$(\tilde{\Sigma}^{k},\tilde{\Gamma}^{k})^{k}$ is a min-max sequence, we have

(7.51) $F_{a}(\tilde{V},\tilde{W})=F_{a}(V,W).$

By construction, it follows that $(\tilde{V},\tilde{W})$ is stationary in
$\operatorname*{An}$ and in $\mathcal{M}\setminus\operatorname*{An}$. We have
only to prove that it is stationary in $\mathcal{M}$. Let us assume, by
contradiction, that $(\tilde{V},\tilde{W})$ is not stationary. We fix
$\eta>0$, a vector field $X\in\mathfrak{X}_{t}(\mathcal{M})$ such that
$\delta_{F_{a}}(\tilde{V},\tilde{W})[X]\leq-\eta$ and we choose a partition of
unity for the cover
$\\{\operatorname*{An}^{\prime},\mathcal{M}\setminus\operatorname*{An}\\}$ of
$\mathcal{M}$: $\varphi\in C_{c}^{\infty}(\operatorname*{An}^{\prime})$ and
$\psi\in C_{c}^{\infty}(\mathcal{M}\setminus\operatorname*{An})$. Hence

(7.52)
$\delta_{F_{a}}(\tilde{V},\tilde{W})[X]=\delta_{F_{a}}(\tilde{V},\tilde{W})[\varphi
X]+\delta_{F_{a}}(\tilde{V},\tilde{W})[\psi X].$

Since $(\tilde{V},\tilde{W})$ is stationary in
$\mathcal{M}\setminus\operatorname*{An}$, it follows that
$\delta_{F_{a}}(\tilde{V},\tilde{W})[\psi X]=0$. Thus

(7.53) $\delta_{F_{a}}(\tilde{V},\tilde{W})[\varphi X]\leq-\eta$

By continuity of first variation there exists $T>0$ such that, if $\Phi_{t}$
is the flow of $\varphi X$, then

(7.54) $\delta_{F_{a}}(\Phi_{t})_{\sharp}(\tilde{V},\tilde{W})[\varphi
X]\leq-\frac{\eta}{2}\qquad\forall t\in[0,T].$

Since
$(\tilde{\Sigma}^{k},\tilde{\Gamma}^{k})\xrightarrow[k]{{}}(\tilde{V},\tilde{W})$,
by continuity of first variation again, there exists $\bar{k}\in\mathbb{N}$
such that

(7.55)
$\delta_{F_{a}}(\Phi_{t})_{\sharp}(\tilde{\Sigma}^{k},\tilde{\Gamma}^{k})[\varphi
X]\leq-\frac{\eta}{4}\qquad\forall t\in[0,T],\forall k>\bar{k}.$

Hence

(7.56)
$F_{a}\bigl{(}(\Phi_{t})_{\sharp}(\tilde{\Sigma}^{k},\tilde{\Gamma}^{k})\bigr{)}\leq
F_{a}(\tilde{\Sigma}^{k},\tilde{\Gamma}^{k})-t\,\frac{\eta}{4}\qquad\forall
t\in[0,T],\,\,\,\forall k>\bar{k}.$

Thus we have a smooth deformation of $(\tilde{\Sigma}^{k},\tilde{\Gamma}^{k})$
with support in $\operatorname*{An}^{\prime}$ which lowers $F_{a}$ of a fixed
amount. But this contradicts, for $k$ large enough, the almost minimizing
property of $(\tilde{\Sigma}^{k},\tilde{\Gamma}^{k})$ in
$\operatorname*{An}^{\prime}$.

∎

######  Remark 7.1.

Since $(\tilde{\Sigma}^{k},\tilde{\Gamma}^{k})$ is a new min-max sequence
which is almost minimizing in each annulus in
$\mathcal{AN}_{\tilde{r}(y)}({y})$, we can repeat the previous arguments to
construct a new replacement of $(\tilde{V},\tilde{W})$ in each annulus in
$\mathcal{AN}_{\tilde{r}(y)}({y})$. In particular, since $\tilde{r}(x)=r(x)$,
we can construct a again a new min-max sequence and a new replacement of
$(\tilde{V},\tilde{W})$ in every annulus in $\mathcal{AN}_{r(x)}({x})$. This
procedure can be in fact repeated an infinite number of times.

##  8\. Regularity of $V$

In this final section we prove that $(V,W)$ is indeed induced by a smooth pair
$(\Sigma,\Gamma)$, thus competing the proof of Theorem 1.1

###  8.1. Integer rectifiability of $V$

######  Proposition 8.1.

Let $(V,W)$ be the pair of varifolds given by Proposition 5.1. Then $V$ and
$W$ are rectifiable and $V$ is integer rectifiable. Moreover, for any
$x\in\partial\mathcal{M}\cap\operatorname{supp}\lVert V\rVert$, any tangent
pair $(C,K)$ to $(V,W)$ at $x$ is such that $C$ and $K$ are half-planes and
$C$ meets $T_{x}\partial\mathcal{M}$ with angle $\theta$.

###### Proof.

We divide the proof in few steps:

$\bullet$_$V$ and $W$ are rectifiable_

If $p\in{\kern 0.0pt\mathcal{M}}^{\mathrm{o}}$, then there exists a
neighborhood $U$ of $p$ such that $V$ is integer rectifiable in $U$, see [5,
Lemma 5.2]. So we fix a point $x\in\partial\mathcal{M}$ such that
$\Theta_{2}(\lVert V\rVert+a\lVert W\rVert,x)>0$ and choose a sequence
$r_{j}\downarrow 0$; the rescaled pairs
$(V_{j},W_{j}):=(D_{{x,r_{j}}})_{\sharp}(V,W)$ converge to a pair $(C,K)$ such
that

  1. (1)

$\operatorname{supp}C\subset T_{x}\mathcal{M}$ and
$\operatorname{supp}K\subset T_{x}\partial\mathcal{M}$;

  2. (2)

$\delta_{F_{a}}(C,K)[X]=0$ for every $X\in\mathfrak{X}_{t}(T_{x}\mathcal{M})$;

  3. (3)

$\rho^{-2}\big{(}\lVert C\rVert(B_{\rho})+a\lVert
K\rVert(B_{\rho})\big{)}=\sigma^{-2}\big{(}\lVert C\rVert(B_{\sigma})+a\lVert
K\rVert(B_{\sigma})\big{)}=\Theta_{2}(\lVert V\rVert+a\lVert W\rVert,x)$ for
every $\rho,\sigma>0$.

All these statements follow by the fact that, by definition, the varifold
$V+aW$ has free boundary (Definition 2.3) and it is indeed stationary for the
mass with respect to all variations tangent to $T_{x}\partial\mathcal{M}$, see
for example the proof of [20, Lemma 5.4].

Since $(V,W)$ has replacement $(\tilde{V}_{j},\tilde{W}_{j})$ in every annulus
$\operatorname*{An}(x,r_{j},2r_{j})$ for $j$ sufficiently large, by Theorem
6.5 the rescaled pairs $(D_{{x,r_{j}}})_{\sharp}(\tilde{V}_{j},\tilde{W}_{j})$
converge to a pair $(\tilde{C},\tilde{K})$ which is a replacement for $(C,K)$
in $\operatorname*{An}(0,1,2)$. By the properties of replacements, we have

(8.1)
$\begin{split}\frac{\lVert\tilde{C}\rVert(B_{1/2})+a\lVert\tilde{K}\rVert(B_{1/2})}{(1/2)^{2}}=&\frac{\lVert{C}\rVert(B_{1/2})+a\lVert{K}\rVert(B_{1/2})}{(1/2)^{2}}\\\
=&\frac{\lVert{C}\rVert(B_{5/2})+a\lVert{K}\rVert(B_{5/2})}{(5/2)^{2}}\\\
=&\frac{\lVert\tilde{C}\rVert(B_{5/2})+a\lVert\tilde{K}\rVert(B_{5/2})}{(5/2)^{2}}.\end{split}$

By the monotonicity identity for varifolds with free boundary (see e.g. Step 5
in the proof of [20, Lemma 5.4]) one gets

(8.2)
$\frac{\lVert\tilde{C}\rVert(B_{\rho})+a\lVert\tilde{K}\rVert(B_{\rho})}{\rho^{2}}=\frac{\lVert\tilde{C}\rVert(B_{\sigma})+a\lVert\tilde{K}\rVert(B_{\sigma})}{\sigma^{2}}=\Theta_{2}(\lVert
V\rVert+a\lVert W\rVert,x)\qquad\forall\rho,\sigma>0.$

This means that for every $\rho>0$ we have
$\operatorname{supp}(\lVert\tilde{C}\rVert+a\lVert\tilde{K}\rVert)\cap\partial
B_{\rho}\neq\emptyset$, because otherwise (8.2) does not hold for
$\rho-\varepsilon,\rho+\varepsilon$ for some small $\varepsilon>0$.

For any $y\in\partial
B_{3/2}\cap\operatorname{supp}(\lVert\tilde{C}\rVert+a\lVert\tilde{K}\rVert)$,
since $\tilde{C}$ and $\tilde{K}$ are regular in $\operatorname*{An}(0,1,2)$,
it follows that
$\Theta_{2}(\lVert\tilde{C}\rVert+a\lVert\tilde{K}\rVert,y)\geq 1/2$. By
monotonicity of density ratios for $\tilde{C}+a\tilde{K}$ given by Proposition
3.3, we have that there exists an universal constant $c>0$ such that

(8.3)
${\lVert\tilde{C}\rVert(B_{1/2}(y))+a\lVert\tilde{K}\rVert(B_{1/2}(y))}\geq
c.$

(8.2) and $B_{1/2}(y)\subset B_{2}$ give the existence of an universal
constant $c^{\prime}>0$ such that

(8.4) $\Theta_{2}(\lVert V\rVert+a\lVert W\rVert,x)>c^{\prime}\qquad\forall
x\in\operatorname{supp}(\lVert V\rVert+a\lVert W\rVert).$

Since $V+aW$ is stationary with respect to $\mathfrak{X}_{t}(\mathcal{M})$,
[20, Theorem 1.1, (1.6)] implies that it has bounded first variation in the
whole space $\mathbb{R}^{3}$ with a uniform bound on the first variation given
by the mass $(\lVert V\rVert+a\lVert W\rVert)(\mathcal{M})$. This, together
with the above uniform lower bound on the density and the Allard
Rectifiability Theorem [25, Theorem 42.4], proves that $V+aW$ is a rectifiable
varifold. Since $W$ is rectifiable because it is the weak-* limit of a
sequence of $L^{\infty}$ functions on $\partial\mathcal{M}$, we have that $V$
is rectifiable.

$\bullet$_$C+aK$ is a rectifiable cone_

We claim that every tangent pair $(C,K)$ to $(V,W)$ is such that $C+aK$ is a
rectifiable cone. Indeed $C+aK$ is obtained as limit of rescalings
$(D_{{x,r_{j}}})_{\sharp}(V+aW)$ which are stationary for
$\mathfrak{X}_{t}\big{(}D_{{x,r_{j}}}(\mathcal{M})\big{)}$; by [20, Theorem
1.1 and (1.6)], we have that $(D_{{x,r_{j}}})_{\sharp}(V+aW)$ have uniformly
bounded first variation, then we can apply the Compactness Theorem for
rectifiable varifolds [25, Theorem 42.7], to get the rectifiability of $C+aK$.
The fact that $C+aK$ is a cone follows by the same arguments in Step 5 in the
proof of [20, Lemma 5.4].

$\bullet$_$C$ and $K$ are half-planes and $C$ intersects
$T_{x}\partial\mathcal{M}$ with angle $\theta$._

Let $(C,K)$ a tangent pair to $(V,W)$ at $x$. For any $\lambda\in(0,1)$,
$(V,W)$ has replacement $(\tilde{V}_{j},\tilde{W}_{j})$ in every annulus
$\operatorname*{An}(x,\lambda r_{j},\lambda^{-1}r_{j})$ for $j$ sufficiently
large and the rescaled pairs
$(D_{{x,r_{j}}})_{\sharp}(\tilde{V}_{j},\tilde{W}_{j})$ converge to a pair
$(\tilde{C},\tilde{K})$ which is a replacement for $(C,K)$ in
$\operatorname*{An}(0,\lambda,\lambda^{-1})$. Moreover, by the above
discussion, $\tilde{C}+a\tilde{K}$ is a rectifiable varifold.

As above, for any $\sigma\in(0,\lambda)$ and for any $\rho>\lambda^{-1}$, we
have

(8.5)
$\begin{split}\frac{\lVert\tilde{C}\rVert(B_{\sigma})+a\lVert\tilde{K}\rVert(B_{\sigma})}{\sigma^{2}}=&\frac{\lVert{C}\rVert(B_{\sigma})+a\lVert{K}\rVert(B_{\sigma})}{\sigma^{2}}\\\
=&\frac{\lVert{C}\rVert(B_{\rho})+a\lVert{K}\rVert(B_{\rho})}{\rho^{2}}\\\
=&\frac{\lVert\tilde{C}\rVert(B_{\rho})+a\lVert\tilde{K}\rVert(B_{\rho})}{\rho^{2}}.\end{split}$

Again this and the same arguments in Step 5 in the proof of [20, Lemma 5.4]
prove that $\tilde{C}+a\tilde{K}$ is a rectifiable cone. The same arguments as
above show that both $\tilde{C},\tilde{K}$ are rectifiable varifolds.

Since $\tilde{C}+a\tilde{K}$ is a cone and $\tilde{C},\tilde{K}$ are smooth
surfaces in $\operatorname*{An}(0,\lambda,\lambda^{-1})$ with integer
multiplicity, it follows that both $\tilde{C}$ and $\tilde{K}$ are cones in
$\operatorname*{An}(0,\lambda,\lambda^{-1})$; indeed, the density of
$\tilde{C}+a\tilde{K}$ takes values in $\mathbb{N}\cup(a+\mathbb{N})$ (since
$\tilde{K}$ has multiplicity 1) and has to be constant on half lines starting
from the origin (because $\tilde{C}+a\tilde{K}$ is a cone), then the densities
of $\tilde{K}$ and $\tilde{C}$ must be constant on lines.

Since $(\tilde{C},\tilde{K})$ is a replacement of $(C,K)$ in
$\operatorname*{An}(0,\lambda,\lambda^{-1})$ and $\tilde{C}$ and $\tilde{K}$
are cones in $\operatorname*{An}(0,\lambda,\lambda^{-1})$, if we consider the
pair of smooth surfaces which induce $(\tilde{C},\tilde{K})$ in
$\operatorname*{An}(0,\lambda,\lambda^{-1})$ and we extend them in a conical
way up to the origin and the infinity, the pair of surfaces $\tilde{\Sigma}$
and $\tilde{\Gamma}$ are stationary and stable in
$\mathbb{R}^{3}\setminus\\{0\\}$, thus we can apply the Bernstein Theorem
(Theorem 6.3) to conclude that $\tilde{\Sigma}$ and $\tilde{\Gamma}$ are half
planes and $\tilde{\Sigma}$ meets $T_{x}\partial\mathcal{M}$ with contact
angle $\theta$.

By the fact that $C+aK$ is a cone which coincides with
$\tilde{\Sigma}+a\tilde{\Gamma}$ in
$\operatorname*{An}(0,\lambda,\lambda^{-1})$, it follows that it coincides
with $\tilde{\Sigma}+a\tilde{\Gamma}$ on the whole $\mathbb{R}^{3}$. This in
particular implies, since $\tilde{\Gamma}$ and $\operatorname{supp}\lVert
K\rVert$ are contained in $T_{x}\partial\mathcal{M}$, that

(8.6) $C\cap{\kern
0.0pt(T_{x}\mathcal{M})}^{\mathrm{o}}=\tilde{\Sigma}\cap{\kern
0.0pt(T_{x}\mathcal{M})}^{\mathrm{o}}.$

This in turn implies that any replacement in an annulus for $(C,K)$ coincide
with $\tilde{\Sigma}$ in the annulus and in ${\kern
0.0pt(T_{x}\mathcal{M})}^{\mathrm{o}}$, that is the surface $\tilde{\Sigma}$
is independent on the choice of $\lambda$.

We claim that $\lVert C\rVert(T_{x}\partial\mathcal{M})=0$. Indeed, if were
not, there would exist $y\in T_{x}\partial\mathcal{M}\setminus\\{0\\}$ and a
small ball $B_{r}(y)$ such that $0\notin B_{r}(y)$ and $\lVert
C\rVert\big{(}B_{r}(y)\cap\partial\mathcal{M}\big{)}>0$. Let us consider
$\lambda$ so small so that
$B_{r}(y)\subset\operatorname*{An}(0,\lambda,\lambda^{-1})$. Let us consider
the replacement $(\tilde{C},\tilde{K})$ of $(C,K)$ in
$\operatorname*{An}(0,\lambda,\lambda^{-1})$. We have

(8.7) $\begin{gathered}\lVert
C\rVert\big{(}B_{\lambda^{-1}}\big{)}=\lVert\tilde{C}\rVert\big{(}B_{\lambda^{-1}})\\\
C\cap B_{\lambda}=\tilde{C}\cap B_{\lambda}\end{gathered}$

This implies that $\lVert
C\rVert\big{(}\operatorname*{An}(0,\lambda,\lambda^{-1})\big{)}=\lVert\tilde{C}\rVert\big{(}\operatorname*{An}(0,\lambda,\lambda^{-1})\big{)}$
but the fact that

(8.8)
$\lVert\tilde{C}\rVert\big{(}T_{x}\partial\mathcal{M})\cap\operatorname*{An}(0,\lambda,\lambda^{-1})\big{)}=\mathcal{H}^{2}\big{(}\tilde{\Sigma}\cap
T_{x}\partial\mathcal{M}\cap\operatorname*{An}(0,\lambda,\lambda^{-1})\big{)}=0$

contradicts (8.6). Hence $\lVert C\rVert(T_{x}\partial\mathcal{M})=0$ and
$C=\tilde{\Sigma}$. In turn, by $C+aK=\tilde{\Sigma}+a\tilde{\Gamma}$, it
follows that $K=\tilde{\Gamma}$. Hence $C$ and $K$ are half-planes and $C$
intersects $T_{x}\partial\mathcal{M}$ with angle $\theta$.

$\bullet$_$V$ is integer rectifiable._

We have proved that, for every tangent pair $(C,K)$ to $(V,W)$ at $x$, it
holds $\lVert C\rVert(T_{x}\partial\mathcal{M})=0$. Since
$(D_{{x,r_{j}}})_{\sharp}V\stackrel{{\scriptstyle\ast}}{{\rightharpoonup}}C$,
we have that

(8.9) $\limsup_{j\to\infty}\frac{\lVert
V\rVert\big{(}B_{r_{j}}(x)\cap\partial\mathcal{M}\big{)}}{r_{j}^{2}}=\limsup_{j\to\infty}\lVert(D_{{x,r_{j}}})_{\sharp}V\rVert\big{(}B_{1}\cap
D_{{x,r_{j}}}(\partial\mathcal{M})\big{)}\leq\lVert
C\rVert\big{(}B_{1}\cap\partial\mathcal{M}\big{)}=0.$

A straightforward covering argument yields $\lVert
V\rVert(\partial\mathcal{M})=0$. Since $V$ is integer rectifiable in the
interior of $\mathcal{M}$, it is integer rectifiable in all $\mathcal{M}$. ∎

###  8.2. Regularity of $V$

We first note that $V$ is regular and stable outside a finite number of point

######  Proposition 8.2.

Let $(V,W)$ be the pair of varifolds given by Proposition 5.1 and let us fix
$x\in\partial\mathcal{M}$. Then there exists $\rho>0$ such that $V$ is induced
in $B_{\rho}(x)\setminus\\{x\\}$ by a regular minimal surface $\Sigma$ which
meets $\partial\mathcal{M}$ with angle $\theta$. Moreover $\Sigma$ is stable
for the capillarity energy.

###### Proof.

The proof is exactly the same as [4, 10.3] but we sketch here the argument for
the reader convenience.

Let us consider $x\in\partial\mathcal{M}$, $r(x)$ so that $(V,W)$ has
replacements in any annulus $\operatorname*{An}\in\mathcal{AN}_{r(x)}({x})$.
Let us fix $\rho>0$ such that $4\rho<r(x)$. By the results in [5], we have
that $V$ is regular in ${\kern 0.0pt\mathcal{M}}^{\mathrm{o}}$. Let us call
$\Sigma_{1},\Sigma_{2},\dots$ the connected components of $V$ in
$\operatorname*{An}(x,\rho,4\rho)\cap{\kern 0.0pt\mathcal{M}}^{\mathrm{o}}$
and let us fix one of them $\Sigma_{i}$ with a regular point $y\in\Sigma_{i}$.
Since $y$ is regular for $\Sigma_{i}$, if $s=|x-y|$ we can find
$\bar{\sigma}>0$ such that

  * •

$\Sigma_{i}$ is regular in $B_{\bar{\sigma}}(y)$;

  * •

$\Sigma_{i}\cap\partial B_{s+\sigma}(x)\neq\emptyset$ (by maximum principle
[5, Theorem 5.1 $(ii)$]) for every $\sigma\in(0,\bar{\sigma})$;

  * •

the intersection between $\Sigma_{i}$ and $\partial B_{s+\sigma}(x)$ is
transversal (by Sard’s Lemma) for almost every $\sigma\in(0,\bar{\sigma})$.

We consider such a $\sigma$, call $t=s+\sigma$ and fix a point
$z\in\Sigma_{i}\cap\partial B_{t}(x)\cap{\kern
0.0pt\mathcal{M}}^{\mathrm{o}}$. We consider a replacement
$(\tilde{V},\tilde{W})$ for $(V,W)$ in $\operatorname*{An}(x,\rho,t)$. Again
by the interior theory in [5], $\tilde{V}$ is regular in ${\kern
0.0pt\mathcal{M}}^{\mathrm{o}}$. Since $V$ and $\tilde{V}$ coincide outside
$B_{t}(x)$, any tangent cone $C$ to $\tilde{V}$ at $z$ must contain a half-
plane $\pi$ (since $z$ is regular for $\Sigma_{i}$); since $C$ is stationary
and stable, it is a plane, hence $C$ coincides with the whole plane $\pi$.
This implies that $z$ is regular for $\tilde{V}$.

If we call $\tilde{\Sigma}$ the connected component of $\tilde{V}$ that
contains $z$, by unique continuation, $\tilde{\Sigma}$ and $\Sigma_{i}$
coincide and, since $\tilde{\Sigma}$ is regular up to $\partial\mathcal{M}$ in
$\operatorname*{An}(x,\rho,t)$ (because $(\tilde{V},\tilde{W})$ is a
replacement for $(V,W)$), also $\Sigma_{i}$ is regular up to
$\partial\mathcal{M}$ in $\operatorname*{An}(x,\rho,t)$. Since the choice of
$y$ is arbitrary, we have that $\Sigma_{i}$ is regular up to
$\partial\mathcal{M}$ in the whole annulus $\operatorname*{An}(x,\rho,4\rho)$.

By monotonicity formula, it follows that there are only finitely many
connected components of $V$ in $\operatorname*{An}(x,2\rho,3\rho)$: $\mkern
1.0mu\overline{\mkern-1.0mu\Sigma\mkern-1.0mu}\mkern 1.0mu_{1},\dots,\mkern
1.0mu\overline{\mkern-1.0mu\Sigma\mkern-1.0mu}\mkern 1.0mu_{N}$ and each of
them is a regular minimal surface which meets $\partial\mathcal{M}$ with angle
$\theta$; let us call $S_{i}$ ($i=1,\dots,N$) the singular sets of the
surfaces $\mkern 1.0mu\overline{\mkern-1.0mu\Sigma\mkern-1.0mu}\mkern
1.0mu_{i}$.

Let us consider a point $y\in\mkern
1.0mu\overline{\mkern-1.0mu\Sigma\mkern-1.0mu}\mkern
1.0mu_{i}\cap\operatorname*{An}(x,2\rho,3\rho)\setminus\bigcup_{j=1}^{N}S_{j}$.
We claim that $y$ is a regular point for $V$; of course, since $y$ is regular
for $\mkern 1.0mu\overline{\mkern-1.0mu\Sigma\mkern-1.0mu}\mkern 1.0mu_{i}$,
to prove this it is sufficient to show that $y$ does not belong to any of the
others $\mkern 1.0mu\overline{\mkern-1.0mu\Sigma\mkern-1.0mu}\mkern
1.0mu_{j}$; if this were the case for a $\mkern
1.0mu\overline{\mkern-1.0mu\Sigma\mkern-1.0mu}\mkern 1.0mu_{j}$, $y$ would be
regular also for $\mkern 1.0mu\overline{\mkern-1.0mu\Sigma\mkern-1.0mu}\mkern
1.0mu_{j}$, and since $\Sigma_{i}\cap\Sigma_{j}\cap{\kern
0.0pt\mathcal{M}}^{\mathrm{o}}=\emptyset$, then $T_{y}\mkern
1.0mu\overline{\mkern-1.0mu\Sigma\mkern-1.0mu}\mkern 1.0mu_{i}=T_{y}\mkern
1.0mu\overline{\mkern-1.0mu\Sigma\mkern-1.0mu}\mkern 1.0mu_{j}$, but this
contradicts the maximum principle.

On the other hand, since $\partial M$ is convex, $\lVert
V\rVert(\partial\mathcal{M})=0$, and any point $y\in\operatorname{supp}\lVert
V\rVert\cap\operatorname*{An}(x,2\rho,3\rho)$ belongs to some $\mkern
1.0mu\overline{\mkern-1.0mu\Sigma\mkern-1.0mu}\mkern 1.0mu_{i}$.

This shows that $V$ is regular up to $\partial\mathcal{M}$ in
$\operatorname*{An}(x,\rho,4\rho)$; since the radius $\rho$ is arbitrary
(except for the fact that $4\rho<r(x)$), this argument proves that $V$ is
regular in $B_{r(x)/4}(x)\setminus\\{x\\}$.

Stability of $\Sigma$ in $B_{r(x)/4}(x)\setminus\\{x\\}$ follows again by
checking the regularity proof of [4, 10.3], since there it is shown that $V$
coincides with each of its replacements in any
$\operatorname*{An}\in\mathcal{AN}_{r(x)}({x})$ ∎

In order to conclude the proof we only need to remove the final (possible)
singular point $x$. Here, the stability of $V$ in a in a (small) punctuerd
ball plays a crucial role. The following Proposition is indeed a
generalization to capillarity surfaces of of the “removable” singularity
theorem for two dimensional stable minimal surfaces first established by
Lawson and Gulliver, [12], see also [24] and [21]. Although the proof of
Lawson and Gulliver can be probably generalized to this setting , we follow
here a different route. To explain the idea, let us assume that
$\mathcal{M}=\\{x_{1}\geq 0\\}$ and that $\Sigma$ is globally stable. Then it
is well known that

$v={\nu_{\Sigma}}\cdot{e_{1}}$

is a jacobi field which is negative at the boundary due to the contact angle
condition. By the stability inequality $v$ should be negative also inside
since otherwise its positive part would induce a negative variation. This
implies that $\Sigma$ can be expressed as a union of graph with respect to
$\\{x_{1}=0\\}$. Each of this graph will be a stationary point point of the
functional

$\mathcal{F}(g)=\int\sqrt{1+|\nabla g|^{2}}+a|\\{g>0\\}|$

and the regularity will follow by the regularity theory of [1] (see also [2])
and more precisely the implementation in [11]). In order to make the proof
rigorous we shall first localize the above argument and then precise in which
sense $g$ is solving a free boundary problem.

######  Proposition 8.3.

Let $(V,W)$ be the pair of varifolds given by Proposition 5.1, let us fix
$x\in\partial\mathcal{M}$ and let $\rho>0$ be the radius given by Proposition
8.2. Then $V$ is regular in the whole $B_{\rho}(x)$, that is is induced in
$B_{\rho}(x)$ by a regular minimal surface which meets $\partial\mathcal{M}$
with angle $\theta$.

###### Proof.

The proof is divided in few steps:

_Step 1_ : First by arguing exactly as in [4, 10.4], we note that $V\llcorner
B_{\rho}(x)$ consists by finitely many connected components $\Sigma_{i}$ with
integer multiplicities $h_{i}$ which are homeomorphic to $2$-balls without the
center $x$.

_Step 2_ : We now fix such a surface $\Sigma^{i}=\Sigma$, remove the
multiplicity $h_{i}$ and we aim to show that it can be extended smoothly
through $x$. We first claim that for $y\in\Sigma\cap\partial\mathcal{M}\cap
B_{\rho}(x)$ the sequence $\Sigma^{y}_{j}:=(D_{{y,r_{j}}})_{\sharp}\Sigma$
converges (up to subsequences) to a plane $\pi_{y}$ which forms the correct
angle with $T_{y}\partial\mathcal{M}$. Note that this is obvious if $y\neq x$
(since the surface is regular there) and we only need to show it for $y=x$.
For, we note that the curvature estimates (Theorem 6.4) yield the existence of
a constant $c>0$ such that

(8.10) $|A_{\Sigma}(y)|\leq\frac{c}{|y-x|}\qquad\forall y\in B_{\rho}(x).$

so that $\Sigma^{x}_{j}$ converges locally smoothly in
$\mathbb{R}^{3}\setminus\\{0\\}$ to an entire surface $\Xi$ which is smooth
and stable outside the origin. By Theorem 6.3, $\Xi$ is a plane through the
origin.

_Step 2_ : We now show that $\Sigma$ is a Lipschitz graph with respect to
$T_{x}\partial\mathcal{M}$. To this end we can assume that

$N_{T_{x}\partial\mathcal{M}}=-e_{1}.$

We now let $r_{j}$ be a sequence such that
$\Sigma_{j}:=(D_{{x,r_{j}}})_{\sharp}\Sigma$ converges to a plane $\Xi$
locally smoothly. Note that the plane might depend on the subsequence, but in
any case we can rotate the coordinates so that

$\nu_{\pi}=-ae_{1}+\sqrt{1-a^{2}}e_{2}.$

Since the convergence of $\Sigma_{j}\to\pi$ is in the $C^{1}$-topology on
annuli, for every $\varepsilon>0$ there exists $J\in\mathbb{N}$ such that

(8.11) $\sup_{y\in B_{r_{j}}(x)\setminus
B_{r_{j}/4}(x)}|\nu_{\Sigma}(y)-\nu_{\pi}|<\varepsilon\qquad\forall j>J,$

where $\nu_{\Sigma}$ is the unit normal vector to $\Sigma$. Hence, for any
$k>j>J$, we have

(8.12) $\sup_{y\in\Sigma\cap(\partial B_{r_{j}}(x)\cup\partial
B_{r_{k}}(x))}|\nu_{\Sigma}(y)-\nu_{\pi}|<\varepsilon.$

Let us consider for $e\in\mathbb{S}^{2}$ and $|e-e_{1}|\leq\delta$ to be fixed
,the function

(8.13) $v(y)=\nu_{\Sigma}(y)\cdot e\quad\forall y\in\Sigma.$

By (8.12), it follows that for every $\varepsilon,\delta>0$ there exists
$J\in\mathbb{N}$ such that

(8.14) $\sup_{y\in\Sigma\cap(\partial B_{r_{j}}(x)\cup\partial
B_{r_{k}}(x))}v(y)\leq-a+\varepsilon+\delta\qquad\forall k>j\geq J.$

Moreover, by the contact angle condition and the regularity of $\Sigma$ up to
$\partial\mathcal{M}$, we get $v(y)=-a$ for every on
$y\in\Sigma\cap\partial\mathcal{M}\cap\operatorname*{An}(x,r_{k},r_{j})$.
Hence

(8.15) $\sup_{y\in\Sigma\cap\big{(}\partial\mathcal{M}\cup\partial
B_{r_{j}}(x)\cup\partial
B_{r_{k}}(x)\big{)}}v(y)\leq-a+\varepsilon+\delta<0\qquad\forall k>j\geq J.$

provided $\delta$ and $\varepsilon$ are chosen sufficiently small but
depending only on $a$.

It is a classical fact that $v$ is a Jacobi field, that is

$\Delta_{\Sigma}v+|A_{\Sigma}|^{2}v=0.$

We now claim that $v$ has constant non-positive sign also in the interior of
the annulus $\operatorname*{An}(x,r_{k},r_{j})$. Indeed, the positive part
$v_{+}:=\max\\{v,0\\}$ is a function with compact support in the interior of
$\operatorname*{An}(x,r_{k},r_{j})$ and we can test the weak form of the
Jacobi equation with the function $v_{+}$ as test function, that is

(8.16) $\begin{split}0=&\int_{\Sigma}\nabla v_{+}\cdot\nabla
v-|A_{\Sigma}|^{2}vv_{+}\mathop{}\\!\mathrm{d}\mathcal{H}^{2}\\\
=&\int_{\Sigma}|\nabla
v_{+}|^{2}-|A_{\Sigma}|^{2}(v_{+})^{2}\mathop{}\\!\mathrm{d}\mathcal{H}^{2}\\\
=&Q(v_{+})\end{split}$

where $Q$ is the stability operator defined in (6.2). Since $Q\geq 0$ (by
stability of $\Sigma$), it follows that $v_{+}$ also is a non negative
solution of the Jacobi equation, a contradiction with the strong minimum
principle. Hence for each $e$ sufficiently close to $e_{1}$,

$\nu_{\Sigma}\cdot e<0\qquad\text{on
$\Sigma\cap\operatorname*{An}(x,r_{k},r_{j})$}$

which is easily seen to imply that
$\Sigma\cap\operatorname*{An}(x,r_{k},r_{j})$ can be represented as the graph
of a Lipschitz function $g:\mathbb{R}^{2}\to\mathbb{R}$

$\Sigma\cap\operatorname*{An}(x,r_{k},r_{j})=\\{x_{1}=g(x_{2},x_{3})\\}\cap\operatorname*{An}(x,r_{k},r_{j})$

and $|\nabla g|\leq 1/\delta$ for $\delta>0$ depending only on $a$. Letting
$k\to\infty$ we deduce that $\Sigma\cap B_{r_{j}}$ is the graph of a Lipschitz
function.

_Step 3_ : We now claim that the function $g$ in the above step is a solution
of a free boundary problem. To this end note first that, up to taking $j$
large and a scaling (and translation) of the coordinates that we can assume
that

$\Sigma\cap B_{1}=\\{x_{1}=g\\}\cap B_{1},\qquad\partial\mathcal{M}\cap
B_{1}=\\{x_{1}=h\\}\cap B_{1}$

for a suitable function $h$ with $h(0)=|\nabla h(0)|=0$. Note that $g\geq h$.
It is now easy to check that $g$ is a viscosity solution of the equation

$\begin{cases}-\operatorname*{div}\Bigl{(}\dfrac{\nabla g}{\sqrt{1+|\nabla
g|^{2}}}\Bigr{)}=0\qquad&\text{on $\\{g>h\\}$}\\\ -\nabla g\cdot{\nabla
h}+1=a\sqrt{1+|\nabla g|^{2}}\sqrt{1+|\nabla h|^{2}}\qquad&\text{on
$\\{g=h\\}$}.\end{cases}$

Indeed the first equation is clearly satisfied classically. As for the
boundary condition we first note that for $x_{0}\in\\{g=h\\}$, the point
$(g(x_{0}),x_{0})\in\Sigma\cap\partial\mathcal{M}$ and thus, thanks to Step
$1$, the functions

$g_{r}(x)=\frac{g(x_{0}+rx)-g(x_{0})}{r}$

converges (up to subsequences) to a linear function $\ell$ with the property
that

$-\nabla\ell\cdot{\nabla h(x_{0})}+1=a\sqrt{1+|\nabla\ell|^{2}}\sqrt{1+|\nabla
h(x_{0})|^{2}}.$

It is easy to verify that this implies that the condition is satisfied in the
viscosity sense, see for instance the computations in [10]. Note that this
also implies that, up to choose $r_{j}$ smaller, we can assume that the free
boundary $\\{g=h\\}$

$\|g-\ell\|_{L^{\infty})B^{\prime}_{1}(0)}<\varepsilon$

where $\ell(x)=x_{2}\sqrt{(1-a^{2})/a^{2}}$ (recall the normalizations at
$x$). One can then apply some straightforward modifications of the arguments
in [11] to show that the free boundary a $C^{1}$ curve. Note here that the
Lipschitz assumption on $g$ ensures the uniform ellipticity of the minimal
surface operator.

_Step 4_ : The previous step implies that each of the $\Sigma^{j}$ are smooth
graphs near $x_{0}$ and that they satisfy the minimal surface equation with
constant capillarity conditions. By Hopf Lemma, they have to coincide, so that
$x$ is a regular point as well.

∎

##  Appendix A Proof of the results of section 3.1

###### Proof of Proposition 3.1.

If $X\in\mathfrak{X}(\mathcal{M})$, there exist $X^{T},X^{\perp}$ such that
$X=X^{T}+X^{\perp}$, with $X^{T}\in\mathfrak{X}_{t}(\mathcal{M})$ and
$X^{\perp}\in\mathfrak{X}_{\perp}(\mathcal{M})$. We have

$\operatorname{div}_{S}X(x)=\operatorname{div}_{S}X^{T}(x)+\operatorname{div}_{S}X^{\perp}(x).$

By Definition 3.1, both $V$ and $V+aW$ have generalized mean curvature
$H\mathbf{1}_{{\kern 0.0pt\mathcal{M}}^{\mathrm{o}}}$ with respect to
$\mathfrak{X}_{0}(\mathcal{M})$; thus by [20, Theorem 1.1], there exists a
positive Radon measure $\sigma_{V}$ on $\partial\mathcal{M}$ and a $\lVert
V\rVert$-measurable vector field $\tilde{H}$ such that

(A.1)
$\displaystyle\begin{split}\int_{G_{2}(\mathcal{M})}\operatorname{div}_{S}X^{\perp}(x)\mathop{}\\!\mathrm{d}V(x,S)=&-\int_{\mathcal{M}}{X^{\perp}}\cdot{H\mathbf{1}_{{\kern
0.0pt\mathcal{M}}^{\mathrm{o}}}+\tilde{H}}\mathop{}\\!\mathrm{d}\lVert
V\rVert+\int_{\partial\mathcal{M}}{X}\cdot{N}\mathop{}\\!\mathrm{d}\sigma_{V}\\\
=&-\int_{\mathcal{M}}{X^{\perp}}\cdot{H+\tilde{H}}\mathop{}\\!\mathrm{d}\lVert
V\rVert+\int_{\partial\mathcal{M}}{X}\cdot{N}\mathop{}\\!\mathrm{d}\sigma_{V},\end{split}$

where the last equality follows because $H$ is assumed to be tangent to
$\partial\mathcal{M}$ on $\partial\mathcal{M}$ (see last part of Definition
3.1). Moreover by the definition of $\tilde{H}$ given in [20, (3.12)], it
follows that in our case
$\tilde{H}=-N(x)\operatorname{div}_{\partial\mathcal{M}}N(x)$ (since on
$\partial\mathcal{M}$, $V$ charges only planes which are tangent to
$\partial\mathcal{M}$, by [20, Lemma 3.1]). The estimates (3.5) and (3.6) on
$\sigma_{V}$ are given by [20, Theorem 1.1].

For the same reason, also $V+aW$ has generalized mean curvature
$H\mathbf{1}_{{\kern 0.0pt\mathcal{M}}^{\mathrm{o}}}$ with respect to
$\mathfrak{X}_{0}(\mathcal{M})$ and satisfies the properties as above with the
same $\tilde{H}$ and $\sigma_{V}$ (because $\sigma_{V}$ depends only of the
behavior of the varifold in the interior of $\mathcal{M}$, see its definition
in [20]).

Thus $W$ has generalized mean curvature $0$ with respect to
$\mathfrak{X}_{0}(\mathcal{M})$ and

(A.2)
$\int_{G_{2}(\mathcal{M})}\operatorname{div}_{\partial\mathcal{M}}X^{\perp}(x)\mathop{}\\!\mathrm{d}\lVert
W\rVert=-\int_{\partial\mathcal{M}}{X^{\perp}}\cdot{\tilde{H}}\mathop{}\\!\mathrm{d}\lVert
W\rVert$

for the same $\tilde{H}$ as above. For what concerns the tangent part $X^{T}$,
Definition 3.1 yields

$\begin{split}\int_{G_{2}(\mathcal{M})}\operatorname{div}_{S}X^{T}(x)\mathop{}\\!\mathrm{d}V(x,S)+a\int_{G_{2}(\mathcal{M})}\operatorname{div}_{S}X^{T}(x)\mathop{}\\!\mathrm{d}\lVert
W\rVert=-\int_{\mathcal{M}}{X^{T}}\cdot{H}\mathop{}\\!\mathrm{d}\lVert
V\rVert.\end{split}$

This shows the conclusion. ∎

In order to prove Proposition 3.2, we first derive a monotonicity inequality
for the masses.

######  Lemma A.1 (Monotonicity inequality).

Suppose
$(V,W)\in\mathcal{V}_{2}(\mathcal{M})\times\mathcal{V}_{2}(\partial\mathcal{M})$
satisfy the contact angle condition $\theta$, with $H\in
L^{p}(\mathcal{M},\lVert V\rVert)$ for some $p\in[1,+\infty)$. Then there
exists a constant $c>0$ that depends only on $p$ and on the second fundamental
form of $\partial\mathcal{M}$ such that, for all
$x_{0}\in\partial\mathcal{M}$, the following inequality holds:

(A.3)
$\begin{split}(1+c\rho)\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}\rho}\bigg{(}\frac{1}{\rho^{2}}&\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert W\rVert)\bigg{)}^{\frac{1}{p}}\\\
\geq&-\rho^{-\frac{2}{p}}\left[\frac{1+\rho}{p}\left(\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}|H+\tilde{H}|^{p}\mathop{}\\!\mathrm{d}\lVert
V\rVert\right)^{\frac{1}{p}}-\frac{a}{p}\left(\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}|\tilde{H}|^{p}\mathop{}\\!\mathrm{d}\lVert
W\rVert\right)^{\frac{1}{p}}\right]\\\
&-c(1+\rho)\left(\frac{1}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert W\rVert)\right)^{\frac{1}{p}}\end{split}$

###### Proof.

Without loss of generality we can suppose $x_{0}=0$. Since for large $\rho$
the statement is obvious, we have to prove it only for
$0<\rho<R(\mathcal{M})$, where $R(\mathcal{M})$ is the minimum radius of
curvature of $\mathcal{M}$. We want to bound from below the following
derivative:

(A.4)
$\begin{split}\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}\rho}&\left(\frac{1}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert W\rVert)\right)^{\frac{1}{p}}\\\
\quad&=\frac{1}{p}\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}\rho}\left(\frac{1}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert
W\rVert)\right)\left(\frac{1}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert W\rVert)\right)^{\frac{1-p}{p}}.\end{split}$

To do so, we want to bound from below the derivative in the right-hand side to
get a differential inequality. We have

$\begin{split}\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}\rho}&\left(\frac{1}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\big{(}\mathop{}\\!\mathrm{d}\lVert
V\rVert(x)+a\lVert W\rVert(x)\big{)}\right)\\\
&\quad=-\frac{1}{\rho^{3}}\int_{G_{2}(\mathcal{M})}\left(2\gamma\Big{(}\frac{|x|}{\rho}\Big{)}+\frac{|x|}{\rho}\gamma^{\prime}\Big{(}\frac{|x|}{\rho}\Big{)}\right)\mathop{}\\!\mathrm{d}(V(x,S)+aW(x,S)).\end{split}$

Let us choose $X(x)=\gamma\Big{(}\frac{|x|}{\rho}\Big{)}x$. Then

$\begin{split}\operatorname{div}_{S}X(x)=2\gamma\Big{(}\frac{|x|}{\rho}\Big{)}+\frac{|x|}{\rho}\gamma^{\prime}\Big{(}\frac{|x|}{\rho}\Big{)}\left|P_{S}\frac{x}{|x|}\right|^{2}.\end{split}$

We use Proposition 3.1: by testing (3.4) with $X$ we get the following
_monotonicity identity_ :

(A.5)
$\begin{split}\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}\rho}\left(\frac{1}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert
W\rVert)\right)=&-\frac{1}{\rho^{3}}\int_{G_{2}(\mathcal{M})}\operatorname{div}_{S}X(x)\mathop{}\\!\mathrm{d}\big{(}V(x,S)+aW(x,S)\big{)}\\\
&-\frac{1}{\rho^{3}}\int_{G_{2}(\mathcal{M})}\frac{|x|}{\rho}\gamma^{\prime}\Big{(}\frac{|x|}{\rho}\Big{)}\left|P_{S^{\perp}}\frac{x}{|x|}\right|^{2}\mathop{}\\!\mathrm{d}\big{(}V(x,S)+aW(x,S)\big{)}\\\
=&\frac{1}{\rho^{3}}\int_{\mathcal{M}}{X}\cdot{H+\tilde{H}}\mathop{}\\!\mathrm{d}\lVert
V\rVert-\frac{1}{\rho^{3}}\int_{\partial\mathcal{M}}{X}\cdot{N}\mathop{}\\!\mathrm{d}\sigma_{V}\\\
&+\frac{a}{\rho^{3}}\int_{\mathcal{M}}{X}\cdot{\tilde{H}}\mathop{}\\!\mathrm{d}\lVert
W\rVert\\\
&+-\frac{1}{\rho^{3}}\int_{G_{2}(\mathcal{M})}\frac{|x|}{\rho}\gamma^{\prime}\Big{(}\frac{|x|}{\rho}\Big{)}\left|P_{S^{\perp}}\frac{x}{|x|}\right|^{2}\mathop{}\\!\mathrm{d}\big{(}V(x,S)+aW(x,S)\big{)}.\end{split}$

We have to estimate from below the last member of the above identity.

Since $\gamma^{\prime}\leq 0$, we can neglect the last integral and, since
$|x|\leq\rho$ by cut-off, we obtain

(A.6)
$\begin{split}\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}\rho}\left(\frac{1}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert
W\rVert)\right)\geq&-\frac{1}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}|H+\tilde{H}|\mathop{}\\!\mathrm{d}\lVert
V\rVert-\frac{a}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}|\tilde{H}|\mathop{}\\!\mathrm{d}\lVert
W\rVert\\\
&-\frac{1}{\rho^{2}}\int_{\partial\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\Bigl{|}{\frac{x}{|x|}}\cdot{N}\Bigr{|}\mathop{}\\!\mathrm{d}\sigma_{V}.\end{split}$

We have now to bound the three terms in the right-hand side of (A.6).

  * •

For the first one, since $H\in L^{p}(\mathcal{M},\lVert V\rVert)$ and
$\tilde{H}\in L^{\infty}(\partial\mathcal{M},\lVert V\rVert)$, also
$H+\tilde{H}\in L^{p}(\mathcal{M},\lVert V\rVert)$. Therefore, by Hölder
inequality we get

(A.7)
$\begin{split}\frac{1}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}|H+\tilde{H}|\mathop{}\\!\mathrm{d}\lVert
V\rVert\leq&\frac{1}{\rho^{2}}\left(\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}|H+\tilde{H}|^{p}\mathop{}\\!\mathrm{d}\lVert
V\rVert\right)^{\frac{1}{p}}\left(\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}\lVert
V\rVert\right)^{1-\frac{1}{p}}.\end{split}$

  * •

For the second one, since $|\tilde{H}|\in L^{\infty}(\partial\mathcal{M})$, by
Hölder inequality and $\gamma\leq 1$ again, we obtain

(A.8)
$\begin{split}\frac{1}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}|\tilde{H}|\mathop{}\\!\mathrm{d}\lVert
W\rVert\leq&\frac{1}{\rho^{2}}\left(\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}|\tilde{H}|^{p}\mathop{}\\!\mathrm{d}\lVert
W\rVert\right)^{\frac{1}{p}}\left(\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}\lVert
W\rVert\right)^{1-\frac{1}{p}}.\end{split}$

  * •

We now move on the estimate of the third integral in the right-hand side of
(A.6). Since $\partial\mathcal{M}$ is of class $C^{2}$ and since
$0\in\partial\mathcal{M}$, there exists a constant $c$ such that

(A.9) $|{\frac{x}{|x|}}\cdot{N(x)}|\leq c|x|.$

This yields

(A.10)
$\frac{1}{\rho^{2}}\int_{\partial\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\big{|}{\frac{x}{|x|}}\cdot{N}\big{|}\mathop{}\\!\mathrm{d}\sigma_{V}\leq\frac{c}{\rho}\int\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}\sigma_{V}.$

We have to further estimate the righ-hand side of this inequality. This is
done by testing (A.1) with $X(x)=-\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\nabla
d(x)$; by standard computation we get

$\begin{split}\frac{1}{\rho}\int\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}\sigma_{V}=&-\frac{1}{\rho}\int_{G_{2}(\mathcal{M})}\gamma^{\prime}\Big{(}\frac{|x|}{\rho}\Big{)}\frac{1}{\rho}{P_{S}\frac{x}{|x|}}\cdot{\nabla
d(x)}\mathop{}\\!\mathrm{d}V(x,S)\\\
&-\frac{1}{\rho}\int_{G_{2}(\mathcal{M})}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\operatorname{div}_{S}\nabla
d(x)\mathop{}\\!\mathrm{d}V(x,S)\\\
&-\frac{1}{\rho}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}{\nabla
d(x)}\cdot{H+\tilde{H}}\mathop{}\\!\mathrm{d}\lVert V\rVert(x).\end{split}$

Since $\gamma^{\prime}(s)=0$ if $s\in(0,1/2)$, we have
$\frac{|x|}{\rho}\geq\frac{1}{2}$; Moreover, using
$|\operatorname{div}_{S}\nabla d|\leq c$ (because $\rho<R(\mathcal{M})$ and
$d$ is of class $C^{2}$ in $\mkern
1.0mu\overline{\mkern-1.0muU_{R(\mathcal{M})}(\partial\mathcal{M})\mkern-1.0mu}\mkern
1.0mu$ ) and (A.7), we obtain

(A.11)
$\begin{split}\frac{1}{\rho}\int\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}\sigma_{V}\leq&-\frac{2}{\rho}\int_{\mathcal{M}}\gamma^{\prime}\Big{(}\frac{|x|}{\rho}\Big{)}\frac{|x|}{\rho^{2}}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert
W\rVert)+\frac{c}{\rho}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert W\rVert)\\\
&+\rho^{1-\frac{2}{p}}\left(\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}|H+\tilde{H}|^{p}\mathop{}\\!\mathrm{d}\lVert
V\rVert\right)^{\frac{1}{p}}\left(\frac{1}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert W\rVert)\right)^{1-\frac{1}{p}}\\\
=&\frac{2}{\rho}\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}\rho}\left(\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert
W\rVert)\right)-\frac{2}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert W\rVert)\\\
&+\frac{2+c\rho}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert W\rVert)\\\
&+\rho^{1-\frac{2}{p}}\left(\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}|H+\tilde{H}|^{p}\mathop{}\\!\mathrm{d}\lVert
V\rVert\right)^{\frac{1}{p}}\left(\frac{1}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert W\rVert)\right)^{1-\frac{1}{p}}.\end{split}$

Concerning the last member, we now observe that

$\begin{split}\frac{2}{\rho}\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}\rho}\left(\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert
W\rVert)\right)-&\frac{2}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert W\rVert)\\\
=&2\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}\rho}\left(\frac{1}{\rho}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert W\rVert)\right).\end{split}$

Substituting in (A.11) we get

$\begin{split}\frac{1}{\rho}\int\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}\sigma_{V}\leq&\,\,2\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}\rho}\left(\frac{1}{\rho}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert W\rVert)\right)\\\
&+\frac{2+c\rho}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert W\rVert)\\\
&+\rho^{1-\frac{2}{p}}\left(\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}|H+\tilde{H}|^{p}\mathop{}\\!\mathrm{d}\lVert
V\rVert\right)^{\frac{1}{p}}\left(\frac{1}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert W\rVert)\right)^{1-\frac{1}{p}}.\end{split}$

Taking into account that

$\begin{split}\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}\rho}&\left(\frac{1}{\rho}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert W\rVert)\right)\\\
&\qquad=\frac{1}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert
W\rVert)+\rho\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}\rho}\left(\frac{1}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert W\rVert)\right),\end{split}$

we obtain

(A.12)
$\begin{split}\frac{1}{\rho}\int\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}\sigma_{V}\leq&\,\frac{4+c\rho}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert
W\rVert)+2\rho\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}\rho}\left(\frac{1}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert W\rVert)\right)\\\
&+\rho^{1-\frac{2}{p}}\left(\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}|H+\tilde{H}|^{p}\mathop{}\\!\mathrm{d}\lVert
V\rVert\right)^{\frac{1}{p}}\left(\frac{1}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert W\rVert)\right)^{1-\frac{1}{p}}.\end{split}$

This complete the estimate of the third term in the right-hand side of (A.6).

Gathering (A.7), (A.8), (A.10) and (A.12) in (A.6), we can estimate (A.4) as
follows:

$\begin{split}\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}\rho}\bigg{(}\frac{1}{\rho^{2}}&\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert W\rVert)\bigg{)}^{\frac{1}{p}}\\\
=&\frac{1}{p}\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}\rho}\left(\frac{1}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert
W\rVert)\right)\left(\frac{1}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert W\rVert)\right)^{\frac{1-p}{p}}\\\
\geq&-\frac{1+\rho}{p}\rho^{-\frac{2}{p}}\left(\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}|H+\tilde{H}|^{p}\mathop{}\\!\mathrm{d}\lVert
V\rVert\right)^{\frac{1}{p}}-\frac{a\rho^{-\frac{2}{p}}}{p}\left(\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}|\tilde{H}|^{p}\mathop{}\\!\mathrm{d}\lVert
W\rVert\right)^{\frac{1}{p}}\\\
&-c(1+\rho)\left(\frac{1}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert W\rVert)\right)^{\frac{1}{p}}\\\
&-c\rho\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}\rho}\left(\frac{1}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert W\rVert)\right)^{\frac{1}{p}},\end{split}$

which is the desired inequality. ∎

###### Proof of Proposition 3.2.

Without loss of generality we can assume $x_{0}=0\in\partial\mathcal{M}$. We
first notice that we have to prove the result only for small $\rho$, since it
is clearly true for large value of $\rho$.

If we call

(A.13)
$f(\rho):=\bigg{(}\frac{1}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert W\rVert)\bigg{)}^{\frac{1}{p}},$

by Lemma A.1, there exists a constant $\Lambda=\Lambda\big{(}c,\lVert
H\rVert_{L^{p}(\mathcal{M})},\mathcal{M},(\lVert V\rVert+a\lVert
W\rVert)(\mathcal{M})\big{)}>0$ (where $c$ is the constant given by Lemma A.1)
such that

(A.14) $f^{\prime}(\rho)+\Lambda f(\rho)+\Lambda\rho^{-\frac{2}{p}}\geq 0.$

Hence, by $p>2$,

(A.15) $f^{\prime}(\rho)+\Lambda
f(\rho)+\Lambda\rho^{-\frac{2}{p}}+\frac{\Lambda^{2}}{1-\frac{2}{p}}\rho^{1-\frac{2}{p}}\geq
0.$

Multiplying the above inequality by $e^{\Lambda\rho}$, we get

(A.16)
$\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}\rho}e^{\Lambda\rho}\left[\left(\frac{1}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert
W\rVert)\right)^{\frac{1}{p}}+\Lambda\rho^{1-\frac{2}{p}}\right]\geq 0.$

Since the constant $c$ of Lemma A.1 does not depend on the choice of $\gamma$
(the estimates are independent on the choice of $\gamma$ unless that
$\gamma^{\prime}(s)=0$ for $s\in(0,1/2)$), letting $\gamma$ increase to
$\mathbf{1}_{[0,1)}$ we have that the function

$\rho\longmapsto
e^{\Lambda\rho}\left[\left(\frac{1}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert
W\rVert)\right)^{\frac{1}{p}}+\Lambda\rho^{1-\frac{2}{p}}\right]$

is monotone increasing. ∎

###### Proof of Proposition 3.3.

Since $H\in L^{\infty}(\mathcal{M},\lVert V\rVert)$, in (A.7) and (A.8) we
simply estimate

(A.17)
$\begin{gathered}\frac{1}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}|H+\tilde{H}|\mathop{}\\!\mathrm{d}\lVert
V\rVert\leq\frac{\lVert
H+\tilde{H}\rVert_{\infty}}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}\lVert
V\rVert,\\\
\frac{1}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}|\tilde{H}|\mathop{}\\!\mathrm{d}\lVert
W\rVert\leq\frac{\lVert\tilde{H}\rVert_{\infty}}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}\lVert
W\rVert.\end{gathered}$

For $\rho^{-1}\int\gamma\mathop{}\\!\mathrm{d}\sigma_{V}$, we have

(A.18)
$\begin{split}\frac{1}{\rho}\int\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}\sigma_{V}\leq&\,\frac{4+c\rho}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert
W\rVert)+2\rho\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}\rho}\left(\frac{1}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert W\rVert)\right)\\\ &+\frac{\lVert
H+\tilde{H}\rVert_{\infty}}{\rho}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}\lVert
V\rVert.\end{split}$

Hence, computations similar to the proof of Lemma A.1 and of Proposition 3.2
yield

(A.19)
$\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}\rho}\bigg{(}\frac{1}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert
W\rVert)\bigg{)}\geq-\frac{\Lambda}{\rho^{2}}\int_{\mathcal{M}}\gamma\Big{(}\frac{|x|}{\rho}\Big{)}\mathop{}\\!\mathrm{d}(\lVert
V\rVert+a\lVert W\rVert)$

for some $\Lambda>0$ which depends only on $\mathcal{M},\lVert
H\rVert_{\infty},\lVert\tilde{H}\rVert_{\infty}$ (which depends on the second
fundamental form of $\partial\mathcal{M}$). By multiplying by
$e^{\Lambda\rho}$ and arguing as in the proof of Proposition 3.2, we obtain
the conclusion. ∎

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