Wulff inequality for minimal submanifolds in Euclidean space

It suffices to show that $\Phi|_{P}$ is the support function of $\text{proj}_{P}\mathcal{W}$, since support functions and Wulff shapes uniquely determine each other (see [CROS16, page 2976]). For any $\nu\in P\cap S^{n+m-1}$ and $x\in\mathcal{W}$, we have $x\cdot\nu=\text{proj}_{P}x\cdot\nu$, hence

(2.1)

$$\Phi|_{P}(\nu)=\sup\{x\cdot\nu:x\in\mathcal{W}\}=\sup\{\text{proj}_{P}x\cdot\nu:x\in\mathcal{W}\}=\sup\{y\cdot\nu:y\in\text{proj}_{P}\mathcal{W}\}.$$

∎

Let $\xi\in\mathcal{W}^{int}$ and we consider the map $w:\Sigma\xrightarrow{}\mathbb{R}$ defined by

(3.4)

$\displaystyle w(x)=u(x)-\langle x,\xi\rangle.$

Denote $\nu(x)$ the unit outward normal vector on the boundary $\partial\Sigma$ relative to $\Sigma$. Then, for any $x\in\partial\Sigma$,

(3.5)

$$\langle\nabla^{\Sigma}w(x),\nu(x)\rangle=\langle\nabla^{\Sigma}u(x),\nu(x)\rangle-\langle\xi,\nu(x)\rangle=\Phi(\nu)-\langle\xi,\nu(x)\rangle>0.$$

Thus $w$ attains its minimum at some interior point $\bar{x}$. It follows that $\nabla^{\Sigma}w(\bar{x})=0$, which implies

(3.6)

$\displaystyle\xi=\nabla^{\Sigma}u(\bar{x})+y$

for some $y\in N_{\bar{x}}\Sigma$. It remains to compute the determinant. Since $w$ achieves its minimum, we have

(3.7)

$$0\leq D^{2}_{\Sigma}w(\bar{x})=D^{2}_{\Sigma}u(x)-\langle\textrm{I\!I}_{\Sigma}(\bar{x}),\xi\rangle=D^{2}_{\Sigma}u(x)-\langle\textrm{I\!I}_{\Sigma}(\bar{x}),y\rangle,$$

where the last equality holds from (3.6). Therefore, we obtain $T(\bar{x},y)=\xi$ and $(\bar{x},y)\in A$. ∎

As shown in [Bre21, Lemma 5], one can show that

(3.9)

$\displaystyle\det DT(x,y)=\det(D^{2}_{\Sigma}u-\langle\textrm{I\!I}_{\Sigma}(x),y\rangle).$

By the definition of $A$, we have $\det(D^{2}_{\Sigma}u-\langle\textrm{I\!I}_{\Sigma}(x),y\rangle)\geq 0$. Thus, by applying the arithmetic-geometric mean inequality, the vanishing mean curvature property of minimal submanifolds, and (3.1), we obtain

(3.10)

$$\det(D^{2}_{\Sigma}u-\langle\textrm{I\!I}_{\Sigma}(x),y\rangle)\leq\left(\frac{\text{tr}(D^{2}_{\Sigma}u-\langle\textrm{I\!I}_{\Sigma}(x),y\rangle)}{n}\right)^{n}=\left(\frac{P_{\Phi}(\Sigma)}{n|\Sigma|}\right)^{n}.$$

∎

We recall

(3.11)

$$f\in\mathcal{F}_{n,m}=\{f\in L^{1}(\mathbb{R}^{n+m}):\textrm{supp}(f)\subset\mathcal{W},f\geq 0,\int_{P}f\leq 1\,\,\text{for}\,\,P\in\textrm{Graff}_{m}(\mathbb{R}^{n+m})\},$$

and by Lemma 3.1 and Lemma 3.2, we have

(3.12)		$\displaystyle\int_{\mathcal{W}}f(\xi)d\xi$	$\displaystyle\leq\int_{\Sigma}\int_{N_{x}\Sigma}f(T(x,y))|\det DT(x,y)|1_{A}(x,y)dydx$
		$\displaystyle\leq\left(\frac{P_{\Phi}(\Sigma)}{n|\Sigma|}\right)^{n}\int_{\Sigma}\int_{N_{x}\Sigma}f(\nabla^{\Sigma}u(x)+y)dydx$
		$\displaystyle\leq\left(\frac{P_{\Phi}(\Sigma)}{n|\Sigma|}\right)^{n}|\Sigma|$
		$\displaystyle=\left(\frac{P_{\Phi}(\Sigma)}{n|\Sigma|^{\frac{n-1}{n}}}\right)^{n}.$

In particular, the above inequality gives

(3.13)

$$\sup_{\mathcal{F}_{n,m}}\int_{\mathcal{W}}f\leq\left(\frac{P_{\Phi}(\Sigma)}{n|\Sigma|^{\frac{n-1}{n}}}\right)^{n},$$

which implies (1.7). It remains to consider the case where $\Sigma$ is disconnected. In that case, we apply the inequality to each individual connected component of $\Sigma$, sum over all connected components, and use the strict inequality

(3.14)

$\displaystyle a^{\frac{n-1}{n}}+b^{\frac{n-1}{n}}>a(a+b)^{-\frac{1}{n}}+b(a+b)^{-\frac{1}{n}}=(a+b)^{\frac{n-1}{n}}$

for all $a,b>0$.

∎

Suppose $\mathcal{W}$ is an ellipsoid given by

(4.4)

$$\frac{x_{1}^{2}}{\lambda_{1}^{2}}+...+\frac{x_{n+1}^{2}}{\lambda_{n+1}^{2}}\leq 1$$

where $0<\lambda_{1}\leq...\leq\lambda_{n+1}$. Denote

(4.5)

$\displaystyle\Lambda=\begin{pmatrix}\lambda_{1}^{-1}&&&&\\ &\lambda_{2}^{-1}&&&\\ &&\ddots&&\\ &&&&\lambda_{n+1}^{-1}\\ \end{pmatrix}.$

Let the function $f(x)$ be defined by

(4.6)

$$f(x)=\frac{1}{\sqrt{1-|\Lambda x|^{2}}}1_{\mathcal{W}}(x).$$

Consider a line $l(t)=t\boldsymbol{\alpha}+\boldsymbol{\omega}$ that passes through the ellipsoid along unit vector $\boldsymbol{\alpha}$ direction. Then the end points $T_{i}\boldsymbol{\alpha}+\boldsymbol{\omega}$ ($i=1,2$) of the chord in the ellipsoid satisfies

(4.7)

$$|\Lambda(T_{i}\boldsymbol{\alpha}+\boldsymbol{\omega})|^{2}=1,\quad i=1,2.$$

This is a quadratic equation

(4.8)

$$-aT_{i}^{2}+bT_{i}+c=0,\quad i=1,2,$$

where

(4.9)

$$a=|\Lambda\boldsymbol{\alpha}|^{2},\quad b=-2\langle\Lambda\boldsymbol{\alpha},\boldsymbol{\omega}\rangle,\quad c=1-|\Lambda\boldsymbol{\omega}|^{2}.$$

Upon solving this, we obtain

(4.10)

$$(\frac{2aT_{i}-b}{\sqrt{b^{2}+4ac}})^{2}=1.$$

Thus, the integral over the chord is given by

	$\displaystyle\int_{T_{1}}^{T_{2}}\frac{1}{\sqrt{1-|\Lambda(t\boldsymbol{\alpha}+\boldsymbol{\omega})|^{2}}}dt$
	$\displaystyle=\int_{T_{1}}^{T_{2}}\frac{1}{\sqrt{-at^{2}+bt+c}}dt$
	$\displaystyle=\left.\frac{1}{\sqrt{a}}\arcsin(\frac{2at-b}{\sqrt{b^{2}+4ac}})\right|_{T_{1}}^{T_{2}}$
(4.11)		$\displaystyle=\frac{\pi}{\sqrt{a}}.$

We notice that

(4.12)

$$\frac{\pi}{\sqrt{a}}=\frac{\pi}{|\Lambda\boldsymbol{\alpha}|}\leq\lambda_{n+1}\pi,$$

and the equality holds if and only if $\boldsymbol{\alpha}=(0,...,0,1)$, that is the line is parallel to the longest axis. We now define the function $\tilde{f}$ as

(4.13)

$$\tilde{f}=\frac{f}{\pi\lambda_{n+1}}\geq 0.$$

It satisfies its integral over any line that passes through the ellipsoid is at most 1, and achieves 1 when the line is parallel to the longest direction of ellipsoid, that is the axis corresponding to $\lambda_{n+1}$. Moreover, Fubini’s Theorem gives

(4.14)

$$\int_{\mathcal{W}}\tilde{f}=\int_{\text{proj}_{\{x_{n+1}=0\}}\mathcal{W}}\int_{\gamma_{x_{n+1}}}\tilde{f}(x)dx^{\prime}dx_{n+1}=|\text{proj}_{\{x_{n+1}=0\}}\mathcal{W}|=|W^{*}|.$$

Thus, $\tilde{f}\in\mathcal{F}_{n,1}$ and satisfies (4.1), and we have completed the proof. ∎

Denote $\mathcal{W}=E_{\lambda_{1},\lambda_{2},\cdots,\lambda_{n+m}}$ the $(n+m)$-dimensional ellipsoid

(4.17)

$\displaystyle\frac{x_{1}^{2}}{\lambda_{1}^{2}}+\frac{x_{2}^{2}}{\lambda_{2}^{2}}+\cdots+\frac{x_{n+m}^{2}}{\lambda_{n+m}^{2}}\leq 1,$

where $\lambda_{1}\leq\lambda_{2}\leq\cdots\leq\lambda_{n+m}$, and $U_{\sigma}=E_{\lambda_{1},\lambda_{2},\cdots,\lambda_{n+m}}\setminus E_{\sigma\lambda_{1},\sigma\lambda_{2},\cdots,\sigma\lambda_{n+m}}$ for some $0<\sigma<1$.

Let $P=\boldsymbol{q}+t_{1}\boldsymbol{r_{1}}+t_{2}\boldsymbol{r_{2}}+\cdots+t_{m}\boldsymbol{r_{m}}$ be an $m$-dimensional affine subspace which intersects with $U_{\sigma}$, where $\{\boldsymbol{r_{j}}\}_{j=1}^{m}$ are orthonormal. The intersection region $U_{\sigma}\cap P$ is the shell of two $m$-dimensional homothetic ellipsoids (if $P$ is very close to the boundary, it would be fully outside of $E_{\sigma\lambda_{1},\cdots,\sigma\lambda_{n+m}}$, so in that case $U_{\sigma}\cap P=E_{\lambda_{1},\cdots,\lambda_{n+m}}\cap P$ is just an ellipsoid). Denote

(4.18)

$\displaystyle\Lambda=\begin{pmatrix}\lambda_{1}^{-1}&&&&\\ &\lambda_{2}^{-1}&&&\\ &&\ddots&&\\ &&&&\lambda_{n+m}^{-1}\\ \end{pmatrix}.$

Note that $\Lambda(E_{\lambda_{1},\cdots,\lambda_{n+m}})=B^{n+m}$, so we can study the intersection between ellipsoid and lower dimensional plane by converting it to an intersection between ball and plane via $\Lambda$, and converting everything back in the end by applying appropriate scaling. (For a detailed computation for $m=1$ and $n=2$, see [Kle12].)

Specifically, the above computation yields that the length of the $i$-th axis of the outer ellipsoid $P\cap E_{\lambda_{1},\cdots,\lambda_{n+m}}$ is

(4.19)

$\displaystyle\frac{\sqrt{1-d}}{|\Lambda\boldsymbol{r_{i}}|},\quad 1\leq i\leq m,$

where $d=|\Lambda\boldsymbol{q}|^{2}-\frac{\langle\Lambda\boldsymbol{q},\Lambda\boldsymbol{r_{1}}\rangle^{2}}{|\Lambda\boldsymbol{r_{1}}|^{2}}-\frac{\langle\Lambda\boldsymbol{q},\Lambda\boldsymbol{r_{2}}\rangle^{2}}{|\Lambda\boldsymbol{r_{2}}|^{2}}-\cdots-\frac{\langle\Lambda\boldsymbol{q},\Lambda\boldsymbol{r_{m}}\rangle^{2}}{|\Lambda\boldsymbol{r_{m}}|^{2}}$ is the squared distance between origin and $\Lambda P$. The length of $i$-th axis of inner ellipsoid $P\cap E_{\sigma\lambda_{1},\cdots,\sigma\lambda_{n+m}}$ is

(4.20)

$\displaystyle\frac{\sqrt{\sigma^{2}-d}}{|\Lambda\boldsymbol{r_{i}}|},\quad 1\leq i\leq m.$

As discussed before, $U_{\sigma}\cap P$ is the shell formed by two homothetic ellipsoids only when $P$ cuts through $E_{\sigma\lambda_{1},\cdots,\sigma\lambda_{n+m}}$, which is the same as $\Lambda P$ cuts through $B^{n}(\sigma)$, or equivalently $d\leq\sigma^{2}$. Hence the intersection volume can be expressed as

(4.21)

$$|U_{\sigma}\cap P|=|B^{m}|[(1-d)^{m/2}-(\sigma^{2}-d)^{m/2}_{+}]\prod_{j=1}^{m}|\Lambda\boldsymbol{r_{j}}|^{-1}.$$

Since $\{\boldsymbol{r_{j}}\}_{j=1}^{m}$ is an orthonormal basis, we have

(4.22)

$$\prod_{j=1}^{m}|\Lambda\boldsymbol{r_{j}}|^{-1}\leq\prod_{j=1}^{m}\lambda_{n+m+1-j}.$$

Then it follows

(4.23)

$$|U_{\sigma}\cap P|\leq|B^{m}|\frac{m}{2}(1-\sigma^{2})\prod_{j=1}^{m}\lambda_{n+m+1-j}=:C_{\sigma}.$$

Now we consider

(4.24)

$$f_{\sigma}=\frac{1}{C_{\sigma}}1_{U_{\sigma}},$$

and it clearly satisfies $\text{supp}(f_{\sigma})\subset\mathcal{W}$ and $f_{\sigma}\geq 0$. For any $m$-dimensional affine subspace $P$ in $\mathbb{R}^{n+m}$,

(4.25)

$$\int_{P}f_{\sigma}=\frac{|U_{\sigma}\cap P|}{C_{\sigma}}\leq 1.$$

On the other hand, the volume of the shell $U_{\sigma}$ is given by the difference of two $(n+m)$-dimensional ellipsoids, so

	$\displaystyle\int_{\mathcal{W}}f_{\sigma}$	$\displaystyle=\frac{|U_{\sigma}|}{C_{\sigma}}=\frac{1}{C_{\sigma}}|B^{n+m}|(1-\sigma^{n+m})\prod_{j=1}^{n+m}\lambda_{j}$
(4.26)			$\displaystyle=\frac{2(1-\sigma^{n+m})}{1-\sigma^{2}}\frac{|B^{n+m}|}{m|B^{m}|}\prod_{j=1}^{n}\lambda_{j}.$

Taking the limit as $\sigma\xrightarrow{}1$, we obtain that

(4.27)

$$\lim_{\sigma\to 1}\int_{\mathcal{W}}f_{\sigma}=\frac{(n+m)|B^{n+m}|}{m|B^{m}|}\prod_{j=1}^{n}\lambda_{j}=\frac{(n+m)|B^{n+m}|}{m|B^{m}||B^{n}|}|W^{*}|,$$

where we used $|W^{*}|=|B^{n}|\prod_{j=1}^{n}\lambda_{j}$ for the ellipsoid. In particular, when $m=2$, the coefficient simplifies to 1, and we have completed the proof of (4.2). ∎

Suppose $\mathcal{W}_{2}\supset\mathcal{W}_{1}$ is obtained via a gluing operation. Then by Definition 1.7, $\mathcal{F}^{\mathcal{W}_{1}}_{n,m}\subset\mathcal{F}^{\mathcal{W}_{2}}_{n,m}$. Moreover, we also have

(4.30)

$\displaystyle|W_{2}^{*}|\geq|W_{1}^{*}|=|\text{proj}_{P}\mathcal{W}_{1}|=|\text{proj}_{P}\mathcal{W}_{2}|\geq|W_{2}^{*}|.$

Thus, we conclude that $|W_{1}^{*}|=|W_{2}^{*}|$, and $\text{proj}_{P}\mathcal{W}_{2}$ is an area-minimizing projection for $\mathcal{W}_{2}$. By assumption, we can thus find a sequence of $f_{\sigma}\in\mathcal{F}^{\mathcal{W}_{1}}_{n,m}\subset\mathcal{F}^{\mathcal{W}_{2}}_{n,m}$ such that

(4.31)

$$\lim_{\sigma\to 1}\int_{\mathcal{W}_{2}}f_{\sigma}=\lim_{\sigma\to 1}\int_{\mathcal{W}_{1}}f_{\sigma}=|W_{1}^{*}|=|W_{2}^{*}|,$$

which finishes the proof for gluing operation.

Now suppose $\mathcal{W}_{2}\subset\mathcal{W}_{1}$ is obtained via a cutting operation. By Definition 1.7, $\text{proj}_{P}\mathcal{W}_{1}\supset\text{proj}_{P}\mathcal{W}_{2}$. For any $x\in\text{proj}_{P}\mathcal{W}_{1}$, denote $Q_{x}=(\text{proj}_{P})^{-1}x$ the $m$-dimensional affine subspace orthogonal to $\text{proj}_{P}\mathcal{W}_{1}$ and passing through $x$. By assumption, we can find a sequence $f_{\sigma}\in\mathcal{F}^{\mathcal{W}_{1}}_{n,m}$ such that

(4.32)

$$\lim_{\sigma\to 1}\int_{\mathcal{W}_{1}}f_{\sigma}=|W_{1}^{*}|.$$

Use the fact that $\text{proj}_{P}\mathcal{W}_{1}$ is an area-minimizing projection, we must have for a.e. $x\in\text{proj}_{P}\mathcal{W}_{1}$,

(4.33)

$$\lim_{{\sigma\to 1}}\int_{Q_{x}\cap\mathcal{W}_{1}}f_{\sigma}=1.$$

On the other hand, for any $x\in\text{proj}_{P}\mathcal{W}_{2}$, by definition of cutting

(4.34)

$$\mathcal{W}_{2}=(\text{proj}_{P})^{-1}(\text{proj}_{P}\mathcal{W}_{2})\cap\mathcal{W}_{1}\supset(\text{proj}_{P})^{-1}x\cap\mathcal{W}_{1}=Q_{x}\cap\mathcal{W}_{1}$$

Certainly $Q_{x}\cap\mathcal{W}_{2}\subset Q_{x}\cap\mathcal{W}_{1}$, so $Q_{x}\cap\mathcal{W}_{2}=Q_{x}\cap\mathcal{W}_{1}$ for any $x\in\text{proj}_{P}\mathcal{W}_{2}$. Therefore, $f_{\sigma}|_{\mathcal{W}_{2}}\subset\mathcal{F}^{\mathcal{W}_{2}}_{n,m}$ satisfies

(4.35)

$$\lim_{{\sigma\to 1}}\int_{\mathcal{W}_{2}}f_{\sigma}=\lim_{{\sigma\to 1}}\int_{\text{proj}_{P}\mathcal{W}_{2}}\int_{Q_{x}\cap\mathcal{W}_{2}}f_{\sigma}=\lim_{{\sigma\to 1}}\int_{\text{proj}_{P}\mathcal{W}_{2}}\int_{Q_{x}\cap\mathcal{W}_{1}}f_{\sigma}=|\text{proj}_{P}\mathcal{W}_{2}|$$

Combining with (1.9), we obtain

(4.36)

$$|W_{2}^{*}|\leq|\text{proj}_{P}\mathcal{W}_{2}|=\lim_{\sigma\to 1}\int_{\mathcal{W}_{2}}f_{\sigma}|_{\mathcal{W}_{2}}\leq\sup_{f\in\mathcal{F}^{\mathcal{W}_{2}}_{n,m}}\int_{\mathcal{W}_{2}}f_{\sigma}\leq|W_{2}^{*}|.$$

Hence $\text{proj}_{P}\mathcal{W}_{2}$ must be area-minimizing projection, which finishes the proof for cutting operation. ∎