a multiplier operator of $L^2(\mathbb{R})$ whose multiplier is: $$m_a :\mathbb{R}\rightarrow\mathbb{R}, \xi\mapsto\frac{\sin(2\pi a\xi)}{2\pi a\xi},$$ i.e.: $$\forall f\in L^2(\mathbb{R}), T_a(f) = \left(\frac{1}{2a}\chi_{[-a,a]}\right)*f = \mathcal{F^{-1}_\xi}\left(\frac{\sin(2\pi a\xi)}{2\pi a\xi}\right)*f \\ = \mathcal{F^{-1}_\xi}\left(m_a(\xi)\right)*\mathcal{F^{-1}_\xi}(\hat{f}) = \mathcal{F^{-1}_\xi}\left(m_a(\xi)\hat{f}(\xi)\right),$$ where the Fourier transform is normalized in the following manner: $$\forall f\in L^1(\mathbb{R}), \forall \xi \in \mathbb{R}, \mathcal{F}(f)(\xi):=\hat{f}(\xi):=\int_{\mathbb{R}}f(t)e^{-2\pi i \xi t}\operatorname{d}t.$$ Then $T_a-I$ is a multiplier operator of $L^2(\mathbb{R})$ whose multiplier is $m_a-1$. Then, remembering that if $T$ is a multiplier operator of $L^2(\mathbb{R})$ whose multiplier is $m$ it holds that $\|T\|_{2\rightarrow2}=\|m\|_\infty,$ we get that: $$\|T_a-I\|_{2\rightarrow2} =\|m_a-1\|_\infty\ge\lim_{\xi\rightarrow\infty}\left|\frac{\sin(2\pi a\xi)}{2\pi a\xi}-1\right| =1,$$ So: $$\liminf_{a\rightarrow0^+}\|T_a-I\|_{2\rightarrow2}\ge1$$ and then: $$\|T_a-I\|_{2\rightarrow2}\not\rightarrow0, a\rightarrow0^+$$

Answer 2

Let $f_a(x)=\begin{cases} \frac{1}{\sqrt{2a}}, & -a\leq x\leq a\\ 0, & others \end{cases},$ then $\|f_a\|_2=1$, $T_af_a(x)=\frac{2a-x}{2a\sqrt{2a}}$ for $x\in [0,2a]$, and $$\|T_a-id\|^2\geq \|T_af_a-f_a\|_2^2\geq \int_{0}^a|T_af_a(x)-f_a(x)|^2dx=\int_{0}^a\frac{x^2}{(2a)^3}dx=\frac{1}{24}$$

Answer 3

A simpler version of the same argument in another answer: For $a>0$ let $$f_a=a^{-1/2}\chi_{(0,a)},$$ so $||f_a||_2=1$..

It's easy to see that $$T_af_a(x)\ge\frac14 a^{-1/2},\quad(-a/2<x<0).$$ Hence $||f_a-T_af_a||_2\ge c>0$.