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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.2012-07-15
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    @copper.hat: Yes, in fact $\|f_\xi\|_\infty = \|\xi\|_2^2$ by the CSB inequality.2012-07-15
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    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., http://www.ms.uky.edu/~rbrown/courses/ma773/notes.pdf2012-07-15
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    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.2012-07-15
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    @LeonidKovalev: :-) What other symbols suffer your disapproval? Surely the curves in $\xi$ are pleasing to the eye?2012-07-15
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    @copper.hat I don't disapprove of $\xi$ per se, but in the context of Fourier transform I expect it to be the time domain variable. Its upper case form is less pleasing to the eye, particularly in formulas like $\frac{\Xi}{\overline{\Xi}}$.2012-07-15
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    @Leonid: I added some clarification about what I'm trying to do. Is there any reason to be more optimistic about this goal?2012-07-15
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    @LeonidKovalev: Jk. No doubt using $j$ for $\sqrt{-1}$ is bad too :-).2012-07-15
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    Regarding your motivation, you might be interested in Young's inequality: http://en.wikipedia.org/wiki/Young's_inequality#Young.27s_inequality_for_convolutions2012-07-15
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    @PZZ: Unfortunately, with $p=r=1/2$ at that article, Young's inequality just tells me something I already know: namely that convolution with any $g \in L^1$ is a bounded linear operator $L^2 \to L^2$.2012-07-15
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    @Mike If you want $g\notin L^1$ to induce a bounded operator, you need some cancellation in $g$, see [singular integrals](http://en.wikipedia.org/wiki/Singular_integral).2012-07-15
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    @Leonid: +1, that is an interesting article. There is also an article specifically on [singular integral operators of convolution type](http://en.wikipedia.org/wiki/Singular_integral_operators_of_convolution_type). Thinking about it, I agree that $g \notin L^1$ forces working with improper integrals at some point. It is possible to choose a sequence of $\xi$ of 2-norm 1 so that the $f_\xi$ are an approximate unit in $C_0(\mathbb{R})$ for instance by taking increasingly shallow normalized Gaussians. So, my formula for the quadratic form above will probably not be bounded unless $g$ is $L^1$.2012-07-16
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    @Mike If the operator is bounded, the associated form is bounded too. Since you are interested in the bilinear form approach, take a look at this [paper](http://arxiv.org/abs/1010.6184) where the authors revisit the basics of the subject from this viewpoint. They avoid PV integrals.2012-07-16

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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.2012-07-19