The near field refractor

Ann. I. H. Poincaré – AN 31 (2014) 655–684

www.elsevier.com/locate/anihpc

The near field refractor

Cristian E. Gutiérrez

, Qingbo Huang

aDepartment of Mathematics, Temple University, Philadelphia, PA 19122, United States bDepartment of Mathematics and Statistics, Wright State University, Dayton, OH 45435, United States

Received 12 October 2012; received in revised form 14 February 2013; accepted 9 July 2013 Available online 16 July 2013

Abstract

We present an abstract method in the setting of compact metric spaces which is applied to solve a number of problems in geometric optics. In particular, we solve the one source near field refraction problem. That is, we construct surfaces separating two homogeneous media with different refractive indices that refract radiation emanating from the origin into a target domain contained in ann−1 dimensional hypersurface. The input and output energy are prescribed. This implies the existence of lenses focusing radiation in a prescribed manner.

©2013 Elsevier Masson SAS. All rights reserved.

1. Introduction

We present in this paper an abstract method, having general interest, that can be applied to prove existence of solutions to a number of problems in geometric optics relative to refraction. The method is formulated in the abstract setting of compact metric spaces and it is based on the ideas of concave and convex mappings and the existence of families of functions that play the role of building blocks satisfying certain properties. The application to show existence of solutions to various problems then consists in selecting the appropriate maps and the appropriate class of building blocks related to the specific problem.

A main application of this method considered in the paper is to solve the one source near field refractor problem, that is, to show existence of surfaces that refract radiation when the output and input intensities are prescribed. More precisely, we have a domainΩ in the unit sphereSⁿ⁻¹and a domainDcontained in ann−1 dimensional surface inRⁿ;Dis referred as the target domain or screen to be illuminated. We also have two homogeneous and isotropic media I and II with refractive indicesn₁andn₂, respectively, and suppose that from a pointOsurrounded by medium I, light emanates with intensityf (x) forx ∈Ω, andD is surrounded by media II. We are also given inD a Radon measureμand the energy conservation equation

Ωf (x) dx=μ(D). We prove the existence of an optical surface Rparameterized byR= {ρ(x)x: x∈Ω}, interface between media I and II, such that all rays refracted byRinto medium II illuminate the objectD, and the prescribed illumination intensity distribution atDisμ. Of course, some

* Corresponding author.

E-mail addresses:[email protected](C.E. Gutiérrez),[email protected](Q. Huang).

1 The author was partially supported by NSF grant DMS-1201401.

2 The author was partially supported by NSF grant DMS-0502045.

0294-1449/$ – see front matter ©2013 Elsevier Masson SAS. All rights reserved.

http://dx.doi.org/10.1016/j.anihpc.2013.07.001

conditions on the relative position ofDand the set of directions inΩare needed to illuminateD. This implies that one can design a lens refracting light beams so that the screenDis illuminated in a prescribed way. The lens is bounded by two optical surfaces, the “outer” surface isRand the “inner” one is a sphere with center at the point from where the radiation emanates.

To implement the application of our abstract result to this case, we determine that the building blocks to construct the surface solution are Descartes ovals. Descartes ovals refract all light rays emanating from a pointO into a fixed pointP; seeFigs. 1 and 2. The solution of the near field refraction problem depends in a essential way of a novel and very delicate study of the eccentricity of the ovals and its approximation properties, Section4. This geometry is much more complicated that the one needed for the solution to the far field problem solved in[5]using mass transport. In particular, existence of solutions for the far field refractor problem can be obtained directly with this abstract method, Section7.

Geometric optics problems – far and near field – received recently attention because of both its applications and their mathematical interest and difficulty. To put our results in perspective, we enumerate some related results. The problem of the far field reflector has been considered by several authors, both mathematicians and engineers. For example, existence and uniqueness up to dilations of solutions for the far field reflector problem were proved by Caffarelli and Oliker in[3]and Wang in[9], and for nonisotropic media by Caffarelli and Huang[2].C¹regularity of solutions for the far field reflector was established in [1]. The near field reflector problem was considered by Kochengin and Oliker in [7], and in recent work by Karakhanyan and Wang [8]. The far field refractor problem was for the first time considered by the authors, and existence and uniqueness up to dilations of solutions were established in [5]. Refraction problems are in general more involved than reflection problems because of physical constraints. A difference between near and far field problems is that the latter can be cast in the frame of optimal transportation. Instead, near field problems are in general not optimal transportation problems which makes their mathematical treatment more difficult. In particular, the presence ofρin the matrixAand on the right hand side of the pde(A.12), indicates the problem is not an optimal transport problem. We mention that some results for the near field refractor problem have been announced in[4].

The plan of the paper is the following. Section2 contains the main results in the paper where we develop the abstract method. Section3describes the Snell law and the physical constraints of refraction. Section4contains our study of Cartesian ovals and estimates that are essential in the application to the near field problem. In Section5we describe the assumptions on the domainsΩ andDand solve the near field problem whenκ <1. Similar existence results whenκ >1 are proved in Section6. In Section7we show further applications of our method: the far field refractor problem and the second boundary value problem for the Monge–Ampère equation.

Finally, the pde governing the near field refractor problem is a fully nonlinear equation of Monge–Ampère type whose derivation is quite complicated and it is included inAppendix A.

2. Continuous mappings and measure equations

LetX, Y be compact metric spaces, andωa Radon measure onX. LetY^Xdenote the class of all set-valued maps Φ fromXtoY such thatΦ(x)is single-valued for a.e.xwith respect to the measureω. We say that the mapΦ∈Y^X is continuous atx₀∈X if givenx_k→x₀andy_k∈Φ(x_k), there exists a subsequencey_kj andy₀∈Φ(x₀)such that yk_j →y0, asj → ∞. LetC(X, Y )denote the class of allΦ∈Y^Xsuch thatΦ is continuous inX; and letCs(X, Y ) denote the class ofΦ∈C(X, Y )such thatΦ(X)=Y.

Lemma 2.1.GivenΦ∈C_s(X, Y ), the set function defined by M_Φ(E)=ω

Φ⁻¹(E) is a Radon measure onY.

Proof. The set C= {E⊂Y: Φ⁻¹(E)isω-measurable}is a σ-algebra containing all Borel sets in Y (Φ⁻¹(E)= {x ∈X: Φ(x)∩E = ∅}). Indeed, Φ⁻¹(∅)= ∅, Φ⁻¹(Y )=X, Φ⁻¹(_∞

i=1E_i)=_∞

i=1Φ⁻¹(E_i), Φ⁻¹(E^c)= (Φ⁻¹(E))^c∪(Φ⁻¹(E)∩Φ⁻¹(E^c)), and ω((Φ⁻¹(E)∩Φ⁻¹(E^c)))=0. If K is compact in Y, then Φ⁻¹(K) is compact inX. In fact, let{x_k} ⊂Φ⁻¹(K), andy_k∈Φ(x_k)∩K. SinceXis compact andΦis continuous, there exist subsequences{x_kj}and{y_kj}such thatx_kj →x₀andy_kj →y₀withy₀∈Φ(x₀), that is,x₀∈Φ⁻¹(K). To show the

σ-additivity, let{E_j}be a disjoint sequence of sets inC. ThenM_Φ(_∞

j=1E_j)=ω(Φ⁻¹(E₁))+_∞

k=2ω(Φ⁻¹(E_k)\ _k−1

i=1Φ⁻¹(E_i))=_∞

k=1M_Φ(E_k). 2

LetC(X)denote the set of continuous functions inXwith the topology of uniform convergence.

Definition 2.2.IfF⊂C(X)andT :F→C_s(X, Y ), we say thatT is continuous at φ∈F, if whenever φ_j ∈F, φ_j→φuniformly inX,x₀∈X, andy_j∈T(φ_j)(x₀), then there exists a subsequencey_j such thaty_j →y₀with y₀∈T(φ)(x₀).

Lemma 2.3.If F⊂C(X), T :F→C_s(X, Y )is continuous at φ∈F, and φ_j →φ inC(X) withφ_j ∈F, then M_T(φj)→M_T(φ)weakly.

Proof. It is enough to show that lim sup

j→∞ M_T(φj)(K)M_T(φ)(K), for allK⊂Y compact, (2.1)

and

lim inf

j→∞ M_T(φj)(G)M_T(φ)(G), for allG⊂Y open. (2.2)

To prove (2.1), we show that lim sup_j→∞T(φj)⁻¹(K)⊂T(φ)⁻¹(K). Let x0∈T(φj)⁻¹(K). So there is yj ∈ T(φj)(x0)∩K, and sinceT is continuous atφ, there exists a subsequenceyj such thatyj →y0withy0∈T(φ)(x0).

Sincey0∈K, we havex0∈T(φ)⁻¹(K). Hence lim sup

j→∞ M_T(φj)(K)=lim sup

j→∞ ω

T(φj)⁻¹(K)

ω

lim sup

j→∞ T(φj)⁻¹(K)

ω

T(φ)⁻¹(K)

=M_T(φ)(K).

To prove(2.2), we show thatT(φ)⁻¹(G)\E0⊂lim infj→∞T(φj)⁻¹(G), withω(E0)=0. Indeed, lim sup

j→∞

T(φ_j)⁻¹(G)c

⊂lim sup

j→∞ T(φ_j)⁻¹(G)c

∪

T(φ_j)⁻¹(G)∩T(φ_j)⁻¹ G^c

=lim sup

j→∞ T(φ_j)⁻¹ G^c

⊂T(φ)⁻¹ G^c

sinceG^cis compact becauseY is compact

=

T(φ)⁻¹(G)c

∪ T(φ)⁻¹(G)∩T(φ)⁻¹ G^c

:=

T(φ)⁻¹(G)c

∪E₀, withω(E₀)=0, sinceT(φ)is single-valued except on a set ofω-measure zero. Therefore

lim inf

j→∞ M_T(φj)(G)=lim inf

j→∞ ω

T(φ_j)⁻¹(G)

ω

lim inf

j→∞ T(φ_j)⁻¹(G) ω

T(φ)⁻¹(G)

=M_T(φ)(G). 2

2.1. Concave case Let

C⁺(X)= {f: f is continuous and positive inX}.

In this subsection, we consider classesF⊂C⁺(X)satisfying the following condition:

(A1) iff₁, f₂∈F, thenf₁∧f₂=min{f₁, f₂} ∈F.

We say that the classF⊂C⁺(X)isT-concave ifF satisfies(A1)and there exists a mapT :F→C_s(X, Y )that is continuous at eachφ∈Fand the following condition holds:

(A2) ifφ₁(x₀)φ₂(x₀), thenT(φ₁)(x₀)⊂T(φ₁∧φ₂)(x₀).

We also introduce the following condition on the classF:

(A3) For eachy0∈Y there exists an interval(αy₀, βy₀)and a family of functions{ht,y₀(x)}α_y0<t <β_y0 ⊂Fsatisfying (a) y₀∈T(h_t,y0)(x)for allx∈X,

(b) h_t,y0h_s,y0 forts,

(c) h_t,y0→0 uniformly ast→α_y0,

(d) h_t,y0 is continuous inC(X)with respect tot, i.e., maxx∈X|h_t,y₀(x)−h_t,y0(x)| →0 ast→t, forα_y0<

t < β_y0.

Remark 2.4.Notice that (A3)(a) implies thatM_T(ht,y

0)=ω(X)δ_y0 forα_y0< t < β_y0. Because ifE⊂Y is a Borel set withy0∈E, then from (A3)(a),X⊂T(h_t,y0)⁻¹(y0)⊂T(h_t,y0)⁻¹(E)⊂Xand soM_T(ht,y

0)(E)=ω(X). Ify0∈/E, theny0∈Y\Eand soM_T(ht,y

0)(Y\E)=ω(X)and thenM_T(ht,y

0)(E)=0.

The following is the main theorem in this section. We solve a measure equation when the given measure inY is discrete.

Theorem 2.5.LetX, Y be compact metric spaces andωis a Radon measure inX. Letp₁, . . . , p_Nbe distinct points inY, andg₁, . . . , g_Nbe positive numbers withN2.

LetF⊂C⁺(X)andT :F→C_s(X, Y )be such thatF isT-concave and(A3)holds.

Assume that ω(X)=

N i=1

g_i, (2.3)

and there exists ρ0=min1iNh_b0

i,pi such thatM_T(ρ0)(pi)gi for 2iN. Then there existbi ∈(αp_i, βp_i), 2iN, such that the function, withb₁=b⁰₁,

ρ(x)= min

1iNh_bi,pi(x) satisfies

M_T(ρ)= N i=1

giδp_i.

Proof. Letb₁=b⁰₁and define the set W=

(b₂, . . . , b_N): α_pi< b_ib_i⁰; M_T(ρ)(p_i)g_i, i=2, . . . , N . By the assumptions,(b⁰₂, . . . , b⁰_N)∈W.

Step 1:There exist constantsL,₀>0 such that for all(b₂, . . . , b_N)∈W we haveb_i α_pi +₀ifα_pi>−∞, andb_i−Lifα_pi= −∞, for 2iN.

To prove this, we first show that the measureM_T(ρ) is supported on{p₁, . . . , p_N}. This follows if we show that the setE=T(ρ)⁻¹(Y\ {p₁, . . . , p_N})⊂N₀, whereN₀:= {x∈X: T(ρ)(x)is not a singleton}hasω-measure zero.

If z₀∈E, thenT(ρ)(z₀)∩(Y \ {p₁, . . . , p_N})= ∅. Therefore there is p∈T(ρ)(z₀)withp=p_i for 1iN. On the other hand,ρ(z₀)=h_bk,pk(z₀)for some 1kN. From (A2) we then haveT(h_bk,pk)(z₀)⊂T(ρ)(z₀), but p_k∈T(h_bk,pk)(z₀)by (A3)(a), and soT(ρ)is not single-valued atz₀.

Consequently, from(2.3)we getM_T(ρ)(p₁)g₁>0, and soω(T(ρ)⁻¹(p₁)) >0. Pickx₀∈T(ρ)⁻¹(p₁)\N₀. We claim that h_b1,p1(x₀)h_bi,pi(x₀)for i2. Otherwise, there is somei2 such that h_bi,pi(x₀) < h_b1,p1(x₀).

Pickj2 such thath_bj,pj(x₀)=min2iNh_bi,pi(x₀). Thenρ(x₀)=h_bj,pj(x₀). Hence from (A2),T(h_bj,pj)(x₀)⊂ T(ρ∧h_bj,pj)(x₀)=T(ρ)(x₀)=p₁. But from (A3)(a),p_j∈T(h_bj,pj)(x₀), a contradiction and the claim is proved.

Sincehb₁,p₁(x0)C >0, we then gethb_i,p_i(x0)C >0 for alli2. From (A3)(c),hb_i,p_i(x0)→0 asbi→αp_i, and therefore Step 1 is proved.

Step 2:W is compact.

It is enough to show that M_T(ρ)(p_i) is continuous in b =(b₂, . . . , b_N) for each 1i N. Let b_m = (b^m₂, . . . , b^m_N)∈W converging to b_∗ =(b₂^∗, . . . , b_N^∗). Then by (A3)(d), ρ_m =(mini=1h_b^m

i,pi)∧ h_b1,p1 →ρ^∗ = (mini=1h_b∗

i,pi) ∧ h_b1,p1 uniformly as m → ∞. Therefore, as in the proof of Lemma 2.3, we obtain lim sup_m→∞M_T(ρm)(p_i)M_T(ρ∗)(p_i) for i=1, . . . , N, and lim infm→∞M_T(ρm)(G)M_T(ρ∗)(G) for each G open. Now choose G open such that p_i ∈G and p_j ∈/G for j =i. Then M_T(ρm)(G)=ω(T(ρ_m)⁻¹(p_i))+ ω(T(ρ_m)⁻¹(G\ {p_i})), but ω(T(ρ_m)⁻¹(G \ {p_i})) ω(T(ρ_m)⁻¹(Y \ {p₁, . . . , p_N})) = 0. We then obtain lim infm→∞M_T(ρm)(p_i)M_T(ρ∗)(p_i)and the continuity follows.

Step 3:Existence of solutions.

The function z=b2+ · · · +bN attains its minimum on W at some point (a2, . . . , aN). We claim that ρ(x)= (min2iNh_ai,pi)∧h_b1,p1 is the desired solution. Otherwise, assume for example that M_T(ρ)(p2) < g2. Let a¯ = (a₂−, a₃, . . . , a_N) andρ(x)¯ =(min2iNh_a¯i,pi)∧h_b1,p1. By continuity M_T(ρ)¯ (p₂) < g₂ for all sufficiently small.

Fori3, we claim thatT(ρ)¯ ⁻¹(p_i)⊂T(ρ)⁻¹(p_i)except on a set ofω-measure zero. Indeed, ifx₀∈T(ρ)¯ ⁻¹(p_i) andT(ρ)(x¯ ₀)is a single point, thenp_i=T(ρ)(x¯ ₀). Notice thatρ(x¯ ₀)=h_ai,pi(x₀). Otherwise, there exists j =i, 1jN, such thatρ(x¯ 0)=ha_¯j,p_j(x0) < ha_i,p_i(x0)(we seta¯1=b1). Then from (A2),T(ha_¯j,p_j)(x0)⊂T(ρ)(x¯ 0), and by (A3)(a)pj∈T(ha_¯j,p_j)(x0)and sopj=pi, a contradiction. From (A3) (b),ha_¯2,p₂ha₂,p₂ soρ(x)¯ ρ(x), and therefore ρ(x¯ 0)=h_ai,pi(x0)=ρ(x0). Hence, and once again from (A2), T(h_ai,pi)(x0)⊂T(ρ∧h_ai,pi)(x0)= T(ρ)(x₀). Thus, p_i ∈T(ρ)(x₀) from (A3)(a), so x₀ ∈T(ρ)⁻¹(p_i), and the claim is proved. We then obtain M_T(ρ)¯ (p_i)M_T(ρ)(p_i)g_ifori3, that is,a¯∈W, a contradiction. 2

Remark 2.6.The existence of the function ρ₀ inTheorem 2.5follows if one assumes that b⁰₁> α_p1 is given and it is sufficiently close to α_p1. In fact, in this case we pick α_pi < τ_i < β_pi for 2iN. Since F⊂C⁺(X), we haveh_τi,pi(x)C_i>0 fori2 andx∈X. Then from (A3)(c), we can choose b₁⁰sufficiently close toα_p1 such thath_τi,pi(x)min2iNC_ih_b0

1,p₁(x). Ifb⁰_i :=τ_i for 2iN, we then selectρ₀(x)=min1iNh_b0

i,p_i(x)= h_b0

1,p₁(x)and so fromRemark 2.4M_T(ρ0)=ω(X)δp₁. Consequently,M_T(ρ0)(pi)=0 for 2iN.

Theorem 2.7.Letρ, ρ^∗be two solutions as inTheorem2.5, withb=(b₁, . . . , b_N), andb^∗=(b^∗₁, . . . , b^∗_N). Assume thatXis connected andω(E) >0for each open setE⊂X. Assume in addition that condition(A3)(b)is replaced by h_t,y0< h_s,y0 fort < s.

(a) Ifb^∗₁b₁, thenb_i^∗b_ifor all1iN. In particular, ifb^∗₁=b₁, thenb^∗_i =b_i for all1iN. (b) Ifρ(x₀)=ρ^∗(x₀)at somex₀∈X, thenρ=ρ^∗.

Proof. (a) Let J= {j: bj< b^∗_j}andI= {i: b_i^∗bi}. Suppose by contradiction thatJ= ∅. We haveI = ∅since 1∈I. For eachj∈J we haveh_bj,pj(x) < h_b∗

j,p_j(x)for allx∈X, sinceb_j< b_j^∗. And alsoh_b∗

i,p_i(x)h_bi,pi(x)for alli∈I and allx∈X.

LetQ= {x∈X:T(ρ^∗)(x)is not a singleton}. From (A2) and (A3)(a), we notice that ifx∈T(ρ^∗)⁻¹(p_i)\Q, for some 1iN, thenρ^∗(x)=h_b∗

i,pi(x).

We next prove that bdyT(ρ^∗)⁻¹(PJ)⊂Q, wherePJ= {p_j:j∈J}. Indeed, letz₀∈bdyT(ρ^∗)⁻¹(PJ)andN_z0be an open neighborhood ofz₀. ThenN_z0∩(T(ρ^∗)⁻¹(PJ))^cis a nonempty open set, since(T(ρ^∗)⁻¹(PJ))is compact.

Thus,N_z0∩(T(ρ^∗)⁻¹(PJ))^c\Qhas a positive measure and therefore is nonempty. We then obtain{z_k}such that z_k→z₀andz_k∈(T(ρ^∗)⁻¹(PJ))^c\Q. So there exists{p_ik}withp_ik =T(ρ^∗)(z_k)andi_k∈I. We may assume that pi_k=pi, for somei∈I. Therefore,ρ^∗(zk)=h_b∗

i,pi(zk). By taking limit,ρ^∗(z0)=h_b∗

i,pi(z0). From (A2) and (A3)(a), this yieldspi∈T(ρ^∗)(z0). SinceT(ρ^∗)(z0)∩PJ= ∅, we obtainz0∈Q.

As a consequence,ω((T(ρ^∗)⁻¹(PJ))^◦)=

j∈Jg_j>0.

Given x₀∈(T(ρ^∗)⁻¹(PJ))^◦, for any open neighborhood N_x0 of x₀, we have that N_x0 ∩(T(ρ^∗)⁻¹(PJ))^◦ is a nonempty open set. Therefore, as in the previous argument, there existsx_k∈(T(ρ^∗)⁻¹(PJ))^◦\Qfork1 such that x_k→x₀. Hence, one may assume that there exists somep_j withj ∈J such thatp_j=T(ρ^∗)(x_k). Soh_b∗

j,pj(x_k)= ρ^∗(x_k)and thenh_b∗

j,pj(x₀)=ρ^∗(x₀)by taking limit. Therefore,h_b∗

j,pj(x₀)h_b∗

i,pi(x₀)for all 1iN. Thus, we obtain forj∈Jthat

h_bj,pj(x₀) < h_b∗

j,pj(x₀)h_b∗

i,pi(x₀)h_bi,pi(x₀) for alli∈I.

Hence by continuity, there existsN_x0 a neighborhood ofx₀such that hb_j,p_j(y) < hb_i,p_i(y) for alli∈I, j ∈J, andy∈Nx₀.

By definition ofρthis implies thatρ(y)=minj∈Jh_bj,pj(y)for ally∈N_x0. Therefore for eachy∈N_x0 there exists j₀∈J, depending ony, such thatρ(y)=h_bj

0,p_j0(y). Hence, once again by (A2) and (A3)(a),p_j0∈T(h_bj

0,p_j0)(y)⊂ T(ρ)(y). That is,y∈T(ρ)⁻¹(p_j0), and therefore

N_x0⊂T(ρ)⁻¹(PJ).

We then have that every pointx∈(T(ρ^∗)⁻¹(PJ))^◦has a neighborhood contained inT(ρ)⁻¹(PJ), that is, T

ρ^∗₋1

(PJ)_◦

⊂

T(ρ)⁻¹(PJ)_◦

=X.

This is a contradiction with the fact that ω

T(ρ)⁻¹(PJ)

=

j∈J

g_j=ω T

ρ^∗₋1

(PJ)_◦ .

(b) If b1=b₁^∗, then bj =b^∗_j for allj >1 by part (a), and we are done. We claim that if b1> b₁^∗, then bj > b^∗_j for all j >1. Indeed, ifb_j=b^∗_j for somej =1, thenb_k =b_k^∗for allk=j by part (a), a contradiction. Therefore ρ^∗(x₀)=minh_b∗

i,pi(x₀) <minh_bi,pi(x₀)=ρ(x₀), a contradiction. 2

Theorem 2.8.Let{μl}be a sequence of discrete Radon measures inY such thatμl→μweakly andμl(Y )=ω(X) forl1. Letρlbe a solution obtained inTheorem2.5corresponding toμl. Assume that there existsR0>0such that R₀∈Range(ρl)forl1. Suppose that

(i) For eachR₁>0withR₁∈Range(ht,y), there existsC_R1>0such thatC⁻_R¹

1 h_t,yC_R1. (ii) For anyC₁> C₀>0, the family{f ∈F: C₀f C₁inX}is compact inC(X).

Then there existsρ∈F satisfyingM_T(ρ)=μ.

Proof. By (ii) andLemma 2.3, it suffices to show{ρ_l}is bounded from below and above. Assumeρ_l(x_l)=R₀for somexl∈X. Then there existshb_l,y_l such thatρlhb_l,y_l andR0=ρl(xl)=hb_l,y_l(xl). By (i),C_R⁻¹

0 hb_l,y_l CR₀

for someCR₀. Therefore,ρlCR₀. To get a lower bound, givenx1∈X, there existsh_b

l,y_l such thatρlh_b l,y_l and ρ_l(x1)=h_b

l,y_l(x1). Hence,R0h_b

l,y_l(x_l). Sinceh_t,y

l is continuous and decreasing to zero ((A3)(b) and (c)), there existsb_l b_l withR₀=h_b

l,y_l(x_l). It follows from (A3)(b) thatρ_l(x₁)h_b

l,y_l(x₁)C_R⁻¹

0. Henceρ_lC_R⁻¹

0. 2

2.2. Convex case

We assume here thatF⊂C⁺(X)and condition (A1) above is replaced by (A1) iff₁, f₂∈F, thenf₁∨f₂=max{f₁, f₂} ∈F.

We say that the classF⊂C⁺(X)isT-convex if Fsatisfies(A1)and there exists a mapT :F→C_s(X, Y )that is continuous at eachφ∈Fand the following condition holds:

(A2) ifφ₁(x₀)φ₂(x₀), thenT(φ₁)(x₀)⊂T(φ₁∨φ₂)(x₀).

Here we substitute condition (A3) by

(A3) for eachy₀∈Y there exists an interval(α_y0, β_y0)and a family of functions{h_t,y0(x)}αy0<t <βy0 ⊂Fsatisfying (a) y₀∈T(h_t,y0)(x)for allx∈X,

(b) h_t,y0h_s,y0 forts,

(c) h_t,y0→0 uniformly ast→β_y0,

(d) h_t,y0 is continuous inC(X)with respect tot, i.e., maxx∈X|h_t,y₀(x)−h_t,y0(x)| →0 ast→t, forα_y0 <

t < β_y0.

Under these assumptions we prove the following theorem.

Theorem 2.9.LetX, Y be compact metric spaces andωis a Radon measure inX. Letp₁, . . . , p_N be distinct points inY, andg₁, . . . , g_N be positive numbers withN2.

LetF⊂C⁺(X)andT :F→C_s(X, Y )be such thatFisT-convex and(A3)holds.

Assume that ω(X)=

N i=1

g_i. (2.4)

Suppose that there exists(b⁰₁, . . . , b⁰_N)withh_b0

1,p₁(x)min2iNh_b0

i,p_i(x)onX. Then there existbi ∈(αp_i, βp_i), 2iN, such that the function, withb₁=b₁⁰,

ρ(x)= max

1iN

h_bi,pi(x) satisfies

M_T(ρ)= N i=1

g_iδ_pi.

Proof. Letb₁=b₁⁰andη=minXh_b1,p1>0. From (A3)(c), there existsτ >0 such that maxX h_βpi−τ,piη

for all 2iN. Therefore, if ρ_τ =h_b1,p1 ∨(max2iNh_βpi−τ,pi)=h_b1,p1, then M_T(ρτ)=ω(X)δ_p1, and so M_T(ρτ)(p_i)=0 for 2iN.

Consider the set W (b₁)=

(b₂, . . . , b_N): α_pi< b_iβ_pi−τ; M_T(ρ)(p_i)g_i, i=2, . . . , N , W (b₁)= ∅, because(β_p2−τ, . . . , β_pN−τ )∈W (b₁).

We claim thatbib_i⁰, for 2iN, for all(b2, . . . , bN)∈W (b1).

To prove the claim, we first show M_T(ρ)(Y \ {p1, . . . , pN})=0;ρ=max1iNhb_i,p_i. Indeed, for z0∈E= T(ρ)⁻¹(Y\ {p₁, . . . , p_N}), there isp∈T(ρ)(z₀)withp=p_i for 1iN. On the other hand,ρ(z₀)=h_bk,pk(z₀) for somek. From (A2) and (A3)(a) we then havep_k∈T(h_bk,pk)(z₀)⊂T(ρ)(z₀). SoT(ρ)is not single-valued at z₀and soω(E)=0. Consequently, from(2.4)we getM_T(ρ)(p₁)g₁>0.

Now suppose by contradiction that b_i < b⁰_i for some 2iN. Sinceb₁=b⁰₁, it follows from the assump- tion and (A3)(b) thath_b1,p1 h_b0

i,p_i h_bi,pi in X. This implies that for eachx₀∈X, there is j =1 such that