Let $f:B_r(0)\rightarrow\mathbb{R}$ with $B_r(0)\subset \mathbb{R}^2$ the open Ball of radius $r>0$. Let us further denote the set of all zeros of $f$ by $$M:=\{x\in B_r(0):\ f(x)=0\}.$$ Let us also assume that the (outer) Lebesgue measure of that set is positiv, i.e. $$\mathcal{L}^2(M)>0.$$ Now my question is the following: If we impose some kind of regularity on $f$, for example continuity, differentiability or Hölder continuity, can we find an inner point of $M$?

Without any regularity this would probably be false, because we could construct something with the fat Cantor set (see e.g. https://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set), but I'm not sure how we could rule out such behaviour.

  • $\begingroup$ If $E$ is any closed set of measure $0$ then $x \to d(x,E)$ is a Holder continuous (hence absolutely continuous) function whose zero set is exactly $E$. I believe we can even construct a $C^{\infty}$ function whose zero set is $E$ but I don't have a proof at this moment. $\endgroup$ – Kavi Rama Murthy Sep 25 '18 at 9:14
  • $\begingroup$ @Kavi Rama Murthy : A very good example, didn't think of that one. By any chance, did you mean that the measure of $E$ is positive, because for the Hölder continuity we should only need that $E$ is closed? $\endgroup$ – humanStampedist Sep 25 '18 at 9:24
  • $\begingroup$ Yes, I meant measure greater than $0$. Sorry for the error. $\endgroup$ – Kavi Rama Murthy Sep 25 '18 at 9:27

The answer is no. First note that for any closed set $E \subset \mathbb{R}^n$ there is a smooth function $f \colon \mathbb{R}^n \rightarrow [0,\infty)$ with zero set $f^{-1}(0) = E$, see e.g. here. Now take a 'fat' cantor set $C$ with positive measure and define $E = C^n$ - by rescaling we may also suppose that $E \subset B_r(0)$. By the previous remark, there exists a smooth function $f$ with zero set $E$. However, $E$ has no inner points.

Only in the case of analytic functions we can deduce that $f \equiv 0$, because the set of zeros is always countable and discret.


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