A two-piece property for free boundary minimal hypersurfaces in the $(n+1)$-dimensional ball

Let $\Sigma$ be as in the hypotheses and denote by $D^{n}$ the equatorial disk that contains $\overline{\Gamma_{I}}$. Let $v\in\operatorname{S}^{n}$ be a vector orthogonal to the disk $D^{n}$ and consider the function $f(x)=\langle x,v\rangle$, $x\in\overline{\Sigma}.$ By hypothesis, we know that $f|_{\Gamma_{I}}\equiv 0$. A standard calculation using that $\Sigma$ is minimal and free boundary yields

Fix $\epsilon>0$ and consider a smooth function $\eta_{\epsilon}:[-1,1]\to[0,1]$ so that

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  $\eta_{\epsilon}(s)=0$ for $|s|<\epsilon$,

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  $\eta_{\epsilon}(s)=1$ for $|s|>2\epsilon$,

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  $|\eta^{\prime}_{\epsilon}|<\displaystyle\frac{C}{\epsilon}$, for some constant $C>0$.

Define $\phi_{0,\epsilon}:\overline{\Sigma}\to[0,1]$ as $\phi_{0,\epsilon}(x)=\eta_{\epsilon}(f(x))$. In particular, we have $|\nabla_{\Sigma}\phi_{0,\epsilon}|<C/\epsilon$ in $\Sigma.$ Observe that the set $S\subset\mathcal{S}_{1}$ where $\phi_{0,\epsilon}$ is not smooth satisfies $\mathcal{H}^{n-2}(S)=0.$

Since $\mathcal{S}_{1}\cap\{|f(x)|\geq\frac{\epsilon}{2}\}$ is compact and $\mathcal{H}^{n-2}\big{(}\mathcal{S}_{1}\big{)}=0$, for any $\epsilon^{\prime}>0$ there exist balls $B_{r_{i}}(p_{i})\subset\mathbb{R}^{n+1},\,i=1,\cdots,m,$ such that

For each $i=1,\cdots,m$, consider a smooth function $\phi_{i}:\overline{\Sigma}\to[0,1]$ such that

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  $\phi_{i}(s)=0$ in $B_{r_{i}}(p_{i})$,

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  $\phi_{i}(s)=1$ in $\mathbb{R}^{n+1}\setminus B_{2r_{i}}(p_{i})$,

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  $|\nabla_{\Sigma}\phi_{i}|<\displaystyle\frac{2}{r_{i}},\,\forall\,x\in\Sigma$.

Define $\phi_{\epsilon},f_{\epsilon}:\overline{\Sigma}\to[0,1]$ by $\phi_{\epsilon}(x)=\displaystyle\min_{0\leq i\leq m}\phi_{i}$, where $\phi_{0}=\phi_{0,\epsilon},$ and $f_{\epsilon}=\phi_{\epsilon}f$. We have that $f_{\epsilon}$ is Lipschitz and $f_{\epsilon}|_{\Gamma_{I}}\equiv 0$, hence (2.2) holds. Moreover

and

since $\Delta_{\Sigma}f\equiv 0$, $\frac{\partial f}{\partial\nu}=f$ and $f|_{\Gamma_{I}}\equiv 0.$ Hence, applying it to (2.2), we get

On the other hand, along $\overline{\Sigma}$ we have $f^{2}\leq 1$. Since $f$ has support away from the singular set, by the classical monotonicity formula at the interior and at the free boundary, there is $C_{\Sigma,\epsilon}>0$ such that

Thus

If we let $\epsilon^{\prime}\to 0$ first and then $\epsilon\to 0$ we obtain

If $|A_{\Sigma}|\equiv 0$ then $\Sigma$ is totally geodesic and we are done. If $|A_{\Sigma}|(x)>0$ for some $x\in\Sigma,$ then we can find a neighborhood $U$ of $x$ in $\Sigma$ such that $|A_{\Sigma}|$ is strictly positive. This implies $\langle y,v\rangle=0$ for any $y\in U$. Therefore, $\Sigma$ is entirely contained in the disk $D^{n};$ in particular, it is totally geodesic. ∎

${\it(i)}$ If the free boundary were empty, we could apply the (interior) maximum principle with the family of hyperplanes parallel to the disk $D^{n}$ and conclude that $\Sigma$ should be contained in the disk $D^{n}$. On the other hand, if the fixed boundary were empty, then we would have a minimal hypersurface entirely contained in a halfball without fixed boundary; hence, we could apply the (interior or free boundary version of) maximum principle with the family of equatorial disks that are rotations of $D^{n}$ around a $(n-1)-$dimensional equatorial disk and conclude that $\Sigma$ should be an equatorial disk.

${\it(ii)}$ Let $D^{n}$ be an equatorial disk and suppose that $\Sigma$ is a (partially) free boundary minimal hypersurface such that $\Sigma\cap D^{n}$ contains a $(n-1)-$dimensional piece $\Upsilon$ of the free boundary of $D^{n}$ in $\operatorname{S}^{n}.$ Assume, without loss of generality, $D^{n}\subset\{x_{n+1}=0\}$.

Observe that since $\Sigma$ is free boundary we know that $\frac{\partial x_{n+1}}{\partial\eta}{\big{|}}_{\Upsilon}=x_{n+1}{\big{|}}_{\Upsilon}=0,$ where $\eta$ is the conormal vector to $\Upsilon;$ and since $\Sigma$ is a minimal hypersurface in $\operatorname{\mathbb{R}}^{n+1}$ we have that $x_{n+1}{\big{|}}_{\Sigma}$ is harmonic.

We will show that $x_{n+1}{\big{|}}_{\Sigma}\equiv 0.$

Consider an extension $\hat{\Sigma}$ of $\Sigma$ along $\Upsilon$ such that $\Upsilon\subset\mbox{int}(\hat{\Sigma})$ and define $\hat{x}_{n+1}$ on $\hat{\Sigma}$ as

Observe that $\hat{x}_{n+1}{\big{|}}_{\Upsilon}=x_{n+1}{\big{|}}_{\Upsilon}\equiv 0$, $\frac{\partial\hat{x}_{n+1}}{\partial\hat{\eta}}{\big{|}}_{\Upsilon}=0$ and $\frac{\partial\hat{x}_{n+1}}{\partial\eta}{\big{|}}_{\Upsilon}=\frac{\partial{x}_{n+1}}{\partial\eta}{\big{|}}_{\Upsilon}=0,$ where $\hat{\eta}$ is the conormal to $\Upsilon$ pointing towards $\Sigma$ and $\eta$ is the conormal to $\Upsilon$ pointing towards $\hat{\Sigma}\setminus\Sigma;$ hence, $\hat{x}_{n+1}$ is $C^{1}$ in a neighborhood of $\Upsilon$ in $\hat{\Sigma}.$

Observe that $\hat{x}_{n+1}$ is a harmonic function on $\hat{\Sigma}\setminus\Upsilon$, so we just need to show the claim in a neighborhood of $\Upsilon.$

Consider a domain $\Omega=\Omega_{1}\cup\Omega_{2}$ where $\partial\overline{\Omega_{i}}=\Gamma_{i}\cup(\overline{\Omega}\cap\Upsilon)$ with $\Omega_{1}\subset\hat{\Sigma}\setminus\Sigma$ and $\Omega_{2}\subset\Sigma$ (see Figure 1), and let $\phi:\hat{\Sigma}\to\operatorname{\mathbb{R}}$ be a smooth function with compact support contained in $\Omega.$

Integration by parts gives us

since supp$(\phi)\subset\subset\Omega,$ where $\nu$ is the outward conormal to $\partial\Omega.$

Then,

We have

where in the first equality we used that $\hat{x}_{n+1}{\big{|}}_{\Omega_{1}}\equiv 0$ and in the second equality we used the fact that $\frac{\partial\hat{x}_{n+1}}{\partial\nu_{1}}{\big{|}}_{\Upsilon}=0,$ where $\nu_{1}$ is the outward conomal to $\Upsilon$ with respect to $\Omega_{1}.$

Analogously, we have

where in the first equality we used that $\hat{x}_{n+1}{\big{|}}_{\Omega_{2}}=x_{n+1}{\big{|}}_{\Omega_{2}}$ is harmonic and in the second equality we used the fact that $\frac{\partial\hat{x}_{n+1}}{\partial\nu_{2}}{\big{|}}_{\Upsilon}=0,$ where $\nu_{2}$ is the outward conomal to $\Upsilon$ with respect to $\Omega_{2}.$

Therefore, the claim follows and, by the Elliptic theory, $\hat{x}_{n+1}$ has to be a (strong) solution to the Laplacian equation. Moreover, since $\hat{x}_{n+1}$ vanishes on an open set, the unique continuation result implies that $\hat{x}_{n+1}\equiv 0$ on $\hat{\Sigma},$ that is, $\Sigma$ is (contained in) the equatorial disk $D^{n}.$ ∎

Let us write

where the union is disjoint and $\mathcal{R}$ consists of the points $x\in\operatorname{spt}T$ such that there is a neighborhood $U\subset\mathbb{R}^{n+1}$ of $x$ where $T\mathbin{\vrule height=6.88889pt,depth=0.0pt,width=0.55974pt\vrule height=0.55974pt,depth=0.0pt,width=5.59721pt}U$ is given by $m$-times ($m\in\mathbb{N}$) integration over an embedded $C^{2}$-hypersurface with boundary. To complete the proof we will prove the following:

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  Regularity at the interior: $(\operatorname{spt}T\setminus\operatorname{spt}\partial T)\cap\mathcal{R}\neq\emptyset$ and $(\operatorname{spt}T\setminus\operatorname{spt}\partial T)\cap\mathcal{S}_{1}$ satisfies (2.9);

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  Regularity at the free-boundary: $\operatorname{spt}\bigl{(}\llbracket\Gamma\rrbracket-\partial T\bigr{)}\cap\mathcal{R}\neq\emptyset$ and $\operatorname{spt}\bigl{(}\llbracket\Gamma\rrbracket-\partial T\bigr{)}\cap\mathcal{S}_{1}$ satisfies (2.9);

The interior regularity is a classical result, see [8, Section 5.3]. Since by Proposition 2 the free part of the boundary is contained in $S\setminus\partial S$, we can use the result by Grüter [16] to conclude the regularity at the free boundary (away from the corners). ∎