TITLE: how wiggly is a generic level set?
QUESTION [7 upvotes]: Typical level sets of smooth real-valued functions are manifolds, so they cannot be fractals. If we coarse grain a bit though, sometimes we get space-filling behavior, eg. every point could be within some epsilon of the level set. Obviously one can find such a level set for any epsilon in any compact region, but for smaller values of epsilon one needs larger derivatives. One way to control the derivatives is to impose a cut-off on the Fourier spectrum. My question asks the relationship between this cut-off and epsilon:
Consider the space of real-valued (smooth) functions $C_\Lambda$ on $\mathbb{R}^n$ whose Fourier components are allowed to be nonzero only up to a frequency $\Lambda$. Let $\epsilon>0$. For which $\Lambda$ can I find an open $U\subset C_\Lambda$ so that the zero set of any element of $U$ gets within $\epsilon$ of any point? (Alexandre Eremenko points out that it's easy to satisfy this in a compact region with arbitrarily small $\Lambda$.)
I'm also interested in this question on the n-torus, where the Fourier decomposition is discrete.
I ask for an open set rather than a single function because I want to study generic functions, so I need some stability. You can use your favorite topology on this space, but it better rule out any neighborhood of the zero function!

REPLY [4 votes]: You can have any epsilon and any $\Lambda$: there is no estimate you ask.
Say in dimension $1$, take a polynomial $p(x)$ whose zeros are $n\epsilon,\; |n|<1/\epsilon$. They are within epsilon of any point of the unit ball. Now
multiply it on any entire function $g$ of exponential type $\delta$, which decreases on the real line faster than this polynomial. Then $pg$ has Fourier spectrum bounded by $\delta$,
and its zero set is $\epsilon$ dense in the unit ball.
To get this in several variables just multiply several such functions depending on one
variable each.
EDIT. In the comment to this answer, you ask about compact manifold, for example a torus.
Then you can estimate your epsilon from below in terms of the upper bound of the spectrum
$\Lambda$. Namely $\epsilon>c/\sqrt{\Lambda}$. Here is a proof suggested by Misha Sodin:
Let $p$ be the point where your function $u$ has maximum modulus. WLOG $u(p)=1$.
Then Bernstein inequality says that $\nabla u\leq c\sqrt{\Lambda}$. It follows that
$u(x)>0$ in a ball of radius at least $c/\sqrt{\Lambda}$.
Bernstein's inequality in the original form applies to trig polynomials in dimension $1$,
and has $c=1$. By applying it to each variable separately we obtain a several-variable version for the torus. But in fact it is true in much more general setting, for balls, spheres,
and any compact manifold, I believe.