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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!

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    $\begingroup$ A simple upper bound is $\Lambda = O(\epsilon^{-1})$. Take $f(x) = \sin (2\pi x_1 \epsilon^{-1})$. It has a single Fourier component of size $\approx \epsilon^{-1}$, and its zero set is within $\epsilon/2$ of any point of $\mathbb{R}^n$. Furthermore since its derivative does not vanish on said level set, for any "sufficiently small" perturbations the level sets move only a tiny bit (implicit function theorem). For the implicit function theorem to apply, you can use a very coarse topology (that of $C^k$ functions for example). I suppose you are asking whether this is sharp? $\endgroup$
    – Willie Wong
    Commented Apr 30, 2015 at 7:49
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    $\begingroup$ BTW, contrary to the question in the title, the level sets of the functions given in my first comment are close to flat. $\endgroup$
    – Willie Wong
    Commented Apr 30, 2015 at 8:32
  • $\begingroup$ Willie Wong is right. Your restriction on the spectrum implies that your function extends to an entire function of exponential type at most $\Lambda$. This helps in proving that $\epsilon$ cannot be much smaller than $c/\Lambda$, where $c$ is a constant which may depend only on dimension. $\endgroup$
    – Alexandre Eremenko
    Commented Apr 30, 2015 at 11:36
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    $\begingroup$ You should have a look at Maria Nastasescu images of level sets in her senior thesis. The problem she investigates are closely related to your question. its.caltech.edu/~mnastase/Senior_Thesis.html $\endgroup$
    – Liviu Nicolaescu
    Commented Apr 30, 2015 at 12:54
  • $\begingroup$ @Christian Remling: Sorry I did not notice the unit ball. But then there is no estimate and the answer to the question is no. $\endgroup$
    – Alexandre Eremenko
    Commented Apr 30, 2015 at 21:15

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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.

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  • $\begingroup$ Thanks. I'd like to ask the nontrivial question then, without reference to the unit ball or perhaps where everything is restricted to a compact region. $\endgroup$
    – lostinloops
    Commented May 1, 2015 at 1:01
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    $\begingroup$ There is no difference between a unit ball, any ball or compact region. $\endgroup$
    – Alexandre Eremenko
    Commented May 1, 2015 at 3:49
  • $\begingroup$ I'm saying use something compact instead of $\mathbb{R}^n$, such as a torus, with discrete Fourier spectrum. $\endgroup$
    – lostinloops
    Commented May 1, 2015 at 16:36
  • $\begingroup$ On compact you have Courant's theorem. Everyting is controlled by $\sqrt{\lambda}$ where $\lambda$ is your largest allowed eigenvalue. $\endgroup$
    – Alexandre Eremenko
    Commented May 1, 2015 at 21:24
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    $\begingroup$ @ragnar: the eigenvalue of $y(x)=\sin(kx)$ is $\lambda=-k^2$. I mean in the sense $y''=\lambda y$. So it is just the question of notation: what you call spectrum, the set of $k$ or the set of $k^2$. $\endgroup$
    – Alexandre Eremenko
    Commented May 7, 2015 at 0:43

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