SYMMETRIC CONVEX SETS WITH MINIMAL GAUSSIAN SURFACE AREA

STEVEN HEILMAN

Abstract. Let Ω ⊆ ℝ^{n+1} have minimal Gaussian surface area among all sets satisfying Ω = −Ω with fixed Gaussian volume. Let A = A_{x} be the second fundamental form of ∂Ω at x, i.e. A is the matrix of first order partial derivatives of the unit normal vector at x ∈ ∂Ω. For any x = (x_{1}, …, x_{n+1}) ∈ ℝ^{n+1}, let γ_{n}(x) = (2π)^{−n/2}e^{−(x_{1}^{2}+···+x_{n+1}^{2})/2}. Let ∥A∥^{2} be the sum of the squares of the entries of A, and let ∥A∥_{2→2} denote the ℓ_{2} operator norm of A.

It is shown that if Ω or Ω^{c} is convex, and if either

∫_{∂Ω}(∥A_{x}∥^{2} − 1)γ_{n}(x)dx > 0  or  ∫_{∂Ω} (∥A_{x}∥^{2} − 1 + 2 sup_{y∈∂Ω} ∥A_{y}∥_{2→2}^{2})γ_{n}(x)dx < 0,

then ∂Ω must be a round cylinder. That is, except for the case that the average value of ∥A∥^{2} is slightly less than 1, we resolve the convex case of a question of Barthe from 2001.

The main tool is the Colding-Minicozzi theory for Gaussian minimal surfaces, which studies eigenfunctions of the Ornstein-Uhlenbeck type operator L = Δ − ⟨x, ∇⟩ + ∥A∥^{2} + 1 associated to the surface ∂Ω. A key new ingredient is the use of a randomly chosen degree 2 polynomial in the second variation formula for the Gaussian surface area. Our actual results are a bit more general than the above statement. Also, some of our results hold without the assumption of convexity.

1. Introduction

In a landmark investigation of mean curvature flow [CM12], Colding and Minicozzi studied a maximal version of the Gaussian surface area of an n-dimensional hypersurface Σ in ℝ^{n+1}. They called this quantity

sup_{a>0,b∈ℝ^{n+1}} ∫_{Σ} a^{− \frac{n}{2}} γ_{n}((x − b)a^{−1/2})dx  (1)

the “entropy” of Σ. The Colding-Minicozzi entropy (1) is of interest since it monotonically decreases under the mean curvature flow. For this reason, [CM12] studied minimizers of (1). Here, with m = n + 1, we define

γ_{n}(x) := (2π)^{−n/2}e^{−∥x∥^{2}/2},  ∥x∥^{2} := ∑_{i=1}^{m} x_{i}^{2},  ∀ x = (x_{1}, …, x_{m}) ∈ ℝ^{m}.

∫_{Σ} γ_{n}(x)dx := lim inf_{ε→0^{+}} \frac{1}{2ε} ∫_{\{x∈ℝ^{n+1} : ∃ y∈Σ, ∥x−y∥<ε\}} γ_{n}(x)dx.

Date: July 9, 2021. 2010 Mathematics Subject Classification. 60E15, 53A10, 60G15, 58E30. Key words and phrases. convex, symmetric, Gaussian, minimal surface, calculus of variations. Supported by NSF Grant DMS 1708908.

In the context of mean curvature flow, the Colding-Minicozzi entropy (1) is an analogue of Perelman’s reduced volume for Ricci flow. It was conjectured in [CIMW13] and ultimately proven in [Zhu20] that, among all compact n-dimensional hypersurfaces Σ ⊆ ℝ^{n+1} with ∂Σ = ∅, the round sphere minimizes the quantity (1).

Mean curvature flow refers to a set of orientable hypersurfaces {Σ_{s}}_{s≥0} such that (d/ds)x = −H(x)N(x), ∀ x ∈ Σ_{s}, ∀ s ≥ 0, where H(x) is the mean curvature of x ∈ Σ and N(x) is the exterior unit normal vector at x ∈ Σ_{s}. (See Section 2.1 for more detailed definitions.)

Influenced by the methods of [CM12], we study minimizers of the Gaussian surface area itself, over symmetric hypersurfaces Σ ⊆ ℝ^{n+1} enclosing a fixed Gaussian volume. We say a hypersurface Σ is symmetric if Σ = −Σ. Without the symmetry assumption, it is well-known that the set Ω ⊆ ℝ^{n+1} of fixed Gaussian volume ∫_{Ω} γ_{n+1}(x)dx and of minimal Gaussian surface area ∫_{∂Ω} γ_{n}(x)dx is a half space. That is, Ω is the set lying on one side of a hyperplane [SC74]. This result has been elucidated and strengthened over the years [Bor85, Led94, Led96, Bob97, BS01, Bor03, MN15a, MN15b, Eld15, MR15, BBJ16]. However, all of these proof methods (with the exception of [MR15, BBJ16]) seem unable to handle the additional constraint that the set Ω is symmetric. i.e. that Ω = −Ω. That is, new methods are needed to find symmetric sets Ω ⊆ ℝ^{n+1} of fixed Gaussian volume and minimal Gaussian surface area. In this work, we demonstrate that the calculus of variations techniques of [CM12, MR15, BBJ16] succeed in this task, where other proof strategies seem insufficient. Informally, the calculus of variations is a “local” proof strategy, whereas other proof strategies such as in [MN15b] or [Eld15] either directly or indirectly use “global” translation invariance of the problem at hand, in the sense that the translation of a half space is still a half space. So, the other methods cannot deal with the constraint Σ = −Σ, since a translation of such a Σ is no longer symmetric.

It was suggested by Barthe in 2001 [Bar01] that the symmetric set Ω of fixed Gaussian volume and minimal Gaussian surface area could be a symmetric strip bounded by two parallel hyperplanes. It was also expressed in [CR11, O’D12] that a Euclidean ball centered at the origin or its complement could minimize Gaussian surface area. A simple calculation demonstrates that the symmetric strip does not minimize Gaussian surface area for certain volume constraints. If t > 0 satisfies γ_{1}([−t, t]) = 1/2, then the Gaussian surface area of [−t, t] is \sqrt{2π}γ_{1}(\{−t\}∪\{t\}) ≈ 1.5932. If r > 0 and if B(0, r) = \{(x_{1}, x_{2}) ∈ ℝ^{2} : x_{1}^{2} +x_{2}^{2} ≤ r^{2}\} satisfies γ_{2}(B(0, r)) = 1/2, then ∫_{∂B(0,r)} γ_{1}(x)dx ≈ 1.4757. Also, the ball B(0, s) in ℝ^{3} with γ_{3}(B(0, s)) = 1/2 satisfies ∫_{∂B(0,s)} γ_{2}(x)dx ≈ 1.4496. So, at least for symmetric sets of Gaussian measure 1/2, the interval or the strip bounded by two hyperplanes does not minimize Gaussian surface area. Moreover, it even appears that the n-dimensional ball of Gaussian measure 1/2 has a decreasing surface area as n increases.

Define S^{n} ⊆ ℝ^{n+1} so that

S^{n} := {(x_{1}, …, x_{n+1}) ∈ ℝ^{n+1} : x_{1}^{2} + ⋯ + x_{n+1}^{2} = 1}.

From the Central Limit Theorem with error bound (also known as the Edgeworth Expansion) [Fel71, XVI.4.(4.1)], for any s ∈ ℝ, the following asymptotic expansion holds as n → ∞:

γ_{n} (B (0, \sqrt{n + s\sqrt{2n}} )) = ∫_{−∞}^{s} e^{−t^{2}/2}dt/\sqrt{2π} + \frac{(1 − s^{2})e^{−s^{2}/2}/\sqrt{2π}}{\sqrt{n}} + o(n^{−1/2}).

Moreover, from the Chain rule, (denoting B(0, r) = {x ∈ ℝ^{n}: ∥x∥ ≤ r}),

\frac{d}{ds}|_{s=0}γ_{n} (B (0, \sqrt{n + s\sqrt{2n}} )) = \frac{d}{dr}|_{r=\sqrt{n}} γ_{n}(B(0, r)) · \frac{d}{ds}|_{s=0}\sqrt{n + s\sqrt{2n}}

= \frac{1}{\sqrt{2π}} + o(n^{−1/2}).

That is,

lim_{n→∞} \frac{d}{dr}|_{r=\sqrt{n}} γ_{n}(B(0, r)) = \frac{1}{\sqrt{π}} ≈ .5642.

So, for symmetric sets of Gaussian measure 1/2, it seems plausible that their Gaussian surface area is at least 1/\sqrt{π}.

Morally, the results of [CIMW13, BW16, Zhu20] imply this result, but we cannot see a formal way of proving this statement.

In summary, it is believed that solid round cylinders (or their complements) minimize Gaussian surface area [Bar01, CR11, O’D12]. We state this as a conjecture below.

Problem 1.1. Fix 0 < c < 1. Minimize

∫_{∂Ω} γ_{n}(x)dx

over all subsets Ω ⊆ ℝ^{n+1} satisfying Ω = −Ω and γ_{n+1}(Ω) = c.

Remark 1.2. If Ω minimizes Problem 1.1, then Ω^{c} also minimizes Problem 1.1, with c replaced by 1 − c.

Conjecture 1.3 ([Bar01, CR11, O’D12]). Suppose Ω ⊆ ℝ^{n+1} minimizes Problem 1.1. Then, after rotating Ω, ∃ r > 0 and ∃ 0 ≤ k ≤ n such that

∂Ω = rS^{k} × ℝ^{n−k}.

Except for the case that the average value of the squared length of the second fundamental form is close to 1, we resolve the case of Conjecture 1.3 where Ω or Ω^{c} is convex. See Theorem 1.11 below.

Besides the relation of this problem to mean curvature flows [CM12, CMP15], recent interest for Gaussian isoperimetric inequalities has developed in theoretical computer science [KKMO07, MOO10, IM12]. A typical application reduces the computational hardness of a discrete computational problem to the solution of a Gaussian isoperimetric inequality. The resolution of a discrete problem using a continuous inequality can be surprising. For example, Borell’s isoperimetric inequality [Bor85, MN15b, Eld15] states that, among all Ω ⊆ ℝ^{n+1} with fixed Gaussian volume, the one with largest noise stability is a half space. Here the noise stability of Ω ⊆ ℝ^{n+1} with parameter 0 < ρ < 1 is

(1 − ρ^{2})^{− \frac{n+1}{2}} (2π)^{−(n+1)} ∫_{Ω} ∫_{Ω} e^{\frac{−∥x∥^{2}−∥y∥^{2}_{2}+2ρ⟨x,y⟩}{2(1−ρ^{2})}} dxdy.

This inequality of Borel generalizes the fact that half spaces minimize Gaussian surface area. And Borell’s inequality implies [KKMO07] that the Goemans-Williamson algorithm for MAX-CUT [GW95] is the best possible polynomial time algorithm, assuming the Unique Games Conjecture [Kho02].

The addition of the constraint that Ω = −Ω has particularly been related to the communication complexity of the Gap-Hamming-Distance problem [CR11, She12, Vid13]. As mentioned above, looking for the symmetric set Ω ⊆ ℝ^{n+1} of fixed Gaussian volume and minimal surface area is intrinsically interesting since previous proof strategies fail to solve this problem, with the exception of recent calculus of variations techniques [CM12, CIMW13, MR15, BBJ16].

The so-called S-inequality of [LaO99] seems superficially related to Problem 1.1. One part of this inequality says: if Ω ⊆ ℝ^{n+1} is a symmetric convex set, and if P is a symmetric strip lying between two hyperplanes such that γ_{n+1}(Ω) = γ_{n+1}(P), then for any t ≥ 1, γ_{n+1}(tΩ) ≥ γ_{n+1}(tP). This result of [LaO99] seems to have a rather different nature than Problem 1.1, since the first step of the proof of [LaO99] reduces to the case n + 1 = 2. As discussed above, Problem 1.1 cannot have such a reduction since symmetric strips do not minimize Problem 1.1 for the measure constraint c = 1/2. One of the difficulties of Problem 1.1 is dealing with higher-dimensional sets.

1.1. Colding-Minicozzi Theory for Mean Curvature Flow. For any x = (x_{1}, …, x_{n+1}) and y = (y_{1}, …, y_{n+1}) ∈ ℝ^{n+1} let ⟨x, y⟩ := ∑_{i=1}^{n+1} x_{i}y_{i} denote their standard inner product, and let ∥x∥ := \sqrt{⟨x, x⟩} denote the corresponding Euclidean norm.

The Colding-Minicozzi theory [CM12, CIMW13] focuses on orientable hypersurfaces Σ with ∂Σ = ∅ satisfying

H(x) = ⟨x, N(x)⟩, ∀ x ∈ Σ. (2)

Here N(x) is the unit exterior pointing normal to Σ at x, and H(x) is the mean curvature of Σ at x. Below, we will often omit the x arguments of H, N for brevity. H(x) is the sum of principal curvatures of Σ at x, or equivalently H(x) = div(N(x)). Here H is chosen so that, if r > 0, then the surface rS^{n} satisfies H(x) = n/r for all x ∈ rS^{n}. A hypersurface Σ satisfying (2) is called a self-shrinker, since it is self-similar under the mean curvature flow. Examples of self-shrinkers include a hyperplane through the origin, the sphere \sqrt{n}S^{n}, or more generally, round cylinders \sqrt{k}S^{k} × S^{n−k}, where 0 ≤ k ≤ n, and also cones with zero mean curvature.

Also, self-shrinkers model singularities of mean curvature flow. And Σ is a self-shrinker if and only if it is a critical point of Gaussian surface area, in the following sense: for any differentiable a: (−1, 1) → ℝ with a(0) = 1, for any differentiable b: (−1, 1) → ℝ^{n+1} with b(0) = 0, and for any normal variation {Σ_{s}}_{s∈(−1,1)} of Σ (as in (6)), we have

\frac{∂}{∂s}|_{s=0} ∫_{Σ_{s}}(a(s))^{− \frac{n}{2}} γ_{n}((x − b(s))(a(s))^{−1/2})dx = 0.

This equivalence was shown in [CM12, Proposition 3.6]. Put another way, self-shrinkers are critical points of Gaussian surface area, if we mod out by translations and dilations. In this paper, we instead study critical points of the Gaussian surface area itself. In this case, ∃ λ ∈ ℝ such that H(x) = ⟨x, N(x)⟩ + λ for all x ∈ Σ if and only if, for any normal variation {Σ_{s}}_{s∈(−1,1)} of Σ,

\frac{∂}{∂s}|_{s=0} ∫_{Σ_{s}} γ_{n}(x)dx = 0.

This fact is well-known, and reproven in Lemma 3.1 below. The special case λ = 0 recovers the self-shrinker equation (2).

The papers [CM12, CIMW13] led to several investigations. On the one hand, it was conjectured that, among all compact hypersurfaces, the round sphere minimizes the entropy (1). This conjecture was studied in [CIMW13] and [BW16] until its ultimate resolution in [Zhu20]. One main technical contribution of [Zhu20] was to extend the Colding-Minicozzi theory to handle perturbations of cones. Also, the main result of [CM12, Theorem 0.12] shows that round cylinders are the only C^{∞} self-shrinkers that locally minimize the entropy (1).

On the other hand, it is natural to study a generalization of surfaces satisfying (2) [MR15, Gua18, CW14, CW18, COW16, BBJ16]. That is, several papers have studied surfaces Σ such that there exists λ ∈ ℝ such that

H(x) = ⟨x, N(x)⟩ + λ,  ∀ x ∈ Σ.  (3)

Surfaces satisfying (3) are called λ-hypersurfaces.

As we just mentioned, the condition (3) is natural in the study of sets minimizing Gaussian surface area, since (3) holds if and only if Σ is a critical point of the Gaussian surface area (see Lemma 3.1).

A key aspect of the Colding-Minicozzi theory is the study of eigenfunctions of the differential operator L, defined for any C^{∞} function f: Σ → ℝ by

Lf(x) := Δf(x) − ⟨x, ∇f(x)⟩ + ∥A_{x}∥^{2} f(x) + f(x),  ∀ x ∈ Σ.  (4)

Here Δ is the Laplacian associated to Σ, ∇ is the gradient associated to Σ, A = A_{x} is the second fundamental form of Σ at x, and ∥A_{x}∥^{2} is the sum of the squares of the entries of the matrix A_{x}. Note that L is an Ornstein-Uhlenbeck-type operator. In particular, if Σ is a hyperplane, then A_{x} = 0 for all x ∈ Ω, so L is exactly the usual Ornstein-Uhlenbeck operator, plus the identity map. (More detailed definitions will be given in Section 2.1 below.)

The work [CM12] made the following crucial observation about the operator L. If (2) holds, then H is an eigenfunction of L with eigenvalue 2:

(2) ⟹ LH = 2H.  (5)

(See (24) below. Note that our definition of L differs from that of [CM12] since our Gaussian measure has a factor of 2, whereas their Gaussian measure has a factor of 4. Consequently, their L operator has different eigenvalues than ours.) The Colding-Minicozzi theory can readily solve Problem 1.1 in the special case that (2) holds (which is more restrictive than (3)). For illustrative purposes, we now sketch this argument, which closely follows [CM12, Theorem 4.30]. In particular, we use the following key insights of [CM12].

• H is an eigenfunction of L with eigenvalue 2. (That is, (5) holds.)

• The second variation formula for Gaussian surface area (17) is a quadratic form involving L.

• If H changes sign, then an eigenfunction of L exists with eigenvalue larger than 2.

Proposition 1.4 (Special Case of Conjecture 1.3). Let Ω ⊆ ℝ^{n+1} minimize Problem 1.1. By Lemma 3.1 below, ∃ λ ∈ ℝ such that (3) holds. Assume λ = 0. Assume also that Σ := ∂Ω is a compact, C^{∞} hypersurface. Then ∃ r > 0 such that ∂Ω = rS^{n}.

Proof. Let H be the mean curvature of Σ. If H ≥ 0, then Huisken’s classification [Hui90, Hui93] [CM12, Theorem 0.17] of compact surfaces satisfying (2) implies that Σ is a round sphere (∃ r > 0 such that Σ = rS^{n}.). So, we may assume that H changes sign. As noted in (5), LH = 2H. Since H changes sign, 2 is not the largest eigenvalue of L, by spectral theory [Zhu20, Lemma 6.5] (e.g. using that (L − ∥A∥^{2} − 2)^{−1} is a compact operator). That is, there exists a C^{2} function g: Σ → ℝ and there exists δ > 2 such that Lg = δg. Moreover, g > 0 on Σ. Since g > 0 and Σ = −Σ, it follows by (4) that g(x) + g(−x) is an eigenfunction of L with eigenvalue δ. That is, we may assume that g(x) = g(−x) for all x ∈ Σ.

Since Σ is not a round sphere, it suffices to find a nearby hypersurface of smaller Gaussian surface area. For any C^{2} function f: Σ → ℝ, and for any s ∈ (−1, 1), consider the hypersurface

Σ_{s} := {x + sN(x)f(x): x ∈ Σ}.  (6)

From the second variation formula, Lemma 3.2 below,

\frac{d^{2}}{ds^{2}} |_{s=0} ∫_{Σ_{s}} γ_{n}(x)dx = − ∫_{Σ} f(x)Lf(x)γ_{n}(x)dx.

So, to complete the proof, it suffices by Lemma 3.2 to find a C^{2} function f such that • f(x) = f(−x) for all x ∈ Σ. (f preserves symmetry.)

• ∫_{Σ} f(x)γ_{n}(x)dx = 0. (f preserves Gaussian volume.)

• ∫_{Σ} f(x)Lf(x)γ_{n}(x)dx > 0. (f decreases Gaussian surface area.) We choose g as above so that Lg = δg, δ > 2 and so that ∫_{Σ}(H(x) + g(x))γ_{n}(x)dx = 0. (Since H changes sign and g ≥ 0, g can satisfy the last equality by multiplying it by an appropriate constant.) We then define f := H + g. Then f satisfies the first two properties. So, it remains to show that f satisfies the last property. Note that, since H and g have different eigenvalues, they are orthogonal, i.e. ∫_{Σ} H(x)g(x)γ_{n}(x)dx = 0. Therefore,

∫_{Σ} f(x)Lf(x)γ_{n}(x)dx =^{(5)} ∫_{Σ}(H(x) + g(x))(2H(x) + δg(x))γ_{n}(x)dx
= 2 ∫_{Σ}(H(x))^{2}γ_{n}(x)dx + δ ∫_{Σ}(g(x))^{2}γ_{n}(x)dx > 0.

(Since H(x) = ⟨x, N(x)⟩ for all x ∈ Σ, and Σ is compact, both ∫_{Σ}(H(x))^{2}γ_{n}(x)dx and ∫_{Σ} H(x)γ_{n}(x)dx exist.) □

Remark 1.5. The case that ∂Ω is not compact can also be dealt with [CM12, Lemmas 9.44 and 9.45], [Zhu20, Proposition 6.11]. Instead of asserting the existence of g, one approximates g by a sequence of Dirichlet eigenfunctions on the intersection of Σ with large compact balls. Since Proposition 1.4 was presented only for illustrative purposes, and since the assumption (2) is too restrictive to resolve Problem 1.1, we will not present the details.

Unfortunately, the proof of Proposition 1.4 does not extend to the more general assumption (3). In order to attack Problem 1.1, we can only assume that (3) holds, instead of the more restrictive (2).

Under the assumption of (3), the proof of Proposition 1.4 breaks in at least two significant ways. First, H is no longer an eigenfunction of L when (3) holds with λ ≠ 0 (see (24) below).

Second, Huisken’s classification no longer holds [Hui90, Hui93]. Indeed, it is known that, for every integer m ≥ 3, there exists λ = λ_{m} < 0 and there exists a convex embedded curve Γ_{m} ⊆ ℝ^{2} satisfying (3) and such that Γ_{m} has m-fold symmetry (and Γ_{m_{1}} ≠ Γ_{m_{2}} if m_{1} ≠ m_{2}) [Cha17, Theorem 1.3, Proposition 3.2]. Consequently, Γ_{m} × ℝ^{n−2} ⊆ ℝ^{n+1} also satisfies (3). That is, Huisken’s classification cannot possibly hold, at least when λ < 0 in (3).

1.2. Our contribution. For any hypersurface Σ ⊆ ℝ^{n+1}, we define (using (4) and (7))

δ = δ(Σ) := sup_{f∈C_{0}^{∞}(Σ)} \frac{∫_{Σ} fLfγ_{n}(x)dx}{∫_{Σ} f^{2}γ_{n}(x)dx}.

The quantity δ(Σ) is denoted −μ_{1} in [CM12, Corollary 5.15].

In Section 7 below, we show that Huisken’s classification does actually hold for surfaces satisfying (3) if λ > 0, if the surface Σ encloses a convex region, and if δ(Σ) < ∞. Due to the following Lemma (see Lemma 5.2 below), we may always assume that δ(∂Ω) < ∞.

Lemma 1.6. If Ω ⊆ ℝ^{n+1} minimizes Problem 1.1, then δ(∂Ω) < ∞.

The assumption that δ(∂Ω) < ∞ is similar to assuming that ∫_{∂Ω} ∥A∥^{2} γ_{n}(x)dx < ∞.

Theorem 1.7 (Huisken-type classification, λ > 0). Let Ω ⊆ ℝ^{n+1} be a convex set. Let λ > 0. Assume that H(x) = ⟨x, N(x)⟩ + λ for all x ∈ ∂Ω. Assume δ(∂Ω) < ∞. Then, after rotating Ω, ∃ r > 0 and ∃ 0 ≤ k ≤ n such that ∂Ω = rS^{k} × ℝ^{n−k}.

Related to Huisken’s classification [Hui90, Hui93] [CM12, Theorem 0.17] are Bernstein theorems. If a hypersurface Σ satisfies (2) and Σ can be written as the graph of a function, then Σ is a hyperplane [EH89] [Wan11]. Also, if a hypersurface Σ satisfies (3), if Σ has polynomial volume growth and if Σ be written as the graph of a function, then Σ is a hyperplane [Gua18, Theorem 1.6] [CW14, Theorem 1.3]. In particular, if Ω = −Ω, if ∂Ω satisfies (3), and if Ω can be separated by a hyperplane into two sets, each of which is the graph of a function, then ∂Ω must consists of two parallel hyperplanes. In this sense, the symmetric strip separated by two parallel hyperplanes (or its complement) are “isolated critical points” in Problem 1.1.

Due to the Bernstein-type theorems of [Gua18, CW14], the main difficulty of Problem 1.1 occurs when Σ is not the graph of a function. Also, by Theorem 1.7, Lemma 1.6 and Lemma 3.1 below, in order to solve the convex case of Problem 1.1, it suffices to restrict to surfaces Σ such that there exists λ < 0 and such that H(x) = ⟨x, N(x)⟩ + λ, for all x ∈ Σ. As discussed above, the case λ < 0 is most interesting, since a Huisken-type classification cannot possibly hold when λ < 0. To deal with the case λ < 0, we use second variation arguments, as in [CM12, CIMW13].

We begin by using the mean curvature minus its mean in the second variation formula for Gaussian surface area.

Theorem 1.8 (Second Variation Using an Eigenfunction of L, λ < 0). Let Ω minimize Problem 1.1 and let Σ := ∂Ω. Then by Lemma 3.1, ∃ λ ∈ ℝ such that H(x) = ⟨x, N(x)⟩+λ, for any x ∈ Σ. Assume λ < 0. If

∫_{Σ}(∥A_{x}∥^{2} − 1)γ_{n}(x)dx > 0,

then, after rotating Ω, ∃ r > 0 and 0 ≤ k ≤ n so that Σ = rS^{k} × ℝ^{n−k}.

Theorem 1.8 follows from a slightly more general inequality in Lemma 6.1 below. In the case that λ < 0 and Ω is convex, the largest eigenvalue of L is at most 2, as shown in Lemma 5.12. With an eigenvalue bound smaller than 2, Lemma 6.1 would improve Theorem 1.8.

To handle the case when the average curvature of Σ is less than 1, we use our intuition about the sphere itself. On the sphere, the mean zero symmetric eigenfunctions of L which maximize the second variation of Gaussian surface area are degree two homogeneous spherical harmonics. This was observed in [Man17]. A similar observation was made in the context of noise stability in [Hei15]. In fact, if v, w ∈ S^{n}, if ⟨v, w⟩ = 0 and if Σ = S^{n} then ⟨v, N⟩⟨w, N⟩ is an eigenfunction of L. So, intuitively, if ∃ λ ∈ ℝ such that H(x) = ⟨x, N(x)⟩ + λ, then ⟨v, N⟩⟨w, N⟩ should also be an eigenfunction of L. Unfortunately, this does not seem to be true. Nevertheless, if we average over all possible choices of v, w ∈ S^{n}, then we can obtain a good bound in the second variation formula. And then there must exist v, w ∈ S^{n} whose second variation exceeds this average value.

If y ∈ Σ, then we define ∥A_{y}∥_{2→2}^{2} := sup_{v∈S^{n}} ∥A_{y}v∥^{2} to be the ℓ_{2} operator norm of A_{y}. Also, Π_{y}: ℝ^{n+1} → ℝ^{n} denotes the linear projection onto the tangent space of y ∈ Σ. (So ∥Π_{y}v∥^{2} = 1 − ⟨N(y), v/ ∥v∥⟩^{2} for any v ∈ ℝ^{n+1} \ {0}.)

As noted in Proposition 1.4, Problem 1.1 reduces to finding functions f: Σ → ℝ such that ∫_{Σ} fLfγ_{n}(x)dx is as large as possible.

Theorem 1.9 (Second Variation Using a Random Bilinear Function). Let Σ ⊆ ℝ^{n+1} be an orientable hypersurface with ∂Σ = ∅. Suppose ∃ λ ∈ ℝ such that H(x) = ⟨x, N(x)⟩ + λ for all x ∈ Σ. Let p := ∫_{Σ} γ_{n}(x)dx.

There exists v, w ∈ S^{n} so that, if m := \frac{1}{p} ∫_{Σ}⟨v, N⟩⟨w, N⟩γ_{n}(y)dy, we have

(n + 1)^{2} ∫_{Σ}(⟨v, N⟩⟨w, N⟩ − m)L(⟨v, N⟩⟨w, N⟩ − m)γ_{n}(x)dx

≥ \frac{1}{p^{2}} ∫_{Σ×Σ×Σ}(1 − ∥A_{x}∥^{2} − 2 ∥A_{y}∥_{2→2}^{2}) ∥Π_{y}(N(z))∥^{2} γ_{n}(x)γ_{n}(y)γ_{n}(z)dxdydz.

Note that convexity is not assumed in Theorem 1.9.

Theorem 1.9 actually follows from a slightly more general statement, Lemma 8.2 below. Theorem 1.9 is sharp for spheres, as observed by [Man17]. If r > 0, and if Σ = rS^{n}, then

∥A_{x}∥^{2} = n/r^{2} and ∥A_{y}∥_{2→2}^{2} = 1/r^{2}, so

1 − ∥A_{x}∥^{2} − 2 ∥A_{y}∥_{2→2}^{2} = \frac{r^{2} − n − 2}{r^{2}}.

If v, w ∈ S^{n} satisfy ⟨v, w⟩ = 0, then m = 0 and ∫_{Σ}⟨v, N⟩⟨w, N⟩L(⟨v, N⟩⟨w, N⟩)γ_{n}(x)dx ≥ 0 if and only if r ≥ \sqrt{n + 2} [Man17, Proposition 1].

Since Theorem 1.9 gives a bound on the second variation, Theorem 1.9 implies the following.

Corollary 1.10 (Second Variation Using a Random Bilinear Function). Let Ω minimize Problem 1.1 and let Σ := ∂Ω, so that ∃ λ ∈ ℝ such that H(x) = ⟨x, N(x)⟩ + λ for all x ∈ Σ. If

∫_{Σ×Σ×Σ}(1 − ∥A_{x}∥^{2} − 2 ∥A_{y}∥_{2→2}^{2}) ∥Π_{y}(N(z))∥^{2} γ_{n}(x)γ_{n}(y)γ_{n}(z)dxdydz > 0,

then, after rotating Ω, ∃ r > 0 and ∃ 0 ≤ k ≤ n so that Σ = rS^{k} × ℝ^{n−k}.

The combination of Remark 1.2, Theorems 1.7 and 1.8 and Corollary 1.10 implies the following.

Theorem 1.11 (Main Result). Let Ω minimize Problem 1.1 and let Σ := ∂Ω. Assume that Ω or Ω^{c} is convex. If

∫_{Σ}(∥A_{x}∥^{2} − 1)γ_{n}(x)dx > 0,

or

∫_{Σ}(∥A_{x}∥^{2} − 1 + 2 sup_{y∈Σ} ∥A_{y}∥_{2→2}^{2})γ_{n}(x)dx < 0,

then, after rotating Ω, ∃ r > 0 and ∃ 0 ≤ k ≤ n so that Σ = rS^{k} × ℝ^{n−k}.

So, except for the case that the average value of ∥A∥^{2} is slightly less than 1, we resolve the convex case of Barthe’s Conjecture 1.3.

In Section 9 we adapt an argument of [CM12] that allows the computation of the second variation of Gaussian volume preserving normal variations, which simultaneously can dilate the hypersurface Σ. When we use the function H −λ in this second variation formula, we get zero. This suggests the intriguing possibility that the fourth variation of H − λ could help to solve Problem 1.1. Instead of embarking on a rather technical enterprise of computing Gaussian volume preserving fourth variations, we instead put the function H − λ + t, t ∈ ℝ into this second variation formula, and we then differentiate twice in t. We then arrive at the following interesting inequality.

Theorem 1.12. Let Ω minimize Problem 1.1 and let Σ := ∂Ω. Assume also that Σ is a compact, C^{∞} hypersurface and Ω is convex. Then

∫_{Σ} ( − ∥A_{x}∥^{2} + H(x)\frac{∫_{Σ} γ_{n}(z)dz}{∫_{Σ}⟨y, N⟩γ_{n}(y)dy} )γ_{n}(x)dx ≥ 0.

This inequality is rather interesting since it is equal to zero exactly when Σ = rS^{n}, for any r > 0, since then ∥A_{x}∥^{2} = n/r^{2}, H(x) = n/r, and ⟨x, N⟩ = r for all x ∈ rS^{n}. So, one might speculate that round spheres are the only compact C^{∞} hypersurfaces, where this quantity is nonnegative, and where ∃ λ ∈ ℝ such that H(x) = ⟨x, N(x)⟩ + λ for all x ∈ Σ

Finally, we show that Theorem 1.11 can be partially generalized to the non-convex case.

Theorem 1.13 (Weak Main Result, Without Convexity). Let Ω minimize Problem 1.1 and let Σ := ∂Ω. From Lemma 3.1 below, ∃ λ ∈ ℝ such that H(x) = ⟨x, N(x)⟩ + λ, ∀ x ∈ Σ. If

∫_{Σ}(∥A_{x}∥^{2} − 1)γ_{n}(x)dx > 0 and − λ ∫_{Σ}⟨x, N(x)⟩γ_{n}(x)dx > 0,

or

∫_{Σ}(∥A_{x}∥^{2} − 1 + 2 sup_{y∈Σ} ∥A_{y}∥_{2→2}^{2})γ_{n}(x)dx < 0,

then, after rotating Ω, ∃ r > 0 and ∃ 0 ≤ k ≤ n so that Σ = rS^{k} × ℝ^{n−k}.

Conjecture 1.3 and our results for it only specify that some cylinder minimizes the Gaussian surface area among all sets of fixed Gaussian volume. That is, the dimension of the cylinder that minimizes Gaussian surface area is not specified. Upon seeing our initial preprint, Frank Morgan suggested the following strengthened version of Conjecture 1.3, which appears to be verified by numerical computations, at least when n ≤ 6. In Conjecture 1.14 below, the cylinder of minimal Gaussian surface area is identified.

Conjecture 1.14 (Morgan’s Conjecture). There exists a sequence of real numbers 1 = a_{0} > a_{1} > a_{2} > ⋯ > 1/2 such that (1/2, 1] = ∪_{k=0}^{∞}(a_{k+1}, a_{k}] and such that the following holds. Fix 1/2 < c < 1. Let n ≥ 0. Suppose Ω_{n+1} ⊆ ℝ^{n+1} minimizes

∫_{∂Ω} γ_{n}(x)dx

over all subsets Ω ⊆ ℝ^{n+1} satisfying Ω = −Ω and γ_{n+1}(Ω) = c. Let k be the unique nonnegative integer such that c ∈ (a_{k+1}, a_{k}]. Then,

∫_{∂Ω_{k+1}} γ_{k}(x)dx = min_{n≥0} ∫_{∂Ω_{n+1}} γ_{n}(x)dx.

Also there exists r = r(c, k) > 0 such that

Ω_{k+1} = {(x_{1}, …, x_{k+1}) ∈ ℝ^{k+1} : x_{1}^{2} + ⋯ + x_{k+1}^{2} ≤ r^{2}}.

That is, the minimum Gaussian surface area of all (measurable) sets of Gaussian measure 1/2 < c < 1 occurs for the ball in ℝ^{k+1} centered at the origin when c ∈ (a_{k+1}, a_{k}]. By Remark 1.2, the above statement holds for any 0 < c < 1/2 by taking complements. That is, the minimum Gaussian surface area of all (measurable) sets of Gaussian measure 0 < c < 1/2 occurs for the complement of the ball in ℝ^{k} centered at the origin when 1 − c ∈ (a_{k+1}, a_{k}].

Lastly, in the case c = 1/2, the ball (or its complement) minimizes Gaussian surface area, asymptotically as n → ∞:

inf_{n≥0} ∫_{∂Ω_{n+1}} γ_{n}(x)dx = lim_{n→∞} ∫_{∂Ω_{n+1}} γ_{n}(x)dx = lim_{n→∞} ∫_{\sqrt{n+1}S^{n}} γ_{n}(x)dx = \frac{1}{\sqrt{π}} ≈ .56419 ….

At present there seems to be no sensible way to analytically find the numbers a_{1}, a_{2}, . . ..

measure constraint c = γ_{n+1}(Ω) Optimal k for ∂Ω = S^{k} × ℝ^{n−k} 1 23 45⋮ ∞  ⋮  ⋮  ⋮⋮ 1 1/2 0  ⋮⋮  a_{1} a_{2} ⋯

Figure 1. Conjecture 1.14 of Morgan.

1.3. Organization.

• Preliminary details are covered in Sections 2 through 4.

• A rather technical section on curvature bounds is given in Section 5.

• Theorem 1.7 is proven in Section 7. Theorem 1.8 is proven in Section 6.

• Theorem 1.9, Corollary 1.10 and Theorem 1.11 are proven in Section 8

• Theorem 1.12 is proven in Section 9.

• Theorem 1.13 is proven in Section 11.

1.4. Discussion. As in the proof of Proposition 1.4, the proof of Theorem 1.11 tries to find a function f: Σ → ℝ with f(x) = f(−x), ∀ x ∈ Σ, ∫_{Σ} f(x)γ_{n}(x)dx = 0, and such that ∫_{Σ} f(x)Lf(x)γ_{n}(x)dx > 0. In Proposition 1.4, this is achieved by letting f be the sum of two distinct eigenfunctions of L with positive eigenvalues. It could occur that L has only one symmetric eigenfunction with a positive eigenvalue, but still we could find a symmetric f with zero mean and ∫_{Σ} f(x)Lf(x)γ_{n}(x)dx > 0. It would be interesting to explore this possibility, since the proof strategy of Proposition 1.4 fails in this case. And indeed, in the proof of the Main Theorem, we have to compensate for the fact that we cannot find explicit eigenfunctions of L. Also, it would be interesting to see if any Ω exists that evades all constraints put upon it by the results in this work (e.g. by Theorems 1.11, 1.13 or Corollary 11.9). To the author’s knowledge, no such Ω is known to exist.

2. Preliminaries

We say that Σ ⊆ ℝ^{n+1} is an n-dimensional C^{∞} manifold with boundary if Σ can be locally written as the graph of a C^{∞} function.

For any (n + 1)-dimensional C^{∞} manifold Ω ⊆ ℝ^{n+1} with boundary, we denote

\begin{aligned}C_{0}^{∞}(Ω; ℝ^{n+1}) & := \{f: Ω → ℝ^{n+1} : f ∈ C^{∞}(Ω; ℝ^{n+1}), f(∂Ω) = 0, \\ & ∃ r > 0, f(Ω ∩ (B(0, r))^{c}) = 0\}. \end{aligned}  (7)

We also denote C_{0}^{∞}(Ω) := C_{0}^{∞}(Ω; ℝ). We let div denote the divergence of a vector field in ℝ^{n+1}. For any r > 0 and for any x ∈ ℝ^{n+1}, we let B(x, r) := {y ∈ ℝ^{n+1} : ∥x − y∥ ≤ r} be the closed Euclidean ball of radius r centered at x ∈ ℝ^{n+1}.

Definition 2.1 (Reduced Boundary). A measurable set Ω ⊆ ℝ^{n+1} has locally finite surface area if, for any r > 0,

sup {∫_{Ω} div(X(x))dx: X ∈ C_{0}^{∞}(B(0, r), ℝ^{n+1}), sup_{x∈ℝ^{n+1}} ∥X(x)∥ ≤ 1} < ∞.

Equivalently, Ω has locally finite surface area if ∇1_{Ω} is a vector-valued Radon measure such that, for any x ∈ ℝ^{n+1}, the total variation

∥∇1_{Ω}∥ (B(x, 1)) := sup_{partitions C_{1},...,C_{m} of B(x,1) m≥1} ∑_{i=1}^{m} ∥∇1_{Ω}(C_{i})∥

is finite [CL12].

If Ω ⊆ ℝ^{n+1} has locally finite surface area, we define the reduced boundary ∂^{∗}Ω of Ω to be the set of points x ∈ ℝ^{n+1} such that

N(x) := − lim_{r→0^{+}} \frac{∇1_{Ω}(B(x, r))}{∥∇1_{Ω}∥ (B(x, r))}

exists, and it is exactly one element of S^{n}.

For more background on the reduced boundary and its regularity, we refer to the discussion in Section 2 of [BBJ16], [AFP00] and [Mag12].

The following argument is essentially identical to [BBJ16, Proposition 1], so we omit the proof.

Lemma 2.2 (Existence). There exists a set Ω ⊆ ℝ^{n+1} minimizing Problem 1.1.

2.1. Submanifold Curvature. Here we cover some basic definitions from differential geometry of submanifolds of Euclidean space.

Let ∇ denote the standard Euclidean connection, so that if X, Y ∈ C_{0}^{∞}(ℝ^{n+1}, ℝ^{n+1}), if Y = (Y_{1}, …, Y_{n+1}), and if u_{1}, …, u_{n+1} is the standard basis of ℝ^{n+1}, then \overline{∇}_{X}Y := ∑_{i=1}^{n+1}(X(Y_{i}))u_{i}. Let N be the outward pointing unit normal vector of an n-dimensional hypersurface Σ ⊆ ℝ^{n+1}. For any vector x ∈ Σ, we write x = x^{T} +x^{N}, so that x^{N} := ⟨x, N⟩N is the normal component of x, and x^{T} is the tangential component of x ∈ Σ. We let ∇^{Σ} := (∇)^{T} denote the tangential component of the Euclidean connection.

Let e_{1}, …, e_{n} be an orthonormal frame of Σ ⊆ ℝ^{n+1}. That is, for a fixed x ∈ Σ, there exists a neighborhood U of x such that e_{1}, …, e_{n} is an orthonormal basis for the tangent space of Σ, for every point in U [Lee03, Proposition 11.17].

Define the mean curvature

H := div(N) = ∑_{i=1}^{n}⟨∇_{e_{i}}N, e_{i}⟩.  (8)

Define the second fundamental form A = (a_{ij})_{1≤i,j≤n} so that

a_{ij} = ⟨∇_{e_{i}}e_{j}, N⟩,  ∀ 1 ≤ i, j ≤ n.  (9)

Compatibility of the Riemannian metric says a_{ij} = ⟨∇_{e_{i}}e_{j}, N⟩ = −⟨e_{j}, ∇_{e_{i}}N⟩ + e_{i}⟨N, e_{j}⟩ = −⟨e_{j}, ∇_{e_{i}}N⟩, ∀ 1 ≤ i, j ≤ n. So, multiplying by e_{j} and summing this equality over j gives

∇_{e_{i}}N = − ∑_{j=1}^{n} a_{ij}e_{j},  ∀ 1 ≤ i ≤ n.  (10)

Using ⟨∇_{N}N, N⟩ = 0,

H =^{(8)} ∑_{i=1}^{n}⟨∇_{e_{i}}N, e_{i}⟩ =^{(10)} − ∑_{i=1}^{n} a_{ii}.  (11)

2.2. First and Second Variation. We will apply the calculus of variations to solve Problem 1.1. Here we present the rudiments of the calculus of variations.

The results of this section are well known to experts in the calculus of variations, and many of these results were re-proven in [BBJ16].

Let Ω ⊆ ℝ^{n+1} be an (n+1)-dimensional C^{2} submanifold with reduced boundary Σ := ∂^{∗}Ω. Let N: ∂^{∗}Ω → S^{n} denote the unit exterior normal to ∂^{∗}Ω. Let X: ℝ^{n+1} → ℝ^{n+1} be a vector field. Unless otherwise stated, we always assume that X(x) is parallel to N(x) for all x ∈ ∂^{∗}Ω. That is,

X(x) = ⟨X(x), N(x)⟩N(x), ∀ x ∈ ∂^{∗}Ω. (12)

Let div denote the divergence of a vector field. We write X in its components as X = (X_{1}, …, X_{n+1}), so that divX = ∑_{i=1}^{n+1} \frac{∂}{∂x_{i}} X_{i}. Let Ψ: ℝ^{n+1} × (−1, 1) → ℝ^{n+1} such that

Ψ(x, 0) = x,  \frac{d}{ds}Ψ(x, s) = X(Ψ(x, s)), ∀ x ∈ ℝ^{n+1}, s ∈ (−1, 1).  (13)

For any s ∈ (−1, 1), let Ω_{s} := Ψ(Ω, s). Note that Ω_{0} = Ω. Let Σ_{s} := ∂^{∗}Ω_{s}.

Definition 2.3. We call {Ω_{s}}_{s∈(−1,1)} as defined above a normal variation of Ω ⊆ ℝ^{n+1}. We also call {Σ_{s}}_{s∈(−1,1)} a normal variation of Σ = ∂Ω.

Lemma 2.4 (First Variation). Let X ∈ C_{0}^{∞}(ℝ^{n+1}, ℝ^{n+1}). Let f(x) = ⟨X(x), N(x)⟩ for any x ∈ ∂^{∗}Ω. Then

\frac{d}{ds}|_{s=0}γ_{n+1}(Ω_{s}) = ∫_{∂^{∗}Ω} f(x)γ_{n+1}(x)dx.  (14)
\frac{d}{ds}|_{s=0} ∫_{∂^{∗}Ω_{s}} γ_{n}(x)dx = ∫_{∂^{∗}Ω}(H(x) − ⟨N(x), x⟩)f(x)γ_{n}(x)dx.  (15)

Lemma 9.8 below (with h = 0) implies (14) and Lemma 9.3 below (with t_{s} = 0 for all s ∈ (−1, 1)) implies (15).

Lemma 2.5 (Second Variation). Let X ∈ C_{0}^{∞}(ℝ^{n+1}, ℝ^{n+1}). Let f(x) = ⟨X(x), N(x)⟩ for all x ∈ ∂^{∗}Ω. Then

\frac{d^{2}}{ds^{2}} |_{s=0}γ_{n+1}(Ω_{s}) = ∫_{∂^{∗}Ω} f(∇_{N}f + f(H − ⟨x, N⟩))γ_{n+1}(x)dx.  (16)

\frac{d^{2}}{ds^{2}} |_{s=0} ∫_{∂^{∗}Ω_{s}} γ_{n}(x)dx = ∫_{∂^{∗}Ω} (−fLf +(H−⟨x, N⟩)(f∇_{N}f +f^{2}(H−⟨x, N⟩)))γ_{n}(x)dx. (17)

Lemma 9.9 (with h = h′ = 0) implies (16) and Lemma 9.6 (with h = h′ = 0 and f′ = f∇_{N}f by Lemma 9.5) implies (17).

3. Variations and Regularity

In this section, we show that a minimizer of Problem 1.1 exists, and the boundary of the minimizer is C^{∞} except on a set of Hausdorff dimension at most n − 7.

Much of this section is a modification of corresponding parts of [BBJ16].

Unless otherwise stated, all sets Ω ⊆ ℝ^{n+1} below are assumed to be measurable sets of locally finite surface area, and such that the Gaussian surface area of Ω, ∫_{∂^{∗}Ω} γ_{n}(x)dx is finite.

Lemma 3.1 (First Variation for Minimizers). Let Ω ⊆ ℝ^{n+1} minimize Problem 1.1. Let Σ := ∂^{∗}Ω. Then there exists λ ∈ ℝ such that, for any x ∈ ∂^{∗}Ω, H(x) − ⟨x, N(x)⟩ = λ.

Proof. Let f: ∂^{∗}Ω → ℝ with ∫_{∂^{∗}Ω} f(x)γ_{n}(x)dx = 0 and f(x) = f(−x) for all x ∈ ∂^{∗}Ω. From Lemma 2.4,

∫_{∂^{∗}Ω}(H(x) − ⟨N(x), x⟩)f(x)γ_{n}(x)dx = 0.

Since Ω = −Ω, this becomes

∫_{(∂^{∗}Ω)∩{x∈ℝ^{n+1} : x_{1}≥0}}(H(x) − ⟨N(x), x⟩)f(x)γ_{n}(x)dx = 0.

This equality is true for any function f such that ∫_{(∂^{∗}Ω)∩{x∈ℝ^{n+1} : x_{1}≥0}} f(x)γ_{n}(x)dx = 0. So, there exists λ ∈ ℝ such that, for any x ∈ ∂^{∗}Ω with x_{1} ≥ 0, H(x) − ⟨N(x), x⟩ = λ. Since Ω = −Ω, we then have H(x) − ⟨N(x), x⟩ = λ for any x ∈ ∂^{∗}Ω with x_{1} ≤ 0. □

Lemma 3.2 (Second Variation for Minimizers). Let Ω ⊆ ℝ^{n+1} minimize Problem 1.1. Let Σ := ∂^{∗}Ω. Then, for any f ∈ C_{0}^{∞}(Σ) such that ∫_{Σ} f(x)γ_{n}(x)dx = 0, and such that f(x) = f(−x) for all x ∈ Σ, we have

∫_{Σ} f(x)Lf(x)γ_{n}(x)dx ≤ 0.

Proof. From Lemma 3.1 there exists λ ∈ ℝ such that, for any x ∈ ∂^{∗}Ω, H(x)−⟨x, N(x)⟩ = λ. Let f: Σ → ℝ satisfy ∫_{Σ} f(x)γ_{n}(x)dx = 0, and such that f(x) = f(−x) for all x ∈ Σ. We extend f to a neighborhood U ⊆ ℝ^{n+1} of Σ (by e.g. Whitney extension [Ste70]), and we denote this extension by f also, so that

∇_{N}f(x) := −λf(x),  ∀ x ∈ Σ.  (18)

Then ∇_{N}f(x) = ∇_{N}f(−x) for all x ∈ Σ.

For any x ∈ ℝ^{n+1}, denote dist(x, ∂^{∗}Ω) := inf{∥y − x∥ : y ∈ ∂^{∗}Ω}. Define the signed distance function d_{Ω} : ℝ^{n+1} → ℝ by

d_{Ω}(x) := \begin{cases}dist(x, ∂^{∗}Ω), & if x ∈ ℝ^{n+1} \ Ω \\ −dist(x, ∂^{∗}Ω), & if x ∈ Ω. \end{cases}

We then define X: ℝ^{n+1} → ℝ^{n+1} so that

X(x) := \begin{cases}f(x)∇d_{Ω}(x) & , if x ∈ U \\ 0 & , if x ∈ ℝ^{n} \ U. \end{cases}  (19)

Let {Ω_{s}}_{s∈(−1,1)} be the normal variation of Ω associated to X.

Since f(x) = f(−x) for all x ∈ Σ and ∇_{N}f(x) = ∇_{N}f(−x) for all x ∈ Σ, sets Ω_{s} are symmetric to first and second order in s near s = 0 (by (51) and Lemma 9.5.)

By (15) and (14), and using the assumption that f has mean zero,

\frac{d}{ds}|_{s=0} ∫_{Σ_{s}} γ_{n}(x)dx = ∫_{Σ_{s}}(H(x) − ⟨x, N(x)⟩)f(x)γ_{n}(x)dx = λ ∫_{Σ_{s}} f(x)γ_{n}(x)dx = 0.

\frac{d}{ds}|_{s=0} ∫_{Ω_{s}} γ_{n+1}(x)dx = ∫_{Σ} f(x)γ_{n+1}(x)dx = 0.

Also, by (16),

\frac{d^{2}}{ds^{2}} |_{s=0} ∫_{Ω_{s}} γ_{n+1}(x)dx = ∫_{Σ} f(x)(∇_{N}f(x) + f(x)(H(x) − ⟨x, N(x)⟩))γ_{n+1}(x)dx

= ∫_{Σ} f(x)(∇_{N}f(x) + λf(x))γ_{n+1}(x)dx =^{(18)} 0.

In summary, the vector field X preserves the symmetry of Ω_{s} to first and second order at s = 0, and the vector field X preserves the Gaussian volume of Ω_{s} to second order at s = 0. Since Ω minimizes Problem 1.1, we must therefore have

\frac{d^{2}}{ds^{2}} |_{s=0} ∫_{∂^{∗}Ω_{s}} γ_{n}(x)dx ≥ 0.

Finally, by (17),

0 ≤ \frac{d^{2}}{ds^{2}} |_{s=0} ∫_{∂^{∗}Ω_{s}} γ_{n}(x)dx

= ∫_{∂^{∗}Ω} ( − fLf + f(H − ⟨x, N⟩)(∇_{N}f + f(H − ⟨x, N⟩)))γ_{n}(x)dx =^{(18)} ∫_{∂^{∗}Ω} −fLfγ_{n}(x)dx.

□

Let g: ℝ^{n+1} → ℝ. We define ∥g∥_{∞} := sup{t ≥ 0: γ_{n+1}(x ∈ ℝ^{n+1} : |g(x)| > t) > 0}. Also, for any 0 < σ < 1, we define

∥g∥_{C^{1,σ}} := ∥g∥_{∞} + ∥∥∇g∥∥_{∞} + max_{i=1,...,n} sup_{x,y∈ℝ^{n+1} : x≠y} \frac{|\frac{∂}{∂x_{i}} g(x) − \frac{∂}{∂x_{i}} g(y)|}{∥x − y∥^{σ}}.

Lemma 3.3 (Existence and Regularity). The minimum value of Problem 1.1 exists.

That is, there exists a set Ω ⊆ ℝ^{n+1} such that Ω achieves the minimum value of Problem 1.1. Also, ∂^{∗}Ω is a C^{∞} manifold. Moreover, if n < 7, then ∂Ω \ ∂^{∗}Ω = ∅, and if n ≥ 7, then the Hausdorff dimension of ∂Ω \ ∂^{∗}Ω is at most n − 7.

Proof. Existence was shown in Lemma 2.2. Now, note that ∂^{∗}Ω is locally the graph of a C^{1,σ} function g, for some 0 < σ < 1. Also, in any neighborhood of x ∈ ∂^{∗}Ω, H(x) can be written as [Eva93]

∑_{i,j=1}^{n+1} (1_{\{i=j\}} − \frac{\frac{∂}{∂x_{i}} g(x)\frac{∂}{∂x_{j}} g(x)}{∥∇g(x)∥^{2}}) \frac{∂^{2}}{∂x_{i}∂x_{j}} g(x).

That is, the equation H(x)−⟨x, N(x)⟩−λ = 0 can locally be written as an elliptic equation. So, “classical Schauder estimates” imply that ∂^{∗}Ω is locally the graph of a C^{∞} function. The final statement concerning Hausdorff dimension follows from the theory of almost minimal surfaces [BBJ16, Proposition 2] [Mag12, Theorem 21.8]. □

4. Eigenfunctions of L

Let e_{1}, …, e_{n} be an orthonormal frame for an orientable n-dimensional hypersurface Σ ⊆ ℝ^{n+1} with ∂Σ = ∅. Let Δ := ∑_{i=1}^{n} ∇_{e_{i}}∇_{e_{i}} be the Laplacian associated to Σ. Let ∇ := ∑_{i=1}^{n} e_{i}∇_{e_{i}} be the gradient associated to Σ. (The symbol ∇_{·}(·) still denotes the Euclidean connection, and the meaning of the symbol ∇ should be clear from context.) For any n × n matrix B = (b_{ij})_{1≤i,j≤n}, define ∥B∥^{2} := ∑_{i,j=1}^{n} b_{ij}^{2}.

For any f ∈ C^{∞}(Σ), define

Lf := Δf − ⟨x, ∇f⟩.  (20)
Lf := Δf − ⟨x, ∇f⟩ + f + ∥A∥^{2} f.  (21)

Note that there is a factor of 2 difference between our definition of L and the definition of L in [CM12].

Below we often remove the x arguments of the functions for brevity. We extend L to matrices so that (LB)_{ij} := L(B_{ij}) for all 1 ≤ i, j ≤ n.

Lemma 4.1 (H is almost an eigenfunction of L [CM15, Proposition 1.2] [Gua18, Lemma 2.1]). Let Σ ⊆ ℝ^{n+1} be an orientable hypersurface. Let λ ∈ ℝ. If

H(x) = ⟨x, N(x)⟩ + λ,  ∀ x ∈ Σ.  (22)

Then

LA = 2A − λA^{2}.  (23)
LH = 2H + λ ∥A∥^{2} .  (24)

Proof. Let 1 ≤ i ≤ n and let x ∈ Σ. Then

\begin{aligned}∇_{e_{i}}H(x) & =^{(22)} ∇_{e_{i}}⟨x, N⟩ = ⟨x, ∇_{e_{i}}N(x)⟩ + ⟨e_{i}, N(x)⟩ = ⟨x, ∇_{e_{i}}N(x)⟩ \\ & = ∑_{j=1}^{n}⟨∇_{e_{i}}N(x), e_{j}⟩⟨x, e_{j}⟩ =^{(10)} − ∑_{j=1}^{n} a_{ij}⟨x, e_{j}⟩. \end{aligned}  (25)

Fix x ∈ Σ. Choosing the frame such that ∇_{e_{k}}^{T}e_{j} = 0 at x for every 1 ≤ j, k ≤ n, we then have ∇_{e_{k}}e_{j} = a_{kj}N at x by (9), so

∑_{j=1}^{n} a_{ij}⟨x, ∇_{e_{k}}e_{j}⟩ = ∑_{j=1}^{n} a_{ij}a_{kj}⟨x, N⟩ =^{(22)} ∑_{j=1}^{n} a_{ij}a_{kj}(H − λ).  (26)

So, ∀ 1 ≤ i, k ≤ n,

\begin{aligned}∇_{e_{k}}∇_{e_{i}}H(x) & =^{(25)} − ∑_{j=1}^{n} ∇_{e_{k}}a_{ij}⟨x, e_{j}⟩ − ∑_{j=1}^{n} a_{ij}⟨e_{k}, e_{j}⟩ − ∑_{j=1}^{n} a_{ij}⟨x, ∇_{e_{k}}e_{j}⟩ \\ & =^{(26)} − ∑_{j=1}^{n} ∇_{e_{k}}a_{ij}⟨x, e_{j}⟩ − ∑_{j=1}^{n} a_{ij}⟨e_{k}, e_{j}⟩ − ∑_{j=1}^{n} a_{ij}a_{kj}(H − λ). \end{aligned}  (27)

Also, for any hypersurface, and for any 1 ≤ i, k ≤ n, (see [Sim83, Lemma B.8] where A has the opposite sign),

(ΔA)_{ik} = − ∥A∥^{2} a_{ik} − ∑_{j=1}^{n} Ha_{ij}a_{kj} − ∇_{e_{i}}∇_{e_{k}}H.

So, using the Codazzi equation (∇_{e_{i}}a_{kj} = ∇_{e_{j}}a_{ki}) and that A is a symmetric matrix,

(ΔA)_{ik} =^{(27)} − ∥A∥^{2} a_{ik} + ∑_{j=1}^{n} ∇_{e_{j}}a_{ki}⟨x, e_{j}⟩ + ∑_{j=1}^{n} a_{ki}⟨e_{i}, e_{j}⟩ − λ ∑_{j=1}^{n} a_{ij}a_{kj}
= − ∥A∥^{2} a_{ik} + ⟨x, ∇a_{ik}⟩ + a_{ik} − λ(A^{2})_{ik}.

Therefore,

LA =^{(21)} 2A − λA^{2}.

Finally, summing the diagonal entries of this equality and applying (11) proves (24).

□

Lemma 4.2 (Linear Eigenfunction of L, [MR15, BBJ16]). Let Σ ⊆ ℝ^{n+1} be an orientable hypersurface. Let λ ∈ ℝ. Suppose

H(x) = ⟨x, N⟩ + λ,  ∀ x ∈ Σ.  (28)

Let v ∈ ℝ^{n+1}. Then

L⟨v, N⟩ = ⟨v, N⟩.

Proof. Let 1 ≤ i ≤ n. Then

∇_{e_{i}}⟨v, N⟩ = ⟨v, ∇_{e_{i}}N⟩ =^{(10)} − ∑_{j=1}^{n} a_{ij}⟨v, e_{j}⟩.  (29)

Fix x ∈ Σ. Choosing the frame such that ∇_{e_{k}}^{T}e_{j} = 0 at x for every 1 ≤ j, k ≤ n, we then have ∇_{e_{k}}e_{j} = a_{kj}N at x by (9), so using also Codazzi’s equation,

\begin{aligned}∇_{e_{i}}∇_{e_{i}}⟨v, N⟩ & = − ∑_{j=1}^{n} ∇_{e_{i}}a_{ij}⟨v, e_{j}⟩ − ∑_{j=1}^{n} a_{ij}⟨v, ∇_{e_{i}}e_{j}⟩ \\ & = − ∑_{j=1}^{n} ∇_{e_{j}}a_{ii}⟨v, e_{j}⟩ − ∑_{j=1}^{n} a_{ij}a_{ij}⟨v, N⟩. \end{aligned}  (30)

Therefore,

Δ⟨v, N⟩ = ∑_{i=1}^{n} ∇_{e_{i}}∇_{e_{i}}⟨v, N⟩ =^{(11)∧(30)} ⟨v, ∇H⟩ − ∥A∥^{2} ⟨v, N⟩.

So far, we have not used any of our assumptions. Using now (25), and that A is symmetric,

⟨v, ∇H⟩ = − ∑_{i,j=1}^{n}⟨x, e_{j}⟩a_{ij}⟨v, e_{i}⟩ =^{(29)} ⟨x, ∇⟨v, N⟩⟩.

In summary,

Δ⟨v, N⟩ = ⟨x, ∇⟨v, N⟩⟩ − ∥A∥^{2} ⟨v, N⟩.

We conclude by (21). □

Remark 4.3. Let f, g ∈ C^{∞}(Σ). Using (21), we get the following product rule for L.

L(fg) = fΔg + gΔf + 2⟨∇f, ∇g⟩ − f⟨x, ∇g⟩ − g⟨x, ∇f⟩ + ∥A∥^{2} fg + fg
= fLg + gLf + 2⟨∇f, ∇g⟩ − ∥A∥^{2} fg − fg.

So, by Lemma 4.2, if v, w ∈ ℝ^{n+1},

L⟨v, N⟩⟨w, N⟩ = ⟨v, N⟩⟨w, N⟩ + 2⟨∇⟨v, N⟩, ∇⟨w, N⟩⟩ − ∥A∥^{2} ⟨v, N⟩⟨w, N⟩.

The following Lemma follows from Stokes’ Theorem.

Lemma 4.4 (Integration by Parts, [CM12, Corollary 3.10], [Zhu20, Lemma 5.4]). Let Σ ⊆ ℝ^{n+1} be an n-dimensional hypersurface. Let f, g: Σ → ℝ. Assume that f is a C^{2} function and g is a C^{2} function with compact support. Then

∫_{Σ} fLgγ_{n}(x)dx = ∫_{Σ} gLfγ_{n}(x)dx = − ∫_{Σ}⟨∇f, ∇g⟩γ_{n}(x)dx.

5. Curvature Bounds

In this rather technical section, we show that the derivatives of the curvature have finite integrals. This will be used later on to justify a more general version of Lemma 4.4. Many of the results of this section are unnecessary if we assume that ∂Ω is a C^{∞} manifold. However, from Lemma 3.3, if Ω minimizes Problem 1.1, it may occur that ∂Ω \ ∂^{∗}Ω is a nonempty set with Hausdorff dimension n − 7. And indeed, due to e.g. the existence of Simons-Lawson cones, this is the best possible. If the singular set ∂Ω \ ∂^{∗}Ω is nonempty, we then have to be careful about integrals of curvature blowing up near the singular set.

For any hypersurface Σ ⊆ ℝ^{n+1}, we define

δ = δ(Σ) := sup_{f∈C_{0}^{∞}(Σ)} \frac{∫_{Σ} fLfγ_{n}(x)dx}{∫_{Σ} f^{2}γ_{n}(x)dx}.  (31)

By the definition of δ,

Σ_{1} ⊆ Σ_{2} ⟹ δ(Σ_{1}) ≤ δ(Σ_{2}). (32)

We say an n-dimensional hypersurface Σ ⊆ ℝ^{n+1} has polynomial volume growth if there exists c > 0 such that, and for any r > 1, ∫_{{x∈Σ: ∥x∥≤r}} dx ≤ cr^{n}.

Lemma 5.1 (Existence of an Eigenfunction, [Zhu20, Lemma 6.5]). Let Σ ⊆ ℝ^{n+1} be a symmetric, connected, orientable hypersurface with polynomial volume growth. Assume that Σ is a C^{∞} hypersurface with possibly nonempty boundary. Assume that δ(Σ) < ∞. Then there exists a positive C^{2} function g on Σ such that Lg = δ(Σ)g, and such that g(x) = g(−x) for all x ∈ Σ.

Proof. Fix x ∈ Σ. Let Σ_{1} ⊆ Σ_{2} ⊆ … be a sequence of compact C^{∞} hypersurfaces such that ∪_{i=1}^{∞}Σ_{i} = Σ. For each i ≥ 1, let g_{i} be a positive Dirichlet eigenfunction of L on Σ_{i} such that Lg_{i} = δ(Σ_{i})g_{i}. By multiplying by a constant, we may assume g_{i}(x) = 1 for all i ≥ 1. Since δ(Σ_{i}) increases to δ(Σ) < ∞ as i → ∞ by (31), the Harnack inequality implies that there exists c = c(Σ_{i}, δ(Σ)) such that 1 ≤ sup_{x∈Σ_{i}} g_{i}(x) ≤ c inf_{x∈Σ_{i}} g_{i}(x) ≤ c. Elliptic theory then gives uniform C^{2,σ} bounds for the functions g_{1}, g_{2}, … on each compact subset of Σ. So, by Arzelà-Ascoli there exists a uniformly convergent subsequence of g_{1}, g_{2}, … which converges to a nonnegative solution of Lg = δ(Σ)g on Σ with g(x) = 1. The Harnack inequality then implies that g > 0 Σ. Finally, the definition of L (21) and symmetry of Σ implies that L(g(x) + g(−x)) = δ(Σ)(g(x) + g(−x)). That is, we may assume that g itself satisfies g(x) = g(−x) for all x ∈ Σ. □

Lemma 5.2. If Ω ⊆ ℝ^{n+1} minimizes Problem 1.1, then δ(∂^{∗}Ω) < ∞.

Proof. We argue by contradiction. Suppose δ(∂^{∗}Ω) = ∞. Then, for any m > 0, there exists a compact symmetric subset Σ_{m} ⊆ ∂^{∗}Ω and there exists a Dirichlet eigenfunction g_{m} > 0 on Σ_{m} (by Lemma 5.1, or by applying spectral theory to the compact operator (L−∥A∥^{2} −2)^{−1} on Σ_{m}) such that Lg_{m} = δ_{m}g_{m}, δ_{m} > m, and such that g_{m}(x) = g_{m}(−x) for all x ∈ Σ_{m}. From Lemma 3.1, ∃ λ ∈ ℝ such that H(x) = ⟨x, N(x)⟩ + λ on ∂^{∗}Ω ⊇ Σ_{m}. To conclude, it suffices by Lemma 3.2 to find a function f on Σ_{m} (extended to be zero on ∂^{∗}Ω \ Σ_{m}) such that ∫_{Σ_{m}} fγ_{n}(x)dx = 0, f(x) = f(−x) for all x ∈ Σ_{m}, and ∫_{Σ_{m}} fLfγ_{n}(x)dx > 0.

Let f := g_{m} + H − λ, where we multiply g_{m} by a constant so that ∫_{Σ_{m}} fγ_{n}(x)dx = 0. (In the case that ∫_{Σ_{m}} g_{m}γ_{n}(x)dx = 0 for some m ≥ 1, we conclude by choosing f = g_{m}. So, we may assume that ∫_{Σ_{m}} g_{m}γ_{n}(x)dx ≠ 0 for all m ≥ 1.) By construction f(x) = f(−x) for all x ∈ Σ_{m}. It remains to bound ∫_{Σ_{m}} fLfγ_{n}(x)dx. From (24) and (21),

L(H − λ) = 2H + λ ∥A∥^{2} − λ ∥A∥^{2} − λ = 2H − λ.  (33)

Integrating by parts with Lemma 4.4, we have

∫_{Σ_{m}} fLfγ_{n}(x)dx = ∫_{Σ_{m}}(g_{m} + H − λ)L(g_{m} + H − λ)γ_{n}(x)dx

= ∫_{Σ_{m}}(g_{m}Lg_{m} + 2g_{m}L(H − λ) + (H − λ)L(H − λ))γ_{n}(x)dx

=^{(33)} ∫_{Σ_{m}}(δ_{m}g_{m}^{2} + 2g_{m}(2H − λ) + (H − λ)(2H − λ))γ_{n}(x)dx

= ∫_{Σ_{m}}(δ_{m}g_{m}^{2} + 2g_{m}(H − λ) + 2Hg_{m} + (H − λ)^{2} + H(H − λ))γ_{n}(x)dx

= ∫_{Σ_{m}}((δ_{m} − 1)g_{m}^{2} + (g_{m} + H − λ)^{2} + 2Hg_{m} + H(H − λ))γ_{n}(x)dx

= ∫_{Σ_{m}}((δ_{m} − 2)g_{m}^{2} + (g_{m} + H − λ)^{2} + (g_{m} + H)^{2} − λH)γ_{n}(x)dx.

By assumption, ∫_{Σ_{m}}(g_{m} + H − λ)γ_{n}(x)dx = 0, so that ∫_{Σ_{m}}(−λH)γ_{n}(x)dx = ∫_{Σ_{m}}(λg_{m} −

λ^{2})γ_{n}(x)dx. From the Cauchy-Schwarz inequality, ∫_{Σ_{m}} g_{m}^{2}γ_{n}(x)dx ≥ \frac{(∫_{Σm} g_{m}γ_{n}(x)dx)^{2}}{∫_{Σm} γ_{n}(x)dx}. So,

∫_{Σ_{m}} fLfγ_{n}(x)dx

≥ ∫_{Σ_{m}} (g_{m}(λ + (δ_{m} − 2)\frac{∫_{Σ_{m}} g_{m}γ_{n}(y)dy}{∫_{Σ_{m}} γ_{n}(z)dz} ) + (g_{m} + H − λ)^{2} + (g_{m} + H)^{2} − λ^{2})γ_{n}(x)dx

= ∫_{Σ_{m}} ( − ⟨x, N⟩(λ + (δ_{m} − 2)\frac{∫_{Σ_{m}} −⟨y, N⟩γ_{n}(y)dy}{∫_{Σ_{m}} γ_{n}(z)dz} )

+ (g_{m} + H − λ)^{2} + (g_{m} + H)^{2} − λ^{2})γ_{n}(x)dx.

In the last line, we used ∫_{Σ_{m}} g_{m}γ_{n}(x)dx = ∫_{Σ_{m}}(λ − H)γ_{n}(x)dx = ∫_{Σ_{m}} −⟨x, N⟩γ_{n}(x)dx. So, letting m → ∞ and using lim_{m→∞} δ_{m} = ∞ concludes the proof, in the case that lim_{m→∞} ∫_{Σ_{m}}⟨x, N⟩γ_{n}(x)dx ≠ 0. (Recall that ∫_{Σ_{m}} g_{m}γ_{n}(x)dx ≠ 0 so ∫_{Σ_{m}}⟨x, N⟩γ_{n}(x)dx ≠ 0 for all m ≥ 1.) It remains to address the case that lim_{m→∞} ∫_{Σ_{m}}⟨x, N⟩γ_{n}(x)dx = 0. That is, it remains to address when ∫_{∂^{∗}Ω}(H − λ)γ_{n}(x)dx = 0. In this case, we use f := H − λ to get

∫_{∂^{∗}Ω}(H − λ)L(H − λ)γ_{n}(x)dx =^{(33)} ∫_{∂^{∗}Ω}(H − λ)(2H − λ)γ_{n}(x)dx

= ∫_{∂^{∗}Ω}(H − λ)(2H − 2λ)γ_{n}(x)dx = 2 ∫_{∂^{∗}Ω}(H − λ)^{2}γ_{n}(x)dx.

The latter quantity is positive, unless H = λ is constant on ∂^{∗}Ω. Then H − λ = ⟨x, N⟩ = 0 for all x ∈ ∂^{∗}Ω. That is, ∂^{∗}Ω is a cone. If λ ≠ 0, this is impossible, since H(tx) = H(x)/t ∀ x ∈ ∂^{∗}Ω, ∀ t > 0. So, it remains to consider the case that H = ⟨x, N⟩ = 0 for all x ∈ ∂^{∗}Ω. This case is eliminated in Lemma 11.5 below. □

• We say that f ∈ L_{2}(Σ, γ_{n}) if ∫_{Σ} |f|^{2} γ_{n}(x)dx < ∞.
• We say that f ∈ W_{1,2}(Σ, γ_{n}) if ∫_{Σ}(|f|^{2} + ∥∇f∥^{2})γ_{n}(x)dx < ∞.
• We say that f ∈ W_{2,2}(Σ, γ_{n}) if ∫_{Σ}(|f|^{2} + ∥∇f∥^{2} + |Lf|^{2})γ_{n}(x)dx < ∞.

Lemma 5.3 ([CM12, Lemma 9.15(2)]). Let Σ ⊆ ℝ^{n+1} be a C^{∞} hypersurface, with possibly nonempty boundary. Assume δ(Σ) < ∞. Suppose g: Σ → ℝ is a C^{2} function with g > 0 and Lg = δg. If ϕ ∈ W_{1,2}(Σ, γ_{n}), then

∫_{Σ} ϕ^{2}(∥A∥^{2} + ∥∇ log g∥^{2})γ_{n}(x)dx ≤ 4 ∫_{Σ}(∥∇ϕ∥^{2} + (δ − 1)ϕ^{2})γ_{n}(x)dx.

Proof.

L log g =^{(20)} ∑_{i=1}^{n} ∇_{e_{i}} (\frac{∇_{e_{i}}g}{g} ) − \frac{⟨x, ∇g⟩}{g} = ∑_{i=1}^{n} \frac{−(∇_{e_{i}}g)^{2}}{g^{2}} + \frac{Lg}{g}

= − ∥∇ log g∥^{2} + \frac{Lg}{g} =^{(21)} − ∥∇ log g∥^{2} + \frac{Lg − ∥A∥^{2} g − g}{g}

=^{(24)} − ∥∇ log g∥^{2} + \frac{δg − ∥A∥^{2} g − g}{g}

= − ∥∇ log g∥^{2} + (δ − 1) − ∥A∥^{2} .

Let η ∈ C_{0}^{∞}(Σ). By Lemma 4.4,

∫_{Σ}⟨∇η^{2}, ∇ log g⟩γ_{n}(x)dx = − ∫_{Σ} η^{2}L log gγ_{n}(x)dx
= ∫_{Σ} η^{2}( ∥∇ log g∥^{2} + (1 − δ) + ∥A∥^{2} )γ_{n}(x)dx.

By the arithmetic mean geometric mean inequality (AMGM),

|⟨∇η^{2}, ∇ log g⟩| ≤ 2 ∥∇η∥^{2} + \frac{1}{2}η^{2} ∥∇ log g∥^{2} .

So,

∫_{Σ} η^{2}(∥A∥^{2} + ∥∇ log g∥^{2})γ_{n}(x)dx ≤ 4 ∫_{Σ} ( ∥∇η∥^{2} + (δ − 1)η^{2})γ_{n}(x)dx.

Letting η approximate ϕ by cutoff functions and applying the monotone convergence theorem completes the proof. □

Lemma 5.4 ([Zhu20, Lemma 6.2]). Let Ω ⊆ ℝ^{n+1} and let Σ := ∂Ω. Assume δ(Σ) < ∞. Suppose g: Σ → ℝ is a C^{2} function with g > 0 and Lg = δg. Assume that the Hausdorff dimension of ∂Ω \ ∂^{∗}Ω is at most n − 7. If ϕ ∈ W_{1,2}(Σ, γ_{n}) and if ∫_{Σ} |ϕ|^{7/3} γ_{n}(x)dx < ∞, then

∫_{Σ} ϕ^{2}(∥A∥^{2} + ∥∇ log |g|∥^{2})γ_{n}(x)dx ≤ 4 ∫_{Σ}(∥∇ϕ∥^{2} + (δ − 1)ϕ^{2})γ_{n}(x)dx.

Corollary 5.5 ([Zhu20, Lemma 6.2]). Let Ω ⊆ ℝ^{n+1} and let Σ := ∂Ω. Assume δ(Σ) < ∞. Suppose g: Σ → ℝ is a C^{2} function with g > 0 and Lg = δg. Assume that the Hausdorff dimension of ∂Ω \ ∂^{∗}Ω is at most n − 7. Then for any k ≥ 0, ∥A∥ ∥x∥^{k} ∈ L_{2}(Σ, γ_{n})

Lemma 5.6 ([CM12, Theorem 9.36]). Let Σ ⊆ ℝ^{n+1} be a connected, orientable C^{∞} hypersurface with polynomial volume growth and with possibly nonempty boundary. Assume ∃ λ ∈ ℝ such that H(x) = ⟨x, N(x)⟩ + λ for all x ∈ Σ. Let δ := δ(Σ). Assume δ(Σ) < ∞. Then

∥∇H∥ ∈ L_{2}(Σ, γ_{n}), ∥A∥ |H| ∈ L_{2}(Σ, γ_{n}).

∫_{Σ} |HLH| γ_{n}(x)dx < ∞.

Proof. As shown in (25), since H(x) = ⟨x, N⟩ + λ, for any 1 ≤ i ≤ n, ∇_{e_{i}}H(x) = − ∑_{j=1}^{n} a_{ij}⟨x, e_{j}⟩. Therefore,

∥∇H∥^{2} ≤ ∥A∥^{2} ∥x∥^{2} , ∥A∥^{2} H^{2} ≤ 2 ∥A∥^{2} (∥x∥^{2} + λ^{2}).

So, Corollary 5.5 implies that ∥∇H∥ ∈ L_{2}(Σ, γ_{n}) and ∥A∥ |H| ∈ L_{2}(Σ, γ_{n}).

For the final assertion, note that LH =^{(21)} LH −∥A∥^{2} H −H =^{(24)} 2H +λ ∥A∥^{2} −∥A∥^{2} H −H.

So,

|HLH| ≤ H^{2} + |λ| |H| ∥A∥^{2} + ∥A∥^{2} H^{2}.

So, the polynomial volume growth (and H^{2} ≤ 2(∥x∥^{2} + λ^{2})) and the above results show that ∫_{Σ} |HLH| γ_{n}(x)dx < ∞. □

The following geometric inequality is essentially shown in [CM12, Lemma 10.8],[Zhu20, Lemma 7.1] and [CW18, Lemma 4.1], and it is inspired by an inequality of Simons [Sim68].

When A, B are n × n matrices, we use the notation ⟨A, B⟩ := ∑_{i,j=1}^{n} a_{ij}b_{ij}. Note that ⟨A, A⟩ = ∥A∥^{2}. Recall that we extend L to matrices so that (LA)_{ij} := L(A_{ij}) for all 1 ≤ i, j ≤ n.

Lemma 5.7 (Simons-type inequality, [Sim68, CM12, CW18, Zhu20]). Let Σ be a C^{∞} orientable hypersurface. Let λ ∈ ℝ. Suppose H(x) = ⟨x, N(x)⟩ + λ, ∀ x ∈ Σ. Then

\begin{aligned}∥A∥ L ∥A∥ & = 2 ∥A∥^{2} − λ⟨A^{2}, A⟩ + ∥∇A∥^{2} − ∥∇ ∥A∥∥^{2} \\ & ≥ 2 ∥A∥^{2} − λ⟨A^{2}, A⟩. \end{aligned}  (34)

Proof. Using for now only (21), we have

L ∥A∥ = L(∥A∥^{2})^{1/2} =^{(21)} ∑_{i=1}^{n} ∇_{e_{i}} (\frac{1}{2} \frac{∇_{e_{i}}(∥A∥^{2})}{∥A∥}) − \frac{1}{2} \frac{⟨x, ∇ ∥A∥^{2}⟩}{∥A∥} + ∥A∥^{3} + ∥A∥

= \frac{1}{2} \frac{∥A∥ Δ ∥A∥^{2} − ∑_{i=1}^{n}[∇_{e_{i}} ∥A∥^{2}][∇_{e_{i}} ∥A∥]}{∥A∥^{2}} − \frac{1}{2} \frac{⟨x, ∇ ∥A∥^{2}⟩}{∥A∥} + ∥A∥^{3} + ∥A∥

= \frac{1}{2} \frac{Δ ∥A∥^{2}}{∥A∥} − \frac{∥∇ ∥A∥∥^{2}}{∥A∥} − \frac{1}{2} \frac{∑_{i=1}^{n}⟨x, e_{i}⟩∇_{e_{i}} ∥A∥^{2}}{∥A∥} + ∥A∥^{3} + ∥A∥

= \frac{⟨A, ΔA⟩ + ∥∇A∥^{2}}{∥A∥} − \frac{∥∇ ∥A∥∥^{2}}{∥A∥} − \frac{∑_{i,j,k=1}^{n}⟨x, e_{i}⟩a_{jk}∇_{e_{i}}a_{jk}}{∥A∥} + ∥A∥^{3} + ∥A∥

=^{(21)} \frac{⟨A, LA⟩}{∥A∥} + \frac{∥∇A∥^{2} − ∥∇ ∥A∥∥^{2}}{∥A∥} .

Now, using (23), we get

∥A∥ L ∥A∥ = 2 ∥A∥^{2} − λ⟨A^{2}, A⟩ + ∥∇A∥^{2} − ∥∇ ∥A∥∥^{2} .

The proof is completed since ∥∇A∥^{2} − ∥∇ ∥A∥∥^{2} ≥ 0, which follows by the Cauchy-Schwarz inequality. □

Lemma 5.8 ([CM12, Lemma 10.2]). Let Σ ⊆ ℝ^{n+1} be any n-dimensional hypersurface. Then

(1 + \frac{2}{n + 1}) ∥∇ ∥A∥∥^{2} ≤ ∥∇A∥^{2} + \frac{2n}{n + 1} ∥∇H∥^{2} .  (35)

The following estimate is adapted from [CM12], which itself was adapted from [SSY75].

Lemma 5.9 ([CM12, Proposition 10.14]). Let Ω ⊆ ℝ^{n+1} be a convex set and let Σ := ∂^{∗}Ω. Assume that the Hausdorff dimension of ∂Ω \ ∂^{∗}Ω is at most n − 4. Assume δ(Σ) < ∞.

Assume ∃ λ ∈ ℝ such that H(x) = ⟨x, N(x)⟩ + λ for all x ∈ Σ. Assume H > 0. Then

∫_{Σ}(∥A∥^{2} + ∥A∥^{4} + ∥∇ ∥A∥∥^{2} + ∥∇A∥^{2})γ_{n}(x)dx < ∞.

Proof. Since H > 0, log H is well-defined, so that

L log H =^{(20)} ∑_{i=1}^{n} ∇_{e_{i}} (\frac{∇_{e_{i}}H}{H} ) − \frac{⟨x, ∇H⟩}{H} = ∑_{i=1}^{n} \frac{−(∇_{e_{i}}H)^{2}}{H^{2}} + \frac{LH}{H}

= − ∥∇ log H∥^{2} + \frac{LH}{H} =^{(21)} − ∥∇ log H∥^{2} + \frac{LH − ∥A∥^{2} H − H}{H}

=^{(24)} − ∥∇ log H∥^{2} + \frac{2H + λ ∥A∥^{2} − ∥A∥^{2} H − H}{H}

= − ∥∇ log H∥^{2} + 1 − ∥A∥^{2} + λ\frac{∥A∥^{2}}{H} .

(36)

Note that ∫_{Σ} ∥A∥^{2} γ_{n}(x)dx < ∞ by Corollary 5.5. Let ϕ ∈ C_{0}^{∞}(Σ). Integrating by parts with Lemma 4.4,

∫_{Σ}⟨∇ϕ^{2}, ∇ log H⟩γ_{n}(x)dx = − ∫_{Σ} ϕ^{2}L log Hγ_{n}(x)dx

=^{(36)} ∫_{Σ} ϕ^{2}( ∥∇ log H∥^{2} − 1 + ∥A∥^{2} − λ\frac{∥A∥^{2}}{H} )γ_{n}(x)dx.

From the AMGM inequality, |⟨∇ϕ^{2}, ∇ log H⟩| ≤ ∥∇ϕ∥^{2} + ϕ^{2} ∥∇ log H∥^{2}, so that

∫_{Σ} ϕ^{2} ∥A∥^{2} γ_{n}(x)dx ≤ ∫_{Σ} [ ∥∇ϕ∥^{2} + ϕ^{2}(1 + λ\frac{∥A∥^{2}}{H} )]γ_{n}(x)dx.  (37)

Let 0 < ε < 1/2 to be chosen later. Using now ϕ := η ∥A∥ in (37), where η ∈ C_{0}^{∞}(Σ), η ≥ 0, and using the AMGM inequality in the form 2ab ≤ εa^{2} + b^{2}/ε, a, b > 0,

∫_{Σ} η^{2} ∥A∥^{4} γ_{n}(x)dx

≤ ∫_{Σ} [η^{2} ∥∇ ∥A∥∥^{2} + 2η ∥A∥ ∥∇ ∥A∥∥ ∥∇η∥ + ∥A∥^{2} ∥∇η∥^{2} + η^{2} ∥A∥^{2} (1 + λ\frac{∥A∥^{2}}{H} )]γ_{n}(x)dx

≤ ∫_{Σ} [(1 + ε)η^{2} ∥∇ ∥A∥∥^{2} + ∥A∥^{2} ∥∇η∥^{2} (1 + 1/ε) + η^{2} ∥A∥^{2} (1 + λ\frac{∥A∥^{2}}{H} )]γ_{n}(x)dx.

(38)

Using the product rule for L, and that L = L − ∥A∥^{2} − 1

\frac{1}{2}L ∥A∥^{2} = ∥∇ ∥A∥∥^{2} + ∥A∥ L ∥A∥ = ∥∇ ∥A∥∥^{2} + ∥A∥ (L ∥A∥ − ∥A∥^{3} − ∥A∥ )

=^{(34)} ∥∇ ∥A∥∥^{2} + 2 ∥A∥^{2} − λ⟨A^{2}, A⟩ + ∥∇A∥^{2} − ∥∇ ∥A∥∥^{2} − ∥A∥^{4} − ∥A∥^{2}

= ∥∇A∥^{2} + ∥A∥^{2} − ∥A∥^{4} − λ⟨A^{2}, A⟩

≥^{(35)} (1 + \frac{2}{n + 1}) ∥∇ ∥A∥∥^{2} − \frac{2n}{n + 1} ∥∇H∥^{2} + ∥A∥^{2} − ∥A∥^{4} − λ⟨A^{2}, A⟩.

Multiplying this inequality by η^{2} and integrating by parts with Lemma 4.4,

− 2 ∫_{Σ} η ∥A∥ ⟨∇η, ∇ ∥A∥⟩γ_{n}(x)dx = −\frac{1}{2} ∫_{Σ}⟨∇η^{2}, ∇ ∥A∥^{2}⟩γ_{n}(x)dx = \frac{1}{2} ∫_{Σ} η^{2}L ∥A∥^{2} γ_{n}(x)dx

≥ ∫_{Σ} η^{2}((1 + \frac{2}{n + 1}) ∥∇ ∥A∥∥^{2} − \frac{2n}{n + 1} ∥∇H∥^{2} − ∥A∥^{4} − λ⟨A^{2}, A⟩)γ_{n}(x)dx.

(We removed the ∥A∥^{2} term since doing so only decreases the quantity on the right.) Rearranging this inequality and then using the AMGM inequality in the form b^{2}/ε − 2ab ≥ −εa^{2},

\begin{matrix}∫_{Σ} (η^{2} ∥A∥^{4} + λη^{2}⟨A^{2}, A⟩ + \frac{2n}{n + 1}η^{2} ∥∇H∥^{2} + \frac{1}{ε}η^{2} ∥A∥^{2} ∥∇η∥^{2} )γ_{n}(x)dx \\ ≥ (1 + \frac{2}{n + 1} − ε) ∫_{Σ} η^{2} ∥∇ ∥A∥∥^{2} γ_{n}(x)dx. \end{matrix}  (39)

Substituting (39) into (38),

∫_{Σ} η^{2} ∥A∥^{4} γ_{n}(x)dx ≤ \frac{1 + ε}{1 + \frac{2}{n+1} − ε} ∫_{Σ} η^{2} ∥A∥^{4} γ_{n}(x)dx

+ 10 ∫_{Σ} [η^{2} ∥∇H∥^{2} + η^{2}(1 + 1/ε)(1 + ∥∇η∥^{2}) ∥A∥^{2} + |λ| η^{2}( |⟨A, A^{2}⟩| + \frac{∥A∥^{4}}{H} )]γ_{n}(x)dx.

Since Ω is convex, A is negative definite with nonpositive diagonal entries, so that H ≥ ∥A∥, i.e. ∥A∥^{4} /H ≤ ∥A∥^{3}. Also |⟨A, A^{2}⟩| ≤ ∥A∥^{3}. Using the AMGM inequality in the form 2a^{3} ≤ a^{4}ε + a^{2}/ε, we then get

∫_{Σ} η^{2} ∥A∥^{4} γ_{n}(x)dx ≤ (10ε |λ| + \frac{1 + ε}{1 + \frac{2}{n+1} − ε}) ∫_{Σ} η^{2} ∥A∥^{4} γ_{n}(x)dx

+ 10 ∫_{Σ} η^{2}( ∥∇H∥^{2} + (1 + (1 + |λ|)/ε)(1 + ∥∇η∥^{2}) ∥A∥^{2} )γ_{n}(x)dx.

Now, choose ε < 1/(20(n + 1)(|λ| + 1)), so that 10 |λ| ε + \frac{1+ε}{1+ \frac{2}{n+1} −ε} < 1. We then can move

the η^{2} ∥A∥^{4} term on the right side to the left side to get some c_{ε} > 0 such that

∫_{Σ} η^{2} ∥A∥^{4} γ_{n}(x)dx ≤ c_{ε} ∫_{Σ} η^{2}[ ∥∇H∥^{2} + ∥A∥^{2} (1 + ∥∇η∥^{2})]γ_{n}(x)dx
≤ c_{ε} ∫_{Σ} η^{2}[ ∥A∥^{2} (1 + ∥x∥^{2} + ∥∇η∥^{2})]γ_{n}(x)dx.

In the last line, we used the inequality ∥∇H∥^{2} ≤ ∥A∥^{2} ∥x∥^{2}. This follows by (25), since H(x) = ⟨x, N⟩ + λ, so for any 1 ≤ i ≤ n, ∇_{e_{i}}H(x) = − ∑_{j=1}^{n} a_{ij}⟨x, e_{j}⟩.

We now choose a sequence of η = η_{r} increasing to 1 as r → ∞ so that the ∥∇η∥^{2} term vanishes. This is possible due to the assumptions that δ < ∞ and the Hausdorff dimension of ∂Ω \ ∂^{∗}Ω is at most n − 4. Such functions are constructed and this estimate is made in [Zhu20, Lemma 6.4]:

∫_{Σ} ∥A∥^{2} ∥∇η_{r}∥^{2} η_{r}^{2}γ_{n}(x)dx ≤ c(δ)(r^{n}e^{−(r−4)^{2}/4} + r^{−1}),  ∀r > 1.  (40)

It therefore follows from Corollary 5.5 applied to Σ = ∂^{∗}Ω that

∫_{Σ} ∥A∥^{4} γ_{n}(x)dx < ∞.  (41)

It then follows from (39) that ∫_{Σ} ∥∇ ∥A∥∥^{2} γ_{n}(x)dx < ∞.

Finally, multiplying the above equality L ∥A∥^{2} = 2 ∥∇A∥^{2} + 2 ∥A∥^{2} − 2 ∥A∥^{4} − 2λ⟨A^{2}, A⟩ by η^{2} and integrating by parts with Lemma 4.4, we get

2 ∫_{Σ} η^{2}(∥∇A∥^{2} − ∥A∥^{4} − λ⟨A^{2}, A⟩)γ_{n}(x)dx ≤ ∫_{Σ} η^{2}L ∥A∥^{2} γ_{n}(x)dx

= −4 ∫_{Σ} η ∥A∥ ⟨∇η, ∇ ∥A∥⟩γ_{n}(x)dx ≤ 2 ∫_{Σ} η^{2} ∥∇ ∥A∥∥^{2} + ∥A∥^{2} ∥∇η∥^{2}⟩γ_{n}(x)dx.

Then the ∥A∥^{4} integral is finite by (41), the ∥∇ ∥A∥∥^{2} integral is finite, the last term has a finite integral by (40), so the integral of ∥∇A∥^{2} is also finite. □

In the following Corollary, the Hausdorff dimension condition is needed to construct functions that converge to 1 while being zero in a neighborhood of the singular set ∂Ω \ ∂^{∗}Ω. For details on this construction, see e.g. [Zhu20, Section 5.2].

Corollary 5.10 (Integration by Parts, Version 2 [Zhu20, Lemma 5.4]). Let Ω ⊆ ℝ^{n+1}. Let f, g: ∂^{∗}Ω → ℝ be C^{2} functions. Suppose the Hausdorff dimension of ∂Ω \ ∂^{∗}Ω is at most n − 7. Assume that

∫_{Σ} ∥∇f∥ ∥∇g∥ γ_{n}(x)dx < ∞,  ∫_{Σ} |fLg| γ_{n}(x)dx < ∞,  ∫_{Σ} ∥f∇g∥^{2} γ_{n}(x)dx < ∞.

Then

∫_{Σ} fLgγ_{n}(x)dx = − ∫_{Σ}⟨∇f, ∇g⟩γ_{n}(x)dx.

Lemma 5.11. Let Σ ⊆ ℝ^{n+1} be a C^{∞} hypersurface, with possibly nonempty boundary. Assume ∃ λ ∈ ℝ such that H(x) = ⟨x, N(x)⟩ + λ for all x ∈ Σ. Let δ := δ(Σ). Assume that δ < ∞ and ∂Σ \ ∂^{∗}Ω has Hausdorff dimension at most n − 7. Let g be the eigenfunction guaranteed to exist by Lemma 5.1. Then for any ε > 0, we have

|∫_{Σ}(HLg − gLH)γ_{n}(x)dx| < ε(∫_{Σ} g^{2}γ_{n}(x)dx)^{1/2}(∫_{Σ} H^{2}γ_{n}(x)dx)^{1/2}.

Proof. Let R > 10 let ε := R^{−3} and let 0 < r_{1}, …, r_{k} < 1, x^{(1)}, …, x^{(k)} ∈ Σ such that ∪_{i=1}^{k}B(x^{(i)}, r_{i}) ⊇ B(0, r) ∩ (∂Σ \ ∂^{∗}Ω) and such that ∑_{i=1}^{k} r_{i}^{n−4} < ε. Such r_{1}, …, r_{k} exist since n − 7 > n − 4, and ∂Σ \ ∂^{∗}Ω has Hausdorff dimension at most n − 7. For each 1 ≤ i ≤ k, we let ϕ_{i}: Σ → [0, 1] such that ϕ_{i} = 1 outside B(x^{(i)}, 3r_{i}), ϕ_{i} = 0 inside B(x^{(i)}, 2r_{i}), and ∥∇ϕ_{i}∥ ≤ 2/r_{i} in B(x^{(i)}, 3r_{i}) \ B(x^{(i)}, 2r_{i}). Let also ψ_{R} so that ψ_{R} = 0 outside B(0, R +2), ψ_{R} = 1 inside B(0, R), and ∥∇ψ_{R}∥ ≤ 2 in B(0, R +2)\B(0, R). Finally, define ϕ = ϕ_{R}: Σ → [0, 1] so that ϕ := min(ψ_{R}, min_{1≤i≤k} ϕ_{i}). Note that ϕ is Lipschitz with compact support and ∥∇ϕ∥ ≤ max(∥∇ψ_{R}∥ , max_{1≤i≤k} ∥∇ψ_{i}∥).

We now integrate by parts with Lemma 4.4 to get

∫_{Σ}(HLg − gLH)γ_{n}(x)dx = ∫_{Σ}(HLg − gLH)γ_{n}(x)dx

= ∫_{Σ} ϕ^{2}(HLg − gLH)γ_{n}(x)dx + ∫_{Σ}(1 − ϕ^{2})(HLg − gLH)γ_{n}(x)dx
= 2 ∫_{Σ} ϕ⟨∇ϕ, H∇g − g∇H⟩γ_{n}(x)dx + ∫_{Σ}(1 − ϕ^{2})(HLg − gLH)γ_{n}(x)dx

The second term is made small by choosing R large, so we focus on the first term. We bound the first by

∫_{Σ} ϕ |⟨∇ϕ, H∇g − g∇H⟩| γ_{n}(x)dx = ∫_{Σ} ϕ |g| |⟨∇ϕ, H \frac{∇g}{g} − ∇H⟩| γ_{n}(x)dx

≤ ∫_{Σ} ϕ |g| ( ∥∇ϕ∥ |H| ∥∇ log g∥ + ∥∇H∥ )γ_{n}(x)dx
≤ (∫_{Σ} g^{2}γ_{n}(x)dx)^{1/2} ∫_{Σ} ϕ^{2} ∥∇ϕ∥^{2} H^{2} ∥∇ log g∥^{2} + ϕ^{2} ∥∇H∥^{2} γ_{n}(x)dx)^{1/2}
≤ (∫_{Σ} g^{2}γ_{n}(x)dx)^{1/2} ∫_{Σ} ϕ^{2} ∥∇ϕ∥^{2} (∥x∥ + |λ|) ∥∇ log g∥^{2} + ϕ^{2} ∥A∥^{2} γ_{n}(x)dx)^{1/2}
≤ (∫_{Σ} g^{2}γ_{n}(x)dx)^{1/2} ∫_{Σ} ϕ^{2} ∥∇ϕ∥^{2} (R + |λ|) ∥∇ log g∥^{2} + ϕ^{2} ∥A∥^{2} γ_{n}(x)dx)^{1/2}.

This term is then small by Lemma 5.3. □

Lemma 5.12. Let Σ ⊆ ℝ^{n+1} be a C^{∞} hypersurface, with possibly nonempty boundary. Assume ∃ λ ∈ ℝ such that H(x) = ⟨x, N(x)⟩ + λ for all x ∈ Σ. Let δ := δ(Σ). Assume that δ < ∞. If H ≥ 0, and if λ < 0, then 1 ≤ δ ≤ 2.

Proof. First, δ ≥ 1 follows from the definition of δ in (31) and Lemma 4.2. More specifically, if ϕ = ϕ_{R} denotes the cutoff function constructed in Lemma 5.11, we have by the product rule (Remark 4.3)

δ ≥ \frac{∫_{Σ} ϕ⟨v, N⟩L(ϕ⟨v, N⟩)γ_{n}(x)dx}{∫_{Σ} ϕ^{2}⟨v, N⟩^{2}γ_{n}(x)dx}

= \frac{∫_{Σ} (ϕ^{2}⟨v, N⟩^{2} + ϕ⟨v, N⟩^{2}Lϕ + 2ϕ⟨v, N⟩⟨∇⟨v, N⟩, ∇ϕ⟩)γ_{n}(x)dx}{∫_{Σ} ϕ^{2}⟨v, N⟩^{2}γ_{n}(x)dx}.

As R → ∞, the last two terms in the top of the integral converge to zero, by construction of ϕ, so that δ ≥ 1. Now, since Lg = δg from Lemma 5.1, by Lemma 5.11,

\begin{aligned}δ ∫_{Σ} Hgγ_{n}(x)dx & = ∫_{Σ} HLgγ_{n}(x)dx = ∫_{Σ} gLHγ_{n}(x)dx + O(ε) \\ & =^{(24)} ∫_{Σ}(2Hg + λg ∥A∥^{2})γ_{n}(x)dx + O(ε). \end{aligned}  (42)

Also, by Corollary 5.10, ∫_{Σ} Lgγ_{n}(x)dx = lim_{R→∞} ∫_{Σ} ϕ_{R}Lgγ_{n}(x)dx = 0, so

δ ∫_{Σ} gγ_{n}(x)dx = ∫_{Σ} Lgγ_{n}(x)dx =^{(21)} ∫_{Σ} g(1 + ∥A∥^{2})γ_{n}(x)dx.

That is,

∫_{Σ} g ∥A∥^{2} γ_{n}(x)dx = (δ − 1) ∫_{Σ} gγ_{n}(x)dx.  (43)

Finally, combining (42) and (43),

(δ − 2) ∫_{Σ} Hgγ_{n}(x)dx = λ ∫_{Σ} g ∥A∥^{2} γ_{n}(x)dx + O(ε) = λ(δ − 1) ∫_{Σ} gγ_{n}(x)dx + O(ε).

So, by our assumptions on λ < 0, g ≥ 0, H ≥ 0, the right side is nonpositive. In order for the left side to be nonpositive, we must have δ ≤ 2. □

6. Perturbations using H or an Eigenfunction

Lemma 6.1 (Perturbation using an Eigenfunction). Let Σ be a symmetric, orientable C^{∞} hypersurface with δ(Σ) < ∞. Assume ∃ λ ∈ ℝ such that H(x) = ⟨x, N(x)⟩ + λ for all x ∈ Σ. Let g: Σ → (0, ∞) from Lemma 5.1 so that Lg = δg. Then

∫_{Σ}(1 + g)L(1 + g)γ_{n}(x)dx = ∫_{Σ} (δ(1 + g)^{2} + ∥A∥^{2} + (1 − δ))γ_{n}(x)dx.

Proof. Integrating by parts with Corollary 5.10,

∫_{Σ}(1 + g)L(1 + g)γ_{n}(x)dx = ∫_{Σ} (gLg + 2Lg + L(1))γ_{n}(x)dx

= ∫_{Σ} (δg^{2} + 2δg + ∥A∥^{2} + 1)γ_{n}(x)dx = ∫_{Σ} (δ(1 + g)^{2} + ∥A∥^{2} + (1 − δ))γ_{n}(x)dx.

□

Corollary 6.2. Let Ω ⊆ ℝ^{n+1} minimize Problem 1.1. Let Σ := ∂^{∗}Ω. Then by Lemma 3.1, ∃ λ ∈ ℝ such that H(x) = ⟨x, N(x)⟩ + λ. Assume that H > 0, λ < 0 and

∫_{Σ}(∥A∥^{2} − 1)γ_{n}(x)dx > 0.

Then, after rotating Ω, ∃ r > 0 and ∃ 0 ≤ k ≤ n such that Σ = rS^{k} × ℝ^{n−k}.

Proof. From Lemma 5.12, 1 ≤ δ ≤ 2. Let g: Σ → (0, ∞) be the eigenfunction guaranteed to exist by Lemma 5.1. If g is not constant, then we can multiply it by a constant as necessary so that ∫_{Σ}(1 + g)γ_{n}(x)dx = 0. Since 1 ≤ δ ≤ 2 and since ∫_{Σ}(∥A∥^{2} − 1)γ_{n}(x)dx > 0, Lemma 6.1 contradicts Lemma 3.2. So, we must assume that g is constant.

If g is constant, then L(1) = g(∥A∥^{2} + 1) = δg. That is, ∥A∥^{2} is equal to a constant c ∈ ℝ. Choose b ∈ ℝ such that ∫_{Σ}(H(x) + b)γ_{n}(x)dx = 0. By (24), LH = 2H + λ ∥A∥^{2} = 2H + λc. So, using Lemma 5.6 and Corollary 5.10, and also using the definition of b,

\frac{1}{2} ∫_{Σ}(H + b)L(H + b)γ_{n}(x)dx = \frac{1}{2} ∫_{Σ}(H + b)(2H + λc + b(c + 1))γ_{n}(x)dx

= ∫_{Σ}(H + b)Hγ_{n}(x)dx = ∫_{Σ} H^{2}γ_{n}(x)dx − \frac{[∫_{Σ} Hγ_{n}(x)dx]^{2}}{∫_{Σ} γ_{n}(y)dy} .

From the Cauchy-Schwarz inequality, −[∫_{Σ} Hγ_{n}(x)dx]^{2} ≥ − ∫_{Σ} H^{2}γ_{n}(x)dx ∫_{Σ} γ_{n}(y)dy, with equality only if H is constant. Therefore

∫_{Σ}(H + b)L(H + b)γ_{n}(x)dx ≥ 0,

with equality only if H is constant. So, if H is not constant, then this inequality contradicts Lemma 3.2. So, we must assume that H is constant. If H is constant and if ∥A∥ is constant, then H/ ∥A∥ is constant. It follows from [CM12, Proof of Theorem 10.1] [Hui93, p.187-188] that Σ is then a round cylinder. (If ∥A∥ = 0 on Σ, then Σ is a rotation of rS^{0} × ℝ^{n} for some r > 0.) □

Remark 6.3. It seems difficult to classify self-shrinkers such that ∥A∥ is constant [Gua17].

7. Huisken-type Classification

The following Lemma is a routine generalization of [CM12, Lemma 10.14] and [Zhu20, Lemma 7.3].

Lemma 7.1. Let Ω ⊆ ℝ^{n+1} such that ∂Ω \ ∂^{∗}Ω has Hausdorff dimension at most n − 7. Let Σ := ∂^{∗}Ω. Let λ ∈ ℝ. Assume δ(Σ) < ∞. Assume H(x) > 0 for all x ∈ Σ and H(x) = ⟨x, N(x)⟩ + λ, ∀ x ∈ Σ. Then

∫_{Σ} ∥ ∥A∥ ∇ log H − ∇ ∥A∥ ∥^{2}γ_{n}(x)dx ≤ λ ∫_{Σ} (\frac{∥A∥^{4}}{H} + ⟨A^{2}, A⟩)γ_{n}(x)dx.

Proof. First, let L := Δ − ⟨x, ∇⟩ as in (20). Using (36) and integrating by parts by Lemma 5.9 and Corollaries 5.10 and 5.5,

− ∫_{Σ}⟨∇ ∥A∥^{2} , ∇ log H⟩γ_{n}(x)dx = ∫_{Σ} ∥A∥^{2} L log Hγ_{n}(x)dx

= ∫_{Σ} ∥A∥^{2} ( − ∥∇ log H∥^{2} + 1 − ∥A∥^{2} + λ\frac{∥A∥^{2}}{H} )γ_{n}(x)dx.

Note that

∥ ∥A∥ ∇ log H − ∇ ∥A∥ ∥^{2} = ∥A∥^{2} ∥∇ log H∥^{2} + ∥∇ ∥A∥∥^{2} − 2 ∥A∥ ⟨∇ log H, ∇ ∥A∥⟩
= ∥A∥^{2} ∥∇ log H∥^{2} + ∥∇ ∥A∥∥^{2} − ⟨∇ log H, ∇ ∥A∥^{2}⟩.

Integrating and combining the above,

∫_{Σ} ∥ ∥A∥ ∇ log H − ∇ ∥A∥ ∥^{2}γ_{n}(x)dx = ∫_{Σ} ( ∥∇ ∥A∥∥^{2} + ∥A∥^{2} − ∥A∥^{4} + λ\frac{∥A∥^{4}}{H} )γ_{n}(x)dx.

(44)

We manipulate the first term on the right. Integrating by parts with Corollary 5.10,

∫_{Σ} ∥∇ ∥A∥∥^{2} γ_{n}(x)dx = − ∫_{Σ} ∥A∥ L ∥A∥ γ_{n}(x)dx

=^{(21)} ∫_{Σ} ( − ∥A∥ L ∥A∥ + ∥A∥^{4} + ∥A∥^{2} )γ_{n}(x)dx

≤^{(34)} ∫_{Σ} ( ∥A∥^{4} + λ⟨A^{2}, A⟩ − ∥A∥^{2} )γ_{n}(x)dx.

Combining this with (44),

∫_{Σ} ∥ ∥A∥ ∇ log H − ∇ ∥A∥ ∥^{2}γ_{n}(x)dx ≤ ∫_{Σ} (λ\frac{∥A∥^{4}}{H} + λ⟨A^{2}, A⟩)γ_{n}(x)dx.

□

Remark 7.2. Repeating the above calculation and replacing log H with log(H − λ) gives

L log(H − λ) = − ∥∇ log(H − λ)∥^{2} + 1 − ∥A∥^{2} + \frac{λ}{H − λ}.

∫_{Σ} ∥ ∥A∥ ∇ log(H − λ) − ∇ ∥A∥ ∥^{2}γ_{n}(x)dx ≤ λ ∫_{Σ} (\frac{∥A∥^{2}}{H − λ} + ⟨A^{2}, A⟩)γ_{n}(x)dx.

Recovering the main result of [CW18, Theorem 4.1] with a slightly different proof.

For curves in the plane, it is known that circles and lines are the only solutions of H(x) = ⟨x, N⟩ + λ when λ > 0 [Gua18, Theorem 1.5] [Cha17, Theorem 1.4]. For convex surfaces, we extend this argument to arbitrary dimensions.

Corollary 7.3 (Huisken-type classification, λ > 0). Let Ω ⊆ ℝ^{n+1} be convex such that ∂Ω \ ∂^{∗}Ω has Hausdorff dimension at most n − 7. Let λ > 0. Assume δ(Σ) < ∞. Suppose H(x) = ⟨x, N(x)⟩ + λ, ∀ x ∈ Σ. Then H/ ∥A∥ is constant on Σ.

Proof. Let B be a positive definite n×n matrix. From logarithmic convexity of the ℓ_{p} norms (or Hölder’s inequality),

(Tr(B^{2}))^{1/2} ≤ (Tr(B))^{1/4}(Tr(B^{3}))^{1/4}.

Since Ω is convex, A is negative definite, and all of the diagonal entries of A are nonpositive. So, taking the fourth power of this inequality,

∥A∥^{4} ≤ H |⟨A^{2}, A⟩| .

That is,

\frac{∥A∥^{4}}{H} ≤ |⟨A^{2}, A⟩| .

So, from Lemma 7.1, λ > 0, and again using that A is negative definite and H > 0,

∫_{Σ} ∥ ∥A∥ ∇ log H − ∇ ∥A∥ ∥^{2}γ_{n}(x)dx ≤ λ ∫_{Σ} (\frac{∥A∥^{4}}{H} + ⟨A^{2}, A⟩)γ_{n}(x)dx ≤ 0.

Therefore, ∇_{e_{i}} log H = ∇_{e_{i}} log ∥A∥ for all 1 ≤ i ≤ n. That is, ∥A∥ /H is constant on Σ. □

Corollary 7.4 (Huisken-type classification, λ > 0). Let Ω ⊆ ℝ^{n+1} be convex such that ∂Ω \ ∂^{∗}Ω has Hausdorff dimension at most n − 7. Let λ > 0. Assume δ(Σ) < ∞. Suppose H(x) = ⟨x, N(x)⟩ + λ, ∀ x ∈ Σ. Then, after rotating Ω, ∃ r > 0 and ∃ 0 ≤ k ≤ n such that Σ = rS^{k} × ℝ^{n−k}.

Proof. Since ∥A∥ /H is constant on Σ, it follows from [CM12, Proof of Theorem 10.1] [Hui93, p.187-188] that Σ is then a round cylinder. □

8. Random Almost-Eigenfunctions

In this section, we let 𝔼 denote the expected value of a random variable.

Lemma 8.1. Let v be a uniformly distributed random vector in S^{n}. Let a, b ∈ S^{n}. Then

𝔼⟨v, a⟩⟨v, b⟩ = \frac{⟨a, b⟩}{n + 1}.

Proof. Let θ := cos^{−1}(⟨a, b⟩). In the case n = 1 we have

𝔼⟨v, a⟩⟨v, b⟩ = \frac{1}{2π} ∫_{0}^{2π} cos(t) cos(t + θ)dt = \frac{1}{2π} ∫_{0}^{2π} cos(t)[cos(t) cos(θ) − sin(t) sin(θ)]dt

= cos θ\frac{1}{2π} ∫_{0}^{2π} cos^{2}(t)dt = cos θ\frac{1}{2π} ∫_{0}^{2π}(1/2)(1 + cos(2t))dt = \frac{cos θ}{2} = \frac{⟨a, b⟩}{2}.

For any n > 1, there then exists c_{n+1} ∈ ℝ such that 𝔼⟨v, a⟩⟨v, b⟩ = c_{n+1}⟨a, b⟩. Choosing a = b, we get c_{n+1} = c_{n+1} ∥a∥^{2} = 𝔼⟨v, a⟩^{2}. And summing over an orthonormal basis of a in ℝ^{n+1} gives (n + 1)c_{n+1} = 𝔼 ∥v∥^{2} = 1. That is, c_{n+1} = 1/(n + 1).

□

Let A_{x} denote the matrix A at the point x. Let Π = Π_{x}: ℝ^{n+1} → ℝ^{n} be the linear projection of ℝ^{n+1} onto the tangent space at x, viewed as ℝ^{n} itself. So, if A is diagonal at x, then there exists a basis u_{1}, …, u_{n+1} of ℝ^{n+1} such that Π_{x}u_{n+1} = 0, Π_{x}u_{i} = u_{i} and A_{x}Π_{x}e_{i} = a_{ii}u_{i} for all 1 ≤ i ≤ n. Let p := ∫_{Σ} γ_{n}(x)dx.

Lemma 8.2 (Independent Bilinear Perturbation). Let Σ be an orientable C^{∞} hypersurface. Assume that there exists λ ∈ ℝ such that for all x ∈ Σ, H(x) = ⟨x, N⟩ + λ. Then there exists v, w ∈ S^{n} so that, if m := \frac{1}{p} ∫_{Σ}⟨v, N⟩⟨w, N⟩γ_{n}(y)dy, then

(n + 1)^{2} ∫_{Σ}(⟨v, N⟩⟨w, N⟩ − m)L(⟨v, N⟩⟨w, N⟩ − m)γ_{n}(x)dx

≥ ∫_{Σ}(1 − ∥A_{x}∥^{2})γ_{n}(x)dx · (1 − \frac{1}{p^{2}} ∫_{Σ} ∫_{Σ}⟨N(y), N(z)⟩^{2}γ_{n}(y)γ_{n}(z)dydz)

− \frac{2}{p} ∫_{Σ} ∫_{Σ} ∥A_{x}Π_{x}(N(y))∥^{2} )γ_{n}(y)γ_{n}(x)dydx.

Proof. Let v, w ∈ S^{n}. From Remark 4.3,

L⟨v, N⟩⟨w, N⟩ = ⟨v, N⟩⟨w, N⟩ + 2⟨∇⟨v, N⟩, ∇⟨w, N⟩⟩ − ∥A∥^{2} ⟨v, N⟩⟨w, N⟩.

So, if we define m := \frac{1}{p} ∫_{Σ}⟨v, N⟩⟨w, N⟩γ_{n}(y)dy, then

L(⟨v, N⟩⟨w, N⟩−m) =^{(21)} ⟨v, N⟩⟨w, N⟩+2⟨∇⟨v, N⟩, ∇⟨w, N⟩⟩−∥A∥^{2} (⟨v, N⟩⟨w, N⟩+m)−m.

Therefore,

∫_{Σ}(⟨v, N⟩⟨w, N⟩ − m)L(⟨v, N⟩⟨w, N⟩ − m)γ_{n}(x)dx
= ∫_{Σ} (⟨v, N⟩^{2}⟨w, N⟩^{2} + 2(⟨v, N⟩⟨w, N⟩ − m)⟨∇⟨v, N⟩, ∇⟨w, N⟩⟩ − 2m⟨v, N⟩⟨w, N⟩
− ∥A∥^{2} (⟨v, N⟩^{2}⟨w, N⟩^{2} − m^{2}) + m^{2})γ_{n}(x)dx
= ∫_{Σ} (⟨v, N⟩^{2}⟨w, N⟩^{2} + 2(⟨v, N⟩⟨w, N⟩ − m)⟨∇⟨v, N⟩, ∇⟨w, N⟩⟩
− ∥A∥^{2} (⟨v, N⟩^{2}⟨w, N⟩^{2} − m^{2}) + m^{2})γ_{n}(x)dx − 2pm^{2}.

From Lemma 8.1, if v, w are uniformly distributed in S^{n}, then

𝔼⟨v, N⟩^{2}⟨w, N⟩^{2} = \frac{1}{(n + 1)^{2}} .

𝔼m^{2} = \frac{1}{p^{2}} ∫_{Σ} ∫_{Σ} 𝔼⟨v, N(y)⟩⟨w, N(y)⟩⟨v, N(z)⟩⟨w, N(z)⟩γ_{n}(y)γ_{n}(z)dydz

= \frac{1}{p^{2}} ∫_{Σ} ∫_{Σ} \frac{⟨N(y), N(z)⟩^{2}}{(n + 1)^{2}} γ_{n}(y)γ_{n}(z)dydz.

Let e_{1}, …, e_{n} ∈ ℝ^{n+1} be an orthonormal frame for Σ (embedded into the tangent space of Σ so that ⟨e_{i}, N(x)⟩ = 0 for all 1 ≤ i ≤ n) such that A is a diagonal matrix at x. Then

⟨∇⟨v, N⟩, ∇⟨w, N⟩⟩ =^{(29)} ⟨∑_{i=1}^{n} a_{ii}⟨v, e_{i}⟩e_{i}, ∑_{j=1}^{n} a_{jj}⟨w, e_{j}⟩e_{j}⟩ = ∑_{i=1}^{n} a_{ii}^{2}⟨v, e_{i}⟩⟨w, e_{i}⟩.

So, using Lemma 8.1 and ∑_{i=1}^{n} a_{ii}^{2} = ∥A_{x}∥^{2},

𝔼⟨v, N(y)⟩⟨w, N(y)⟩⟨∇⟨v, N(x)⟩, ∇⟨w, N(x)⟩⟩ = ∑_{i=1}^{n} a_{ii}^{2}𝔼⟨v, N(y)⟩⟨w, N(y)⟩⟨v, e_{i}⟩⟨w, e_{i}⟩

= ∑_{i=1}^{n} a_{ii}^{2} \frac{⟨e_{i}, N(y)⟩^{2}}{(n + 1)^{2}} = \frac{∥A_{x}Π_{x}(N(y))∥^{2}}{(n + 1)^{2}} .

In particular, if x = y, we have Π(N(y)) = 0, so that

𝔼⟨v, N(x)⟩⟨w, N(x)⟩⟨∇⟨v, N(x)⟩, ∇⟨w, N(x)⟩⟩ = 0.

Also,

𝔼m⟨∇⟨v, N(x)⟩, ∇⟨w, N(x)⟩⟩ = \frac{1}{p} ∫_{Σ} 𝔼⟨v, N(y)⟩⟨w, N(y)⟩⟨∇⟨v, N(x)⟩, ∇⟨w, N(x)⟩⟩γ_{n}(y)dy

= \frac{1}{p} ∫_{Σ} \frac{∥A_{x}Π_{x}(N(y))∥^{2}}{(n + 1)^{2}} γ_{n}(y)dy.

Combining the above calculations,

𝔼 ∫_{Σ}(⟨v, N⟩⟨w, N⟩ − m)L(⟨v, N⟩⟨w, N⟩ − m)γ_{n}(x)dx

= 𝔼 ∫_{Σ} (⟨v, N⟩^{2}⟨w, N⟩^{2} + 2(⟨v, N⟩⟨w, N⟩ − m)⟨∇⟨v, N⟩, ∇⟨w, N⟩⟩

− ∥A∥^{2} (⟨v, N⟩^{2}⟨w, N⟩^{2} − m^{2}) + m^{2})γ_{n}(x)dx − 2p𝔼m^{2}

= \frac{1}{(n + 1)^{2}} ∫_{Σ} (1 − \frac{2}{p} ∫_{Σ} \frac{∥A_{x}Π(N(y))∥^{2}}{(n + 1)^{2}} γ_{n}(y)dy

− ∥A_{x}∥^{2} + \frac{(1 + ∥A_{x}∥^{2})}{p^{2}} ∫_{Σ} ∫_{Σ}⟨N(y), N(z)⟩^{2}γ_{n}(y)γ_{n}(z)dydz)γ_{n}(x)dx

− \frac{2}{(n + 1)^{2}} \frac{1}{p} ∫_{Σ} ∫_{Σ}⟨N(y), N(z)⟩^{2}γ_{n}(y)γ_{n}(z)dydz.

Simplifying a bit using the definition of p, we get

(n + 1)^{2}𝔼 ∫_{Σ}(⟨v, N⟩⟨w, N⟩ − m)L(⟨v, N⟩⟨w, N⟩ − m)γ_{n}(x)dx

= ∫_{Σ}(1 − ∥A_{x}∥^{2})γ_{n}(x)dx + ∫_{Σ}(∥A_{x}∥^{2} − 1)γ_{n}(x)dx\frac{1}{p^{2}} ∫_{Σ} ∫_{Σ}⟨N(y), N(z)⟩^{2}γ_{n}(y)γ_{n}(z)dydz

− \frac{2}{p} ∫_{Σ} ∫_{Σ} ∥A_{x}Π(N(y))∥^{2} )γ_{n}(y)γ_{n}(x)dydx.

Therefore, ∃ v, w ∈ S^{n} such that ∫_{Σ}(⟨v, N⟩⟨w, N⟩ − m)L(⟨v, N⟩⟨w, N⟩ − m)γ_{n}(x)dx exceeds or equals the above expected value. □

Remark 8.3. If we repeat the proof of Lemma 8.2 for v, w which are conditioned to satisfy ⟨v, w⟩ = 0, then the result is the same.

As above, let A_{x} denote the matrix A at the point x. Let Π = Π_{x}: ℝ^{n+1} → ℝ^{n} be the linear projection of ℝ^{n+1} onto the tangent space at x, viewed as ℝ^{n} itself. Let p := ∫_{Σ} γ_{n}(x)dx.

Corollary 8.4. Let Σ be an orientable C^{∞} hypersurface. Assume that there exists λ ∈ ℝ such that for all x ∈ Σ, H(x) = ⟨x, N⟩ + λ. Then there exists v, w ∈ S^{n} so that, if m := \frac{1}{p} ∫_{Σ}⟨v, N⟩⟨w, N⟩γ_{n}(y)dy, then

(n + 1)^{2} ∫_{Σ}(⟨v, N⟩⟨w, N⟩ − m)L(⟨v, N⟩⟨w, N⟩ − m)γ_{n}(x)dx

≥ \frac{1}{p^{2}} ∫_{Σ×Σ×Σ}(1 − ∥A_{x}∥^{2} − 2 ∥A_{y}∥_{2→2}^{2}) ∥Π_{y}(N(z))∥^{2} γ_{n}(x)γ_{n}(y)γ_{n}(z)dxdydz.

Proof. From Lemma 8.2, there exists v, w ∈ S^{n} such that

(n + 1)^{2} ∫_{Σ}(⟨v, N⟩⟨w, N⟩ − m)L(⟨v, N⟩⟨w, N⟩ − m)γ_{n}(x)dx

≥ ∫_{Σ}(1 − ∥A_{x}∥^{2})γ_{n}(x)dx · (1 − \frac{1}{p^{2}} ∫_{Σ} ∫_{Σ}⟨N(y), N(z)⟩^{2}γ_{n}(y)γ_{n}(z)dydz)

− \frac{2}{p} ∫_{Σ} ∫_{Σ} ∥A_{z}Π_{z}(N(y))∥^{2} )γ_{n}(y)γ_{n}(z)dydz.

Now, using the definition of p and Π_{z},

(1 − \frac{1}{p^{2}} ∫_{Σ} ∫_{Σ}⟨N(y), N(z)⟩^{2}γ_{n}(y)γ_{n}(z)dydz)

= \frac{1}{p^{2}} ∫_{Σ} ∫_{Σ}(1 − ⟨N(y), N(z)⟩^{2})γ_{n}(y)γ_{n}(z)dydz

= \frac{1}{p^{2}} ∫_{Σ} ∫_{Σ} ∥Π_{z}N(y)∥^{2} γ_{n}(y)γ_{n}(z)dydz.

Recall that ∥A_{z}∥_{2→2}^{2} = sup_{u∈S^{n}} ∥A_{z}u∥^{2}. Combining the above with ∥A_{z}Π_{z}(N(y))∥^{2} ≤ ∥A_{z}∥_{2→2}^{2} ∥Π_{z}(N(y))∥^{2} gives

(n + 1)^{2} ∫_{Σ}(⟨v, N⟩⟨w, N⟩ − m)L(⟨v, N⟩⟨w, N⟩ − m)γ_{n}(x)dx

≥ ∫_{Σ}(1 − ∥A_{x}∥^{2})γ_{n}(x)dx · \frac{1}{p^{2}} ∫_{Σ} ∫_{Σ} ∥Π_{z}N(y)∥^{2} γ_{n}(y)γ_{n}(z)dydz.

− (∫_{Σ} γ_{n}(x)dx)\frac{2}{p^{2}} ∫_{Σ} ∥A_{z}∥_{2→2}^{2} ∫_{Σ} ∥Π_{z}(N(y))∥^{2} γ_{n}(y)γ_{n}(z)dydz

= \frac{1}{p^{2}} ∫_{Σ} ∫_{Σ}(1 − ∥A_{x}∥^{2} − 2 ∥A_{z}∥_{2→2}^{2}) ∫_{Σ} ∥Π_{z}(N(y))∥^{2} γ_{n}(y)γ_{n}(x)γ_{n}(z)dydxdz.

□

We now prove the Main Theorem, Theorem 1.11

Proof of Theorem 1.11. Combine Lemma 3.2 with Corollaries 7.3, 8.4 and 6.2 (note that if Ω is convex, then H ≥ 0). □

9. Normal Variations with Dilations

Lemma 9.1 ([CM12, Lemma 3.20], [CW18, Lemma 3.1]). Let Σ ⊆ ℝ^{n+1} be a C^{∞} hypersurface. Let λ ∈ ℝ. Assume that

H(x) = ⟨x, N⟩ + λ,  ∀ x ∈ Σ,  (45)

then

\frac{1}{2}Δ ∥x∥^{2} = −H⟨N, x⟩ + n.  (46)
\frac{1}{2}L ∥x∥^{2} = n − ∥x∥^{2} − λ⟨x, N⟩ = n − ∥x∥^{2} − λ(H − λ).  (47)

Proof. Using that any hypersurface Σ satisfies Δx = −HN,

\frac{1}{2}Δ ∥x∥^{2} = ⟨Δx, x⟩ + ∥∇x∥^{2} = −H⟨N, x⟩ + n.

Also,

\frac{1}{2}L ∥x∥^{2} =^{(20)} \frac{1}{2}Δ ∥x∥^{2} − \frac{1}{2}⟨x, ∇ ∥x∥^{2}⟩ = \frac{1}{2}Δ ∥x∥^{2} − ∥x^{T} ∥_{2}^{2}

=^{(46)} −H⟨N, x⟩ + n − ∥x^{T} ∥_{2}^{2} =^{(45)} n − ⟨N, x⟩^{2} − ∥x^{T} ∥_{2}^{2} − λ⟨x, N⟩

= n − ∥x∥^{2} − λ⟨x, N⟩. =^{(45)} n − ∥x∥^{2} − λ(H − λ).

□

Lemma 9.2 ([CM12, Lemma 3.25] [CW18, Lemma 3.3]). Let Σ ⊆ ℝ^{n+1} be a C^{∞} hypersurface with ∂Σ = ∅. Let λ ∈ ℝ. Assume that H(x) = ⟨x, N⟩ + λ ∀ x ∈ Σ, and that Σ has polynomial volume growth. Then

∫_{Σ} (n − ∥x∥^{2} − λH + λ^{2})γ_{n}(x)dx = 0.  (48)

∫_{Σ} ((n + 2) ∥x∥^{2} − ∥x∥^{4} − λ ∥x∥^{2} (H − λ) − 2(H − λ)^{2}) = 0.  (49)

∫_{Σ}(∥x∥^{2} − n)^{2}γ_{n}(x)dx = ∫_{Σ} (2n + (H − λ)( − 2H + λ(n − ∥x∥^{2})))γ_{n}(x)dx.  (50)

Proof. Integrating by parts with Corollary 5.10 proves (48) as follows.

0 = ∫_{Σ} (L ∥x∥^{2} )γ_{n}(x)dx =^{(47)} ∫_{Σ} (n − ∥x∥^{2} − λ(H − λ))γ_{n}(x)dx.

Using Corollary 5.10 again,

−4 ∫_{Σ} ∥x^{T} ∥^{2} γ_{n}(x)dx = − ∫_{Σ} ∥∇ ∥x∥^{2}∥^{2} γ_{n}(x)dx = ∫_{Σ} ∥x∥^{2} L ∥x∥^{2} γ_{n}(x)dx

=^{(47)} 2 ∫_{Σ} (n ∥x∥^{2} − ∥x∥^{4} − λ ∥x∥^{2} (H − λ)).

Rearranging and using ∥x^{T} ∥_{2}^{2} = −∥x^{N}∥_{2}^{2} + ∥x∥^{2} =^{(45)} −(H − λ)^{2} + ∥x∥^{2}, we get

∫_{Σ} ((n + 2) ∥x∥^{2} − ∥x∥^{4} − λ ∥x∥^{2} (H − λ) − 2(H − λ)^{2}) = 0.

To prove (50), we write

∫_{Σ}(∥x∥^{2} − n)^{2}γ_{n}(x)dx = ∫_{Σ} ( ∥x∥^{4} − 2n ∥x∥^{2} + n^{2})γ_{n}(x)dx

=^{(49)} ∫_{Σ} ((n + 2 − 2n) ∥x∥^{2} − λ ∥x∥^{2} (H − λ) − 2(H − λ)^{2} + n^{2})γ_{n}(x)dx

= ∫_{Σ} ((−n + 2) ∥x∥^{2} − λ ∥x∥^{2} (H − λ) − 2(H − λ)^{2} + n^{2})γ_{n}(x)dx

=^{(48)} ∫_{Σ} ((−n + 2)[n − λ(H − λ)] − λ ∥x∥^{2} (H − λ) − 2(H − λ)^{2} + n^{2})γ_{n}(x)dx

= ∫_{Σ} 2n + (H − λ)(nλ − 2λ − λ ∥x∥^{2} − 2H + 2λ)γ_{n}(x)dx

= ∫_{Σ} 2n + (H − λ)( − 2H + λ(n − ∥x∥^{2}))γ_{n}(x)dx.

□

Recall that if X: ℝ^{n+1} → ℝ^{n+1} is a given vector field, then we define Ψ: ℝ^{n+1} ×(−1, 1) → ℝ^{n+1} so that Ψ(x, 0) = x and such that \frac{d}{ds}|_{s=0}Ψ(x, s) = X(Ψ(x, s)) ∀ x ∈ ℝ^{n+1} and ∀ s ∈ (−1, 1) as in (13). And for any s ∈ (−1, 1), we define Ω_{s} := Ψ(Ω, s) and Σ_{s} := ∂Ω_{s}.

Define Z: ℝ^{n+1} → ℝ^{n+1} so that Z(x) := \frac{d^{2}}{ds^{2}} |_{s=0}Ψ(x, s) is the “acceleration” vector field. Suppose we write Ψ in its Taylor expansion (with respect to s) as

Ψ(x, s) = x + sX(x) + \frac{s^{2}}{2} Z(x) + o(s^{2}),  ∀ x ∈ ℝ^{n+1}, s ∈ (−1, 1).  (51)

Note that Z(x) := \frac{d^{2}}{ds^{2}} |_{s=0}Ψ(x, s) and (13) imply that

Z(x)_{i} = ∑_{j=1}^{n+1} X_{j}(x)\frac{∂}{∂x_{j}} X_{i}(x),  ∀ x ∈ ℝ^{n+1}, ∀ 1 ≤ i ≤ n + 1.  (52)

Let Jac denote the Jacobian determinant in ℝ^{n+1}. As shown in [Hei15, Lemma 12.2][CS07],

JacΨ(x, 0) = 1.  (53)
(d/ds)JacΨ|_{s=0} = div(X).  (54)
\frac{d^{2}}{ds^{2}} JacΨ(x, s)|_{s=0} = div((div(X))X).  (55)

For any t_{s} > 0, define

F_{t_{s}} := (2πt_{s})^{− \frac{n}{2}} ∫_{Σ_{s}} e^{− \frac{∥x∥^{2}}{2ts}} dx = (2π)^{− \frac{n}{2}} ∫_{Σ_{s}/\sqrt{t_{s}}} e^{− \frac{∥x∥^{2}}{2}} dx.

Lemma 9.3 (First Variation of Surface area [CM12, Lemma 3.1]). Let Σ ⊆ ℝ^{n+1} be a C^{∞} hypersurface. Denote f := ⟨X, N⟩. Then

\frac{∂}{∂s}F_{t_{s}} = ∫_{Σ_{s}} [f(H − ⟨x, N⟩/t_{s}) + \frac{\frac{∂}{∂s}t_{s}}{2} (\frac{∥x∥^{2}}{t_{s}^{2}} − \frac{n}{t_{s}} )](2πt_{s})^{− \frac{n}{2}} e^{− \frac{∥x∥^{2}}{2ts}} dx.

Proof. We use logarithmic differentiation.

\frac{∂}{∂t_{s}} log ((2πt_{s})^{− \frac{n}{2}} e^{− \frac{∥x∥^{2}}{2ts}} ) = −\frac{n}{2} \frac{1}{t_{s}} + \frac{∥x∥^{2}}{2t_{s}^{2}} .

Taking the s derivative and applying the chain rule, using x′ = fN and (dx)′ = fHdx,

\frac{∂}{∂s} ((2πt_{s})^{− \frac{n}{2}} e^{− \frac{∥x∥^{2}}{2ts}} dx)

= (fH − \frac{f}{t_{s}} ⟨x, N⟩ + (\frac{∂}{∂s}t_{s})( − \frac{n}{2} \frac{1}{t_{s}} + \frac{∥x∥^{2}}{2t_{s}^{2}} )) (2πt_{s})^{− \frac{n}{2}} e^{− \frac{∥x∥^{2}}{2ts}} dx.

□

Remark 9.4. Let Σ be a C^{∞} hypersurface. If ∃ λ ∈ ℝ such that H = ⟨x, N⟩ + λ ∀ x ∈ Σ, and if h := \frac{∂}{∂s}|_{s=0}t_{s}, then

\frac{∂}{∂s}|_{s=0}F_{t_{s}} = ∫_{Σ}(fλ + (h/2)(∥x∥^{2} − n))γ_{n}dx

=^{(48)} ∫_{Σ}[fλ + (h/2)(−λ)(H − λ)]γ_{n}dx = λ ∫_{Σ}[f − (h/2)⟨x, N⟩]γ_{n}dx.

Lemma 9.5. Let Σ ⊆ ℝ^{n+1} be a C^{∞} hypersurface. Then

f(x)⟨∇f(x), N(x)⟩ = ⟨Z(x), N(x)⟩, ∀ x ∈ Σ.

Proof. Let div denote the divergence on ℝ^{n+1}, and let div_{τ} denote the divergence on Σ. Let DN denote the matrix of partial derivatives of N. Using (52) and div(N(x)) = div_{τ}(N(x)) ∀ x ∈ Σ to get

\begin{matrix}⟨N(x), Z(x)⟩ = (f(x))^{2}⟨(DN(x))N(x), N(x)⟩ + f(x)⟨∇f(x), N(x)⟩ \\ = (f(x))^{2}(div(N(x)) − div_{τ}(N(x))) + f(x)⟨∇f(x), N(x)⟩ = f(x)⟨∇f(x), N(x)⟩. \end{matrix} (56)
□

Lemma 9.6 (Second Variation of Surface Area, with Dilations [CM12, Theorem 4.1]). Let Σ ⊆ ℝ^{n+1} be a C^{∞} hypersurface with ∂Σ = ∅. Let t_{0} := 1, h := \frac{∂}{∂s}|_{s=0}t_{s}, f := ⟨X, N⟩.

\frac{∂^{2}}{∂s^{2}} |_{s=0}F_{t_{s}} = ∫_{Σ} ( − fLf + 2hf⟨x, N⟩ − h^{2}( ∥x∥^{2} − \frac{n}{2})

+ [f(H − ⟨x, N⟩) + \frac{h}{2}(∥x∥^{2} − n)]^{2}

+ f′(H − ⟨x, N⟩) + \frac{h′}{2} (∥x∥^{2} − n))γ_{n}(x)dx.

Proof. We let ′ denote \frac{∂}{∂s}|_{s=0}. Using Lemma 9.3 and (ab)′′ = ab[log(ab)]′′ + ab([log(ab)]′)^{2},

\frac{∂^{2}}{∂s^{2}} |_{s=0}F_{t_{s}} = ∫_{Σ} (f(H − ⟨x, N⟩/t_{s})′ + \frac{h}{2}(\frac{∥x∥^{2}}{t_{s}^{2}} − \frac{n}{t_{s}} )′

+ [f(H − ⟨x, N⟩/t_{0}) + \frac{h}{2}(\frac{∥x∥^{2}}{t_{0}^{2}} − \frac{n}{t_{0}} )]^{2}

+ f′(H − ⟨x, N⟩) + \frac{h′}{2} (\frac{∥x∥^{2}}{t_{0}^{2}} − \frac{n}{t_{0}} ))γ_{n}(x)dx.

We use H′ = −Δf − ∥A∥^{2} f, N′ = −∇f, x′ = fN [CM12, A.3, A.4] to get

⟨x, N⟩′ = f − ⟨x, ∇f⟩.

Using (t_{s}^{−1})′ = −ht_{0}^{−2} and t_{0} = 1,

(H − ⟨x, N⟩/t_{s})′ = −Δf − ∥A∥^{2} f − \frac{(f − ⟨x, ∇f⟩)}{t_{0}} + h\frac{⟨x, N⟩}{t_{0}^{2}} = =^{(21)} −Lf + h⟨x, N⟩.

Using (t_{s}^{−2})′ = −h2t_{0}^{−3}, x′ = fN and t_{0} = 1,

(\frac{∥x∥^{2}}{t_{s}^{2}} − \frac{n}{t_{s}} )′ = 2\frac{⟨x, fN⟩}{t_{0}^{2}} − 2h\frac{∥x∥^{2}}{t_{0}^{3}} + \frac{hn}{t_{0}^{2}} = 2f⟨x, N⟩ − 2h( ∥x∥^{2} − \frac{n}{2}).

Combining the above completes the proof. □

Corollary 9.7. Let Σ be a C^{∞} hypersurface with ∂Σ = ∅. Let λ ∈ ℝ. Assume that H(x) = ⟨x, N⟩ + λ for all x ∈ Σ. Then

\frac{∂^{2}}{∂s^{2}} |_{s=0}F_{t_{s}} = ∫_{Σ} ( − fLf + 2hf(H − λ) + h^{2}(H − λ)(λ − H/2 + λ(n − ∥x∥^{2})/4)

+ f^{2}λ^{2} + λhf(∥x∥^{2} − n) + λf′ + \frac{h′}{2} (−λ(H − λ)))γ_{n}(x)dx.

Proof. Using (48) and (50) in Lemma 9.6,

\frac{∂^{2}}{∂s^{2}} |_{s=0}F_{t_{s}} = ∫_{Σ} ( − fLf + 2hf⟨x, N⟩ − h^{2}(−λ(H − λ) + n/2) + f^{2}λ^{2} + λhf(∥x∥^{2} − n)

+ \frac{h^{2}}{4} [2n + (H − λ)(−2H + λ(n − ∥x∥^{2}))] + λf′ + \frac{h′}{2} (−λ(H − λ)))γ_{n}(x)dx

= ∫_{Σ} ( − fLf + 2hf⟨x, N⟩ + h^{2}(H − λ)(λ − H/2 + λ(n − ∥x∥^{2})/4)

+ f^{2}λ^{2} + λhf(∥x∥^{2} − n) + λf′ + \frac{h′}{2} (−λ(H − λ)))γ_{n}(x)dx.

□

For any t_{s} > 0, define

G_{t_{s}} := (2πt_{s})^{− \frac{n+1}{2}} ∫_{Ω_{s}} e^{− \frac{∥x∥^{2}}{2ts}} dx = (2π)^{− \frac{n+1}{2}} ∫_{Ω_{s}/\sqrt{t_{s}}} e^{− \frac{∥x∥^{2}}{2}} dx.

Repeating much of the reasoning of Lemma 9.3 (see e.g. [BBJ16, Eq. (20)]), we get

Lemma 9.8 (First Variation of Volume). Let Σ be a C^{∞} hypersurface. Let Jac denote the Jacobian determinant in ℝ^{n+1}. Let t_{0} := 1, h := \frac{∂}{∂s}|_{s=0}t_{s}, f := ⟨X, N⟩.

\frac{∂}{∂s}G_{t_{s}} = (2πt_{s})^{− \frac{n+1}{2}} ∫_{Ω} e^{− \frac{∥Ψ(x,s)∥^{2}}{2ts}} JacΨ(x, s)

[\frac{h}{2}(\frac{∥x∥^{2}}{t_{s}^{2}} − \frac{n + 1}{t_{s}}) + \frac{[JacΨ(x, s)]′}{JacΨ(x, s)} − \frac{⟨Ψ(x, s), Ψ′(x, s)⟩}{t_{s}}]dx.

\frac{∂}{∂s}|_{s=0}G_{t_{s}} = (2πt_{s})^{− \frac{n+1}{2}} ∫_{Ω} e^{− \frac{∥x∥^{2}}{2}} [\frac{h}{2}(\frac{∥x∥^{2}}{t_{s}^{2}} − \frac{n + 1}{t_{s}}) + div(X) − ⟨x, X⟩]dx

= ∫_{Ω} div(( − \frac{xh}{2} + X)γ_{n+1}(x)) = ∫_{Σ} ( − \frac{h}{2}⟨x, N⟩ + ⟨X, N⟩)γ_{n+1}(x)dx

= ∫_{Σ} ( − \frac{h}{2}⟨x, N⟩ + f)γ_{n+1}(x)dx.

Lemma 9.9 (Second Variation of Volume). Let Σ ⊆ ℝ^{n+1} be a C^{∞} hypersurface with ∂Σ = ∅. Let h := \frac{∂}{∂s}|_{s=0}t_{s}, f := ⟨X, N⟩

\frac{∂^{2}}{∂s^{2}} |_{s=0}G_{t_{s}} = ∫_{Σ} (f(∇_{N}f + fH − f⟨x, N⟩) + \frac{1}{2}(h^{2} + h′)⟨x, N⟩
+ h(f − (h/4)⟨x, N⟩)(∥x∥^{2} − [n + 1]))γ_{n+1}(x)dx.

Proof. For any 1 ≤ i ≤ n + 1, let Z_{i} := ∑_{j=1}^{n+1} X_{j} \frac{∂}{∂x_{j}} X_{i}. We let ′ denote \frac{∂}{∂s}|_{s=0}. Using (ab)′′ = ab[log(ab)]′′ + ab([log(ab)]′)^{2} in Lemma 9.8, with Ψ(x, 0) = x (51), (53) and (54),

\frac{∂^{2}}{∂s^{2}} |_{s=0}G_{t_{s}}

= ∫_{Ω} (\frac{h}{2}(\frac{∥Ψ(x, s)∥^{2}}{t_{s}^{2}} − \frac{n + 1}{t_{s}})′ + [JacΨ(x, s)]′′ − ⟨x, Ψ′′(x, s)⟩ − ⟨Ψ′(x, s), Ψ′(x, s)⟩

+ [\frac{h}{2}(\frac{∥x∥^{2}}{t_{0}^{2}} − \frac{n + 1}{t_{0}}) + div(X) − ⟨x, X⟩]^{2} − [(JacΨ(x, s))′]^{2} + ⟨x, X⟩h

+ \frac{h′}{2} (n + 1 − ∥x∥^{2}))γ_{n+1}(x)dx.

Using (51), (t_{s}^{−1})′ = −ht_{0}^{−2}, (t_{s}^{−2})′ = −h2t_{0}^{−3} and t_{0} = 1,

(\frac{∥Ψ(x, s)∥^{2}}{t_{s}^{2}} − \frac{n + 1}{t_{s}})′ = 2\frac{⟨x, X⟩}{t_{0}^{2}} − 2h\frac{∥x∥^{2}}{t_{0}^{3}} + \frac{h(n + 1)}{t_{0}^{2}} = 2⟨x, X⟩ − 2h( ∥x∥^{2} − \frac{n + 1}{2}).

Combining the above with (51), (54), (55) and t_{0} = 1,

\frac{∂^{2}}{∂s^{2}} |_{s=0}G_{t_{s}} = ∫_{Ω} (⟨x, X⟩h − h^{2}( ∥x∥^{2} − \frac{n + 1}{2}) + div(Xdiv(X)) − ⟨Z, x⟩ − ⟨X, X⟩

+ [\frac{h}{2}(\frac{∥x∥^{2}}{t_{0}^{2}} − \frac{n + 1}{t_{0}}) + div(X) − ⟨x, X⟩]^{2} − (div(X))^{2} + ⟨x, X⟩h

+ \frac{h′}{2} (n + 1 − ∥x∥^{2}))γ_{n+1}(x)dx

= ∫_{Ω} (2⟨x, X⟩h − h^{2}( ∥x∥^{2} − \frac{n + 1}{2}) + \frac{h^{2}}{4} (∥x∥^{2} − [n + 1])^{2}

+ h[div(X) − ⟨x, X⟩](∥x∥^{2} − [n + 1]) + \frac{h′}{2} (n + 1 − ∥x∥^{2})

+ div(Xdiv(X)) − ⟨Z, x⟩ − ⟨X, X⟩ − 2⟨x, X⟩div(X) + ⟨x, X⟩^{2})γ_{n+1}(x)dx.

Then using Lemma 9.12, and also using

div(X(∥x∥^{2} − [n + 1])γ_{n+1}(x)) = (∥x∥^{2} − [n + 1])div(Xγ_{n+1}(x)) + 2⟨x, X⟩γ_{n+1}(x)

= (∥x∥^{2} − [n + 1])[div(X) − ⟨x, X⟩]γ_{n+1}(x) + 2⟨x, X⟩γ_{n+1}(x).

div(x(∥x∥^{2} − [n + 1])γ_{n+1}(x)) = (∥x∥^{2} − [n + 1])div(xγ_{n+1}(x)) + 2 ∥x∥^{2} γ_{n+1}(x)
= −(∥x∥^{2} − [n + 1])^{2}γ_{n+1}(x) + 2 ∥x∥^{2} γ_{n+1}(x).

We get

\frac{∂^{2}}{∂s^{2}} |_{s=0}G_{t_{s}}

= ∫_{Ω} div(Xdiv(Xγ_{n+1}(x))) + (2⟨x, X⟩h − h^{2}( ∥x∥^{2} − \frac{n + 1}{2}) + \frac{h^{2}}{4} (∥x∥^{2} − [n + 1])^{2}

+ h[div(X) − ⟨x, X⟩](∥x∥^{2} − [n + 1]) + \frac{h′}{2} (n + 1 − ∥x∥^{2}))γ_{n+1}(x)dx

= ∫_{Ω} div(Xdiv(Xγ_{n+1}(x))) + (h^{2}[div(xγ_{n+1}(x)) − [n + 1]/2]

+ \frac{h^{2}}{4} [−div(x(∥x∥^{2} − [n + 1])γ_{n+1}(x))] + \frac{h^{2}}{2} ∥x∥^{2} γ_{n+1}(x)

+ h[div(X(∥x∥^{2} − [n + 1])γ_{n+1}(x))] + \frac{h′}{2} div(xγ_{n+1}(x)))dx.

Applying the divergence theorem,

\frac{∂^{2}}{∂s^{2}} |_{s=0}G_{t_{s}} = ∫_{Σ} f(div(Xγ_{n+1}(x)))

+ (h^{2}⟨x, N⟩ − (h^{2}/4)⟨x, N⟩(∥x∥^{2} − [n + 1]) + hf(∥x∥^{2} − [n + 1]) + (h′/2)⟨x, N⟩)γ_{n+1}(x)dx + ∫_{Ω} ( − (h^{2}/2)(n + 1) + (h^{2}/2) ∥x∥^{2} )γ_{n+1}(x)dx = ∫_{Σ} f(div(Xγ_{n+1}(x))) + (h^{2}⟨x, N⟩ − (h^{2}/4)⟨x, N⟩(∥x∥^{2} − [n + 1])

+ hf(∥x∥^{2} − [n + 1]) + (h′/2)⟨x, N⟩)γ_{n+1}(x)dx − ∫_{Ω} \frac{h^{2}}{2} div(xγ_{n+1}(x))dx

= ∫_{Σ} f(fH + ∇_{N}f − f⟨x, N⟩) + (h^{2}/2)⟨x, N⟩ + ( − (h^{2}/4)⟨x, N⟩(∥x∥^{2} − [n + 1]) + hf(∥x∥^{2} − [n + 1]) + (h′/2)⟨x, N⟩)γ_{n+1}(x)dx. Above we used

div(Xγ_{n+1}(x)) = div(fNγ_{n+1}(x)) = (∇_{N}f + fH − f⟨x, N⟩)γ_{n+1}(x).

□

For technical reasons, we restrict the following Lemma to f > 0. Lemma 9.10 (Volume Preserving Second Variation of the Surface area). Let Σ ⊆ ℝ^{n+1} be a ℂ^{∞} hypersurface. Let λ ∈ ℝ. Suppose H(x) = ⟨x, N⟩ + λ for all x ∈ Σ. Given any f: Σ → ℝ with f > 0, ∃ h, h′, f′ so that \frac{∂}{∂s}|_{s=0}G_{t_{s}} = \frac{∂^{2}}{∂s^{2}} |_{s=0}G_{t_{s}} = \frac{∂}{∂s}|_{s=0}F_{t_{s}} = 0, and

\frac{∂^{2}}{∂s^{2}} |_{s=0}F_{t_{s}} = ∫_{Σ} ( − fLf + 2hf⟨x, N⟩ − \frac{h^{2}}{2} ⟨x, N⟩^{2} + λ\frac{h^{2}}{4} ⟨x, N⟩)γ_{n}(x)dx.

Proof. As above, we let′ denote \frac{∂}{∂s}|_{s=0}. Using N′ = −∇f [CM12, A.3], and that X is parallel to N on Σ, and Lemma 9.5,

f′ = ⟨X, N⟩′ = ⟨Z, N⟩ = f∇_{N}f.

Using this fact, we let h′ := 0, we choose h so that \frac{∂}{∂s}|_{s=0}G_{t_{s}} = 0 in Lemma 9.8. That is, we choose h so that

∫_{Σ} fγ_{n}(x)dx = \frac{h}{2} ∫_{Σ}⟨x, N⟩γ_{n}(x)dx.  (57)

By Remark 9.4, \frac{∂}{∂s}|_{s=0}F_{t_{s}} = 0 as well. We then choose f∇_{N}f in Lemma 9.9 so that \frac{∂^{2}}{∂s^{2}} G_{t_{s}} = 0. That is, we choose ∇_{N}f so that

f∇_{N}f := −f^{2}H + f^{2}⟨x, N⟩ − (h^{2}/2)⟨x, N⟩ − h(f − (h/4)⟨x, N⟩)(∥x∥^{2} − [n + 1]).

We then substitute these choices of h, h′, f′ into Lemma 9.6.

\frac{∂^{2}}{∂s^{2}} |_{s=0}F_{t_{s}}

= ∫_{Σ} ( − fLf + 2hf⟨x, N⟩ − h^{2}( ∥x∥^{2} − \frac{n}{2}) + [fλ + \frac{h}{2}(∥x∥^{2} − n)]^{2} + f′λ)γ_{n}(x)dx

= ∫_{Σ} ( − fLf + 2hf⟨x, N⟩ − h^{2}( ∥x∥^{2} − \frac{n}{2}) + [fλ + \frac{h}{2}(∥x∥^{2} − n)]^{2}

+ λ[ − f^{2}H + f^{2}⟨x, N⟩ − (h^{2}/2)⟨x, N⟩ − h(f − (h/4)⟨x, N⟩)(∥x∥^{2} − [n + 1])])γ_{n}(x)dx.

Applying (48) and (50), along with repeated use of H = ⟨x, N⟩ + λ gives

\frac{∂^{2}}{∂s^{2}} |_{s=0}F_{t_{s}} = ∫_{Σ} ( − fLf + 2hf⟨x, N⟩ − h^{2}( ∥x∥^{2} − n + (n/2))

+ f^{2}λ^{2} + λfh(∥x∥^{2} − n) + \frac{h^{2}}{4} (∥x∥^{2} − n)^{2}

+ λ[ − f^{2}H + f^{2}⟨x, N⟩ − (h^{2}/2)⟨x, N⟩ − h(f − (h/4)⟨x, N⟩)(∥x∥^{2} − [n + 1])])γ_{n}(x)dx

= ∫_{Σ} ( − fLf + 2hf⟨x, N⟩ − h^{2}λ(λ − H) − h^{2}(n/2)) + f^{2}λ^{2} + λfh(∥x∥^{2} − n)

+ \frac{h^{2}}{4} (2n + ⟨x, N⟩(−2H + λ(n − ∥x∥^{2})))

+ λ[ − f^{2}H + f^{2}⟨x, N⟩ − (h^{2}/2)⟨x, N⟩ − h(f − (h/4)⟨x, N⟩)(∥x∥^{2} − [n + 1])])γ_{n}(x)dx

= ∫_{Σ} ( − fLf + 2hf⟨x, N⟩ + h^{2}λ⟨x, N⟩ − h^{2}(n/2)) + f^{2}λ^{2} + λfh(∥x∥^{2} − n)

+ \frac{h^{2}}{4} (2n + ⟨x, N⟩(−2H + λ(n − ∥x∥^{2})))

+ λ[ − f^{2}H + f^{2}⟨x, N⟩ − (h^{2}/2)⟨x, N⟩ − h(f − (h/4)⟨x, N⟩)(∥x∥^{2} − [n + 1])])γ_{n}(x)dx

= ∫_{Σ} ( − fLf + 2hf⟨x, N⟩ + h^{2}λ⟨x, N⟩ − h^{2}(n/2)) + f^{2}λ^{2} + λfh

+ \frac{h^{2}}{4} (2n + ⟨x, N⟩(−2H + λ(n − ∥x∥^{2})))

+ λ[ − f^{2}H + f^{2}⟨x, N⟩ − (h^{2}/2)⟨x, N⟩ + (h^{2}/4)⟨x, N⟩(∥x∥^{2} − [n + 1])])γ_{n}(x)dx

= ∫_{Σ} ( − fLf + 2hf⟨x, N⟩ − h^{2}(n/2)) + f^{2}λ^{2} + λfh

+ \frac{h^{2}}{4} (2n + ⟨x, N⟩(−2H + λ(n − ∥x∥^{2})))

+ λ[ − f^{2}H + f^{2}⟨x, N⟩ + (h^{2}/4)⟨x, N⟩(∥x∥^{2} − [n − 1])])γ_{n}(x)dx

= ∫_{Σ} ( − fLf + 2hf⟨x, N⟩ − h^{2}(n/2)) + f^{2}λ^{2} + λfh

+ \frac{h^{2}}{4} (2n + ⟨x, N⟩(−2H)) + λ[ − f^{2}H + f^{2}⟨x, N⟩ + (h^{2}/4)⟨x, N⟩])γ_{n}(x)dx

= ∫_{Σ} ( − fLf + 2hf⟨x, N⟩) + f^{2}λ^{2} + λfh

− \frac{h^{2}}{2} ⟨x, N⟩H + λ[ − f^{2}H + f^{2}⟨x, N⟩ + (h^{2}/4)⟨x, N⟩])γ_{n}(x)dx

= ∫_{Σ} ( − fLf + 2hf⟨x, N⟩) + λfh − \frac{h^{2}}{2} ⟨x, N⟩H + λ(h^{2}/4)⟨x, N⟩)γ_{n}(x)dx.

Applying H = ⟨x, N⟩ + λ one more time

\frac{∂^{2}}{∂s^{2}} |_{s=0}F_{t_{s}} = ∫_{Σ} ( − fLf + 2hf⟨x, N⟩ + λfh − \frac{h^{2}}{2} ⟨x, N⟩^{2} − λ\frac{h^{2}}{4} ⟨x, N⟩)γ_{n}(x)dx.

Finally, by (57), we then get

\frac{∂^{2}}{∂s^{2}} |_{s=0}F_{t_{s}} = ∫_{Σ} ( − fLf + 2hf⟨x, N⟩ − \frac{h^{2}}{2} ⟨x, N⟩^{2} + λ\frac{h^{2}}{4} ⟨x, N⟩)γ_{n}(x)dx.

□

Using Lemma 9.10 for H − λ + t, and then differentiating the resulting expression twice at t = 0, we get the following interesting inequality. Lemma 9.11. Let Σ ⊆ ℝ^{n+1} be a ℂ^{∞} hypersurface. Let λ ∈ ℝ. Suppose H(x) = ⟨x, N⟩ + λ for all x ∈ Σ. Let t ∈ ℝ. Let f := H − λ + t. Define h := 2\frac{∫_{Σ} fγ_{n+1}(x)dx}{∫_{Σ}⟨x,N⟩γ_{n+1}(x)dx} as in (57). Then

∫_{Σ} ( − fLf + 2hf⟨x, N⟩ − \frac{h^{2}}{2} ⟨x, N⟩^{2} + λ\frac{h^{2}}{4} ⟨x, N⟩)γ_{n+1}(x)dx

≤ t^{2} ∫_{Σ} ( − ∥A∥^{2} + H \frac{∫_{Σ} γ_{n+1}(x)dx}{∫_{Σ}⟨x, N⟩γ_{n+1}(x)dx})γ_{n+1}(x)dx.

Proof. From (24), and (21), L(H − λ) = 2H + λ ∥A∥^{2} − λ ∥A∥^{2} − λ = 2H − λ.

Also, by the definition of H and f, and using H − λ = ⟨x, N⟩,

h := 2\frac{∫_{Σ} fγ_{n+1}(x)dx}{∫_{Σ}⟨x, N⟩γ_{n+1}(x)dx} = 2\frac{∫_{Σ}(⟨x, N⟩ + t)γ_{n+1}(x)dx}{∫_{Σ}⟨x, N⟩γ_{n+1}(x)dx} = 2 + 2\frac{∫_{Σ} tγ_{n+1}(x)dx}{∫_{Σ}⟨x, N⟩γ_{n+1}(x)dx}.

For brevity, we define

s := 2\frac{∫_{Σ} tγ_{n+1}(x)dx}{∫_{Σ}⟨x, N⟩γ_{n+1}(x)dx}.

Then h = 2 + s, and

∫_{Σ} ( − fLf + 2hf⟨x, N⟩ − \frac{h^{2}}{2} ⟨x, N⟩^{2} + λ\frac{h^{2}}{4} ⟨x, N⟩)γ_{n+1}(x)dx

= ∫_{Σ} ( − (H − λ + t)(2H − λ + t ∥A∥^{2} + t) + 2(2 + s)(⟨x, N⟩ + t)⟨x, N⟩

− \frac{(2 + s)^{2}}{2} ⟨x, N⟩^{2} + λ\frac{(2 + s)^{2}}{4} ⟨x, N⟩)γ_{n+1}(x)dx

= ∫_{Σ} ( − (H − λ + t)^{2} − (⟨x, N⟩ + t)(H + t ∥A∥^{2}) + (4 + 2s)⟨x, N⟩^{2} + (4 + 2s)t⟨x, N⟩

− \frac{(2 + s)^{2}}{2} ⟨x, N⟩^{2} + λ\frac{(2 + s)^{2}}{4} ⟨x, N⟩)γ_{n+1}(x)dx

= ∫_{Σ} ( − (⟨x, N⟩ + t)^{2} − (⟨x, N⟩ + t)(H + t ∥A∥^{2}) + (4 + 2s)⟨x, N⟩^{2} + (4 + 2s)t⟨x, N⟩

− \frac{(2 + s)^{2}}{2} ⟨x, N⟩^{2} + (H − ⟨x, N⟩)\frac{(2 + s)^{2}}{4} ⟨x, N⟩)γ_{n+1}(x)dx

= ∫_{Σ} ( − ⟨x, N⟩^{2} − 2t⟨x, N⟩ − t^{2} − H⟨x, N⟩ − tH − (⟨x, N⟩ + t)t ∥A∥^{2}

+ 4⟨x, N⟩^{2} + 2s⟨x, N⟩^{2} + (4 + 2s)t⟨x, N⟩ − 2⟨x, N⟩^{2} − 2s⟨x, N⟩^{2} − \frac{s^{2}}{2} ⟨x, N⟩^{2}

− ⟨x, N⟩^{2} + H⟨x, N⟩ + s(H − ⟨x, N⟩)⟨x, N⟩ + \frac{s^{2}}{4} (H − ⟨x, N⟩)⟨x, N⟩)γ_{n+1}(x)dx

= ∫_{Σ} ( − 2t⟨x, N⟩ − t^{2} − H⟨x, N⟩ − tH − (⟨x, N⟩ + t)t ∥A∥^{2}

+ 2s⟨x, N⟩^{2} + (4 + 2s)t⟨x, N⟩ − 2s⟨x, N⟩^{2} − \frac{s^{2}}{2} ⟨x, N⟩^{2}

+ H⟨x, N⟩ + s(H − ⟨x, N⟩)⟨x, N⟩ + \frac{s^{2}}{4} (H − ⟨x, N⟩)⟨x, N⟩.

= ∫_{Σ} ( − 2t⟨x, N⟩ − t^{2} − tH − (⟨x, N⟩ + t)t ∥A∥^{2}

+ (4 + 2s)t⟨x, N⟩ − \frac{s^{2}}{2} ⟨x, N⟩^{2} + s(H − ⟨x, N⟩)⟨x, N⟩ + \frac{s^{2}}{4} (H − ⟨x, N⟩)⟨x, N⟩.

Let F (t) be the above expression, as a function of t. Then F′(0) = ∫_{Σ} ( − 2⟨x, N⟩ − H − ⟨x, N⟩ ∥A∥^{2} + 4⟨x, N⟩

+ 2\frac{∫_{Σ} γ_{n+1}(x)dx}{∫_{Σ}⟨x, N⟩γ_{n+1}(x)dx}(H − ⟨x, N⟩)⟨x, N⟩)γ_{n+1}(x)dx

= ∫_{Σ} (2⟨x, N⟩ − H − (H − λ) ∥A∥^{2} + 2\frac{∫_{Σ} γ_{n+1}(x)dx}{∫_{Σ}⟨x, N⟩γ_{n+1}(x)dx}(H − ⟨x, N⟩)⟨x, N⟩)γ_{n+1}(x)dx.

Integrating by parts to get ∫_{Σ} LHγ_{n}(x)dx = 0 by Corollary 5.10,

∫_{Σ} 2Hγ_{n}(x)dx =^{(24)} ∫_{Σ}(LH − λ ∥A∥^{2})γ_{n}(x)dx = ∫_{Σ}(H(∥A∥^{2} + 1) − λ ∥A∥^{2})γ_{n}(x)dx.

So,

∫_{Σ} H ∥A∥^{2} γ_{n}(x)dx = ∫_{Σ} Hγ_{n}(x)dx + λ ∫_{Σ} ∥A∥^{2} γ_{n}(x)dx.  (58)

From (58), ∫_{Σ}((H − λ) ∥A∥^{2} − H)γ_{n+1}(x)dx = 0. So,

F′(0) = ∫_{Σ} (2⟨x, N⟩ − 2H + 2\frac{∫_{Σ} γ_{n+1}(x)dx}{∫_{Σ}⟨x, N⟩γ_{n+1}(x)dx}(H − ⟨x, N⟩)⟨x, N⟩)γ_{n+1}(x)dx

= ∫_{Σ} ( − 2λ + 2\frac{∫_{Σ} γ_{n+1}(x)dx}{∫_{Σ}⟨x, N⟩γ_{n+1}(x)dx}λ⟨x, N⟩)γ_{n+1}(x)dx = 0.

F′′(0) = ∫_{Σ} ( − 2 − 2 ∥A∥^{2} + 8\frac{∫_{Σ} γ_{n+1}(x)dx}{∫_{Σ}⟨x, N⟩γ_{n+1}(x)dx}⟨x, N⟩ − 4\begin{matrix}[∫_{Σ} γ_{n+1}(x)dx] & 2 \\ [∫_{Σ}⟨x, N⟩γ_{n+1}(x)dx] & 2 \end{matrix} ⟨x, N⟩^{2}

+ 2\begin{matrix}[∫_{Σ} γ_{n+1}(x)dx] & 2 \\ [∫_{Σ}⟨x, N⟩γ_{n+1}(x)dx] & 2 \end{matrix} (H − ⟨x, N⟩)⟨x, N⟩)γ_{n+1}(x)dx

= ∫_{Σ} (6 − 2 ∥A∥^{2} + 2λ\frac{∫_{Σ} γ_{n+1}(x)dx}{∫_{Σ}⟨x, N⟩γ_{n+1}(x)dx} − 4\begin{matrix}[∫_{Σ} γ_{n+1}(x)dx] & 2 \\ [∫_{Σ}⟨x, N⟩γ_{n+1}(x)dx] & 2 \end{matrix} ⟨x, N⟩^{2})γ_{n+1}(x)dx.

Then using the Cauchy-Schwarz inequality,

F′′(0) ≤ ∫_{Σ} (6 − 2 ∥A∥^{2} + 2λ\frac{∫_{Σ} γ_{n+1}(x)dx}{∫_{Σ}⟨x, N⟩γ_{n+1}(x)dx} − 4\frac{∫_{Σ} γ_{n+1}(x)dx}{∫_{Σ}⟨x, N⟩^{2}γ_{n+1}(x)dx}⟨x, N⟩^{2})γ_{n+1}(x)dx

= 2 ∫_{Σ} (1 − ∥A∥^{2} + λ\frac{∫_{Σ} γ_{n+1}(x)dx}{∫_{Σ}⟨x, N⟩γ_{n+1}(x)dx})γ_{n+1}(x)dx

= 2 ∫_{Σ} ( − ∥A∥^{2} + H \frac{∫_{Σ} γ_{n+1}(x)dx}{∫_{Σ}⟨x, N⟩γ_{n+1}(x)dx})γ_{n+1}(x)dx.

□

Proof of Theorem 1.12. Combine Lemmas 9.10 and 9.11. □

Lemma 9.12. Let X ∈ C_{0}^{∞}(ℝ^{n+1}, ℝ^{n+1}). Then for any x ∈ ℝ^{n+1},

div(X(x)div(X(x)γ_{n+1}(x))) = (div(X(x)divX(x)) − ⟨X(x), x⟩div(X(x)) − ⟨X(x), X(x)⟩

− ⟨x, Z(x)⟩ − ⟨x, X(x)⟩div(X) + ⟨X(x), x⟩^{2})γ_{n+1}(x).

Proof. Below, we let ∇ denote the gradient on ℝ^{n+1}.

div(X(x)div(X(x)γ_{n+1}(x)))

= div(X(x)div(X(x))γ_{n+1}(x) − X(x)⟨X(x), x⟩γ_{n+1}(x)))

= (div(X(x)div(X(x))) − ⟨X(x), x⟩div(X(x)))γ_{n+1}(x) − div(X(x)⟨X(x), x⟩γ_{n+1}(x))

= (div(X(x)div(X(x))) − ⟨X(x), x⟩div(X(x)) − div(X(x))⟨X(x), x⟩

+ ⟨X(x), x⟩^{2} − ⟨X(x), ∇⟨X(x), x⟩⟩)γ_{n+1}(x)

= (div(X(x)div(X(x))) − ⟨X(x), x⟩div(X(x)) − div(X(x))⟨X(x), x⟩

+ ⟨X(x), x⟩^{2} − ⟨X(x), X(x)⟩ − ∑_{i,j=1}^{n+1} x_{i}X_{j}(x)\frac{∂}{∂x_{j}} X_{i}(x))γ_{n+1}(x).

We then conclude by (52). □

10. Open Questions

• If Ω is convex, is it possible to improve the eigenvalue bound of Lemma 5.12 in certain specific cases? If so, then Lemma 6.1 would give an improvement to Theorem 1.8.

• Is it possible to classify convex Ω ⊆ ℝ^{n+1} with Σ = ∂Ω such that ∃ λ ∈ ℝ so that H(x) = ⟨x, N(x)⟩ + λ, and such that Σ satisfies the inequality in Theorem 1.12?

11. Comments on the Non-Convex Case

As mentioned previously, the second condition of the Main Theorem, Theorem 1.11, holds without the assumption of convexity. However, the first condition of Theorem 1.11 requires Ω or Ω^{c} to be convex. In the current section, we therefore try to find a result similar to the first part of Theorem 1.11 without the assumption of convexity.

Lemma 11.1 (Perturbation using H). Let Ω ⊆ ℝ^{n+1} such that ∂Ω \ ∂^{∗}Ω has Hausdorff dimension at most n − 7. Let Σ := ∂^{∗}Ω. Suppose δ(Σ) < ∞ and

H(x) = ⟨x, N⟩ + λ,  ∀ x ∈ Σ.  (59)

Let b ∈ ℝ so that ∫_{Σ}(H(x) + b)γ_{n}(x)dx = 0. If H is not constant, then

\begin{matrix}∫_{Σ}(H + b)L(H + b)γ_{n}(x)dx \\ = ∫_{Σ} (2(H + b)^{2} + (b + λ)^{2} ∥A∥^{2} )γ_{n}(x)dx + b ∫_{Σ}⟨x, N⟩γ_{n}(x)dx \\ = ∫_{Σ} (2(H + b)^{2} + (b + λ)^{2}[∥A∥^{2} − 1])γ_{n}(x)dx − λ ∫_{Σ}⟨x, N⟩γ_{n}(x)dx \end{matrix}  (60)

Proof. Using Lemma 5.6, Corollary 5.10 and (21)

∫_{Σ}(H + b)L(H + b)γ_{n}(x)dx = ∫_{Σ}(HLH + 2HLb + b^{2}L(1))γ_{n}(x)dx

=^{(24)∧(21)} ∫_{Σ} (H(2H + λ ∥A∥^{2}) + 2Hb(∥A∥^{2} + 1) + b^{2}(∥A∥^{2} + 1))γ_{n}(x)dx.

From (24) and Corollary 5.10 again,

∫_{Σ}(2H + λ ∥A∥^{2})γ_{n}(x)dx = ∫_{Σ} LHγ_{n}(x)dx = ∫_{Σ} HL(1)γ_{n}(x)dx = ∫_{Σ} H(∥A∥^{2} + 1)γ_{n}(x)dx.

So, ∫_{Σ} H ∥A∥^{2} γ_{n}(x)dx = ∫_{Σ}(H + λ ∥A∥^{2})γ_{n}(x)dx. In summary,

∫_{Σ}(H + b)L(H + b)γ_{n}(x)dx
= ∫_{Σ} (2H^{2} + (λ + 2b)(H + λ ∥A∥^{2}) + 2Hb + b^{2}(∥A∥^{2} + 1))γ_{n}(x)dx
= ∫_{Σ} (2(H + b)^{2} − 2b^{2} + Hλ + λ^{2} ∥A∥^{2} + 2bλ ∥A∥^{2} + b^{2}(∥A∥^{2} + 1))γ_{n}(x)dx
= ∫_{Σ} (2(H + b)^{2} − b^{2} − bλ + (λ + b)^{2} ∥A∥^{2} )γ_{n}(x)dx
= ∫_{Σ} (2(H + b)^{2} + (b + λ)^{2} ∥A∥^{2} )γ_{n}(x)dx − b(b + λ) ∫_{Σ} γ_{n}(x)dx.

In the penultimate line, we used the definition of b, so that ∫_{Σ} Hγ_{n}(x)dx = − ∫_{Σ} bγ_{n}(x)dx.

Finally, ∫_{Σ}(b + λ)γ_{n}(x)dx = ∫_{Σ}(−H + λ)γ_{n}(x)dx =^{(59)} − ∫_{Σ}⟨x, N⟩γ_{n}(x)dx. The first part of (60) is proven. The second follows by writing b = b + λ − λ. □

Lemmas 11.1, 3.2 and 5.2 have the following corollary.

Corollary 11.2. Let Ω ⊆ ℝ^{n+1} minimize Problem 1.1. Let Σ := ∂^{∗}Ω. From Lemma 3.1, ∃ λ ∈ ℝ such that H(x) = ⟨x, N(x)⟩ + λ for all x ∈ Σ.

If either

(i) − ∫_{Σ} H(x)γ_{n}(x)dx · ∫_{Σ}⟨x, N⟩γ_{n}(x)dx > 0, or (ii) ∫_{Σ}(∥A∥^{2} − 1)γ_{n}(x)dx > 0 and −λ ∫_{Σ}⟨x, N⟩γ_{n}(x)dx > 0,

then H must be constant on ∂^{∗}Ω.

By Corollary 11.2(ii), if we want a condition resembling the second condition of Theorem 1.11 to hold without the assumption of convexity, we must investigate the case that H is constant in Problem 1.1.

11.1. The case of constant mean curvature.

Proposition 11.3. Let Ω ⊆ ℝ^{n+1} minimize Problem 1.1. Let Σ := ∂^{∗}Ω. Assume H is constant on Σ. Then either Σ is a round cylinder, or H = ⟨x, N⟩ = 0 for all x ∈ Σ.

Proof. By (24) and (21),

LH = H(∥A∥^{2} + 1) = 2H + λ ∥A∥^{2} .

So, ∥A∥^{2} (H − λ) = H. If H ≠ 0, then λ ≠ H. So, if H ≠ 0, then ∥A∥ is constant. So, if H ≠ 0, it follows from [CM12, Proof of Theorem 10.1] [Hui93, p.187-188] that Σ is a round cylinder.

If H = 0, then either λ = H = 0 or ∥A∥ = 0. If λ = 0, then H = ⟨x, N⟩ = 0 by Lemma 3.1. If ∥A∥ is constant, then it follows from [CM12, Proof of Theorem 10.1] [Hui93, p.187-188] that Σ is a round cylinder. □

Remark 11.4. By Lemma 3.3, if n ≤ 6 and if H is constant on Σ, then in the setting of Corollary 11.2, it cannot occur that H = ⟨x, N⟩ = 0 ∀ x ∈ Σ.

By Proposition 11.3, the only remaining case to consider in Corollary 11.2 is when H = ⟨x, N⟩ = 0 for all x ∈ Σ. That is, we must consider when Σ is a cone with mean curvature zero. This case is eliminated by the following Lemma.

Lemma 11.5. Let Ω ⊆ ℝ^{n+1} minimize Problem 1.1. Let Σ := ∂^{∗}Ω. Then it cannot occur that H(x) = ⟨x, N(x)⟩ = 0 for all x ∈ Σ.

Proof. At every point x ∈ Σ, suppose we label the n eigenvalues of A in order as λ_{1}(x) ≤ ⋯ ≤ λ_{n}(x). Except on a set of Hausdorff dimension at most n−1 on Σ, these eigenvalues are C^{∞} functions [Kat66, Theorem II.5.4, p. 111]. For any 1 ≤ i ≤ n, we claim that Lλ_{i} = 2λ_{i}. This follows from (23) with λ = 0, since A can be diagonalized in a neighborhood of any x ∈ Σ (except on a set of x of Hausdorff dimension at most n − 1.) We claim that there exists a function h: Σ → ℝ and there exist constants c_{1}, …, c_{n} ∈ ℝ such that

λ_{i}(x) = c_{i}h(x),  ∀ x ∈ Σ.  (61)

To see this, let α, β be any two distinct eigenvalues of A. Let t ∈ ℝ such that ∫_{Σ}(α + tβ)γ_{n}(x)dx = 0. Let f := α + tβ. Then ∫_{Σ} f(x)γ_{n}(x)dx = 0, f(x) = f(−x) for all x ∈ Σ, and Lf = 2f. So, ∫_{Σ} f(x)Lf(x)γ_{n}(x) = 2 ∫_{Σ}(f(x))^{2}γ_{n}(x). From Lemma 3.2, we conclude that f(x) = 0 for all x ∈ Σ. Equation (61) follows. Since all eigenvalues of A are multiples of the same function, ∥A∥ is also an eigenfunction of L, with eigenvalue 2. Then Lemma 5.7 with λ = 0 says that

2∥A∥^{2} = ∥A∥L∥A∥ = 2∥A∥^{2} + ∥∇A∥^{2} − ∥∇∥A∥∥^{2}.

That is, ∥∇A∥^{2} = ∥∇∥A∥∥^{2} for all x ∈ Σ. As shown in [CM12, Eq. (10.33)], if A is diagonal at x ∈ Σ and if ∥∇A∥^{2} = ∥∇∥A∥∥^{2}, then ∇_{e_{n}}a_{ii} = 0 for all 1 ≤ i ≤ n − 1. Since Σ is a cone, if we choose e_{n} such that e_{n} is invariant under a dilation of the cone, we have |∇_{e_{n}}a_{ii}| = |a_{ii}/ ∥x∥| at x ∈ Σ. That is, a_{ii} = 0 for all 1 ≤ i ≤ n − 1. And a_{nn} = 0 as well, by the choice of e_{n}. We conclude that A = 0 everywhere, so that Σ is a plane through the origin. But this finally contradicts that Ω is a symmetric set and Σ = ∂^{∗}Ω. We conclude that H = ⟨x, N⟩ = 0 cannot occur on a set of positive n-dimensional Hausdorff measure. □

We can now finally improve the conclusion of Corollary 11.2.

Corollary 11.6. Let Ω ⊆ ℝ^{n+1} minimize Problem 1.1. Let Σ := ∂^{∗}Ω. From Lemma 3.1, ∃ λ ∈ ℝ such that H(x) = ⟨x, N(x)⟩ + λ for all x ∈ Σ.

If either

(i) − ∫_{Σ} H(x)γ_{n}(x)dx · ∫_{Σ}⟨x, N⟩γ_{n}(x)dx > 0, or (ii) ∫_{Σ}(∥A∥^{2} − 1)γ_{n}(x)dx > 0 and −λ ∫_{Σ}⟨x, N⟩γ_{n}(x)dx > 0,

Then, after rotating Ω, ∃ r > 0 and ∃ 0 ≤ k ≤ n such that Σ = rS^{k} × ℝ^{n−k}.

Proof. Let Ω minimize Problem 1.1. From Proposition 11.3, Σ is either a round cylinder, or H(x) = ⟨x, N(x)⟩ = 0 for all x ∈ Σ. The latter case is eliminated by Lemma 11.5. □

Proof of Theorem 1.13. Combine Corollaries 1.10 and 11.6. □

11.2. Infinitesimal Rotations. Corollary 11.6 can eliminate e.g. star-shaped sets with λ < 0 such as the interior of a hyperboloid. Below we present an argument that also eliminates sets with “lumpy” boundary as candidates for minimizers in Problem 1.1. This argument is a modification of one from [HMRR02], which was itself inspired by the standard proof of the Courant Nodal Domain Theorem.

Lemma 11.7 (Infinitesimal Rotations as Eigenfunctions of L). Let Σ ⊆ ℝ^{n+1} be an orientable hypersurface. Let λ ∈ ℝ. Assume that

H(x) = ⟨x, N(x)⟩ + λ,  ∀ x ∈ Σ.  (62)

Let Q be an (n + 1) × (n + 1) real antisymmetric matrix with Q^{∗} = −Q. Then

L⟨Qx, N⟩ = 0,  ∀ x ∈ Σ.  (63)

Proof. Let 1 ≤ i ≤ n and let x ∈ Σ. Then

\begin{aligned}∇_{e_{i}}⟨Qx, N⟩ & = ⟨Qe_{i}, N⟩ + ⟨Qx, ∇_{e_{i}}N⟩ \\ & = ⟨Qe_{i}, N⟩ + ∑_{j=1}^{n}⟨∇_{e_{i}}N, e_{j}⟩⟨Qx, e_{j}⟩ =^{(10)} ⟨Qe_{i}, N⟩ − ∑_{j=1}^{n} a_{ij}⟨Qx, e_{j}⟩. \end{aligned}  (64)

∇H = ∑_{i=1}^{n} ∇_{e_{i}}He_{i} =^{(22)} ∑_{i=1}^{n} ∇_{e_{i}}⟨x, N⟩e_{i} = ∑_{i=1}^{n}⟨x, ∇_{e_{i}}N⟩e_{i} =^{(10)} − ∑_{i,j=1}^{n} a_{ij}⟨x, e_{j}⟩e_{i}.  (65)

Using that A is a symmetric matrix,

\begin{aligned}⟨∇H, Qx⟩ & =^{(65)} − ∑_{i,j=1}^{n} a_{ij}⟨x, e_{j}⟩⟨e_{i}, Qx⟩ = − ∑_{i,j=1}^{n} a_{ij}⟨x, e_{i}⟩⟨e_{j}, Qx⟩ \\ & =^{(64)} ⟨∇⟨Qx, N⟩, x⟩ − ∑_{i=1}^{n}⟨x, e_{i}⟩⟨Qe_{i}, N⟩. \end{aligned}  (66)

Writing x = ⟨x, N⟩N + ∑_{i=1}^{n}⟨x, e_{i}⟩e_{i}, so ⟨Qx, N⟩ = ⟨x, N⟩⟨QN, N⟩ + ∑_{i=1}^{n}⟨x, e_{i}⟩⟨Qe_{i}, N⟩,

⟨∇H, Qx⟩ =^{(66)} ⟨∇⟨Qx, N⟩, x⟩ + ⟨x, N⟩⟨QN, N⟩ − ⟨Qx, N⟩.  (67)

Choose the frame such that ∇_{e_{k}}^{T}e_{j} = 0 at x for every 1 ≤ j, k ≤ n, we then have ∇_{e_{k}}e_{j} = a_{kj}N at x by (9). So, ∀ 1 ≤ i ≤ n,

∇_{e_{i}}∇_{e_{i}}⟨Qx, N⟩

=^{(64)} ⟨Q∇_{e_{i}}e_{i}, N⟩ + ⟨Qe_{i}, ∇_{e_{i}}N⟩ − ∑_{j=1}^{n} ∇_{e_{i}}a_{ij}⟨Qx, e_{j}⟩ − ∑_{j=1}^{n} a_{ij}⟨Qe_{i}, e_{j}⟩ − ∑_{j=1}^{n} a_{ij}⟨Qx, ∇_{e_{i}}e_{j}⟩

=^{(65)} +a_{ii}⟨QN, N⟩ − ∑_{j=1}^{n} a_{ij}⟨Qe_{i}, e_{j}⟩ − ∑_{j=1}^{n} ∇_{e_{i}}a_{ij}⟨Qx, e_{j}⟩ − ∑_{j=1}^{n} a_{ij}⟨Qe_{i}, e_{j}⟩ − ∑_{j=1}^{n} a_{ij}^{2}⟨Qx, N⟩.

(68)

So, using the Codazzi equation (∇_{e_{i}}a_{kj} = ∇_{e_{j}}a_{ki} ∀ 1 ≤ i, j, k ≤ n),

Δ⟨Qx, N⟩ = ∑_{i=1}^{n} ∇_{e_{i}}∇_{e_{i}}⟨Qx, N⟩

=^{(68)∧(11)} −H⟨QN, N⟩ − 2 ∑_{i,j=1}^{n} a_{ij}⟨Qe_{i}, e_{j}⟩ − ∑_{i,j=1}^{n} ∇_{e_{j}}a_{ii}⟨x, e_{j}⟩ − ∑_{ij=1}^{n} a_{ij}^{2}⟨Qx, N⟩

=^{(11)} −H⟨QN, N⟩ − 2 ∑_{i,j=1}^{n} a_{ij}⟨Qe_{i}, e_{j}⟩ + ⟨∇H, Qx⟩ − ∑_{ij=1}^{n} a_{ij}^{2}⟨Qx, N⟩

=^{(67)} (−H + ⟨x, N⟩)⟨QN, N⟩ − 2 ∑_{i,j=1}^{n} a_{ij}⟨Qe_{i}, e_{j}⟩ + ⟨∇⟨Qx, N⟩, x⟩ − ⟨Qx, N⟩ − ∥A∥^{2} ⟨Qx, N⟩.

In summary,

L⟨Qx, N⟩ =^{(21)∧(62)} −λ⟨QN, N⟩ − 2 ∑_{i,j=1}^{n} a_{ij}⟨Qe_{i}, e_{j}⟩.  (69)

It remains to show that the right side of (69) is zero. Note that we have not yet used any property of Q. Since Q^{∗} = −Q, for any v, w ∈ ℝ^{n+1}, we have

⟨Qv, w⟩ = w^{∗}Qv = (w^{∗}Qv)^{∗} = v^{∗}Q^{∗}w = −v^{∗}Qw = −⟨Qw, v⟩.  (70)

In particular, choosing v = w = N, we get

⟨Qv, v⟩ = 0,  ∀ v ∈ ℝ^{n+1}.  (71)

So, the first term on the right of (69) is zero. Lastly, using that A is a symmetric matrix,

∑_{i,j=1}^{n} a_{ij}⟨Qe_{i}, e_{j}⟩ =^{(70)} − ∑_{i,j=1}^{n} a_{ij}⟨Qe_{j}, e_{i}⟩ = − ∑_{i,j=1}^{n} a_{ij}⟨Qe_{i}, e_{j}⟩.

So, the last term on the right of (69) is zero. That is, (63) holds. □

Lemma 11.8. Let Ω ⊆ ℝ^{n+1} with −Ω = Ω. Let Σ := ∂^{∗}Ω. Let Q be a real antisymmetric (n + 1) × (n + 1) matrix. Then

∫_{Σ}⟨Qx, N(x)⟩γ_{n}(x)dx = 0.

Proof. By the divergence theorem,

∫_{Σ}⟨Qx, N(x)⟩γ_{n}(x)dx = ∫_{Ω}(−⟨Qx, x⟩ + Tr(Q))γ_{n}(x)dx =^{(71)} 0.

Here Tr denotes the trace of a matrix. □

We define the number of nodal domains of a function f: Σ → ℝ to be the number of connected components of the set {x ∈ Σ: f(x) ≠ 0}.

Corollary 11.9. Let Ω ⊆ ℝ^{n+1} with −Ω = Ω. Let Σ := ∂^{∗}Ω. Suppose there exists a real antisymmetric (n + 1) × (n + 1) matrix Q with Q^{∗} = −Q such that the function f: Σ → ℝ defined by f(x) := ⟨Qx, N(x)⟩ ∀ x ∈ Σ has more than four nodal domains. Then Ω does not minimize Problem 1.1.

Remark 11.10. Since Ω = −Ω and ∫_{Σ} f(x)γ_{n}(x)dx = 0 by Lemma 11.8, f cannot have exactly two nodal domains. If Σ is a non-spherical ellipsoid aligned with the coordinate axes,

and if we choose Q to have all zero entries other than the upper left corner of (\begin{matrix}0 & 1 \\ −1 & 0 \end{matrix}), then

f has four connected components. And if Σ is a curve in the plane with many oscillations, then f has many connected components. So, the assumption of the theorem implies that Ω has a “lumpy” boundary.

Proof. Label two of the nodal domains as D_{1}, D_{2} ⊆ Σ, so that D_{1} ≠ D_{2}, D_{1} ≠ −D_{2}. Let α ∈ ℝ. Define g: Σ → ℝ so that g(x) := f(x) for any x ∈ D_{1} ∪ (−D_{1}), g(x) := αf(x) for any x ∈ D_{2} ∪ (−D_{2}), and g(x) := 0 otherwise. Choose α ∈ ℝ, such that ∫_{Σ} g(x)γ_{n}(x)dx = 0. Then g(x) = g(−x) for all x ∈ Σ, g vanishes on an open subset of Σ, and g vanishes on the set where ∇g is discontinuous. Also, Lg = 0 on Σ \ [(∂D_{1}) ∪ (∂(−D_{1})) ∪ (∂D_{2}) ∪ (∂(−D_{2}))]. Also ∫_{Σ} gLgγ_{n}(x)dx = 0 by Lemma 11.7.

Assume for the sake of contradiction that Ω minimizes Problem 1.1. From Lemma 3.2, if h: Σ → ℝ is any C^{∞} function such that ∫_{Σ} h(x)γ_{n}(x)dx and h(x) = h(−x) for all x ∈ Σ, then for any t ∈ ℝ,

∫_{Σ}(g + th)L(g + th)γ_{n}(x)dx ≤ 0.

Since this holds for all t ∈ ℝ and ∫_{Σ} gLgγ_{n}(x)dx = 0, we conclude that

∫_{Σ}(gLh + hLg)γ_{n}(x)dx = 0.

Integrating by parts with Lemma 5.10 (which is valid since g vanishes on the set where ∇g is discontinuous and g ∈ W_{1,2}(Σ, γ_{n}), as defined before Lemma 5.3),

2 ∫_{Σ} hLgγ_{n}(x)dx = 0.

Since this equation holds for any mean zero symmetric C^{∞} function h, we conclude that Lg = 0 in the distributional sense. By elliptic regularity, Lg = 0 on all of Σ. By the unique continuation property, since g vanishes on an open subset of Σ, we conclude that g = 0 on Σ. This contradicts the existence of more than one nodal domain of f. We conclude that Ω does not minimize Problem 1.1. □

Acknowledgement. Thanks to Vesa Julin for helpful discussions, especially concerning volume preserving extensions of a function. Thanks also to Domenico La Manna for sharing his preprint [Man17].

References

[AFP00] Luigi Ambrosio, Nicola Fusco, and Diego Pallara, Functions of bounded variation and free discontinuity problems, Oxford Mathematical Monographs, The Clarendon Press, Oxford University Press, New York, 2000. MR 1857292

[Bar01] Franck Barthe, An isoperimetric result for the Gaussian measure and unconditional sets, Bulletin of the London Mathematical Society 33 (2001), 408–416.

[BBJ16] Marco Barchiesi, Alessio Brancolini, and Vesa Julin, Sharp dimension free quantitative estimates for the Gaussian isoperimetric inequality, to appear, Annals of Probability, 2016.

[Bob97] S. G. Bobkov, An isoperimetric inequality on the discrete cube, and an elementary proof of the isoperimetric inequality in gauss space, Ann. Probab. 25 (1997), no. 1, 206–214.

[Bor85] Christer Borell, Geometric bounds on the Ornstein-Uhlenbeck velocity process, Z. Wahrsch. Verw. Gebiete 70 (1985), no. 1, 1–13. MR 795785 (87k:60103)

[Bor03] , The Ehrhard inequality, C. R. Math. Acad. Sci. Paris 337 (2003), no. 10, 663–666. MR 2030108 (2004k:60102)

[BS01] A. Burchard and M. Schmuckenschlger, Comparison theorems for exit times, Geometric & Functional Analysis GAFA 11 (2001), no. 4, 651–692 (English).

[BW16] Jacob Bernstein and Lu Wang, A sharp lower bound for the entropy of closed hypersurfaces up to dimension six, Invent. Math. 206 (2016), no. 3, 601–627. MR 3573969

[Cha17] Jui-En Chang, 1-dimensional solutions of the λ-self shrinkers, Geometriae Dedicata (2017), 1–16.

[CIMW13] Tobias Holck Colding, Tom Ilmanen, William P. Minicozzi, II, and Brian White, The round sphere minimizes entropy among closed self-shrinkers, J. Differential Geom. 95 (2013), no. 1, 53–69. MR 3128979

[CL12] Marco Cicalese and Gian Paolo Leonardi, A selection principle for the sharp quantitative isoperimetric inequality, Arch. Ration. Mech. Anal. 206 (2012), no. 2, 617–643. MR 2980529

[CM12] Tobias H. Colding and William P. Minicozzi, II, Generic mean curvature flow I: generic singularities, Ann. of Math. (2) 175 (2012), no. 2, 755–833. MR 2993752

[CM15] Tobias Holck Colding and William P. Minicozzi, II, Uniqueness of blowups and łojasiewicz inequalities, Ann. of Math. (2) 182 (2015), no. 1, 221–285. MR 3374960

[CMP15] Tobias Holck Colding, William P. Minicozzi, II, and Erik Kjær Pedersen, Mean curvature flow, Bull. Amer. Math. Soc. (N.S.) 52 (2015), no. 2, 297–333. MR 3312634

[COW16] Qing-Ming Cheng, Shiho Ogata, and Guoxin Wei, Rigidity theorems of λ-hypersurfaces, Comm. Anal. Geom. 24 (2016), no. 1, 45–58. MR 3514553

[CR11] Amit Chakrabarti and Oded Regev, An optimal lower bound on the communication complexity of gap Hamming distance, Proc. 43rd Annual ACM Symposium on the Theory of Computing, 2011, pp. 51–60.

[CS07] Rustum Choksi and Peter Sternberg, On the first and second variations of a nonlocal isoperimetric problem, J. Reine Angew. Math. 611 (2007), 75–108. MR 2360604 (2008j:49062)

[CW14] Qing-Ming Cheng and Guoxin Wei, The gauss image of λ-hypersurfaces and a bernstein type problem, Preprint, arXiv:1410.5302, 2014.

[CW18] Qing-Ming Cheng and Guoxin Wei, Complete λ-hypersurfaces of weighted volume-preserving mean curvature flow, Calc. Var. Partial Differential Equations 57 (2018), no. 2, Art. 32, 21. MR 3763110

[EH89] Klaus Ecker and Gerhard Huisken, Mean curvature evolution of entire graphs, Ann. of Math. (2) 130 (1989), no. 3, 453–471. MR 1025164

[Eld15] Ronen Eldan, A two-sided estimate for the gaussian noise stability deficit, Inventiones mathematicae 201 (2015), no. 2, 561–624 (English).

[Eva93] Lawrence C. Evans, Convergence of an algorithm for mean curvature motion, Indiana Univ. Math. J. 42 (1993), no. 2, 533–557. MR 1237058 (95d:58023)

[Fel71] William Feller, An introduction to probability theory and its applications. Vol. II., Second edition, John Wiley & Sons, Inc., New York-London-Sydney, 1971. MR 0270403

[Gua17] Qiang Guang, Self-shrinkers with second fundamental form of constant length, Bull. Aust. Math. Soc. 96 (2017), no. 2, 326–332. MR 3703914

[Gua18] , Gap and rigidity theorems of λ-hypersurfaces, Proc. Amer. Math. Soc. 146 (2018), no. 10, 4459–4471. MR 3834671

[GW95] Michel X. Goemans and David P. Williamson, Improved approximation algorithms for maximum cut and satisfiability problems using semidefinite programming, J. Assoc. Comput. Mach. 42 (1995), no. 6, 1115–1145. MR 1412228 (97g:90108)

[Hei15] Steven Heilman, Low correlation noise stability of symmetric sets, to appear, Journal of Theoretical Probability. Preprint, arXiv:1511.00382, 2015.

[HMRR02] Michael Hutchings, Frank Morgan, Manuel Ritoré, and Antonio Ros, Proof of the double bubble conjecture, Ann. of Math. (2) 155 (2002), no. 2, 459–489. MR 1906593 (2003c:53013)

[Hui90] Gerhard Huisken, Asymptotic behavior for singularities of the mean curvature flow, J. Differential Geom. 31 (1990), no. 1, 285–299. MR 1030675

[Hui93] , Local and global behaviour of hypersurfaces moving by mean curvature, Differential geometry: partial differential equations on manifolds (Los Angeles, CA, 1990), Proc. Sympos. Pure Math., vol. 54, Amer. Math. Soc., Providence, RI, 1993, pp. 175–191. MR 1216584

[IM12] Marcus Isaksson and Elchanan Mossel, Maximally stable Gaussian partitions with discrete applications, Israel J. Math. 189 (2012), 347–396. MR 2931402

[Kat66] Tosio Kato, Perturbation theory for linear operators, Die Grundlehren der mathematischen Wissenschaften, Band 132, Springer-Verlag New York, Inc., New York, 1966. MR 0203473

[Kho02] Subhash Khot, On the power of unique 2-prover 1-round games, Proceedings of the Thirty-Fourth Annual ACM Symposium on Theory of Computing (New York), ACM, 2002, pp. 767–775 (electronic). MR MR2121525

[KKMO07] Subhash Khot, Guy Kindler, Elchanan Mossel, and Ryan O’Donnell, Optimal inapproximability results for MAX-CUT and other 2-variable CSPs?, SIAM J. Comput. 37 (2007), no. 1, 319–357. MR 2306295 (2008d:68035)

[LaO99] Rafał Latał a and Krzysztof Oleszkiewicz, Gaussian measures of dilatations of convex symmetric sets, Ann. Probab. 27 (1999), no. 4, 1922–1938. MR 1742894

[Led94] Michel Ledoux, Semigroup proofs of the isoperimetric inequality in Euclidean and Gauss space, Bull. Sci. Math. 118 (1994), no. 6, 485–510. MR 1309086 (96c:49061)

[Led96] , Isoperimetry and Gaussian analysis, Lectures on probability theory and statistics (Saint-Flour, 1994), Lecture Notes in Math., vol. 1648, Springer, Berlin, 1996, pp. 165–294. MR 1600888 (99h:60002)

[Lee03] John M. Lee, Introduction to smooth manifolds, Graduate Texts in Mathematics, vol. 218, Springer-Verlag, New York, 2003. MR 1930091 (2003k:58001)

[Mag12] Francesco Maggi, Sets of finite perimeter and geometric variational problems, Cambridge Studies in Advanced Mathematics, vol. 135, Cambridge University Press, Cambridge, 2012, An introduction to geometric measure theory. MR 2976521

[Man17] Domenico Angelo La Manna, Local minimality of the ball for the Gaussian perimeter, Preprint,, 2017.

[MN15a] Elchanan Mossel and Joe Neeman, Robust dimension free isoperimetry in gaussian space, The Annals of Probability 43 (2015), no. 3, 971–991.

[MN15b] , Robust optimality of Gaussian noise stability, J. Eur. Math. Soc. (JEMS) 17 (2015), no. 2, 433–482. MR 3317748

[MOO10] Elchanan Mossel, Ryan O’Donnell, and Krzysztof Oleszkiewicz, Noise stability of functions with low influences: invariance and optimality, Ann. of Math. (2) 171 (2010), no. 1, 295–341. MR 2630040 (2012a:60091)

[MR15] Matthew McGonagle and John Ross, The hyperplane is the only stable, smooth solution to the isoperimetric problem in Gaussian space, Geom. Dedicata 178 (2015), 277–296. MR 3397495

[O’D12] Ryan O’Donnell, Open problems in analysis of boolean functions, Preprint, arXiv:1204.6447v1, 2012.

[SC74] V. N. Sudakov and B. S. Cirel′son, Extremal properties of half-spaces for spherically invariant measures, Zap. Naučn. Sem. Leningrad. Otdel. Mat. Inst. Steklov. (LOMI) 41 (1974), 14–24, 165, Problems in the theory of probability distributions, II. MR 0365680 (51 #1932)

[She12] Alexander A. Sherstov, The communication complexity of gap hamming distance, Theory of Computing 8 (2012), no. 1, 197–208.

[Sim68] James Simons, Minimal varieties in riemannian manifolds, Ann. of Math. (2) 88 (1968), 62–105. MR 0233295

[Sim83] Leon Simon, Lectures on geometric measure theory, Proceedings of the Centre for Mathematical Analysis, Australian National University, vol. 3, Australian National University, Centre for Mathematical Analysis, Canberra, 1983. MR 756417

[SSY75] R. Schoen, L. Simon, and S. T. Yau, Curvature estimates for minimal hypersurfaces, Acta Math. 134 (1975), no. 3-4, 275–288. MR 0423263

[Ste70] Elias M. Stein, Singular integrals and differentiability properties of functions, Princeton Mathematical Series, No. 30, Princeton University Press, Princeton, N.J., 1970. MR 0290095 (44 #7280)

[Vid13] Thomas Vidick, A concentration inequality for the overlap of a vector on a large set, with application to the communication complexity of the gap-hamming-distance problem, Chicago J. Theor. Comput. Sci. 2013 (2013).

[Wan11] Lu Wang, A Bernstein type theorem for self-similar shrinkers, Geom. Dedicata 151 (2011), 297–303. MR 2780753

[Zhu20] Jonathan J. Zhu, On the entropy of closed hypersurfaces and singular self-shrinkers, J. Differential Geom. 114 (2020), no. 3, 551–593. MR 4072205

Department of Mathematics, University of Southern California, Los Angeles, CA 90089-2532

E-mail address: stevenmheilman@gmail.com