Murat Akman. The Brunn-Minkowski Inequality and a Minkowski Problem for Nonlinear Capacities. March 10

The Brunn-Minkowski Inequality and a Minkowski Problem for Nonlinear Capacities Murat Akman March 10 Postdoc in HA Group and Postdoc at the University of Connecticut

Minkowski Addition of Sets Let E 1 and E 2 be convex bodies (compact convex sets with non-empty interiors). Minkowski addition of E 1 + E 2 is defined as E 1 + E 2 := {a + b a E 1, b E 2 } = b E 2 E 1 + {b}. y sa := {sa a A}. E 2 E 1 + E 2 E 1 x

Minkowski sum of a square and a disk A + ɛb = A + ɛb radius = ɛ length = l Therefore, A + ɛb = A + 4lɛ + ɛb A + 2 πlɛ + ɛb = A + 2 A ɛb + ɛb.

Minkowski sum of a square and a disk A + ɛb = A + ɛb radius = ɛ length = l Therefore, A + ɛb = A + 4lɛ + ɛb A + 2 πlɛ + ɛb = A + 2 A ɛb + ɛb. Hence A + ɛb 1/2 A 1/2 + ɛb 1/2. Of course this is not a coincidence!

The Brunn-Minkowski inequality Theorem (Minkowski in 1887 and Brunn in 1896) Let A and B two convex sets in R n and let λ [0, 1]. Then (1 λ)a + λb 1 n (1 λ) A 1 n + λ B 1 n. Moreover, equality holds iff A is a homothetic copy of B. The function 1/n is a concave function in the class of convex sets in R n under the Minkowski addition. It also holds for bounded measurable sets in R n. [Due to Lusternik in 1935, Hadwiger and Ohmann in 1956.] The Brunn-Minkowski Inequality implies the isoperimetric inequalities. Connections with: Sobolev inequality, Poincaré inequality, Young inequality, Prékopa-Leindler inequality, etc. *: i.e. are equal up to translation and dilation.

Equivalent forms of the Brunn-Minkowski inequality Let A, B be convex bodies in R n and let λ [0, 1]. Then TFAE. Classic (1 λ)a + λb 1 n (1 λ) A 1 n + λ B 1 n. Elegant A + B 1 n A 1 n + B 1 n. Multiplicative (1 λ)a + λb A 1 λ B λ. Minimal (1 λ)a + λb min{ A, B }.

Is it special to Lebesgue measure? Let K = {Convex bodies in R n }. 0 is a homogeneous of degree n; tk = t n K whenever t 0 and K K.

Is it special to Lebesgue measure? Let K = {Convex bodies in R n }. 0 is a homogeneous of degree n; tk = t n K whenever t 0 and K K. Let F : K R be a functional such that F(K) 0 whenever K K. F is α-homogeneous (α 0); F(tK) = t α F(K), whenever t 0 and K K. Is there any other homogeneous of degree α functional F which verifies the Brunn-Minkowski inequality for convex bodies? [F((1 λ)a + λb)] 1 α (1 λ) [F(A)] 1 α + λ [F(B)] 1 α. What happens in the case of equality?

Inequalities of Brunn-Minkowski type Torsional rigidity. α = n + 2. (Borell in 1985, Colesanti in 2005). Eigenvalue of the Monge-Ampère equation. α = 2n. (Salani in 2005). Homogeneous Minkowski-concave function of degree m. α = m + n (Knothe in 1957). Affine quermassintegral of order i. α = n i. (Lutwak in 1984). p-minkowski addition with order i. α = p/(n i). (Firey in 1962). Nilpotent Brunn-Minkowski in simply connected nilpotent Lie group of dimension n. α = n. (Leonardi and Mansou in 2005). Newtonian and Logarithmic capacity. α = n 2 and α = 1 (Borell in 1983, Caffarelli, Jerison and Lieb in 1996, Colesanti and Cuoghi in 2005).... (See Gardner s survey article).

Brunn-Minkowski inequality for p-capacity p-capacity, 1 < p < n, of a set convex body K is defined as { } Cap p (K) = inf v p dx, v Cc (R n ) : v 1 on K. R n

Brunn-Minkowski inequality for p-capacity p-capacity, 1 < p < n, of a set convex body K is defined as { } Cap p (K) = inf v p dx, v Cc (R n ) : v 1 on K. R n If u is the minimizer then p u = ( u p 2 u) = 0 in R n \ K, u = 1 on K, lim x u(x) = 0. Cap p (K) = R n \K p u is a nonlinear generalization of the Laplace equation u = 0. Cap( ) is homogeneous of degree n p. Cap( ) satisfies the Brunn-Minkowski inequality for convex bodies; [ Capp ((1 λ)a + λb) ] 1 n p (1 λ) [Cap(A) p ] 1 n p + λ [Cap(B) p ] 1 n p. Equality holds iff A is a homothetic copy of B. u(x) p dx.

A harmonic PDEs More general nonlinear elliptic PDEs; Let p be fixed, 1 < p < n. Let A = (A 1,..., A n ) : R n \ {0} R n. Suppose also that A = A(η) has continuous partial derivatives in η k, 1 k n, on R n \ {0}. We say that the function A belongs to the class M p (α) if the following conditions are satisfied whenever ξ R n and η R n \ {0}: (i) α 1 η p 2 ξ 2 n i,j=1 (ii) A(η) = η p 1 A(η/ η ). A i η j (η)ξ i ξ j α η p 2 ξ 2, (1)

Brunn-Minkowski inequality for nonlinear Capacities Given convex compact set K with H n p (K) =. Then there is u with A( u) = 0 in R n \ K, u = 1 on K, Define Cap A (K) := A( u), u dy. lim x u(x) = 0. Theorem (A., J. Gong, J. Hineman, J. Lewis, A. Vogel) Let E 1, E 2 R n be compact convex sets with H n p (E i ) = then [Cap A ((1 λ)e 1 + λe 2 )] 1 n p (1 λ) [Cap A (E 1 )] 1 n p +λ [Cap A (E 2 )] 1 n p. If equality holds and (i) There exists 1 Λ < such that Ai η j (η) Ai (η ) Λ η η η p 3 whenever 0 < η 2 η and 1 i n, (ii) A i (η) = f η i, 1 i n, where f (tη) = t p f (η), when t > 0, η R n \ {0}, then E 2 is a homothetic copy of E 1. η j

Why does it work for general elliptic PDEs?

Why does it work for general elliptic PDEs? Step 1: If u is capacitary function for convex body E then u(x) lim x G(x) = Cap A (E) 1 p 1 where G(x) is the fundamental solution to the corresponding PDE. Let E 1, E 2 be convex bodies and let u 1, u 2 be capacitary functions. Let u be the capacitary function for E 1 + E 2. Fix λ (0, 1), define { } u x = λy + (1 λ)z, (x) = sup min{u 1 (y), u 2 (z)} ; λ [0, 1], y, z R n. Step 2: prove that u (x) u(x) in R n. Use Step 2; u(x) u (x) min(u 1 (x), u 2 (x)) u(x) ( G(x) min u1 (x) G(x), u ) 2(x). G(x)

Minkowski Problem for Polyhedron Fact: A convex polygon in R 2 is uniquely determined (up to translation) by the unit normals n 1,..., n m of the faces and lengths l 1,..., l m of its edges. n 2 n 1 n 3 l 3 l 2 l 1 How about in R 3 or in R n? Problem (Minkowski Problem for Polyhedron) Let unit normal vectors n 1,..., n m and positive numbers A 1,..., A m be given.

Necessary conditions for Minkowski problem for Polyhedron Condition for Normals: The set of unit normals n 1,..., n m can not live in any single closed hemisphere;

Necessary conditions for Minkowski problem for Polyhedron Condition for Normals: The set of unit normals n 1,..., n m can not live in any single closed hemisphere; n 4 n 2 n 1 nm hemisphere n 3 The unit sphere S 2

Necessary conditions for Minkowski problem for Polyhedron Condition for Normals: The set of unit normals n 1,..., n m can not live in any single closed hemisphere; w n 4 n 2 Otherwise, there exists a unit vector w s.t. w n i = 0 for every i = 1,..., m. n 1 nm hemisphere n 3 Then Polyhedron will not be closed in the w direction. The unit sphere S 2

Necessary conditions for Minkowski problem for Polyhedron Both sides equal condition: A 1 n 1 +... + A m n m = 0. There are n equations here so this imposes n conditions on A i and n i.

Necessary conditions for Minkowski problem for Polyhedron Both sides equal condition: A 1 n 1 +... + A m n m = 0. There are n equations here so this imposes n conditions on A i and n i. If w S n 1 then the area of the projection of ith face to w is A i (n i w). It is positive for faces on the w side of the Polyhedron and it is negative for those of the other side. Therefore, for every w S n 1, m A i (n i w) = A i (n i w) + A i (n i w) = 0. i=1 n i w>0 n i w<0

Minkowski Problem for Polyhedron Theorem (Solution to Minkowski Problem for Polyhedron) Suppose that n 1,..., n m S n 1 spans R n and positive numbers A 1,..., A m are given. Then there exists a Polyhedron P whose faces have unit normals n 1,..., n m and surface areas A 1,..., A m if and only if A 1 n 1 +... + A m n m = 0. Moreover, this Polyhedron is unique up to translation. Due to Minkowski (1903).

Surface area measure on S n 1 Let g K = g : K S n 1 be the Gauss map; x n x. K is convex. K g K

Surface area measure on S n 1 Let g K = g : K S n 1 be the Gauss map; x n x. K is convex. K g K Consider the measure dµ K = g (dh n 1 ) defined on the unit sphere by µ K (E) = dh n 1 whenever E S n 1 is Borel. g 1 (E) This is well-defined almost everywhere with respect to H n 1. When K = P is convex Polyhedron with unit normals n 1,..., n m S n 1 and surface areas A 1,..., A m then m dµ P = A i δ ni where δ ni is Dirac point mass measure at n i. i=1

Surface area measure on S n 1 If K C 2 and has positive Gauss curvature everywhere then dσ K (X) = 1 κ(x) dhn 1 (X) on S n 1 where κ(x) is the Gauss curvature at the point of K where X is the outer unit normal to K. Gauss curvature κ is the Jacobian determinant of the Gauss map g.

Minkowski Problem General Case Problem (Minkowski Problem) Given a positive Borel measure µ on S n 1, does there exists a convex body K in R n with surface area measure µ K such that µ K = µ?

Minkowski Problem General Case Problem (Minkowski Problem) Given a positive Borel measure µ on S n 1, does there exists a convex body K in R n with surface area measure µ K such that µ K = µ? Necessary conditions: C1: the centroid of the measure µ is at the origin; S n 1 w dµ(w) = 0. This is due to µ is translation invariant. C2: µ is not supported on any equator of S n 1 ; S n 1 θ w dµ(w) > 0 for every θ S n 1. This is due to the convex body has non-empty interior.

Minkowski Problem General Case Problem (Minkowski Problem) Given a positive Borel measure µ on S n 1, does there exists a convex body K in R n with surface area measure µ K such that µ K = µ? Necessary conditions: (sufficient conditions as well) C1: the centroid of the measure µ is at the origin; S n 1 w dµ(w) = 0. This is due to µ is translation invariant. C2: µ is not supported on any equator of S n 1 ; S n 1 θ w dµ(w) > 0 for every θ S n 1. This is due to the convex body has non-empty interior.

Minkowski Problem General Case Problem (Minkowski Problem) Given a positive Borel measure µ on S n 1, does there exists a convex body K in R n with surface area measure µ K such that µ K = µ? Necessary conditions: (sufficient conditions as well) C1: the centroid of the measure µ is at the origin; S n 1 w dµ(w) = 0. This is due to µ is translation invariant. C2: µ is not supported on any equator of S n 1 ; S n 1 θ w dµ(w) > 0 for every θ S n 1. This is due to the convex body has non-empty interior. K is unique up to translation. These results are due to Minkowski, Alexandrov, Fenchel-Jessen.

Minkowski Problem - General case Theorem Let µ be a non-negative Borel measure on S n 1 satisfying w dµ(w) = 0 S n 1 and θ w dµ(w) > 0 S n 1 for every θ S n 1. Existence: There is a convex body K with non-empty interior such that µ K = µ. Uniqueness: K is unique up to translation. Regularity: If dµ K = 1 κ dhn 1 for some strictly positive function κ C k,α (S n 1 ) for some k N and α (0, 1) then K is C k+2,α. Existence is solved by Minkowski (1903) for the case of polyhedron, in general Alexandrov (1937-1938), Fenchel and Jessen (1938). C regularity is proved by Lewy (1938), Pogorelov (1953), Nirenberg (1953), Cheng and Yau (1976). The precise gain of two derivatives and the treatment of small values of k due to Caffarelli (1990).

The support function of a convex body The support function h K of a convex domain K in R n is defined as h K : S n 1 R, h K (X) = sup{ X, x ; x K}

The support function of a convex body The support function h K of a convex domain K in R n is defined as h K : S n 1 R, h K (X) = sup{ X, x ; x K} i.e.: h K (X) is the signed distance of supporting hyperplane at a point on K from the origin whose outer unit normal is X. X h K (X) -h K is homogeneous of degree 1. -Any non-empty closed convex set K is uniquely determined by h K. K 0 For every convex bodies K, L and constants α, β 0; h αk+βl = αh K + βh L where A + B := {a + b a A, b B}.

The Hadamard Variational Formula If K is convex body with support function h K and L is any other convex body with support function h L. Then K + ɛl K M(K, L) := lim = h L (X) dµ K (X) = h L (g(x)) dh n 1. ɛ 0 + ɛ S n 1 K A A + ɛb = A + 4lɛ + ɛb. + ɛb radius = ɛ = A + ɛb lim ɛ 0 A + ɛb A ɛ = 4lɛ + ɛ2 B ɛ = 4l length = l

The Hadamard Variational Formula If K is convex body with support function h K and L is any other convex body with support function h L. Then K + ɛl K M(K, L) := lim = h L (X) dµ K (X) = h L (g(x)) dh n 1. ɛ 0 + ɛ S n 1 K A A + ɛb = A + 4lɛ + ɛb. + ɛb radius = ɛ = A + ɛb A + ɛb A lim = 4lɛ + ɛ2 B ɛ 0 ɛ ɛ = 4l length = l The first variation M(K, L) of the Lebesgue measure is the surface area measure µ K.

A representation formula for volume K + ɛl K M(K, L) = lim = ɛ 0 + ɛ S n 1 h L (X) dµ K (X). If we take L = K in M(K, L) then K + ɛk K (1 + ɛ)k K h K (X) dµ K (X) = lim = lim ɛ 0 + ɛ ɛ 0 + ɛ S n 1 (1 + ɛ) n 1 = K lim = n K. ɛ 0 + ɛ

A representation formula for volume K + ɛl K M(K, L) = lim = ɛ 0 + ɛ S n 1 h L (X) dµ K (X). If we take L = K in M(K, L) then K + ɛk K (1 + ɛ)k K h K (X) dµ K (X) = lim = lim ɛ 0 + ɛ ɛ 0 + ɛ S n 1 Hence (1 + ɛ) n 1 = K lim = n K. ɛ 0 + ɛ K = 1 h K (X) dµ K (X) = n S n 1 K h K (g(x)) dh n 1 = 1 M(K, K) n

Proof of Minkowski Problem: Existence Existence: Given positive finite Borel measure µ on S n 1, find convex set K with non-empty interior such that µ K = µ.

Proof of Minkowski Problem: Existence Existence: Given positive finite Borel measure µ on S n 1, find convex set K with non-empty interior such that µ K = µ. Define a functional from set of convex bodies K; F : K R by F(L) = h L (X) dµ(x) for L K. S n 1 Consider the following minimization problem inf{f(l) subject to the constraint L 1}. Say K is minimizer then the Lagrange multiplier implies λ R df( K) = λd K. Use the fact that the first variation of the Volume is µ K ; df( K) = dµ and d K = dµ K = dµ = λdµ K.

Proof of Minkowski Problem: Uniqueness Uniqueness: Let K, L be two convex domains with surface measures µ K, µ L which are solution to Minkowski problem, i.e., µ K = µ L = µ given µ.

Proof of Minkowski Problem: Uniqueness Uniqueness: Let K, L be two convex domains with surface measures µ K, µ L which are solution to Minkowski problem, i.e., µ K = µ L = µ given µ. Define The Brunn-Minkowski inequality F(t) = tk + (1 t)l 1/n, t [0, 1]. tk + (1 t)l 1/n t K 1/n + (1 t) L 1/n for t [0, 1] implies F(t) is concave in [0, 1]. The variational formula for volume and µ K = µ L gives F F(t) F(0) (0) = lim = 1 n 1 d L n t 0 + t n dt ( tk + (1 t)l ) t=0 = 1 n 1 L n (h K (X) h L (X)) dµ L (X) n S n 1 = 1 n 1 L n [ K L ] = [F(0)] 1 n [F(1) n F(0) n ]. n

Proof of Minkowski Problem: Uniqueness Now F (0) = [F(0)] 1 n [F(1) n F(0) n ] and concavity of F F (0) F(1) F(0) F(1) n 1 F(0) n 1. That is, F(t) = tk + (1 t)l 1/n, K 1/n L 1/n,

Proof of Minkowski Problem: Uniqueness Now F (0) = [F(0)] 1 n [F(1) n F(0) n ] and concavity of F F (0) F(1) F(0) F(1) n 1 F(0) n 1. That is, F(t) = tk + (1 t)l 1/n, K 1/n L 1/n, reverse roles of K and L; K 1/n = L 1/n. This gives L 1/n = F(0) = F(1) = K 1/n and F (0) = 0. Hence F(t) = constant for t [0, 1]; F(1/2) = 1 2 K + L ]1/n = 1 2 [F(0) + F(1)] = 1 2 [ L 1/n + K 1/n ] gives us equality in the Brunn-Minkowski inequality.

Proof of Minkowski Problem: Uniqueness Now F (0) = [F(0)] 1 n [F(1) n F(0) n ] and concavity of F F (0) F(1) F(0) F(1) n 1 F(0) n 1. That is, F(t) = tk + (1 t)l 1/n, This gives K 1/n L 1/n, reverse roles of K and L; K 1/n = L 1/n. L 1/n = F(0) = F(1) = K 1/n and F (0) = 0. Hence F(t) = constant for t [0, 1]; F(1/2) = 1 2 K + L ]1/n = 1 2 [F(0) + F(1)] = 1 2 [ L 1/n + K 1/n ] gives us equality in the Brunn-Minkowski inequality. Therefore K and L are homothetic. µ K = µ L implies K and L have same perimeter K is at most translation of L.

Minkowski type problems for other measures L p Minkowski problem (L 0 is the usual Minkowski problem) due to Andrews in 1999, Chou and Wang in 2006, Hug, Lutwak, Yang, and Zhang in 2005, Ludwig in 2011, Lutwak and V. Oliker in 1995. First eigenvalue of the Laplace operator with Dirichlet boundary conditions; existence due to Jerison in 1996, uniqueness due to Brascamp and Lieb in 1976 and Colesanti in 2005. The torsional rigidity Fimiani and Colesanti in 2008.

Capacity of a convex body If K R n is convex body, n 3, (Newtonian) capacity of K is { } Cap(K) = inf v 2 dx, v Cc (R n ) : v 1 on K. R n

Capacity of a convex body If K R n is convex body, n 3, (Newtonian) capacity of K is { } Cap(K) = inf v 2 dx, v Cc (R n ) : v 1 on K. R n If u is the minimizer then u is called the capacitary function for K; u = 0 in R n \ K, u = 1 on K Hence Cap(K) = u 2 dx = γ. lim x u(x) = 0. R n \K γ is known as the electrostatic capacity of K. Indeed, u has the asymptotic expansion u(x) = γa n x 2 n + O( x 1 n ) as x. Moreover, G(x) = x 2 n is the fundamental solution to u = 0; G(x) = δ 0.

Minkowski problem for electrostatic capacity Classical Minkowski problem; dµ K = g (dh n 1 ) defined on the unit sphere by µ K (E) = dh n 1 whenever E S n 1 is Borel. g 1 (E)

Minkowski problem for electrostatic capacity Classical Minkowski problem; dµ K = g (dh n 1 ) defined on the unit sphere by µ K (E) = dh n 1 whenever E S n 1 is Borel. g 1 (E) What if we consider dµ 2 K = g ( u 2 dh n 1 ) on S n 1 defined by µ 2 K(E) = u 2 dh n 1 whenever E S n 1 is Borel. g 1 (E) dµ 2 K is well defined surface area measure on S n 1 as (due to Dahlberg) u is defined almost everywhere on K and H n 1 u 2 H n 1 H n 1 on K.

The Minkowski problem for electrostatic capacity Theorem (Jerison in 1996) Let µ be a non-negative Borel measure on S n 1. Then there exists a convex body K such that µ = µ 2 K if and only if w dµ(w) = 0 S n 1 and θ w dµ(w) > 0 S n 1 for every θ S n 1. K is unique up to translation when n 4 (and is unique up to translation and dilation when n = 3). Jerison observed the resemblance between Cap(K) = 1 h K u K 2 dh n 1 and K = 1 n 2 n K K h K dh n 1.

Sketch of Uniqueness As in the classical Minkowski problem, existence relies on The Hadamard variational formula for capacity. The Brunn-Minkowski inequality for capacity. Use these two to show that m(t) is constant for t [0, 1] where m(t) = (Cap((1 t)e 1 + te 2 )) 1 n 2.

Sketch of Existence Let n 1,..., n m be unit normals which spans R n and let c 1,..., c m be m positive numbers such that c i n i = 0 µ = m i=1 c iδ ni. i=1

Sketch of Existence n 2 n 1 Given q = (q 1,..., q m ) R m with q i 0. n 3 q 2 q 1 q 3 q 6 q 5 q 4 n 6 m P(q) := {x R n ; x, n i q i }. i=1 n 4 n 5 Let n 1,..., n m be unit normals which spans R n and let c 1,..., c m be m positive numbers such that c i n i = 0 µ = m i=1 c iδ ni. i=1 Consider the following minimization problem { m } inf c i q i ; P(q) Θ where Θ = {P(q); Cap(P(q)) 1}. i=1 Prove that there exists q R m which is a minimizer.

The Minkowski problem for p-capacitary surface measures Let u be the p-capacitary function (1 < p < n) for convex body K p u = 0 in R n \ K, u = 1 on K, Hence Cap p (K) = u p dx. lim x u(x) = 0. R n \K What if we consider dµ p K = g ( u p dh n 1 ) on S n 1 defined by µ p K (E) = u p dh n 1 whenever E S n 1 is Borel. g 1 (E) dµ p K is well defined surface area measure on Sn 1 as (due to Nyström and Lewis) H n 1 u p H n 1 H n 1 on K.

The Minkowski problem for p-capacitary measure Theorem (A., Gong, Hineman, Lewis, Vogel) Let µ be a non-negative Borel measure on S n 1. Then there exists a bounded convex domain K such that µ = µ p K if and only if w dµ(w) = 0 S n 1 and θ w dµ(w) > 0 S n 1 for every θ S n 1. K is unique up to translation when p n 1 (and is unique up to translation and dilation when p = n 1). This problem has been considered by Colesanti, Nyström, Salani, Xiao, Yang, Zhang under additional assumptions on µ with for 1 < p < 2. We prove the same result for capacitary surface measures associated to A harmonic PDEs.

T HAN KS!

Regularity of Minkowski Problem - Monge-Ampère equation Let µ be a positive measure on S n 1 with dh n 1 = dµ whenever E S n 1 is Borel. g 1 (E) E Suppose that dµ = (1/κ(ξ))dξ for some integrable κ C 0 > 0. Let φ denote the convex Lipschitz function defined on an open subset O of R n 1 whose graph {(x, φ(x)) : x O} is a portion of K. Then φ satisfies the Monge-Ampère equation det(d 2 φ(x)) ( 1, φ(x)) = κ(ξ) where ξ = (1 + φ(x) 2 ). (n+1)/2 1 + φ(x) 2

Equivalent statement of Minkowski problem: Find a closed, convex hypersurface K whose Gaussian curvature prescribed as a positive function defined on S n 1. Regularity of Minkowski Problem - Monge-Ampère equation Let µ be a positive measure on S n 1 with dh n 1 = dµ whenever E S n 1 is Borel. g 1 (E) E Suppose that dµ = (1/κ(ξ))dξ for some integrable κ C 0 > 0. Let φ denote the convex Lipschitz function defined on an open subset O of R n 1 whose graph {(x, φ(x)) : x O} is a portion of K. Then φ satisfies the Monge-Ampère equation det(d 2 φ(x)) ( 1, φ(x)) = κ(ξ) where ξ = (1 + φ(x) 2 ). (n+1)/2 1 + φ(x) 2 This is a Monge-Ampère equation with right hand side f = κ(ξ)(1 + φ 2 ) (n+1)/2. This corresponds to imposing that the Gauss curvature of the graph of φ at the point (x, φ(x)) is κ(ξ).