I am faced with the following task:
a) Let $T_t$ be the operator $T_t:\phi(x) \mapsto \phi(x + t)$ on $L^2(\mathbb{R})$. What is the norm of $T_t$? To what operator does $T_t$ converge as $t \to \infty$ and in what topology?
b) Answer the same question for $T_t$ if the Hilbert space is $L^2(\mathbb{R},e^{-x^2}dx)$.
For a), it is clear to me that $\Vert T_t \Vert = 1$ and, based on the result for $\ell^2$, I would assume that $T_t \overset{w}{\to} T$, i.e. $T_n$ converges to the zero element in the weak operator topology. However, I do not know how to show this formally as I do not know how to transfer the arguments from $\ell^2$, which are based on the Schwarz inequality using $$| \langle y , T_nx \rangle | = | \langle S_ny , T_nx \rangle | \leq \Vert S_ny \Vert \Vert T_nx \Vert = \Vert S_ny \Vert \Vert x \Vert \to 0,$$ where $T_n$ is the right shift operator and $S_n$ the operator that sets the first $n$ elements to zero. And I don't know what the additional requirement in b) on the square-integrability under - essentially - a standard normal distribution changes. Can any of you please help me in these questions, please?
Here is an answer to (a). $L^2$ functions vanish at infinity in the following sense: for all $\epsilon>0$ there is $M>0$ such that $$ \int_{\mathbb R \setminus [-M,+M]} \phi^2 \le \epsilon. $$ Now take $\psi\in C_c(\mathbb R)$. Then the support of $\psi$ is contained in some intervall $[-R,R]$.
Take $\phi \in L^2$, $\epsilon>0$ and $M$ as above. Then $$ \int T_t\phi \psi \le \epsilon \|\psi\|_{L^2} $$ for all $t>M+R$. So $\int T_t\phi\psi\to0$ for all $\psi\in C_c(\mathbb R)$. Since these smooth functions are dense in $L^2(\mathbb R)$ this implies $T_t\phi \rightharpoonup 0$.