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