Q:

Convergence of the translation operator in $L^2(\mathbb{R}, e^{-x^2}dx)$

For $t \ge 0$, let $T_t : L^2(\mathbb{R}, e^{-x^2}dx) \to L^2(\mathbb{R}, e^{-x^2}dx)$ be the translation operator given by $f(x) \mapsto f(x + t)$. I would like to:

Find the norm of $T_t$.
Determine whether $T_t$ converges to an operator in the weak, strong, or uniform topologies, as $t \to \infty$.

This is a homework problem, so I would appreciate suggestions and hints only.
I have used this previous post, Convergence of translation operator, to help me answer this question when we consider $T_t$ as an operator $L^2(\mathbb{R}, dx) \to L^2(\mathbb{R}, dx)$. In this situation, $||T_t|| = 1$ and $T_t$ just converges weakly. 
But now I am finding the modified problem considerably harder. In fact, I'm even having trouble showing $T_t$ is bounded (i.e., $||f(x + t)||_{L^2(\mathbb{R}, e^{-x^2}dx)} \le C || f(x)||_{L^2(\mathbb{R}, e^{-x^2}dx)}$). Is there some clever way to rewrite 
$$\int_{\mathbb{R}}|f(x + t)|^2 e^{-x^2} dx$$
in terms of just $\int_{\mathbb{R}}|f(x)|^2 e^{-x^2} dx$? I think this is what I need to accomplish to the show the boundedness and get a handle on the norm of $T_t$. Also, any hints about convergence in the various topologies is greatly appreciated.  

A:

Operators $T_t$ are unbounded for $t\neq 0.$ 
We first show that $T_t, t>0$ is not defined on the whole Hilbert space. 
Write
$||T_tf||^2=\int_{\mathbb{R}}|f(x + t)|^2 e^{-x^2} dx=\int_{\mathbb{R}}|f(x)|^2 e^{-(x-t)^2} dx=e^{-t^2}\int_{\mathbb{R}}e^{2tx}|f(x)|^2 e^{-x^2} dx.$
Then $\int_{\mathbb{R}}|f(x)|^2 e^{-x^2} dx<\infty$ does not imply $\int_{\mathbb{R}}e^{2tx}|f(x)|^2 e^{-x^2} dx<\infty$
Take e.g. $f(x)=e^{-tx}e^{x^2/2},x>0,\ f(x)=0,\ x\leq 0.$ Then $f\in L^2(\mathbb R,e^{-x^2}dx)$ but is not in the domain of $T_t.$
Now consider the sequence $f_n(x)=f(x)$ on $(0,n)$ and $0$ otherwise. Then $f_n$ is norm bounded and $||T_tf_n||\to\infty, n\to\infty.$

