The Gauss Image Problem

KÁROLY J. BÖRÖCZKY

Central European University and Rényi Institute ERWIN LUTWAK

Courant Institute DEANE YANG Courant Institute GAOYONG ZHANG

Courant Institute AND YIMING ZHAO

Massachusetts Institute of Technology Dedicated to the memory of Professor Delin Ren

1 Introduction

The Brunn-Minkowski theory and the dual Brunn-Minkowski theory are two core theories in convex geometric analysis that center on the investigation of global geometric invariants and geometric measures associated with convex bodies. The two theories display an amazing conceptual duality that involves many dual con- cepts in both geometry and analysis such as dual spaces in functional analysis, polarity in convex geometry, and projection and intersection in geometric tomog- raphy; see Schneider [49, p. 507] for a lucid explanation.

In the conceptual duality, a central role is assumed by theradial Gauss image _K (defined immediately below) of a convex body K in euclideann-space, Rⁿ. The radial Gauss image is a map on the unit sphere, S^{n 1}, of Rⁿ whose values are subsets of the unit sphere. It is known thatAleksandrov’s integral curvature onS^{n 1}and spherical Lebesgue measure are “linked” via the radial Gauss image, and so are the classicalsurface area measureof Aleksandrov-Fenchel-Jessen and Federer’s.n 1/^thcurvature measure(see Schneider [49, theorem 4.2.3] and [27]).

The importance of the radial Gauss image was made more evident in the recent work [27], in which the long-soughtdual curvature measures(the dual counterparts of Federer’s curvature measures) were unveiled. In [27] new links were established between the Brunn-Minkowski theory and the dual Brunn-Minkowski theory by making critical use of the radial Gauss image. Motivated by the manner in which these new geometric measures are defined via the radial Gauss image, it becomes

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natural to introduce a general new concept—theGauss image measureassociated with a convex body. Among other things, this concept bridges the classical and the recently discovered geometric measures of convex bodies.

In light of the role that the radial Gauss image plays in connecting various spher- ical Borel measures, a central question regarding Gauss image measures is: Given two spherical Borel measures, under what conditions does there exist a convex body so that one measure is the Gauss image measure of the other? We call this theGauss image problemand state it more precisely immediately below.

LetKⁿdenote the set of convex bodies (compact, convex subsets with nonempty interior) inn-dimensional euclidean space, Rⁿ, with Kⁿ_o denoting the bodies that contain the origin in their interiors.

IfK 2Kⁿ_o andx 2 @K is a boundary point, then thenormal coneofK atxis defined by

N.K; x/ D fv 2 S^{n 1}W .y x/ v 0for ally 2 Kg;

where.y x/ vdenotes the standard inner product ofy x andvinRⁿ. The radial mapr_K W S^{n 1} ! @KofKis defined foru 2 S^{n 1}byr_K.u/ D ru 2 @K, wherer > 0. For! S^{n 1}, theradial Gauss imageof!is defined by

_K.!/ D [

x2rK.!/

N.K; x/ S^{n 1}:

The radial Gauss image is the composite of the multivalued Gauss map and the radial map. It is well-known (see Schneider [49]) that for a Borel measurable

! S^{n 1}, the set_K.!/ S^{n 1} is spherically Lebesgue measurable but not necessarily Borel measurable.

Recall (see, e.g., [30, p. 1117]) that a submeasure differs from a measure in that the countable additivity in the definition of a measure is replaced by countable subadditivity. (See Section 3 for precise definitions.)

DEFINITION. Supposeis a submeasure defined on spherical Lebesgue measur- able subsets ofS^{n 1}, andK 2K_oⁿ. Then.K; /, theGauss image measure of viaK, is the submeasure onS^{n 1}defined by

.K; !/ D ._K.!//

for each Borel! S^{n 1}.

When we write that a Borel measureon S^{n 1} isabsolutely continuous, we shall always mean that it is absolutely continuous with respect to spherical Le- besgue measure. Obviously, the completion of an absolutely continuous Borel measure is defined on all spherically Lebesgue measurable subsets ofS^{n 1}. When we speak of Borel measures onS^{n 1}, we shall always assume them to be finite, nonnegative, and nonzero.

As will be shown, whenis an absolutely continuous Borel measure onS^{n 1}, andK 2Kⁿ_o, then.K; /, is a Borel measure onS^{n 1}. Whenis Lebesgue mea- sure onS^{n 1}, then.K; /is simplyAleksandrov’s integral curvatureof the body

K (see, e.g., [2]). Moreover, the classical surface area measures of Aleksandrov- Fenchel-Jessen [1, 49], and the recently discovered, in [27], dual curvature mea- sures are all Gauss image measures. This makes the Gauss image measure an object of significant interest that requires extensive study.

It is the aim of this work to introduce and attack theGauss image problem:

The Gauss image problem.Supposeis a submeasure defined on the Lebesgue measurable subsets ofS^{n 1}, andis a Borel submeasure onS^{n 1}. What are the necessary and sufficient conditions, onand, so that there exists a convex body K 2Kⁿ_o such that

(1.1) .K; / D

on the Borel subsets ofS^{n 1}? And if such a body exists, to what extent is it unique?

When is spherical Lebesgue measure, the Gauss image problem is just the classical Aleksandrov problem. Note that since obviouslyK.S^{n 1}/ D S^{n 1}, a solution to (1.1) is only possible ifjj D jj; i.e.,.S^{n 1}/ D .S^{n 1}/.

Purely as an aside, we note that for the special case in whichis a measure that has a density, sayf, andis a measure that has a density, sayg, the geometric problem (1.1) is the equation of Monge-Ampère type,

(1.2) g

rh C h jrh C hj

jrh C hj ⁿhdet.r²h C hI / D f;

where hW S^{n 1} ! .0; 1/is the unknown function. In (1.2), I is the standard Riemannian metric onS^{n 1}, the mapW S^{n 1} ! S^{n 1} is the identity, whilerh andr²hare, respectively, the gradient and the Hessian ofhwith respect toI.

The focus of this work will be on solving the general question posed by (1.1).

Special cases, such as (1.2), shall be ignored. Our approach in attacking equation (1.1) uses convex geometric methods of a variational nature. What will be needed are delicate estimates for geometric invariants in order to solve an associated max- imization problem. The techniques developed in this work in order to obtain these critical estimates are new and different from those developed in [16, 27].

IfK 2Kⁿ_o, then its radial function_KW Rⁿnf0g ! Ris defined, for eachx ¤ 0, by_K.x/ D maxfr > 0 W rx 2 Kg. Ifis a Borel measure onS^{n 1}, then for a realq ¤ 0, define theq^thdual volume ofK with respect toby

_q.K/ D 1

jj Z

S^{n 1}_K^q.u/ d.u/

¹_q :

Recall thatq.K/is monotone nondecreasing and continuous inq. Define thelog- volumeofK with respect toby₀.K/ Dlim_q!0q.K/. Ifis the spherical Lebesgue measure, then the dual volume_q.K/is just the normalized classicalq^th dual volume. Dual volumes associated with the spherical Lebesgue measure are fundamental geometric invariants. Their connections to dual curvature measures and the dual Minkowski problem were discovered in [27]. Surprisingly, as will be seen, log-volumes are closely related to the Gauss image problem.

ForQ 2 Kⁿ_o, letQ D fx 2 Rⁿ W x y 1for ally 2 Kgdenote thepolar ofQ. As will be shown, the solutions of the Gauss image problem are closely tied to the following:

Maximizing the log-volume-product.If; are Borel measures onS^{n 1}of the same total mass, what are the necessary and sufficient conditions onandso that there exists a convex bodyK 2K_oⁿsuch that

sup

Q2Kⁿ_o

0.Q/0.Q/ D0.K/0.K/‹

If ! S^{n 1} is contained in a closed hemisphere, then the polar set ! is defined by

(1.3) !D fv 2 S^{n 1} W u v 0for allu 2 !g D \

u2!

fv 2 S^{n 1} W u v 0g:

A critical new concept introduced here is that of two Borel measures onS^{n 1} beingAleksandrov related.

DEFINITION. Two Borel measuresandonS^{n 1}are calledAleksandrov related if

.S^{n 1}/ D .S^{n 1}/ > .!/ C .!/

for each compact, spherically convex set! S^{n 1}.

This relationship is easily seen to be symmetric since! D ! for each com- pact, spherically convex set! S^{n 1}. If is Aleksandrov related to spherical Lebesgue measure, then the measureis said tosatisfy the Aleksandrov condition, which is an important well-known notion.

The following solution to a critical case of the Gauss image problem will be presented:

THEOREM1.1. Supposeandare Borel measures onS^{n 1}andis absolutely continuous. Ifandare Aleksandrov related, then there exists a bodyK 2 Kⁿ_o such that D .K; /.

It will be shown that when the measureis strictly positive on nonempty open sets, the requirement that the measures be Aleksandrov related is also necessary.

Moreover, it will be shown that the convex body in the solution is unique up to dilation.

When the measure is spherical Lebesgue measure, Theorem 1.1 is origi- nally due to Aleksandrov. New proofs were presented by Oliker [47] and later by Bertrand [10]. The approach taken below is different from these.

It will be shown that in an important case, the Gauss image problem and the problem of maximizing the log-volume-product are equivalent.

THEOREM 1.2. Suppose and are Borel measures on S^{n 1}, and is both absolutely continuous and strictly positive on nonempty open sets. Ifjj D jj, then the following statements are equivalent:

(1) There exists a bodyK 2Kⁿ_osuch that.K; / D . (2) There exists a bodyK 2Kⁿ_osuch that

sup

Q2Kⁿ_o

0.Q/0.Q/ D0.K/0.K/:

(3) andare Aleksandrov related.

Moreover, if the convex bodyKexists, then it is unique up to dilation.

It can be shown that two even Borel measures with the same total mass are always Aleksandrov related (a Borel measure is even if its value is the same for each Borel set and its antipode).

THEOREM1.3. Supposeis an even Borel measure onS^{n 1}that is not concen- trated on any great hypersphere, andis an even Borel measure onS^{n 1} that is absolutely continuous and strictly positive on nonempty open sets. Ifjj D jj, then there exists an origin-symmetric convex bodyK 2 Kⁿ_o, unique up to dilation, such that

(1) .K; / D , and

(2) the maximum of0.Q/0.Q/overQ 2Kⁿ_o is attained atK.

It is necessary to contrast the Gauss image problem with the various Minkowski problems and dual Minkowski problems that have been extensively studied (see, e.g., [14, 16–18, 27–29, 37–42, 45–47, 54, 56–59]). A good way to do that is to contrast the Gauss image problem with a specific Minkowski problem, say the log- Minkowski problem. The cone volume measure of a convex body has been of considerable recent interest (see, e.g., [9, 12, 13, 16, 25, 26, 43, 44, 54]). Thecone- volume measureV_K of a convex body K is a Borel measure on the unit sphere, defined for Borel! S^{n 1}as then-dimensional Lebesgue measure of the cone

ftx W 0 t 1andx 2 @KwithN.K; x/ \ ! ¤ ¿g:

Thelog-Minkowski problemasks: Given a Borel measure, does there exist a convex body K such that D V_K? And if the body exists, to what extent is it unique? (For recent work on this, see, e.g., [6–8, 16, 25, 26].) It is precisely here that we can see the difference between Minkowski problems and the Gauss image problem. In the Gauss image problem, a pair of submeasures is given and it is asked if there exists a convex body “linking” them via its radial Gauss image. Thus, we need to construct a convex body whose radial Gauss image “links” the two given submeasures. On the other hand, in a Minkowski problem, only one measure is given, and the question asks if this measure is a specific geometric measure of a convex body, such as the cone-volume measure of a convex body. To solve a Minkowski problem, we are attempting to construct a convex body for a specific geometric measure of convex bodies. However, the Gauss image problem could be a Minkowski problem. For example, ifis spherical Lebesgue measure, then .K; /is just Aleksandrov’s integral curvature ofK. Here we are dealing with a Minkowski problem, namely, the Minkowski problem for Aleksandrov’s integral

curvature: Given a Borel measure, does there exist a convex bodyK such that D .K; /; i.e., does there exist a convex body K whose integral curvature is the given measure? And if the body exists, to what extent is it unique? In this sense, the Gauss image problem broadens the study of Minkowski problems. But the essence of the problem is an attempt at a deeper understanding of the Gauss image map.

2 Preliminaries Forx 2 Rⁿ, letjxj D p

x xbe the euclidean norm ofx. Forx 2 Rⁿn f0g, define xx D x=jxj. For a subset E Rⁿ, let E D fxx W x 2 E n f0gg. Thex origin-centered unit ballfx 2 RⁿW jxj 1gis always denoted byB.

Lebesgue measure inRⁿis denoted byV, which is also called “volume.” Write

!nfor the volume ofB. We shall writeH^{n 1}for.n 1/-dimensional Hausdorff measure.

For the set of continuous functions defined onS^{n 1}, writeC.S^{n 1}/, and for f 2 C.S^{n 1}/, write kf k₁ D max_v2Sn 1jf .v/j. We shall viewC.S^{n 1}/ as endowed with the topology induced by thismax-norm. We writeC^C.S^{n 1}/ for the set of strictly positive functions inC.S^{n 1}/, andC_e^C.S^{n 1}/for the set of even functions inC^C.S^{n 1}/.

LetKⁿ denote the set of compact, convex subsets of Rⁿ. For K 2 Kⁿ, the support functionhK W Rⁿ ! RofK is defined byhK.x/ Dmaxfx y W y 2 Kg for x 2 Rⁿ. The support function is convex and homogeneous of degree 1. A compact convex subset ofRⁿis uniquely determined by its support function. The setKⁿis viewed as endowed with theHausdorff metric. So, the distance between K; L 2 Kⁿ is simplyd.K; L/ D kh_K h_Lk₁. IfAis a compact subset ofRⁿ, then convA, theconvex hullofA, is the smallest convex set that containsA. It is easily seen that its support function is given by

(2.1) hconvA.x/ Dmaxfx y W y 2 Ag;

forx 2 Rⁿ.

Aconvex bodyinRⁿis a compact convex set with nonempty interior. Denote by intK the interior of the convex body K. Denote by K_eⁿ the class of origin- symmetric convex bodies inRⁿ. Obviously,Kⁿ_o is a subspace ofKⁿ, andKⁿ_e is a subspace ofKⁿ_o.

The radial function_KW S^{n 1}! Rof a compact setK that is star-shaped, with respect to the origin, is defined byK.x/ D maxfa W au 2 Kgforu 2 S^{n 1}. A compact star-shaped set with respect to the origin is uniquely determined by its radial function. The radial function of a convex body inK_oⁿis continuous and positive. IfK 2Kⁿ_o, then obviously

@K D f_K.u/u W u 2 S^{n 1}g:

Theradial metric defines the distance betweenK; L 2 K_oⁿask_{K L}k₁. We shall use the well-known fact that onKⁿ_o, the Hausdorff metric and radial metric are topologically equivalent.

For a Borel measureonS^{n 1}, define _p.f / D

1 jj

Z

S^{n 1}f^pd _1=p

; p ¤ 0;

and

0.f / Dexp 1

jj Z

S^{n 1}logf d

for eachf 2 C^C.S^{n 1}/. Whenf D _K, for someK 2Kⁿ_o, then_p.f /will be written as_p.K/. Whenis spherical Lebesgue measure, then the_p.K/are the normalized dual volumes from the dual Brunn-Minkowski theory—a theory that played a critical role in the ultimate solution of the Busemann-Petty problem (see, e.g., [19, 21, 31, 32, 55]).

IfK 2K_oⁿ, then it is easily seen that the radial function and the support function ofKare related by

h_K.v/ Dmax_u2Sn 1.u v/ _K.u/; v 2 S^{n 1}; and

1=_K.u/ Dmax_v2Sn 1.u v/=h_K.v/; u 2 S^{n 1}: From the definition of the polar body, we see that

(2.2) _K D 1=h_K and h_K D 1=_K

onS^{n 1}.

ForK; L 2Kⁿand reala; b 0, theMinkowski combination,aK C bL 2Kⁿ, is the compact convex set defined by

aK C bL D fax C by W x 2 Kandy 2 Lg;

and its support function is given by

h_aKCbLD ah_KC bh_L:

Suppose  S^{n 1} is closed and not contained in any closed hemisphere of S^{n 1}. For a functionf W  ! .0; 1/, definehf ito be the convex hull inRⁿ,

hf i Dconvff .u/u W u 2 g:

Since f is strictly positive and is not contained in any closed hemisphere of S^{n 1}, it follows thathf i 2 Kⁿ_o. Note thathaf i D ahf ifora > 0. From (2.1), we see that the support function ofhf iis given by

(2.3) h_{hf i}.x/ Dmax_u2.x u/f .u/;

forx 2 Rⁿ. We shall make use of the fact that iff0; f1; : : : 2 C^C.S^{n 1}/, then

(2.4) lim

k!1f_k D f₀uniformly onS^{n 1} H) hf_ki ! hf₀iinKⁿ_o:

See, e.g., [27, p. 345] for a proof.

If! S^{n 1}, define cone!, thecone generated by!, as cone! WD ftu W t 0andu 2 !g and define€!, therestricted cone generated by!, as

€

! D ftu W 0 t 1andu 2 !g:

A subset ! S^{n 1} is spherically convex, if cone! is a nonempty proper con- vex subset ofRⁿ.This definition implies that a spherically convex set onS^{n 1} is nonempty and is always contained in a closed hemisphere ofS^{n 1}. A spherically convex set! S^{n 1}is said to bestrongly spherically convexif it is contained in an open hemisphere.

If!is a compact spherically convex set inS^{n 1}, then!is strongly spherically convex if and only if! \ . !/ D ¿, or equivalently! does not contain a pair of antipodal points. Indeed, when! S^{n 1} is compact spherically convex and

! \ . !/ D ¿, then conv!and conv. !/, the convex hulls inRⁿ, are disjoint. If this were not the case, then this would immediately imply that the origin belongs to conv!. But, to see that this is impossible write the origin as a convex combination ofu1; : : : ; ur 2 ! with strictly positive coefficients. This would imply that the point u₁ 2 cone!, and since u₁ 2 S^{n 1}, it would follow that u₁ 2 !, thus contradicting the fact thatu1 2 !. Since conv! and conv. !/are disjoint compact convex sets inRⁿthat do not contain the origin, the hyperplane separation theorem tells us that conv! and conv. !/ are contained in the opposite open sides of a hyperplane passing through the origin. Thus,!is contained in an open hemisphere.

For a subset! S^{n 1} that is contained in a closed hemisphere, itspolar set

!is defined by

!D fv 2 S^{n 1}W u v 0for allu 2 !g:

Thespherical convex hull,h!i, of!is defined by h!i D S^{n 1}\conv.cone!/:

The polar set!is always convex and

(2.5) !D h!i:

For recent work on spherical convex bodies, see Besau and Werner [11].

As is well-known, the Hausdorff metric can be extended to the set of all non- empty compact subsets ofRⁿ. IfK andLare nonempty compact subsets ofRⁿ, then the Hausdorff distance between them can be defined by

max

x2Ksup inf

y2Ljx yj; sup

y2L inf

x2Kjx yj :

LetO^{n 1} denote the set of spherically compact convex sets ofS^{n 1} endowed with the topology of the Hausdorff metric. It is easily verified that a sequence

!_i 2O^{n 1}converges to! 2O^{n 1}if and only if€!_i converges to€!.

Let  S^{n 1} be a closed set that is not contained in a closed hemisphere ofS^{n 1}. Let f W  ! R be continuous and > 0. Let h_tW  ! .0; 1/be a continuous function defined for eacht 2 . ; /by

loght Dlogh C tf C o.t; /;

whereo.t; /W  ! R is continuous and lim_t!0o.t; /=t D 0 uniformly on. Denote by

Œh_t D fx 2 RⁿW x v h_t.v/for allv 2 g

the Wulff shape determined byh_t. We shall callŒh_talogarithmic family of Wulff shapes formed by .h; f /. On occasion, we shall write Œht asŒh; f , and, if h happens to be the support function of a convex body K, perhaps as ŒK; f  or

ŒK; f; torŒK; f; o; t, if required for clarity. We callŒK; f a logarithmic family of Wulff shapes formed by.K; f /.

LetgW  ! Rbe continuous and > 0. Let_tW  ! .0; 1/be a continuous function defined for eacht 2 . ; /by

logt Dlog C tg C o.t; /;

where againo.t; /W  ! Ris continuous and lim_t!0o.t; /=t D 0uniformly on

. Denote by

hti Dconvft.u/u W u 2 S^{n 1}g

the convex hull generated by_t. We will callh_tialogarithmic family of convex hulls generated by.; g/. On occasion, we shall writeh_tiash; g; ti, and if happens to be the radial function of a convex bodyK 2K_oⁿashK; giorhK; g; ti orhK; g; o; ti, if required for clarity. We callhK; gia logarithmic family of convex hulls generated by.K; g/.

From [27] we will use the easily established fact that ifK 2Kⁿ_oandf W  ! R is continuous, where S^{n 1} is a closed set that is not contained in a closed hemisphere ofS^{n 1}, then

(2.6) hK; f iD ŒK; f :

It will be important to recall the fact that every Borel measure that is absolutely continuous vanishes on the boundaries of spherically convex subsets of the sphere.

Schneider’s book [49] is our standard reference for the basics regarding convex bodies. The books [20, 22] are also good references.

3 The Gauss Image Measure

LetKbe a convex body inRⁿ. For eachv 2 S^{n 1}, the hyperplane H_K.v/ D fx 2 RⁿW x v D h_K.v/g

is called the supporting hyperplane to K with unit normal v. For @K, the spherical image of is defined by

_K./ D fv 2 S^{n 1}W x 2 H_K.v/for somex 2 g S^{n 1}:

For S^{n 1}, thereverse spherical image ofis defined by

x_K./ D fx 2 @K W x 2 H_K.v/for somev 2 g @K:

Let_K @K be the set consisting of all x 2 @K for which the set_K.fxg/, abbreviated as_K.x/, contains more than a single element. The points in@K n _K are calledregularpoints of@K. It is well-known (Schneider [49, p. 84]) that the .n 1/-dimensional Hausdorff measure of the set of singular (i.e., nonregular) points of a convex body is0; i.e.,H^{n 1}.K/ D 0. The function

_KW @K n _K ! S^{n 1};

defined by letting_K.x/be the unique element in_K.x/for eachx 2 @K n _K, is called thespherical image map(also known as the Gauss map) ofKand is known to be continuous (see lemma 2.2.12 of Schneider [49]).

The set_K S^{n 1}consisting of allv 2 S^{n 1}for which the setx_K.v/contains more than a single element is ofH^{n 1}-measure0(see theorem 2.2.11 of Schneider [49]). The function

x_KW S^{n 1}n _K ! @K;

defined, for eachv 2 S^{n 1}n_K, by lettingx_K.v/be the unique element inx_K.v/, is called thereverse spherical image map. The vectors inS^{n 1}n _K are called the regular normal vectorsofK. Thus,v 2 S^{n 1} is a regular normal vector ofK if and only if@K \ H_K.v/consists of a single point. The functionx_Kis well-known to be continuous (see lemma 2.2.12 of Schneider [49]).

ForK 2K_oⁿ, define theradial mapofK,

rKW S^{n 1} ! @K by rK.u/ D K.u/u 2 @K

foru 2 S^{n 1}. Note that the mappingr_K¹W @K ! S^{n 1} is just the restriction of the mapxW Rⁿn f0g ! S^{n 1}to the set@K. The radial map is bi-Lipschitz.

For! S^{n 1}, define theradial Gauss image of!by K.!/ D K.rK.!// S^{n 1}: Thus, foru 2 S^{n 1},

_K.fug/ D fv 2 S^{n 1}W r_K.u/ 2 H_K.v/g:

We will need the fact that_Kmaps closed sets ofS^{n 1}into closed sets ofS^{n 1}. LEMMA3.1. If! S^{n 1}is closed, then_K.!/is also closed.

PROOF. Suppose the pointsvi 2 _K.!/are such thatvi ! v0. We will show thatv₀2 _K.!/. Nowv_i 2 _K.r_K.!//means thatv_i is a unit outer normal toK atrK.ui/for someui 2 !; i.e.,

(3.1) x v_i r_K.u_i/ v_i for allx 2 K:

Since ! S^{n 1} is compact, ui 2 ! has a convergent subsequence, which we will again denote byu_i, that is,u_i ! u₀ 2 !. Sincer_K is a continuous function,

r_K.u_i/ ! r_K.u₀/, and together withv_i ! v₀and (3.1) gives x v₀ r_K.u₀/ v₀ for allx 2 K:

Hence,v₀2 _K.r_K.!// D _K.!/.

Define theradial Gauss mapof the convex bodyK 2Kⁿ_{o K}W S^{n 1}n !_K ! S^{n 1} by_KD _K r_K;

where !_K D x_K D r_K¹._K/. Since r_K¹ D x is a bi-Lipschitz map between the spaces @K andS^{n 1}, it follows that !_K has spherical Lebesgue measure 0.

Observe that ifu 2 S^{n 1}n !_K, then _K.fug/contains only the element_K.u/. Note that since both_Kandr_Kare continuous,_Kis continuous.

From [27] Lemma 2.2, ifK0; K1; : : : 2K_oⁿ, then

(3.2) Ki ! K0 H) _Ki ! _K0;

almost everywhere, with respect to spherical Lebesgue measure.

For S^{n 1}, define thereverse radial Gauss imageofby (3.3) _K./ D r_K¹.x_K.// Dx_Kx./:

Thus,

(3.4) _K./ D fxx W x 2 @Kwherex 2 HK.v/for somev 2 g:

Define thereverse radial Gauss mapof the convex bodyK 2Kⁿ_o, _KW S^{n 1}n _K ! S^{n 1} by _K D r_K¹ x_K: Note that since bothr_K¹andx_Kare continuous,_K is continuous.

If S^{n 1}is a Borel set, then_K./ DxKx./ S^{n 1}is spherical Lebesgue measurable. This fact is lemma 2.2.14 of Schneider [49]; an alternate proof was given in [27]. It was shown in [27] that ifv … _Kand! S^{n 1}, then

(3.5) v 2 K.!/ if and only if _K.v/ 2 !:

Hence (3.5) holds for almost all v 2 S^{n 1}, with respect to spherical Lebesgue measure. It was also shown in [27] that ifK 2 Kⁿ_o, then the reverse radial Gauss image ofKand the radial Gauss image of the polar body,K, are identical; i.e.,

(3.6) _K./ D _K./;

for each S^{n 1}. It follows that for K 2 Kⁿ_o, the set _K./ is spherical Lebesgue measurable whenever S^{n 1} is a Borel set. SinceK D K, this shows that_K.!/is spherically Lebesgue measurable whenever! S^{n 1} is a Borel set andK 2K_oⁿ. From (3.6) we also see that forK 2Kⁿ_o,

(3.7) _K D _K

almost everywhere onS^{n 1}, with respect to spherical Lebesgue measure.

If K₀; K₁; : : : 2 Kⁿ_o are such that K_i ! K₀, then K_i ! K₀. This and (3.2) give us_Ki ! _K0 almost everywhere with respect to spherical Lebesgue measure. Now (3.7) allows us to conclude that

(3.8) K_i ! K₀ H) _Ki ! _K0;

almost everywhere, with respect to spherical Lebesgue measure.

ForK 2Kⁿ_o, Aleksandrov’sintegral curvature,C₀.K; /, is a Borel measure on S^{n 1}defined, for Borel! S^{n 1}, by

(3.9) C0.K; !/ DH^{n 1}._K.!//I

i.e., C₀.K; !/ is the spherical Lebesgue measure of_K.!/. The total measure C₀.K; S^{n 1}/of integral curvature of each convex bodyKisn!_n, the surface area of the unit sphereS^{n 1}inRⁿ.

Thesolid-angle measure Cz₀.K; /, also known as the 0^th dual curvature mea- sure, introduced in [27], can be defined by

(3.10) n zC₀.K; / DH^{n 1}._K.//

for each Borel S^{n 1}. From (3.9), (3.10), and (3.6), we have C₀.K; / D n zC₀.K; /:

The .n 1/^th area measure S_{n 1}.K; / is the classical surface area measure S.K; /, which is defined, for each Borel S^{n 1}, by

(3.11) S_{n 1}.K; / DH^{n 1}.x_K.//:

Federer’s.n 1/^thcurvature measureCn 1.K; /onS^{n 1}can be defined, for each Borel! S^{n 1}, by

(3.12) Cn 1.K; !/ DH^{n 1}.r_K.!//:

From (3.11) and (3.12), and the definition (3.3) that_K D r_K¹ x_K, we see that the.n 1/^thcurvature measureC_{n 1}.K; /onS^{n 1}and the.n 1/^tharea measure Sn 1.K; /onS^{n 1}are related by

(3.13) C_{n 1}.K; _K.// D S_{n 1}.K; / for each Borel S^{n 1}. See Schneider [49, theorem 4.2.3].

The following lemma establishes a fundamental property of the radial Gauss image.

LEMMA3.2. LetK 2Kⁿ_o. If! S^{n 1}is a spherically convex set, then

(3.14) _K.!/ S^{n 1}n !;

and furthermore the set.S^{n 1}n !/ n _K.!/has interior points.

PROOF. Consider an arbitrary u 2 ! and an arbitraryv 2 _K.u/; i.e.,v is an outer unit normal ofK atr_K.u/. From the definition of the support function, we see that

₀ h_K.v/ D _K.u/u v ₁u v;

which implies

(3.15) u v 0=1;

where₀is the minimum of_KonS^{n 1}and₁is the maximum of_K onS^{n 1}. The definition of!and the fact thatu 2 !now give us thatv … !, which yields (3.14).

Now (3.14) is just _K.!/ \ ! D ¿. When! is spherically convex, ! is nonempty. However, (3.15) implies that if we choose02 .0; 0=1/, then the set

!⁰₀ D \

u2!

fv 2 S^{n 1}W v u < 0g n !

is disjoint from_K.!/. Note that!⁰₀ has nonempty interior. Therefore, the set .S^{n 1}n !/ n _K.!/

has interior points.

Aspherical submeasureWB! Œ0; 1/, defined on a-algebraBof subsets of S^{n 1}, is a function that satisfies the following:

(1) .¿/ D 0.

(2) IfA; B 2Bare such thatA B, then.A/ .B/.

(3) IfA1; A2; : : : 2B, then.S₁

1 Ai/ P₁

1 .Ai/.

Our interest will be limited to spherical Lebesgue submeasures and spherical Borel submeasures, whereB is the collection of spherical Lebesgue measurable subsets ofS^{n 1}and spherical Borel subsets ofS^{n 1}, respectively.

Supposeis a spherical Lebesgue submeasure andK 2Kⁿ_o. TheGauss image measure.K; /ofviaKis the spherical Borel submeasure defined by

(3.16) .K; !/ D .K.!//

for each Borel set! S^{n 1}. To see that.K; /is indeed a submeasure, we recall the basic properties of the Gauss image_Kof a bodyK 2Kⁿ_o:

(1) _K.¿/ D ¿.

(2) If!; !⁰ S^{n 1}are such that! !⁰, then_K.!/ _K.!⁰/. (3) If!₁; !₂; : : : S^{n 1}, then_K.S₁

1 !_i/ DS₁

1 _K.!_i/.

(4) If !₁; !₂; : : : S^{n 1} are pairwise disjoint, then up to a set of spherical Lebesgue measure0, the sets_K.!₁/; _K.!₂/; : : : are pairwise disjoint as well.

Properties (1) and (2) are completely trivial, while Property (3) follows directly from the trivial lemma 2.3 in [27] together with (3.6). Property (4) is lemma 2.4

in [27]. Thereverse Gauss image measure.K; /ofviaK is the Borel sub- measure onS^{n 1}defined by

(3.17) .K; !/ D ._K.!// D ._K.!//

for each Borel set! S^{n 1}. Note that the second identity in (3.17) is from (3.6).

Since fora > 0obviously_aK D _K and_aK D _K, it follows, from their definitions, that

.aK; / D .K; / and .aK; / D .K; /

for alla > 0; i.e., the Gauss image measure and the reverse Gauss image measure of a convex body are invariant under dilations of the convex body. From (3.16), (3.17), and (3.6), we immediately obtain

(3.18) .K; / D .K; /:

When is spherical Lebesgue measure H^{n 1}

S^{n 1}, it follows from (3.9) and (3.10) that the Gauss image measure.K; /is integral curvature and the reverse Gauss image measure.K; /isntimes the solid-angle measure, i.e.,

DH^{n 1}j_Sn 1 H) .K; / D C₀.K; / and .K; / D n zC₀.K; /:

Ifis the curvature measureCn 1.K; /of a convex bodyK, then, by (3.13), the reverse Gauss image measure.K; /is the surface measureS_{n 1}.K; /, i.e.,

D C_{n 1}.K; / H) .K; / D S_{n 1}.K; /:

Whenis an absolutely continuous Borel measure, the Gauss image measure is a Borel measure, for which we have the following integral representation.

LEMMA3.3. Ifis an absolutely continuous Borel measure andK 2Kⁿ_o, then (3.19)

Z

S^{n 1}f .u/d.K; u/ D Z

S^{n 1}f ._K.v//d.v/

for each bounded Borelf W S^{n 1}! R.

PROOF. Let be a simple function onS^{n 1}given by DX

i

c_i1_!i

wherec_i 2 R, where!_i S^{n 1} are Borel sets, and where 1_!i is the indicator function of!_i. Since_Khas spherical Lebesgue measure0, we can conclude from (3.5) that

(3.20) 1_K.!i/.v/ D 1_!i._K.v//;

for almost allv 2 S^{n 1}, with respect to spherical Lebesgue measure. Sinceis absolutely continuous, (3.20) gives

(3.21)

Z

S^{n 1}1_K.!i/.v/ d.v/ D Z

S^{n 1}1_!i._K.v// d.v/:

We now use (3.16) and (3.21), and get Z

S^{n 1}.u/d.K; u/ D Z

S^{n 1}

X

i

c_i1_!i.u/d.K; u/

DX

i

ci.K; !i/ DX

i

c_i._K.!_i//

D Z

S^{n 1}

X

i

c_i1_K.!i/.v/d.v/

D Z

S^{n 1}

X

i

c_i1_!i._K.v//d.v/

D Z

S^{n 1}._K.v//d.v/:

This establishes (3.19) for simple functions. Given a bounded Borel f, we now choose a sequence of simple functions _k ! f uniformly. Then _{k K} converges tof _K a.e. with respect to spherical Lebesgue measure, and thus a.e.

with respect to. Sincef is a Borel function onS^{n 1}and the inverse radial Gauss map_K is continuous onS^{n 1}n K, the composite functionf _K is a Borel function onS^{n 1}n _K. Since_k ! f uniformly andf is bounded, the functions _kare uniformly bounded. Note that bothand.K; /are finite measures. By the dominated convergence theorem, we take the limitk ! 1to establish (3.19).

When the measure is an absolutely continuous Borel measure, we can (and will) speak of its Gauss image measure (as opposed to submeasure). The Gauss image measure as a functional from the spaceKⁿ_o to the space of Borel measures onS^{n 1}is weakly convergent with respect to the Hausdorff metric.

LEMMA 3.4. If is an absolutely continuous Borel measure on S^{n 1} and the bodiesK₀; K₁; : : : 2Kⁿ_oare such thatK_i ! K₀, then.K_i; / ! .K; /weakly.

PROOF. SinceK_i ! K₀, from (3.8) we see that_Ki ! _K0almost everywhere with respect to spherical Lebesgue measure. Then for each continuous functionf onS^{n 1}, we havef _Ki ! f _K0almost everywhere with respect to spherical Lebesgue measure, and thus almost everywhere with respect to. Sincejf _Kij is obviously bounded by max_v2Sn 1jf .v/j, we have

Z

S^{n 1}f ._Ki.v//d.v/ ! Z

S^{n 1}f ._K0.v//d.v/:

This and Lemma 3.3 show that Z

S^{n 1}f .u/d.K_i; u/ ! Z

S^{n 1}f .u/d.K₀; u/

for each continuousf W S^{n 1}! R. Thus,.K_i; / ! .K₀; /weakly.

LEMMA 3.5. If is an absolutely continuous Borel measure onS^{n 1}, then for eachK 2 Kⁿ_o, the Gauss image measure .K; / is absolutely continuous with respect to the surface area measureS.K; /of the polar bodyKofK.

PROOF. Since the polar of the polar is the original body, from (3.18) we see that all we need show is that the reverse Gauss image measure.K; /is absolutely continuous with respect to the surface area measureS.K; /ofK.

Suppose S^{n 1} is such that S.K; / D 0. Then from the definition of S.K; /we know thatH^{n 1}.x_K.// D 0. But since the map xW @K ! S^{n 1} is bi-Lipschitz, we haveH^{n 1}. xx_K.// D 0. This, in turn, can be rewritten using the definition (3.3) of_K, as

H^{n 1}._K.// D 0:

This, (3.17), and the fact thatis absolutely continuous imply

.K; / D ._K.// D 0:

Takingto be spherical Lebesgue measure in Lemma 3.5 and using definition (3.9) give the following:

COROLLARY3.6. The integral curvatureC₀.K; /ofK is absolutely continuous with respect to the surface area measureS.K; /of the polar bodyKofK.

The following lemma shows that an absolutely continuous Borel measurethat is positive on nonempty open subsets ofS^{n 1}and its Gauss image measure.K; / are always Aleksandrov related. As will be seen, this turns out to be a critical property.

LEMMA3.7. Supposeis an absolutely continuous Borel measure that is strictly positive on nonempty open subsets ofS^{n 1}. If K 2 Kⁿ_o, then the Gauss image measure.K; /satisfies

(3.22) .K; !/ < .S^{n 1}n !/

for each spherically convex set! S^{n 1}.

PROOF. Lemma 3.2 tells us that _K.!/ S^{n 1} n ! for each convex set

! S^{n 1}, and that.S^{n 1}n !/ n _K.!/has interior points. Thus, ._K.!// .S^{n 1}n !/;

and sinceis strictly positive on open sets, we also know that ..S^{n 1}n !/ n _K.!// > 0:

Thus,

._K.!// < .S^{n 1}n !/:

This and (3.16), the definition of the Gauss image measure,.K; /, immediately

yield (3.22).

The following lemma establishes uniqueness, up to dilation, for the Gauss image measure. The proof below is in the spirit of Aleksandrov’s proof for the case of integral curvature.

We shall use the fact that if the convex bodiesK andL have parallel support hyperplanes at the pointsrK.u/andrL.u/whenever both points are regular, then KandLare dilates (of one another).

LEMMA3.8. Supposeis an absolutely continuous Borel measure onS^{n 1}that is strictly positive on open sets. IfK; L 2 Kⁿ_o are such that.K; / D .L; /, thenKandLare dilates(of one another).

PROOF. We will show thatKandLhave parallel support hyperplanes at points r_K.u/andr_L.u/that are regular. Assume that there exists a u₀ 2 S^{n 1} so that rK.u0/andrL.u0/are regular and the support hyperplane ofKatrK.u0/and the support hyperplane ofLatr_L.u₀/are not parallel; i.e., _K.u₀/ ¤ _L.u₀/. Let c > 0be such thatcr_K.u₀/ D r_L.u₀/, and letK⁰D cK. Define the regular point x₀D r_K⁰.u₀/ D r_L.u₀/.

Define the disjoint decompositionS^{n 1}D !⁰[ ! [ !₀by letting

!⁰D fu 2 S^{n 1}W _K⁰.u/ > _L.u/g;

! D fu 2 S^{n 1}W _K⁰.u/ < _L.u/g;

!0 D fu 2 S^{n 1}W _K⁰.u/ D _L.u/g:

Supposeu 2 !⁰andLis a support hyperplane ofLatrL.u/. Obviously,rK⁰.!⁰/ is not completely contained in the half-space containing L that is generated by _L. Thus, there is a support hyperplane_K⁰ ofK⁰at some point ofr_K⁰.!⁰/that is parallel to_L. This implies that

(3.23) _L.!⁰/ _K⁰.!⁰/ D _K.!⁰/;

from which follows

(3.24) .L; !⁰/ .K; !⁰/:

To obtain the contradiction, we shall show that the inequality (3.24) is strict.

The continuity of the radial function and the definitions of!and!⁰show that the sets! [ !₀and!⁰[ !₀are closed, and thus by Lemma 3.1 the Gauss images K⁰.! [ !0/andL.!⁰[ !0/are closed as well. ThusS^{n 1}n K⁰.! [ !0/and S^{n 1}n _L.!⁰[ !₀/are open. Observe that, from the definitions of!,!₀, and!⁰ and the definition of the Gauss image, we have

(3.25) S^{n 1}n _K⁰.! [ !₀/ _K⁰.!⁰/ and

(3.26) .S^{n 1}n L.!⁰[ !0// \ L.!⁰/ D ¿:

Let

D .S^{n 1}n _K⁰.! [ !₀// \ .S^{n 1}n _L.!⁰[ !₀//:

Thenis an open set, and from (3.26) and (3.25) we obviously have (3.27) \ _L.!⁰/ D ¿ and _K⁰.!⁰/:

Let₀⁰ be the support hyperplane ofK⁰at the regular pointx₀D r_K⁰.u₀/ 2 @K⁰ with outer unit normal_K.u₀/, and let₀ be the support hyperplane ofL at the regular pointx₀ D r_L.u₀/ 2 @Lwith outer unit normal_L.u₀/. Recall that we assumed that the pointu0 is such that 0 ¤ ₀⁰. Note that _K.u0/ and_L.u0/ cannot be opposite of each other, since both K and L contain the origin in the interior.

Consider the hyperplaneP that is orthogonal to

v₁D ._K.u₀/ C _L.u₀//=j_K.u₀/ C _L.u₀/j

and passes through the pointx0. Note thatv1 K.u0/ > 0andv1 L.u0/ > 0. LetP_Cbe the half-space defined by

P_C D fx 2 RⁿW x v₁> x₀ v₁g:

Sincex0is a regular point for bothK⁰andL, the intersectionsPC\K⁰andPC\L must be nonempty.

Observe that ifr_K⁰.u/ 2 H_K⁰.v₁/thenu 2 !⁰. To see this, note that

(3.28) r_K⁰.u/ D x⁰C cv1

for somex⁰2 P andc > 0. By definition of support function, x_{0 K}⁰.u₀/ D h_K⁰._K⁰.u₀// r_K⁰.u/ _K⁰.u₀/

D x⁰ K⁰.u0/ C cv1 K⁰.u0/:

Sincev_{1 K}⁰.u₀/ > 0, we havex⁰ _K⁰.u₀/ < x_{0 K}⁰.u₀/. This, combined with the fact thatx⁰ v₁ D x₀ v₁(sincex⁰; x₀ 2 P) and the definition ofv₁, implies (3.29) x⁰ _L.u0/ > x0 L.u0/:

By (3.28), (3.29), and the fact thatv_{1 L}.u₀/ > 0,

r_K⁰.u/ _L.u₀/ D x⁰ _L.u₀/ C cv_{1 L}.u₀/ > x_{0 L}.u₀/ D h_L._L.u₀//:

This implies thatrK⁰.u/ … L, which in turn givesK⁰.u/ > L.u/oru 2 !⁰. This implies thatv_{1… K}⁰.! [ !₀/.

The same argument givesv_{1… L}.!⁰[ !₀/. Hence,v₁ 2 . Therefore,is a nonempty open set. Sinceis by hypothesis positive on open sets,./ > 0.

From (3.23) and (3.27),

(3.30) L.!⁰/ D L.!⁰/ n K⁰.!⁰/ n : Thus, (3.30) and./ > 0, give

.L; !⁰/ D ._L.!⁰// ._K⁰.!⁰/ n /

< ._K⁰.!⁰/ n / C ./ D ._K⁰.!⁰// D .K; !⁰/;

which contradicts.L; / D .K; /.

It is easily seen that the integral curvature of a convex body is not concentrated in any closed hemisphere, and the total measure of the integral curvature of a convex body is the surface area of the unit sphere. Then it is natural to find a complete set of properties that characterize the integral curvature. The following result shows that, when is an absolutely continuous Borel measure, then the Gauss image measure as a functional from the spaceKⁿ_o of convex bodies to the space of Borel measures is a valuation. The theory of valuations has seen explosive growth in the last quarter century (see, e.g., [3–5, 15, 23, 24, 33–36, 50–53], and the references therein). It would be interesting to characterize this valuation.

PROPOSITION3.9. Ifis an absolutely continuous Borel measure onS^{n 1}, then the Gauss image measure ofis a valuation;i.e., forK; L 2K_oⁿ,

.K; / C .L; / D .K [ L; / C .K \ L; /;

wheneverK [ L 2K_oⁿ.

PROOF. Sincer_K andr_Lare bijections betweenS^{n 1} and@Kand@L, respec- tively, we have the following disjoint partition ofS^{n 1} D ₀[ _L[ _K, where

0D r_K¹.@K \ @L/ D r_L¹.@K \ @L/ D fu 2 S^{n 1}W K.u/ D L.u/g;

LD r_K¹.@K \intL/ D r_L¹..Rⁿn K/ \ @L/ D fu 2 S^{n 1}W K.u/ < L.u/g;

K D r_K¹.@K \ .Rⁿn L// D r_L¹.intK \ @L/ D fu 2 S^{n 1}W K.u/ > L.u/g:

SinceK [ Lis a convex body, we have, forH^{n 1}-almost allu 2 _K, _K.u/ D _K[L.u/ and _L.u/ D _K\L.u/I

forH^{n 1}-almost allu 2 _L,

_K.u/ D _K\L.u/ and _L.u/ D _K[L.u/I and forH^{n 1}-almost allu 2 0,

_K.u/ D _L.u/ D _K\L.u/ D _K[L.u/:

Sinceis absolutely continuous, for a Borel set! S^{n 1}, we have .K; ! \ _K/ D ._K.! \ _K// D ._K[L.! \ _K//

D .K [ L; ! \ _K/;

and also

.L; ! \ K/ D .K \ L; ! \ K/:

Adding the last two, we obtain

.K; ! \ K/ C .L; ! \ K/ D .K [ L; ! \ K/ C .K \ L; ! \ K/:

Similarly, we have

.K; ! \ _L/ C .L; ! \ _L/ D .K [ L; ! \ _L/ C .K \ L; ! \ _L/;

.K; ! \ 0/ C .L; ! \ 0/ D .K [ L; ! \ 0/ C .K \ L; ! \ 0/:

Summing up the last three gives the desired valuation property.

4 Variational Formulas for the Log-Volumes of Convex Bodies Letbe a Borel measure on S^{n 1}. The log-volume₀.K/ of a convex body K 2Kⁿ_o with respect tois defined by

(4.1) ₀.K/ Dexp

1 jj

Z

S^{n 1}log_K.v/d.v/

: We require the following lemma established in [27].

LEMMA4.1. Suppose S^{n 1}is a closed set that is not contained in any closed hemisphere ofS^{n 1}. Suppose₀W  ! .0; 1/andgW  ! Rare continuous. If htiis a logarithmic family of convex hulls of.0; g/, then

(4.2) lim

t!0

logh_hti.v/ logh_h0i.v/

t D g._h0i.v//

for allv 2 S^{n 1}n _h0i;i.e., for all regular normalsvofh₀i. Hence(4.2)holds a.e. with respect to spherical Lebesgue measure. Moreover, there exists0 > 0and M > 0so that

jlogh_hti.v/ logh_h0i.v/j M jtj for allv 2 S^{n 1}and allt 2 . 0; 0/.

We require the following lemma. When the measure is spherical Lebesgue mea- sure, it was established in [27].

LEMMA4.2. Supposeis an absolutely continuous Borel measure onS^{n 1};the bodyK 2 Kⁿ_o and f; gW S^{n 1} ! Rare continuous. If hK; giis a logarithmic family of convex hulls generated by.K; g/, then

(4.3) d

dt log0.hK; g; ti/

tD0D 1 jj

Z

S^{n 1}g.u/d.K; u/:

IfŒK; f is a logarithmic family of Wulff shapes formed by.K; f /, then

(4.4) d

dt log₀.ŒK; f; t/

tD0D 1 jj

Z

S^{n 1}f .v/d.K; v/:

PROOF. Write_t D _KCtgCo.t; /. Note thathK; g; ti D h_ti. In particular, ₀ D _Kandh₀i D K.

From Lemma 4.1, the dominated convergence theorem, and (3.19), we have d

dt log₀.hK; g; ti/

tD0D lim

t!0

1 jj

Z

S^{n 1}

logh_hti.v/ logh_h0i.v/

t d.v/

D 1

jj Z

S^{n 1}g._h0i.v//d.v/

D 1

jj Z

S^{n 1}g.u/d.K; u/: