zero set of a bounded variation function on $\mathbb{R}^3$

62 Views Asked by At

Suppose $f(t,s,u)$ is a continuous function on $[0,1]^3$. Suppose $f$ is monotone with respect to $t$, and monotone with respect to $u$. Suppose for each $t,u$ fixed, $f$ is of bounded variation $\leq M$ with respect to $s$. Assume the zero set $Z$ of $f$ is a hypersurface. What can we say about the zero set of $f$? Can we say anything about the 2-dimensional Hausdorff measure of its zero set? Are there any related theorems?

The reason I ask this is that, if $f$ has a zero set equal to a hyperplane, then we can conclude that the Whitney decomposition of $[0,1]^3\backslash Z =\cup B_r$ satisfies that the number of Whitney cubes of sidelength $2^{-k}$ is $\sim 2^{2k}$, and therefore we have $$ \iiint_{[0,1]^3} \frac{1}{\text{dist}\,(x,Z)^\sigma}\,dx \sim \sum \iiint_{B_r} \frac{1}{r^\sigma} = \sum_k \sum_{r=2^{-k}} r^{3-\sigma} \sim \sum_k 2^{2k} 2^{-3k+\sigma k} =\sum_k 2^{-(1-\sigma)k} \lesssim_\sigma 1, $$ for any $\sigma <1$. I want to achieve a similar (possibly weaker) result for a zero set that is a hypersurface (possibly under extra conditions).

The Whitney decomposition theorem that I read is on Page 167-170 in Singular Integrals and Differentiability Properties of Functions by Stein.