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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.

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It must be in $L_{\infty}$ since continuous and vanishing. – copper.hat Jul 15 '12 at 22:23
@copper.hat: Yes, in fact $\|f_\xi\|_\infty = \|\xi\|_2^2$ by the CSB inequality. – Mike F Jul 15 '12 at 22:27
I refuse to use $\xi$ for a function. Applying Fourier transform, we get something like $f_g(t)=\int \hat g e^{it}\bar {\hat g}$ which is something like $\widehat{|\hat g|^2}$. (I refuse to keep track of the signs and conjugates, too.) So your question becomes a question about the decay of the Fourier transform of an $L^1$ function. "It is a difficult and unsolved problem to describe the image of $L^1$ under the Fourier transform" E.g., – user31373 Jul 15 '12 at 22:32
Also note that $f_{\xi}(-x)$ is the convolution of $\xi(x)$ and $\xi(-x)$. If you had other assumptions, you could use Young's inequality. – Yury Jul 15 '12 at 22:47
@Mike If you want $g\notin L^1$ to induce a bounded operator, you need some cancellation in $g$, see singular integrals. – user31373 Jul 15 '12 at 23:22
up vote 1 down vote accepted

It can decay arbitrarily slowly. That is, let $g$ be any decreasing continuous function on $[0,\infty)$ with $\lim_{s \to \infty} g(s) = 0$. Then there is $f \in L^1({\mathbb R})$ such that $|\hat{f}(s)| \ge g(|s|)$ for all real $s$.

By scaling we may assume $g(0) = 1$. Take $s_n$ so that $g(s_n) = 1/n^2$. Thus $s_1 = 0$, and on $[s_n, s_{n+1}]$ we have $g(s) < 1/n^2$. Consider $f_n(x) = \dfrac{s_{n+1}}{n^2} \exp(-s_{n+1} |x|)$ which is in $L^1$ with $\| f_n \|_1 = 2/n^2$ and has $\widehat{f_n}(s) = \dfrac{2}{n^2} \dfrac{s_{n+1}^2}{s_{n+1}^2+s^2}$. So $\widehat{f_n}(s) > 0$ everywhere and $\widehat{f_n}(s) \ge 1/n^2 \ge g(|s|)$ for $s_n \le |s| \le s_{n+1}$. Now take $f = \sum_{n=1}^\infty f_n$.

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Thank you @Robert, this confirms my suspicion that I was on a wild goose chase. – Mike F Jul 19 '12 at 3:23

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