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 $m \not \in [-L,L]$. 



extremal kernels for functions on the torus
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,](https://en.wikipedia.org/wiki/Jackson%27s_inequality) 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 $L\gtrsim 1/\delta$.


