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.
I. INTRODUCTION
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:
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,
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,
Define the Neumann and Dirichlet portions of the boundary, and , by
The original Friedland–Hayman inequality is the case with . 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 . 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 as N → ∞. Using this device, we will prove the following key proposition:
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
Because Γ is convex, F is a generalized convex function, that is, a convex function allowing for the value +∞. Put another way, e−F 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:
(Ref. 8 ). Let μ, ν be positive measures on and with finite second order moments. Suppose also that μ is absolutely continuous with respect to the Lebesgue measure. Then, there exists a convex function such that transports μ onto cν. That is, μ(T−1(E)) = cν(E), with the normalizing constant .
The variant of Caffarelli’s contraction theorem that we require can be stated as follows:
Caffarelli’s theorem is the case . 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.
Our second application is to the case of equality of a Poincaré–Wirtinger type inequality: For a given convex domain , define μ1(K) by
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
In Ref. 5, Brandolini, Chiacchio, Henrot, and Trombetti showed that if has a C2-smooth boundary, then
In Ref. 6, this inequality is shown to hold for any convex planar domain, and if 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.
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 . 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).
II. PROOF OF THE CAFFARELLI CONTRACTION THEOREM
Recall from Theorem I.4 that T = ∇φ transports the Gaussian measure onto the measure cν, where ν = e−F(x)1Kμ for a convex domain and a convex function F on K. Here, φ(x) is a convex function on , and the constant c is chosen so that μ and cν 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.
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 .
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 , this part of the proof requires a small modification of Caffarelli’s proof, and so we write it out in detail.
III. CONSEQUENCES OF THE CONTRACTION THEOREM
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.
A. Poincaré–Wirtinger inequality
B. The Friedland–Hayman inequality
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 and (∂/∂ν)u+ = 0 (in the weak sense) on . 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 to be constant in the extra variable,
Denote the spherical cross section of the cone by
where m = n + N. Let dσm, ∇m, and Δm denote the spherical measure, gradient, and Laplace–Beltrami operator on the unit (m − 1) sphere in . Then, setting α = α+, uN is homogeneous of degree α in the product cone in , and so by separation of variables,
Integrating by parts, using the boundary conditions, we get
We rewrite this as
The integrand in the numerator, on |x|2 + |y|2 = 1, is given by
and the formula for the integral of a function f on Σm that depends only on x is
Therefore,
Note that the left-hand side is α + O(1/N) as N → ∞.
Next, change variables by setting , , to obtain
Because
this nearly completes the proof. The additional property we need to check is that as N → ∞,
An estimate like (9) is required because fN is not a suitable test function. Although fN does vanish on as required, it does not vanish on the outer boundary . The simplest truncation is by a radial cut-off function of slope 1 on a band of unit width , 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 . In detail, set
Then, since u+ is homogeneous of degree α,
By Eq. (8) for m = n, N = 0, we have
Thus, since ∇u+ is homogeneous of degree α − 1,
Note that the denominator Nα is the same in both formulas because of the choice of change of variable .
With these formulas for the numerator and denominator, one sees that (9) is valid. Indeed, setting ,
tends to zero very fast since in the range , whereas
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 .
For half spaces, the corresponding eigenfunction is a function of a single variable. Therefore, we can apply the results of Beckner, Kenig, and Pipher.
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 , we will use Theorems I.5 and I.6 to compare the Gaussian eigenvalues on Γ± to those on complementary subsets of . 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 to cn,Kν, with (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
T is Lipschitz, with the Lipschitz constant bounded by 1, and so
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 . Applying Theorems III.2 and III.3 thus implies that
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 of positive measures for which λj(D2φ(x)) ≤ 1 − ɛ for all x ∈ U1, 1 ≤ j ≤ n. 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
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 , where Γ′ does not contain a line. In particular, by Theorem I.6, we have T(x) = (x1 − p1, …, xk − pk, T′(xk+1, …, xn+1)) for some fixed and with |∇T′| < 1 on a set of positive measures U2 in . Assuming that has positive measure in [where for some ], 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+(x − p) for some p ∈ Rn, 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′, for a half-space . This, in particular, ensures that also has positive measure in , and so for equality, w− depends only on the variables x1, …, xk. We have therefore reduced to the setting of Theorem III.3 in , and so for equality, after a rotation, we have Γ+ = {x1 > 0} ∩ Γ and Γ− = {x1 < 0} ∩ Γ.
IV. REMARKS ON TRANSPORT MAPS ON THE SPHERE
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 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 for a c-concave function ψ, where one defines a function ψ to be c-concave if for
Now let μ be the uniformly distributed probability measure on a hemisphere in , the appropriate multiple of the volume form of the round metric. Define the measure ν by
Here, W is a (geodesically) convex subset of and f is a convex function on W such that ν is a probability measure. It is then natural to pose the following questions:
Does there exist a transport map from μ to ν for a c-concave function ψ such that T is Lipschitz on with the Lipschitz constant bounded by 1?
If, at almost every point of , 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 and that f is independent of the x1-variable?
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.
ACKNOWLEDGMENTS
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).
AUTHOR DECLARATIONS
Conflict of Interest
The authors have no conflicts to disclose.
DATA AVAILABILITY
Data sharing is not applicable to this article as no new data were created or analyzed in this study.