Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Say I have a compactly supported $C^1$ function $f:\mathbb{R} \to \mathbb{R}$. Let $R>0$. Let $\nu$ be some reasonable measure on $\mathbb{R}$ -- take, for instance, (a) $d\nu(t)=dt$ or (b) $d\nu(t)=e^{-t}$ for $t>0$ and $d\nu(t)=0$ for $t\leq 0$.

Let $\delta(R)$ be the minimum of $|f-\widehat{g}|_2 = \left( \int_\mathbb{R} |f(t)-\widehat{g}(t)|^2 d\nu(t)\right)^{1/2}$ over all functions $g:\mathbb{R} \to \mathbb{C}$ supported on $\lbrack -R,R\rbrack$.

What is $\delta(R)$? How fast does it decrease as $R\to \infty$? Given $R$, can one construct a $g$ that attains the minimum?

(A variation on the same question: allow measures, not just functions $g$, supported on $\lbrack -R,R\rbrack$.)

Update: for $d\nu(t) = t$ this is very easy by isometry, as mentioned below; the minimum is attained for $g$ equal to the restriction of $\widehat{f}$ to $\lbrack -R,R\rbrack$ -- and so, if $f$ is in $C^k$, $\delta(R)$ decreases at least as fast as $1/R^{k-1}$ as $R\to \infty$. I am really more interested in the answers for the measure $\nu$ given in (b) above.

share|improve this question
2  
Really? $\widehat g$ is entire if $g$ is compactly supported, so we do not have any choice at all as to what to take for $g$ and most of the time we have no possibility to find such $g$ with any $R$. I hope you meant something else or I just misunderstood what is written... –  fedja Nov 4 '12 at 14:05
    
Hm, I guess I really had (b) in mind. Let me rewrite this... –  H A Helfgott Nov 4 '12 at 15:55
    
All right, rewritten. Thanks. –  H A Helfgott Nov 4 '12 at 16:15
2  
Maybe I'm missing something, but in case (a) (Lebesgue measure), $|f-\widehat{g}|_2 = |\widehat{f}-g|_2$ by Plancherel, so shouldn't you just take $g$ to be the restriction of $\widehat{f}$ to $[-R,R]$? –  Henry Cohn Nov 4 '12 at 17:36
    
I was about to say, yes, for (a) this is trivial by isometry. But what can one say about (b), say? –  H A Helfgott Nov 4 '12 at 20:10
show 11 more comments

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.