What is the operator norm of the left shift operator $T_t:\phi(x) \mapsto \phi(x+t)$ in $L^2(\mathbb{R},e^{-x^2}dx)$?

Question

My argumentation would be that $$\Vert T_t \Vert = \sup_{f \in L^2(\mathbb{R},e^{-x^2}dx), f \neq 0} \frac{\Vert T_tf \Vert}{\Vert f \Vert} = \left( \frac{\int_{\mathbb{R}} |f(x+t)|^2 e^{-x^2}dx}{\int_{\mathbb{R}} |f(x)|^2 e^{-x^2}dx} \right) = \infty$$ because $L^2(\mathbb{R},e^{-x^2}dx)$ contains all functions whose square does not increase faster than the exponentiated square decreases, so we can essentially choose a just slightly less quickly increasing function, which - when shifted - blows up the $e^{-x^2}$ term, which leads the integral to diverge. Is this a correct (albeit vague and not very formal) mathematical argument? Does this mean that the $T_t$ operator does not converge to any operator even in the weak topology?

Answer 1

As far as I can see $T_t$ is not even a map on $L^{p}(e^{-x^{2}}dx)$. If $f(x)=\frac 1 {\sqrt {1+x^{2}}}e^{x^{2}/2}$ then $f \in L^{2}(e^{-x^{2}}dx)$ but $f(x+t) \notin L^{2}(e^{-x^{2}}dx)$ for any $t >0$.