Let $f: S^{n-1}\to \mathbb{R}$ be a continuous function ($S^{n-1}\subset \mathbb{R}^n$ is the unit sphere), $f(a)>0$ and $f(b)<0$ for certain points $a,b\in S^{n-1}$. By continuity these inequalities hold in $B_r(a)\cap S^{n-1}$ and $B_r(b)\cap S^{n-1}$ for a small radius $r$. Let $v:=\mathcal{H}^{n-2}(\partial (B_r(a)\cap S^{n-1}))=\mathcal{H}^{n-2}(\partial (B_r(b)\cap S^{n-1}))$ (taking the boundary with respect to the $S^{n-1}$-topology; the $n-2$-dimensional Hausdorff measure $v$ is roughly the surface area of the $n-2$-sphere of radius $r$ for small $r$).

Is it true that $\mathcal{H}^{n-2}(\{f=0\}) \ge v$, and how would one prove this? It seems geometrically obvious; the picture I have in mind looks as follows: Imagine the 2-sphere and two points $a,b$ on it. The zero set of $f$ is at least a curve which surrounds both $a$ and $b$, and the least length for such a curve is $v$, which is the length of the boundary of the neighbourhoods of $a$ and $b$ where $f$ is known to be positive/negative.


2 Answers 2


Yes. Here is an elementary proof.

Construct a distance non-increasing retraction $p:S^{n-1}\setminus B_r(a)\setminus B_r(b)\to \partial B_r(a)$. (For example: divide the sphere into two-hemispheres by the hyperplane $H$ of symmetry between $a$ and $b$. In the hemisphere containing $a$, let $p$ be the radial projection, namely $p(x)$ is the intersection of the geodesic segment $[ax]$ with $\partial B_r(a)$. In the other hyperplane, let $p$ be the composition of the reflection in $H$ and the above radial projection. This map is distance non-increasing since the radial projection to $\partial B_r(a)$ is distance non-increasing away from the $r$-pall centered at the point opposite to $a$ on the sphere, and this ball is contained in the second hemisphere.)

For each $x\in \partial B_r(a)$, the pre-image $p^{-1}(x)$ contains a continuous path connecting $\partial B_r(a)$ to $\partial B_r(b)$ and therefore a point of the set $\{f=0\}$. Hence $p(\{f=0\})=\partial B_r(a)$. Since $p$ does not decrease distances, it does not increase the Hausdorff measure and the desired inequality follows.

  • $\begingroup$ Very nice argument. $\endgroup$
    – Rbega
    Aug 3, 2011 at 23:51

The answer is yes. For simplicity take n=3 (the argument is the same regardless but this makes things more straightforward). Let us also assume that $\lbrace f=0\rbrace$ is rectifiable (this allows us to ignore technical issues).

We have Levy's isoperimetric inequality for a domain $\Omega$ $$(\mathcal{H}^1(\partial\Omega))^2\geq \mathcal{H}^2(\Omega)(4\pi-\mathcal{H}^2(\Omega))$$. with equality when and only when $\Omega$ is a geodesic ball.

Let $\Omega_a$ be the component of $S^{2}\backslash \lbrace f=0\rbrace$ that contains $a$ and $\Omega_b$ be the component containing $b$. By construction we have that $B_r(a)\subset \Omega_a$ and $B_r(b)\subset \Omega_b$. Notice that up to a relabelling we may take $\mathcal{H}^2(B_r(a))=A\leq \mathcal{H}^2(\Omega_a)\leq 2\pi$.

By the isoperimetric inequality we have $$ v^2=A(4\pi-A )\leq \mathcal{H}^2(\Omega_a)(4\pi-\mathcal{H}^2(\Omega_a))\leq (\mathcal{H}^1(\partial\Omega_a))^2$$ where we have used that $x < y\leq 2\pi$ implies a $x (4\pi-x) \leq y(4\pi-y)$ and the isoperimetric inequality.

Finally, we note that $\partial \Omega_a\subset \lbrace f=0 \rbrace$ so taking square roots gives $$ v\leq \mathcal{H}^1(\lbrace f=0 \rbrace) $$ as desired.


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