Let $n>1$, and let $f:\mathbb{R}^n \to \mathbb{R}$ be a real-analytic function which is not identically zero.
Does $\dim_{\mathcal H}(f^{-1}(0)) \le n-1$? here $\dim_{\mathcal H}$ refers to the Hausdorff dimension. (I have read this claim in a paper, but there was no reference).
I know that $f^{-1}(0)$ has Lebesgue measure zero.
If this is false, is it true then that $\dim_{\mathcal H}(f^{-1}(0)) < n$?
Any reference would be appreciated.
I would like to propose an elementary approach based on the implicit function theorem.
Consider $F_0=f^{-1}(0)$. By the implicit function theorem if $\nabla f(x)\neq 0$ for some $x\in F_0$ then $F_0$ is a graph locally around $x$ and thus of dimension $n-1$ (actually a smooth submanifold).
Consider then the exceptional set $F_1=\{x:f(x)=0\wedge \nabla f(x)=0\}$. If $\nabla^2 f(x)\neq 0$ for some $x\in F_1$ then in particular $\nabla (\partial_i f)(x)\neq 0$ for some $i\in \{1,\ldots ,n\}$ and thus $F_1\subset \{x:\partial_if(x)=0\}$ which is a graph locally around $x$ and thus of dimension $n-1$.
The exceptional set is now $F_2=\{x:f(x)=0\wedge \nabla f(x)=0\wedge \nabla^2 f(x)=0\}$. Continuing this way the exceptional set $F_k$ where all derivatives up to order $k$ vanish is contained in a smooth submanifold of dimension $n-1$. What remains is the set of those points where all derivatives of all orders vanish, which is empty by assumption (otherwise by analyticity the function is identically zero).
Yes, this is true and follows, e.g. from Łojasiewicz's stratification theorem: Every real-analytic subset of $R^n$ is a locally finite (hence, countable) union of pairwise disjoint smooth real-analytic submanifolds. Take a look for instance here for references and generalizations:
A. Parusinsky, Lipschitz stratification of subanalytic sets Annales scientifiques de l’É.N.S. 4e série, tome 27, no 6 (1994), p. 661-696
Having problems typing today, I'm going to write $Z=f^{-1}(0)$.
The answer is yes if $n=1$, since then $Z$ is discrete.
Hence in general $Z\cap L$ has dimension $0$ for every line $L$. Does that imply $\dim(Z)\le n-1$? Surely that implication is well known if true -- I can imagine it could follow from Frostman's lemma maybe...
