The Friedland–Hayman inequality is a sharp inequality concerning the growth rates of homogeneous, harmonic functions with Dirichlet boundary conditions on complementary cones dividing Euclidean space into two parts. In this paper, we prove an analogous inequality in which one divides a convex cone into two parts, placing Neumann conditions on the boundary of the convex cone and Dirichlet conditions on the interface. This analogous inequality was already proved by us jointly with Sarah Raynor. Here, we present a new proof that permits us to characterize the case of equality. In keeping with the two-phase free boundary theory introduced by Alt, Caffarelli, and Friedman, such an improvement can be expected to yield further regularity in free boundary problems.

The Friedland–Hayman inequality19 is a sharp inequality concerning the growth rates of homogeneous harmonic functions with Dirichlet boundary conditions on complementary cones dividing Euclidean space into two parts. It plays a crucial role in the interior regularity theory of two-phase free boundary problems, as developed by Alt, Caffarelli, and Friedman.1 In this paper, we prove an analogous inequality in which one divides a convex cone into two parts, placing Neumann conditions on the boundary of the convex cone and Dirichlet conditions on the interface. This analogous inequality was already proved in Ref. 3 and leads to regularity of two-phase free boundaries at points near a fixed boundary with Neumann conditions in convex domains. Here, we present a new proof of independent interest that leads, in addition, to the characterization of the case of equality. In keeping with the theory introduced in Ref. 1 (see also Ref. 12), such an improvement should ultimately yield further regularity properties of the free boundary.

The Friedland–Hayman inequality can be stated as follows:

Theorem I.1
(Ref.19 ). Letu1andu2be non-negative, Hölder continuous functions defined onRn, withu1u2 ≡ 0, and harmonic where they are positive, that is, Δui(x) = 0 wheneverui(x) > 0. Ifuiis homogeneous of degreeαi, then
α1+α22.

Equality holds if and only if the two functions are (up to constant multiples) the positive and negative parts of a linear function. In other words, after rigid motion,

u1(x)=c1x1+,u2(x)=c2x1,c1>0,c2>0,

with x1 being the first coordinate of x and u±(x): = max(0, ±u(x)). This was proved in Ref. 1 (Lemma 6.6) in dimension 2. In addition, in Remark 6.1 of that paper, the authors show that the characterization of equality is valid in all dimensions, provided one knows that the case of equality in a rearrangement theorem on the sphere due to Sperner25 is acheived only for rotationally symmetric caps. This case of equality in Sperner’s rearrangement theorem was subsequently proved by Brothers and Ziemer.9 

To state our main theorem, consider a convex, open cone Γ and two nonempty, disjoint, open, connected, conic subsets Γ±, that is,

Γ̄+Γ̄Γ̄,Γ+Γ=,Γ±.

Define the Neumann and Dirichlet portions of the boundary, γN± and γD±, by

γN±=ΓΓ±andγD±:=ΓΓ±so thatΓ±=γN±γD±.

Theorem I.2
(Friedland–Hayman inequality for convex cones). Suppose thatΓRnis an open, convex cone containing two disjoint, open, connected cones Γ±as above. Letu±denote the unique (up to constant multiples) positive, harmonic function on Γ±that is homogeneous of positive degree and satisfies the mixed boundary conditionsu± = 0 onγD±and (∂/∂ν)u± = 0 onγN±in the weak sense. If the degree ofu±is denotedα±, then
α++α2.
Moreover, if equality holds, then after rotation, there is an open, convex coneΓRn1such that
Γ=R×Γ,Γ±={±x1>0}×Γandu±(x)=c±x1±
for some constantsc± > 0.

The original Friedland–Hayman inequality is the case Γ=Rn with γN±=. If Γ is a half space, then the result also follows from the original Friedland–Hayman inequality by an argument using reflection. The proof of the inequality stated in Theorem I.2 in our joint work with Beck et al.3 uses the Lévy–Gromov isoperimetric inequality on Ricci non-negative manifolds. As we mentioned earlier, the new proof here will permit us to characterize the case of equality.

The exponents α± can be expressed in terms of the lowest eigenvalues of a mixed boundary problem on the spherical cross sections Γ±Sn1. The relationship is given by (8). Those eigenvalues have a variational characterization, which is important to the applications to the free boundary regularity, that is, to the monotonicity formulas of Refs. 1 and 3. Variational characterizations are also important to this proof, as we shall see.

By rearrangement, the Friedland–Hayman inequality reduces to a family of one-dimensional problems parameterized by the dimension n. On the other hand, the inequality in dimension n + 1 implies the inequality in dimension n, as one can see by considering the product of a cone with a line. Thus, the one-dimensional inequality gets harder to prove as n increases. Beckner, Kenig, and Pipher4 gave a more conceptual proof of the original Friedland–Hayman theorem that relies on this rearrangement but circumvents the Brothers–Ziemer result. They took the limit as n tends to infinity and identified and fully analyzed the problem one obtains in the limit, an extremal eigenvalue problem on the real line with a Gaussian weight. The limiting “infinite-dimensional” problem dominates all the finite dimensional ones, and its extremal sets are the half lines x1 > 0, x1 < 0. See Sec. 12.2 in Ref. 12 for the details of this approach.

Rearrangement cannot be used to solve the problem on convex domains. However, it is possible to take the dimension to infinity by considering cones of the form Γ±×RN as N → ∞. Using this device, we will prove the following key proposition:

Proposition I.3.
Ifα+is the characteristic exponent associated with Γ+in Theorem I.2, then
α+inffY+Γ+f(x)2e|x|2/2dxΓ+f(x)2e|x|2/2dx,
withY+={fC0(Rn):f(x)=0forxγD+}. Evidently, the same result holds with + replaced by.

Note that the weight in the variational expression on the right side of the inequality in Proposition I.3 is the restriction to Γ+ of

e|x|2/2eF(x)dx,F(x)=0,xΓ,,xRn\Γ.

Because Γ is convex, F is a generalized convex function, that is, a convex function allowing for the value +∞. Put another way, eF is a generalized log-concave function.

Because our measure is “more log concave” than the Gaussian, we will be able to invoke a variant of Caffarelli’s contraction theorem for the Brenier optimal transport mapping. Recall that Brenier’s mapping can be characterized as follows:

Theorem I.4

(Ref.8 ).Letμ, νbe positive measures onRnand with finite second order moments. Suppose also thatμis absolutely continuous with respect to the Lebesgue measure. Then, there exists a convex functionφ:RnRsuch thatT=φ:RnRntransportsμonto. That is,μ(T−1(E)) = (E), with the normalizing constantc=μ(Rn)/ν(Rn).

The variant of Caffarelli’s contraction theorem that we require can be stated as follows:

Theorem I.5.
LetKbe a convex subset ofRn, and letFbe a convex function onRn. Set
μ=e|x|2/2dx,ν=eF1Kμ.
Then, the Brenier mappingT = ∇φis a contraction: |T(x) − T(y)| ≤ |xy|. Put another way,D2φhas eigenvalues bounded above by 1.

Caffarelli’s theorem is the case K=Rn. The Friedland–Hayman inequality will follow from Proposition I.3 and Theorem I.5 with K being a convex cone and F ≡ 0 on K.

De Philippis and Figalli16 characterized the case of equality in the eigenvalue formulation of Caffarelli’s theorem. This result will permit us to characterize the case of equality in the Friedland–Hayman inequality, as stated in Theorem I.2. Their theorem is stated here in our generalized setting.

Theorem I.6.
Letφbe as in Theorem I.5, and let
0λ1(D2φ(x))λn(D2φ(x))1
be the eigenvalues ofD2φ(x). If for somem, 1 ≤ mn, λnm+1(D2φ(x)) = 1 for allxRn, then, after translation and rotation, there is a convex setKRnmand a convex functionG on K′ such that
K=Rm×K,ν=e|x|2/2|y|2/2G(y)1K(y)dxdy
withx=(x,y)Rm×Rnm.

Our second application is to the case of equality of a Poincaré–Wirtinger type inequality: For a given convex domain KRn, define μ1(K) by

μ1(K)=inffKf(x)2e|x|2/2dxK|f(x)|2e|x|2/2dx:Kf(x)e|x|2/2dx=0.
(1)

Here, the infimum is taken over function for which the expressions are finite. The value μ1(K) can also be viewed as the first non-zero eigenvalue of

dive|x|2/2u=μe|x|2/2uinK,un=0onK.

In Ref. 5, Brandolini, Chiacchio, Henrot, and Trombetti showed that if KRn has a C2-smooth boundary, then

μ1(K)μ1(Rn)=μ1(R)=1.
(2)

In Ref. 6, this inequality is shown to hold for any convex planar domain, and if KR2 is contained in a strip, then the equality in (2) holds precisely when K is itself a strip. Here, we use Theorems I.5 and I.6 to prove the theorem in the non-smooth case and describe fully the case of equality.

Theorem I.7.
For any convex domainKRnandμ1(K) as in (1), we have
μ1(K)μ1(R)=1.
Moreover, (up to a rotation) equality holds precisely whenKis of the formK=R×K for a convex domain KRn1.

Let us make a few remarks about the existing literature. Our Proof of Theorem I.5 will follow Caffarelli’s proof, exploiting the fact that φ satisfies a Monge–Ampère equation and that the second difference of φ(x) is well-behaved as |x| tends to infinity. Other proofs, variants, and extensions of this theorem have also been given in Refs. 18, 21, 22, and 26. The theorem of De Philippis and Figalli (Theorem 1.2 of Ref. 16 identifying the case of equality in Caffarelli’s original theorem) has an alternative proof due to Cheng and Zhou. That proof involves the first non-zero eigenvalue of a Laplacian with drift on a complete smooth metric space with a lower bound on the Bakry–Émery Ricci curvature (see Theorem 2 in Ref. 14).

The rest of this paper is structured as follows: In Sec. II, we prove the version of Caffarelli’s contraction theorem and the equality case (Theorems I.5 and I.6). We then show how our two applications follow from these theorems in Sec. III. We first prove the Poincaré–Wirtinger type inequality and the case of equality, which is a direct consequence of Theorems I.5 and I.6. Our version of the Friedland–Hayman inequality requires Proposition I.3. This converts our problem to the one about Gaussian eigenvalues of a convex domain to which Theorems I.5 and I.6 apply in Rn. We emphasize here that our argument is inspired by the argument of Beckner, Kenig, and Pipher. Moreover, because our method reduces the case of the convex cone to the case of the entire Euclidean space, it depends on the original Friedland–Hayman theorem and does not replace it. We end by discussing a possible, natural variant of Caffarelli’s contraction theorem for geodesically convex subsets of spheres. This variant would provide an alternative path to the main theorem (Theorem I.2).

Recall from Theorem I.4 that T = ∇φ transports the Gaussian measure μ=e|x|2/2dx onto the measure , where ν = eF(x)1Kμ for a convex domain KRn and a convex function F on K. Here, φ(x) is a convex function on Rn, and the constant c is chosen so that μ and have the same total measure. To prove Theorem I.5, we need to show that the eigenvalues of D2φ(x) are bounded above by 1.

Proof of Theorem I.5.
We follow the proof Caffarelli used to prove Theorem 11 in Ref. 10. In particular, rather than studying D2φ(x) directly, we work with second differences of φ(x). That is, we fix h > 0, and for each unit direction e, xRn, we set
δeφ(x)=φ(x+he)+φ(xhe)2φ(x).
To prove the theorem, we need to show that
0δeφ(x)h2
(3)
for all x, e, and h > 0, since then letting h → 0 gives the desired upper bound. To prove (3), it is sufficient to study δeφ(x) at a point where it achieves its maximum in both x and e, together with its behavior as |x| tends to infinity. However, in the case where the convex set K is not smooth, strictly convex, and bounded, then the behavior of φ(x) at infinity can be more complicated. Therefore, we form a sequence of smooth, strictly convex, and bounded sets KjK, which converges to K in Hausdorff distance on compact sets.24 (By strictly convex, we mean that each tangent plane to Kj touches ∂Kj at a unique point.) We also obtain corresponding Brenier maps Tj = ∇φj transporting μ=e|x|2/2dx onto cjνj, with νj=eF(x)1Kjμ. Here, the constant cj > 0 is chosen to ensure that cjνj and μ have the same total measure.
As F is a convex function on Rn, it is, in particular, in L(KBR) for all R > 0. Since the sets K and Kj are convex, this ensures that φ and φj are C2 (see Theorem 1 in Ref. 15). Therefore, φj is a classical solution to the Monge–Ampére equation
detD2φj(x)=e|x|2/2exp|φj(x)|2/2F(φj(x)),
(4)
and φ satisfies the analogous equation. Moreover, since Kj converges to K on compact sets, the measures eF(x)1Kjμ converge strongly to ν as j tends to infinity. Therefore, φj(x) converges to φ(x) uniformly on compact sets [see Theorem 3 in Ref. 10 and also Ref. 27, (5.23)]. To show (3) and complete the proof of the theorem, it is thus sufficient to establish
0δeφj(x)h2
(5)
for all x, e, and h > 0 and j fixed.

Note that the lower bound is guaranteed since φj is convex. To prove the upper bound, suppose first that δeφj(x) attains its maximum in x and e at x = x0 and e = e0. Then, as shown in the Proof of Theorem 11 in Ref. 10 (see also Ref. 11), using the fact that φj(x) satisfies the Monge–Ampère equation in (4), it satisfies a maximum principle ensuring that δe0φj(x0)h2.

To complete the proof, we therefore need to study the behavior of δeφj(x) as |x| tends to infinity. This part of the proof is the reason for using the smooth, strictly convex approximating sets Kj and also why we work with the second difference rather than the second derivative directly. Since the sets Kj will not be balls centered at the origin when K is a proper subset of Rn, this part of the proof requires a small modification of Caffarelli’s proof, and so we write it out in detail.

Lemma II.1.
Suppose that |x| tends to infinity, withx|x|converging to a directionn. Then,
φj(x)yn,
with uniform convergence in the directionn. Here, ynis the unique point onKjwith the outward unit normal pointing in the directionn.

Proof of Lemma II.1.
The uniqueness of the point yn follows immediately from the strict convexity and smoothness of Kj. Given xRn, let y = ∇φj(x) ∈ Kj. Define Γy(θ) to be the cone of vertex y pointing in the direction of x, with angle θ (here θ is fixed, with 0<θ<π2), so that
Γy(θ)={yRn:angle(x,yy)θ}.
By the cyclical monotonicity of the optimal transport mapping, if y′ = ∇φj(x′), then the inner product of x′ − x and y′ − y is non-negative. Therefore, the pre-image of Γy(θ) under ∇φj is contained in the cone
Γx(θ)={xRn:angle(x,xx)θ+π2}.
There exists a constant aθ > 0 such that the complement of Γx(θ) contains the ball of radius aθ|x| centered at the origin. In particular, as |x| tends to infinity, for θ fixed, the Gaussian measure of Γx(θ) tends to zero. Since the map ∇φj is measure preserving and the measure νj is bounded from below on Kj, this means that the Lebesgue measure of Γy(θ) ∩ Kj tends to 0. Therefore, given η > 0 and 0<θ<π2, there exists Mθ > 0 such that if |x| > Mθ, then the Lebesgue measure of Γy(θ) ∩ Kj is less than η. This, in particular, ensures dist(y, ∂Kj) < C1η, where C1 is a constant depending only on the Lipschitz bound for the convex set Kj. As θ tends to π2, the cone Γy(θ) approaches the half-plane passing through y in the direction x|x|, and this forces y to approach yn, the unique point on ∂Kj with outward unit normal n. More precisely, for any η > 0, if |nm| < η, then |ynym| < C2η, for a constant C2 depending only on the strict convexity of Kj (but not n). Therefore, given η > 0, we can choose δ > 0 and θ*<π2 such that if x|x|n<δ, with |x|>Mθ*, then |yyn| < ɛ. This proves that y converges to yn uniformly in the direction n.□
To conclude the proof of the theorem, we note that
0h2δeφj(x)φj(x+he)eφj(xhe)e2h,
and for h > 0 and e fixed, x+he|x+he|xhe|xhe| tends to 0 as |x| tends to infinity. Therefore, Lemma II.1 implies that δeφj(x) tends to 0 as |x| tends to infinity, and hence, δeφj(x) ≤ h2 for all x, e, and h > 0 as required.□

Proof of Theorem I.6.
Now that we have proved Theorem I.5, after ordering the eigenvalues of D2φ(x), we can ensure that
0λ1(D2φ(x))λn(D2φ(x))1.
To deal with the case of equality, where for some k ≥ 1, λnk+1(D2φ(x)) = 1 for all xRn, the proof of De Philippis and Figalli in Theorem 1.2 in Ref. 16 (used to deal with the case of stability in Caffarelli’s original contraction theorem) still applies and so we just briefly summarize their proof: Defining the convex function Ψ(x)=12|x|2φ(x), the assumption of the theorem implies that det(D2Ψ)(x) = 0. Subtracting a linear function from Ψ(x) (which only translates ν), we can assume that Ψ(x) ≥ Ψ(0) = 0. Combining det(D2Ψ)(x) = 0 with an Alexandrov estimate, de Philippis and Figalli showed that the set Σ = {Ψ = 0} does not have an exposed point. Therefore, the set Σ must contain a line (see Ref. 2, Lemma 3.5 in Chap. 2). Rotating so that this line is Re1 and combining this with the convexity of Ψ implies that e1Ψ(x)0. Therefore, we can view Ψ as a function of the variables x′ = (x2, …, xn) and write the mapping T = ∇φ as T(x)=x1,xΨ(x). This means that the measure ν can be written as ν=ex12/2dx1ν1, where T1(x′) = x′ − ∇Ψ(x′) transports the (n − 1)-dimensional Gaussian measure e|x|2/2 onto 1. Moreover, we can write ν1 as e|x|2/2G1(x)1K1dx for a convex set K1 and convex function G1, since these properties are preserved under taking marginals (see Theorem 4.3 in Ref. 7). This proves the theorem for k = 1, and by recursively applying this argument, the theorem holds.□

In this section, we use Theorems I.5 and I.6 to study the Friedland–Hayman and Poincaré–Wirtinger type inequalities discussed in the Introduction. The Poincaré–Wirtinger result follows as a direct application, so we will prove it first.

Proof of Theorem I.7.
Let u(x) be the eigenfunction corresponding to minimizing the quantity μ1(K) given in (1). Applying Theorem I.5 with F(x) ≡ 0 (so that ν=e|x|2/21Kdx), we obtain a transport map T = ∇φ from μ=e|x|2/2dx to such that T is Lipschitz, with the Lipschitz constant bounded by 1. We define v(x) by v(x) = u(T(x)), and since T is a transport map, we have
Rnv(x)e|x|2/2dx=Rnu(T(x))e|x|2/2dx=cKu(x)dν=0.
Therefore, v(x) is an admissible test function for μ1(Rn), and so
μ1(R)=μ1(Rn)Rnv(x)2e|x|2/2dxRn|v(x)|2e|x|2/2dx.
(6)
Moreover, since the Lipschitz constant of T is bounded by 1, we have
v(x)(u)(T(x))
(7)
for all xRn. Combining this with the fact that T is a transport map, we can use (6) to obtain
μ1(R)Ku(x)2e|x|2/2dxK|u(x)|2e|x|2/2dx=μ1(K).
To deal with the equality case μ1(K)=μ1(R)=1, we use Theorem I.6. If K contains a line (say Re1), then we can set u(x) = x1 to obtain μ1(K) = 1. Now, suppose that K does not contain a line and again let u(x) be the eigenfunction corresponding to μ1(K). Then, by Theorem I.6, there exist ɛ > 0 and a set URn of positive measures for which λj(D2φ(x)) ≤ 1 − ɛ for all xU, 1 ≤ jn. The image of U under T also has a positive measure. Setting v(x) = u(T(x)), we thus have
Uv(x)2e|x|2/2dx(1ε)Uu(T(x))2e|x|2/2dx.
Using this inequality in the above argument in place of (7), we obtain μ1(K) > 1, and so we cannot have equality unless K contains a line.□

In this section, we prove Theorem I.2. Recall that u+ is a harmonic function on the cone Γ+ of homogeneous degree α+ satisfying boundary conditions u+ = 0 on γD+ and (∂/∂ν)u+ = 0 (in the weak sense) on γN+. The first step is to prove Proposition I.3, a lower bound on the characteristic exponent α+ in terms of the lowest Gaussian eigenvalue on the cone with the same boundary conditions. The convexity of the cone Γ is not used in this proof.

Fix an integer N ≥ 0, and extend the function u+ to the product cone Γ+×RN to be constant in the extra variable,

uN(x,y)=u+(x),(x,y)Γ+×RN.

Denote the spherical cross section of the cone Γ+×RNRm by

Σm={(x,y)Γ+×RN:|x|2+|y|2=1},

where m = n + N. Let m, ∇m, and Δm denote the spherical measure, gradient, and Laplace–Beltrami operator on the unit (m − 1) sphere in Rm. Then, setting α = α+, uN is homogeneous of degree α in the product cone in Rm, and so by separation of variables,

ΔmuN=α(α+m2)uNonΣm.

Integrating by parts, using the boundary conditions, we get

Σm|muN|2dσm=α(α+m2)ΣmuN2dσm.
(8)

We rewrite this as

α(α+N+n2)=Σm|muN|2dσmΣmuN2dσm.

The integrand in the numerator, on |x|2 + |y|2 = 1, is given by

|muN|2=|x,yuN(x,y)|2|(x,y)x,yuN(x,y)|2=|u+(x))|2|xu+(x)|2=|u+(x)|2(αu+(x))2,

and the formula for the integral of a function f on Σm that depends only on x is

Σmf(x)dσm(x,y)=cNΓ+{|x|<1}f(x)(1|x|2)N/21dx.

Therefore,

α(α+N+n2)N=1NΓ+{|x|<1}[|u+(x))|2(αu+(x))2](1|x|2)N/21dxΓ+{|x|<1}u+(x)2(1|x|2)N/21dx.

Note that the left-hand side is α + O(1/N) as N → ∞.

Next, change variables by setting fN(z)=u+(z/N), |z|<N, to obtain

α=Γ+{|z|<N}|fN(z)|21N|αfN(z)|21|z|2NN/21dzΓ+{|z|<N}fN(z)21|z|2NN/21dz+O(1/N)=Γ+{|z|<N}|fN(z)|21|z|2NN/21dzΓ+{|z|<N}fN(z)21|z|2NN/21dz+O(1/N).

Because

1|z|2NN/21e|z|2/2asN,

this nearly completes the proof. The additional property we need to check is that as N → ∞,

Γ+{N1<|z|<N}[fN(z)2+|fN(z)|2]1|z|2NN/21dzΓ+{|z|<N}fN(z)21|z|2NN/21dz0.
(9)

An estimate like (9) is required because fN is not a suitable test function. Although fN does vanish on γD+ as required, it does not vanish on the outer boundary |z|=N. The simplest truncation is by a radial cut-off function of slope 1 on a band of unit width N1<|z|<N, which gives a legitimate test function and proves our proposition, assuming (9) holds.

Finally, (9) follows from the fact that fN and ∇fN grow like powers of |z|, whereas the weight resembles e|z|2/2. In detail, set

A=Σnu+2dσn.

Then, since u+ is homogeneous of degree α,

Γ+{|z|<N}fN(z)2w(|z|)dz=A0Nr2αNαw(r)rn1dr.

By Eq. (8) for m = n, N = 0, we have

Σn|u+|2dσn=α(α+n2)A.

Thus, since ∇u+ is homogeneous of degree α − 1,

Γ+{|z|<N}|fN(z)|2w(|z|)dz=α(α+n2)A0Nr2α2Nαw(r)rn1dr.

Note that the denominator Nα is the same in both formulas because of the choice of change of variable z/N.

With these formulas for the numerator and denominator, one sees that (9) is valid. Indeed, setting w(r)=1r2/NN/21,

N1N(r2α+r2α2)(1r2/N)N/21rn1dr

tends to zero very fast since (1r2/N)N/21er2/2eN/2 in the range N1<r<N, whereas

0Nr2α(1r2/N)N/21rn1dr

tends to a positive limit. This concludes the Proof of Proposition I.3.

We are now ready to prove the Friedland–Hayman inequality in the original case with Γ=Rn.

Definition III.1.
ForΩRnopen, with the Gaussian measure satisfyingΩe|x|2/2dx<1, defineλg(Ω) by
λg(Ω)=infwH01(Ω)Ωw(x)2e|x|2/2dxΩw(x)2e|x|2/2dx.
Alternatively, λg(Ω) is the first eigenvalue of
(/xi)e|x|2/2u/xi=λe|x|2/2u(x)in  Ω,u(x)=0on  Ω.
Through Gaussian symmetrization, the following theorem holds:

Theorem III.2
(Ref.17, Proposition 2.3 and Ref.13, Theorem 3). Let Ω andλgbe as in Definition III.1. LetHbe a half-space with the same Gaussian measure as Ω, then
λg(Ω)λg(H),
with equality if and only if Ω is equal toH, up to arotation.

For half spaces, the corresponding eigenfunction is a function of a single variable. Therefore, we can apply the results of Beckner, Kenig, and Pipher.

Theorem III.3
(Ref.4and Sec. 12.2 of Ref.12). LetH±be complementary half-spaces. Then,
λg(H+)+λg(H)2,
with equality if and only ifH+e|x|2/2dx=He|x|2/2dx.

In particular, combining these two theorems with Proposition I.3 establishes the original Friedland–Hayman inequality and classifies the case of equality. For the general case where Γ is a proper, convex subset of Rn, we will use Theorems I.5 and I.6 to compare the Gaussian eigenvalues on Γ± to those on complementary subsets of Rn. In particular, the proof is now similar to that of Theorem I.7 with some small modifications coming from the fact that we have to consider eigenvalues on complementary subsets.

We first apply Theorem I.5 with K = Γ and F(x) ≡ 0. This gives a transport map T = ∇φ from μ=e|x|2/2dx to cn,Kν, with ν=e|x|2/21Γdx (and cn,K being a constant ensuring cn,Kν and μ have the same total measure), such that T is Lipschitz with the Lipschitz constant bounded above by 1. Let w±, defined on Γ±, respectively, be admissible test functions for the infimum given in Proposition I.3. Extending w± by zero so that they are defined on Γ, we set v+(x) = w+(T(x)), v(x) = w(T(x)). Since T is a transport map, we have

Rnv±(x)2e|x|2/2dx=Rnw±(T(x))2e|x|2/2dx=cΓwi(x)2dν.
(10)

T is Lipschitz, with the Lipschitz constant bounded by 1, and so

v±(x)w±(T(x))T(x)w±(T(x)).
(11)

Combining (10) and (11) gives

Γ+w+(x)2e|x|2/2dxΓ+w+(x)2e|x|2/2dxRnv+(x)2e|x|2/2dxRnv+(x)2e|x|2/2dx
(12)

and the same for Γ, w, and v. Since w1 and w2 have disjoint supports in the cone Γ, so do the functions v1 and v2 in Rn. Applying Theorems III.2 and III.3 thus implies that

Rnv+(x)2e|x|2/2dxRnv+(x)2e|x|2/2dx+Rnv(x)2e|x|2/2dxRnv(x)2e|x|2/2dx2,

and by Proposition I.3, this proves the desired inequality in Theorem I.2.

We now turn to the case of equality. Suppose that the cone Γ does not contain a line. Then, by Theorem I.6, there exist ɛ > 0 and a set U1Rn of positive measures for which λj(D2φ(x)) ≤ 1 − ɛ for all xU1, 1 ≤ jn. The image of U1 under T is also of positive measure, and so without loss of generality, assume that T(U1) ∩ Γ+ has positive measure. Letting w+ be the minimizer for the infimum in Proposition I.3 for Γ+, we therefore have

v+(x)(1ε)w+(T(x))onT1T(U1)Γ+.

Inserting this strict inequality into the argument above ensures that we cannot have the equality α+ + α = 2.

Now suppose that we have equality. Then, we can write the cone Γ in the form Γ=Rk×Γ, where Γ′ does not contain a line. In particular, by Theorem I.6, we have T(x) = (x1p1, …, xkpk, T′(xk+1, …, xn+1)) for some fixed (p1,,pk)Rk and with |∇T′| < 1 on a set of positive measures U2 in Rnk. Assuming that T(U2)ΠnkΓ+ has positive measure in Rnk [where ΠnkΓ+={xRnk:(x1,,xk,x)Γ+ for some (x1,,xk)Rk}], in order to have equality, the minimizer w+ for Γ+ in Proposition I.3 must depend only on the variables x1, …, xk. Therefore, v+(x) = w+(xp) for some pRn, and by Theorem III.2, w+ is an explicit ordinary differential equation (ODE) solution depending only on one variable and Γ+ equals the intersection of Γ with a half-space. That is, we can write Γ+ = H × Γ′, Γ=Rk\H×Γ for a half-space HRk. This, in particular, ensures that T(U2)ΠnkΓ also has positive measure in Rnk, and so for equality, w depends only on the variables x1, …, xk. We have therefore reduced to the setting of Theorem III.3 in Rk, and so for equality, after a rotation, we have Γ+ = {x1 > 0} ∩ Γ and Γ = {x1 < 0} ∩ Γ.

To prove the Friedland–Hayman type inequality (Theorem I.2), we used Proposition I.3 to obtain a lower bound on the characteristic exponents α± in terms of Gaussian eigenvalues. This then allowed us to apply the Caffarelli contraction theorem (Theorem I.5). A possible alternative to this is to work directly on the sphere Sn1 and prove an analogous version of the Caffarelli contraction theorem on the sphere. In Ref. 23, McCann showed the following: Let (M, g) be a smooth, compact manifold without boundary, with the distance function d(x, y), and let μ1, μ2 be probability measures on M. There exists a unique mapping T transporting μ1 onto μ2, minimizing the total cost with respect to the cost function c(x, y) = d(x,y)2/2. The mapping T(x)=expxψ(x) for a c-concave function ψ, where one defines a function ψ to be c-concave if ψ=ψcc for

ψc(y)=infxMc(x,y)ψ(x).

Now let μ be the uniformly distributed probability measure on a hemisphere S+n1 in Sn1, the appropriate multiple of the volume form of the round metric. Define the measure ν by

ν=efμW.

Here, W is a (geodesically) convex subset of S+n1 and f is a convex function on W such that ν is a probability measure. It is then natural to pose the following questions:

  1. Does there exist a transport map T(x)=expxψ(x) from μ to ν for a c-concave function ψ such that T is Lipschitz on S+n1 with the Lipschitz constant bounded by 1?

  2. If, at almost every point of S+n1, there is a direction in which T is not a strict contraction, then is it true that, after a rotation, W contains the antipodal points (±1, 0, …, 0) in S+n1 and that f is independent of the x1-variable?

  3. If there is a pair of points x and y for which d(x, y) = d(T(x), T(y)), then does W contain the full geodesic through x and y, all the way to the antipodal points? And does T split in that direction as in question (2)? Is there an infinitesimal version of this phenomenon at a single point?

The tools of Fathi et al.18 seem appropriate to study these questions. As well as giving a version of the Caffarelli contraction theorem on the sphere, answering these questions in the affirmative would also give a different Proof of Theorem I.2 while working directly on the spherical cross sections of the cones Γ±. Finally, we suggest that, more broadly, sharp global inequalities like ours should have consequences for the behavior of level sets of solutions to semilinear elliptic equations in the context of manifolds with lower bounds on Ricci curvature and with log-concave weights. We refer the reader to Ref. 20 in which Jerison discusses, in particular, isoperimetric surfaces and Neumann eigenfunctions.

This paper is offered in memory of Jean Bourgain. T.B. was supported, in part, by NSF Grant No. 2042654. D.J. was supported, in part, by NSF Grant No. 1500771, a Simons Fellowship, a Guggenhein Fellowship, and a Simons Foundation Grant (Grant No. 601948).

The authors have no conflicts to disclose.

Data sharing is not applicable to this article as no new data were created or analyzed in this study.

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