Given $C \geq 1$ and $\epsilon > 0$, is there a number $N = N(C,\, \epsilon)$ such that the following holds:


For every set $S \subseteq S^1$ of cardinality $C$, there is a function $f: S^1 \to \mathbb{R}$ such that:


1). $f(z) = \sum\_{n = -N}^N a\_n z^n$ (i.e. $f$ is a Laurent polynomial of in degrees $[-N,\, N]$).


2). $f \geq 0$ on $S^1$


3). $f(z) \geq 1$ for each $z \in S$


4). $\|f\|\_{L^1} < \epsilon$



Degree of polynomial approximating characeristic function of finte set
So it suffices to do this for $S=\{1\}$. If you can find a degree $N$ Laurent polynomial doing the job for $\{1\}$ with $L^1$ distance $\epsilon$, $f\_1$ say, then $f\_\zeta(z):=f\_1(\bar \zeta z)$ is another Laurent polynomial of the same degree doing the job for $\{\zeta\}$ with the same $\epsilon$. Suppose you can find $N(1,\epsilon)$ for each $\epsilon>0$, then there is a trivial bound $N(C,\epsilon)\le N(1,\epsilon/C)$: just run the argument above to get an $f\_\zeta$ for each $\zeta$ in $C$ and add them together. The sum has the right lower bounds, and the $L^1$ bound is trivial. 


Finally how to get an $f\_1$ with $L^1$ norm less than $\epsilon$. No doubt there are much better ways than I'm about to suggest, but anyway: take a continuous function $g$ of $L^1$ norm less than $2\epsilon/3$ that is bounded below by $\epsilon/3$ and satisfies $g(1)>1+\epsilon/3$. Using the Féjér kernel, obtain a trigonometric polynomial of some degree approximating $g$ uniformly within $\epsilon/3$. This is your $f\_1$. *Et voilà!*


