Let $H$ be the Hilbert space $L^2(\mathbb{R})$. For $t \in \mathbb{R}$, let $\lambda_t \in B(H)$ be the unitary operator which translates by $t$, that is $(\lambda_t \xi)(s) = \xi(-t +s)$.
For $\xi \in H$, fixed but arbitrary, define $f_\xi$ by $f_\xi(t) = \langle \xi, \lambda_t \xi \rangle.$
It's pretty easy to see that $f_\xi$ is continuous (since the inner product is continuous and the map $t \mapsto \lambda_t:\mathbb{R} \to B(H)$ is strongly continuous) and vanishes at infinity (just translate $\xi$ by some large $t$ where the integral of $|\xi|^2$ over the complement of $[-2t,2t]$ is small). I am curious what else can be said about $f_\xi$. Does it decay quickly enough to be in any $L^p$ space for example?
Motivation: I am wondering for which sort of functions $g : \mathbb{R} \to \mathbb{C}$ does convolution with $g$ define a bounded operator $\kappa_g \in B(H)$. For example, this is the case for $g \in L^1(\mathbb{R})$. The first thing I thought to do was look to see when the quadratic form $\xi \mapsto \langle \xi , \kappa_g \xi \rangle$ might make sense and be bounded. Formally, we have $\langle \xi ,\kappa_g \xi\rangle = \langle \xi, g * \xi \rangle = \int \overline{ \xi(s)} \int g(t)\xi(-t+s) dt ds = \int g(t) \int \overline{\xi(s)} \xi(-t +s) ds dt = \int g(t) \langle \xi,\lambda_t \xi \rangle dt = \int g(t) f_\xi(t) dt$ so it is mandatory for $g$ to be integrable against every function in the collection $\{f_\xi : \xi \in H \}$. It was at this point I became interested in the decay properties of these functions.