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*}