0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.