Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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., ms.uky.edu/~rbrown/courses/ma773/notes.pdf –  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
1  
@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

1 Answer 1

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 $f = \sum_{n=1}^\infty f_n$.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.