Weak Convergence of Shift Operator on $L^2(\mathbb{R})$

Question

Let $T_t$ be the operator $T_t: \varphi(x) = \varphi(x+t)$ on $L^2(\mathbb{R})$. To what operator does $T_t$ converge as $t \to \infty$ and in what topology?

I believe that $T_t$ should converge weakly to $T = 0$. I have outlined my proof below, but I am unsure if the steps are logical (particularly the limits).

For each functional $\ell \in L^2(\mathbb{R})^*$, the Riesz Representation Theorem implies that there exists $\psi \in L^2(\mathbb{R})$ such that $\ell(\phi) = (\psi, \phi)$ for all $\phi \in L^2(\mathbb{R})$. Then, \begin{align*} |\ell(T\varphi) - \ell(T_n\varphi)| &= |(\psi, T\varphi) - (\psi, T_n\varphi)| \\ &= |(\psi, 0) - (\psi, T_n\varphi)| \\ &= |(\psi, T_n\varphi)| \\ &= \left|\int_{-\infty}^{\infty} \overline{\psi(x)} \phi(x+n) dx\right| & (\text{Defintion of Inner Product}) \\ &= \left|\int_{-\infty}^{-m} \overline{\psi(x)} \phi(x+n) dx\right| + \left|\int_{-m}^{\infty} \overline{\psi(x)} \phi(x+n) dx\right| \\ &\leq \int_{-\infty}^{-m} \left|\overline{\psi(x)} \phi(x+n)\right| dx + \int_{-m}^{\infty} \left|\overline{\psi(x)} \phi(x+n)\right| dx & (\text{Triangle Inequality}) \\ &\leq \left[\int_{-\infty}^{-m} \left|\overline{\psi(x)} \phi(x+n)\right|^2 dx\right]^{1/2} + \left[\int_{-m}^{\infty} \left|\overline{\psi(x)} \phi(x+n)\right|^2 dx\right]^{1/2} \\ &\leq \left[\int_{-\infty}^{-m} \left|\overline{\psi(x)}\right|^2 \|\phi\|^2 dx\right]^{1/2} + \left[\int_{-m}^{\infty} \|\psi\|^2\left|\phi(x+n)\right|^2 dx\right]^{1/2} & (\text{$\psi, \phi$ bounded}) \\ &= \|\phi\| \left[\int_{-\infty}^{-m} \left|\overline{\psi(x)}\right|^2 dx\right]^{1/2} + \|\psi\| \left[\int_{-m+n}^{\infty} \left|\phi(x)\right|^2 dx\right]^{1/2} \\ &\xrightarrow[n \to \infty]{} \|\phi\| \left[\int_{-\infty}^{-m} \left|\overline{\psi(x)}\right|^2 dx\right]^{1/2} + \|\psi\| \cdot 0 \\ &\xrightarrow[m \to \infty]{} \|\phi\| \cdot 0 + \|\psi\| \cdot 0 \\ &= 0. \end{align*}

Answer 1

I think your guess that $T_t$ converges to $0$ in some topology (weak operator topology) is correct.

Note: You say that $T_t$ converges weakly to $0$, but some people call this (what you wanted to show in your proof) weakly operator convergent to $0$.

But in my opinion your proof is not correct: you have to show $\lim_{n\to\infty}|\ell(T\varphi)-\ell(T_n\varphi)|=0$, but your expression also involves the variable $m$ (this variable should be explained somewhere in my opinion). You finish by taking a limit with respect to $m$. I do not see how this directly relates to a proof that $\lim_{n\to\infty}|\ell(T\varphi)-\ell(T_n\varphi)|=0$.

Maybe you can try to avoid the variable $m$ somehow?

In general, if you are in doubt over limits, then you can always go back to the $\forall \varepsilon>0\exists N\in\Bbb N\dots$-style definition of the limit.

Answer 2

To be precise, show that, for every $f\in L^2(\mathbb R)$, given $\varepsilon>0$, there is sufficiently large $t_o$ such that, for every $t\ge t_o$, $\Big|\int f(x)\cdot f(x+t)\;dx\Big|<\varepsilon$. To do so, you'll need to prove that the "tails" of $f$ get small, in a precise sense.