Let $\delta, \epsilon>0$ and $f: \mathbb{T} \rightarrow \mathbb{C}$ such that (i) $f(0)=1$ and (ii) $|f(x)| \le \epsilon$ for all $|x| > \delta$. What is the smallest $L = L(\delta, \epsilon)$ (asymptotically) such that there exists such a $f$ with $\widehat{f}(m) =0$ for all $m \not \in [-L,L]$.
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$\begingroup$ Do you want any other assumptions such as positivity of $f$, or $\int |f(x)| dx = 1$? $\endgroup$– Yemon ChoiCommented Mar 15, 2016 at 22:49
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$\begingroup$ No, I am happy with any complex valued $f$ (and I do not require its $\ell_1$ norm to be 1). I just want to know what kind of $L(\delta, \epsilon)$ might be achievable. $\endgroup$– Anindya DeCommented Mar 15, 2016 at 23:24
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It suffices to approximate uniformly a function $g$ with $g(0)=1$, $g(x)=0$ for $|x|>\delta$ by a trigonometric polynomial $p_L$ of degree $\le L$, with error $\le \epsilon/2$. By Jackson's Theorem, the error can be made $$ \|g-p_L\|_{\infty} \le \frac{C_n}{L^n} \|g^{(n)}\|_{\infty} . $$ For the $g$ we are interested in, we will have that $\|g^{(n)}\|_{\infty}\lesssim \delta^{-n}$. It follows that there will be an approximation with the desired properties if $$ L \gtrsim \frac{1}{\delta \epsilon^{1/n}} , $$ and here $n\ge 1$ is at our disposal (the implied constant will of course depend on $n$; more precise information on this is in principle available).
The argument was probably too simple to give an optimal bound, but on the other hand, we're not off by very much because the uncertainty principle suggests that you will have to take $L\gtrsim 1/\delta$.
answered Mar 16, 2016 at 0:37
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$\begingroup$ Thanks! Yeah, I was worried about the dependence on $\epsilon$, but it seems like by pushing n to be large, one can get a fairly good dependence on $\epsilon$. $\endgroup$ Commented Mar 16, 2016 at 16:07