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Let $f : \mathbb R^n\to \mathbb R$ continuous, for which there exist $x,y\in\mathbb R^n$, such that $f(x)f(y)<0$.

Is it true that the Hausdorff dimension of the zero set of $f$ is at least $n-1$?

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  • $\begingroup$ In fact the zero set has topological dimension at least $n-1$, and therefore Hausdorff dimension at least $n-1$. $\endgroup$
    – Gerald Edgar
    Commented Dec 11, 2013 at 22:41

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Yes, it is true. If the zero set has Hausdorff dimension $<n-1$ then almost every line in the direction of a coordinate axis will not intersect the set. This easily follows from the definition of the Hausdorff dimension. Then you connect your $x$ and $y$ by a piecevise linear curve which does not intersect the zero-set.

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IT seems easy to adapt in your case the nice construction of Sergei Ivanov of this related question inside the sphere to your case. And the answer would be yes

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