Algebra univers. 49 (2003) 305–319 0002-5240/03/030305 – 15 DOI 10.1007/s00012-003-1719-2 c Birkh¨auser Verlag, Basel, 2003 Algebra Universalis

Support functions of general convex sets

Dug-Hwan Choi and Jonathan D. H. Smith

Dedicated to the Memory of Gian-Carlo Rota

1. Introduction

The concept of the support of anon-emptycompact convex was introduced by Minkowski at the end of the 19th century [3, pp. 106, 144, 231]. Since then it has played a vital rˆole in many of the applications of convexity, from optimization theory to the geometry of numbers. Support functions of non-empty compact convex subsets of a finite-dimensional Euclidean space Rd are character- ized as positively homogeneous convex real-valued functions on Rd,andnon-empty compact convex sets are determined uniquely by their support functions. Work of Romanowska and the second author [5], [7, §11] extended the concept of the support function from compact non-empty convex sets to general bounded non- empty convex sets, thus to convex sets that are not necessarily closed. The method was to find a suitable Dd,replacing the codomain R of Minkowski’s support functions, so that non-empty bounded convex subsets of Rd are determined d uniquely by their Dd-valued support functions defined on R .Conditions were also found, analogous to Minkowski’s positive homogeneity and convexity, characterizing d the support functions amongst all the Dd-valuedfunctions on R .Thesewerethe so-called G-conditions of [5, Definition 4.11], [7, Definition 11.2]. The purpose of the current paper is to produce support functions (Definition 7.1) specifying general convex subsets ofafinite-dimensional Euclidean space Rd, essentially removing the restriction to the bounded, non-empty case that was im- ± posed in [5]. A codomain Dd is constructed so that arbitrary convex subsets of d ± d R are determined uniquely by their Dd -valued support functions defined on R . ± d These support functions are characterized amongst all Dd -valuedfunctions on R by satisfaction of a set of conditions called the G±-conditions (Section 6 below), analogues in turn of the G-conditions of [5] and of Minkowski’s positive homogeneity and convexity conditions.

Presented by J. Kung. Received September 2, 1999; accepted in final form December 14, 2000. 2000 Mathematics Subject Classification: 52A20; 06F99. Key words and phrases: Support function, , entropic algebra, mode, modal, barycen- tric algebra, semilattice, lexicographic order, ordinal product, ordinal sum, Plonka sum.

305 306 D.-H. ChoiandJ.D.H. Smith Algebra univers.

The correspondences between convex sets and their support functions go fur- ther. Algebraic structure on R induces algebraic structure on the set of real-valued functions on Rd.Thesetofsupport functions of non-empty, compact convex sets should be closed under this algebraic structure, and should reflect comparable alge- braic structure on the set of non-empty compact convex subsets of Rd.Theusual linear algebraic structure on R is unsuitable here: for example, the negative of a is no longer convex.Thekeyalgebraic structure on R for use in the context of support functions comprises convex combinations forming a barycentric algebra (see [4]) and the maximum operation forming a join semilattice, the convex combinations distributing over the join so that the two structures combine to form a modal in the sense of [4]. The support functions then form a submodal of the induced modal structure on the full set of functions, and the modal structure on the support functions reflects exactly the modal structure on the compact convex sets given by convex combinations and convex hulls of unions. (See Section 3 below for an outline, and [4, 3.7] for details.) Given this algebraic approach to Minkowski’s support functions, and the fact that the set of all convex subsets of Rd forms a modal under convex combinations and convex hulls of unions, it is then shown that ± the codomain Dd carries a modal structure (Proposition 5.1) such that support functions form a submodal (Corollary 8.6) of the induced modal of functions from d ± R to Dd ,thissubmodal being isomorphic to the modal (3.3) of all convex subsets of Rd (Theorem 8.7).

2. Algebraic preliminaries

An algebra (A, Ω) of type τ :Ω→ N is entropic if each operation ω is a homo- morphism ωτ ω :(A , Ω) → (A, Ω) ; (a1,...,aωτ ) → a1 ···aωτ ω. (2.1) An algebra is idempotent if each singleton subset is a subalgebra. An algebra is a mode if it is idempotent and entropic. An algebra (A, +, Ω) is a modal if (A, +) is ajoinsemilattice, (A, Ω) is a mode, and for each operation ω in Ω, the distributive law a1 ···(ai + ai) ···aωτ ω = a1 ···ai ···aωτ ω + a1 ···ai ···aωτ ω (2.2) holds for 1 ≤ i ≤ ωτ. For a real number r,define a binary operation r on R by r : R2 → R ;(x, y) → x(1 − r)+yr. (2.3) Let I◦ =[0, 1]◦ denote the interior of the unit interval I.Let D =(R, +,I◦)(2.4) Vol. 49, 2003 Support functions of general convex sets 307 denote the algebra structure on R given by the join semilattice operation x + y = max{x, y} and the binary operations p of (2.3) for p in I◦.ThenD is a modal. Now let R±∞ = R ∪{−∞, ∞} be the ordered set of (doubly) extended reals.Foreach element r of the set R+ of strictly positive reals, a binary operation r will be defined on R±∞ to yield an algebra (R±∞, R+). Define xy1 = y.Forreal arguments x and y,define xyr using (2.3), so that (R, R+)isasubalgebra of (R±∞, R+). For {x, y} ⊆ R and p ∈ I◦,define xyp = if −∞ ∈ x, y then −∞ else ∞ (2.5) For {x, y} ⊆ R and q ∈ (1, ∞), define xyq = if x<∞ and y = ∞ then ∞ else −∞ (2.6) Thus {−∞} is also a subalgebra of (R±∞, R+), and a sink of (R±∞,I◦). Bearing in mind that (R, R+)isasubalgebra of (R±∞, R+), one may summarize the algebraic structure on the doubly extended reals by the tables I◦ −∞ R ∞ −∞ −∞ −∞ −∞ R −∞ R ∞ (2.7) ∞ −∞ ∞ ∞ and (1, ∞) −∞ R ∞ −∞ −∞ −∞ ∞ R −∞ R ∞ (2.8) ∞ −∞ −∞ −∞.

3. Extended real valued support functions

For a finite dimension d,letRd denote the Rd equipped with the Euclidean inner product Rd × Rd → R ;(x, y) → (x|y). (3.1) The vector space Rd also supports a mode structure (Rd, R+)oftypeR+ ×{2} given as the d-th power of the mode (R, R+)using(2.3). [The d-th power semilattice structure on Rd will not be needed, so that + will denote addition on Euclidean spaces Rd,including R1,while (R, +) is the semilattice of (2.4).] Note that the convex subsets of Rd are precisely thesubalgebras of (Rd,I◦). Let RdT denote the set of all convex subsets of Rd.Thissetbecomes a mode (RdT,I◦) under the complex products XYp = {xyp | x ∈ X, y ∈ Y }. (3.2) 308 D.-H. ChoiandJ.D.H. Smith Algebra univers.

Furthermore, (RdT,+) is a join semilattice under the containment relation ⊆;for convex subsets X and Y of Rd,thejoinX + Y is the of X ∪ Y .One thus obtains a modal structure (RdT,+,I◦)(3.3) on the set of convex subsets of Rd [7, §7].

Definition 3.1. Consider the function H : RdT × Rd → R±∞ ;(X, y) → sup{(x|y) | x ∈ X}. (3.4) For a fixed convex subset X of Rd,the(extended real valued) support function of X is the function d ±∞ HX : R → R ; y → H(X, y). (3.5) For an element y of Rd,thesupporting hyperplane of X in the y-direction is    d  z ∈ R (x|y)=HX (y) . (3.6)

Note that (3.6) will be empty if HX (y)isnotfinite. Also, note

HX (0) = if X = ∅ then −∞ else 0(3.7) Thus the supporting hyperplane of ∅ in any direction is empty. For a non-empty convex subset of Rd,thesupporting hyperplane in the 0-direction is Rd.

Proposition 3.2. For a convex subset X of Rd,theclosure X of X is given by     d  X = z ∈ R (z|y) ≤ HX (y) . (3.8) y∈Rd

d Proof. For any element y of R ,onehas H∅(y)=−∞,sothat(3.8) holds for empty X.Nowsuppose that X is non-empty. For each element (x, y)ofX × Rd, one has (x|y) ≤ sup{(z|y) | z ∈ X} = HX (y). Thus X is contained in the closed set on the right hand side of (3.8). Conversely, suppose that t does not lie in X.Then 0doesnotlie in the translated closed convex set X − t.Itfollows that there is a vector y with 0 >HX−t(y)=sup{(x − t|y) | x ∈ X} =sup{(x|y) | x ∈ X}−(t|y), whence (t|y) > sup{(x|y) | x ∈ X}≥sup{(x|y) | x ∈ X} = HX (y)andt does not lie in the right hand side of (3.8). 

Afunction f : Rd → R±∞ is said to be positively homogeneous if f(0) ∈{0, −∞} and ∀r ∈ R+, ∀y ∈ Rd,f(0yr)=f(0)f(y)r. (3.9)

d Proposition 3.3. For eachconvex subset X of R ,thesupport function HX is positively homogeneous. Vol. 49, 2003 Support functions of general convex sets 309

Proof. In view of (3.7), it suffices to verify

HX (yr)=HX (0yr)=HX (0)HX (y)r (3.10) for all y in Rd and r in R+.IfX is empty, then all the terms of (3.10) are −∞, ±∞ + since {−∞} is a subalgebra of (R , R ). Otherwise, suppose that HX (y)=∞, so that HX (yr)=∞ also. Since X is non-empty, HX (0) = 0, and the right hand side of (3.10) reduces to 0∞r = ∞.Finally, if HX (y)isfinite, then verification of (3.10) amounts to straightforward linear algebra.  Afunction f :(A, Ω) → (B,Ω) between algebras of type τ :Ω→ N,withordered codomain, is said to be convex if its epigraph epi f = {(a, b) ∈ A × B | f(a) ≤ b} (3.11) is a subalgebra of (A×B,Ω) [4,(316)] [7, (5.3)]. For a function f :(Rd,I◦) → (D, I◦) with D as in (2.4), note that this definition of convexity agrees with the usual one given by “Jensen’s Inequality”. Proposition 3.4. For eachconvex subset X of Rd,theextended real valued support d ◦ ±∞ ◦ function HX :(R ,I ) → (R ,I ) is convex. Proof. Note that d ±∞ epi HX = {(y,s) ∈ R × R |∀x ∈ X, (x|y) ≤ s}. (3.12) Suppose that (x|y) ≤ s and (x|z) ≤ t for all x in X.Then∀x ∈ X, (x|yzp)= (x|y)(x|z)p ≤ stp,the latterinequality holding by the Monotonicity Lemma [4, 315] [7, 5.3] in the modal D of (2.4). 

4. Directed derivatives of convex functions

d ±∞ By Propositions 3.3 and 3.4, HX : R → R ,theextended real valued support function of a convex subset X of Rd,ispositively homogeneous and convex. In this section, directed derivatives of convex functions are defined, and are shown to preserve positive homogeneity and convexity of their primitives. Lemma 4.1. Let f : Rd → R±∞ be convex. Then f −1{−∞} is either ∅ or Rd. Proof. Suppose f(x)=−∞ for some x in Rd.Bytheconvexity of f,onehas f(0) = f(x(−x)1/2) ≤ f(x)f(−x)1/2 = −∞,whence f(0) = −∞.Foreach element y of Rd,onethen has f(y)=f(0(2y)1/2) ≤ f(0)f(2y)1/2 = −∞,whence f(y)=−∞.  Proposition 4.2. Let f : Rd → R±∞ be convex. Then for all x, y in Rd,thelimit lim f(x)f(xy1/r)r exists. r→∞ 310 D.-H. ChoiandJ.D.H. Smith Algebra univers.

Proof. If f(x)=−∞,thenbyLemma 4.1 f(xy 1/r)=−∞ for all r in R+,whence lim f(x)f(xy 1/r)r = −∞.Iff(x)=∞,thenf(x)f(xy 1/r)r = −∞ for all r>1 r→∞ by the bottom line of (2.8), whence lim f(x)f(xy 1/r)r = −∞.Otherwise,f(x) r→∞ is finite. It will then be shown that 1

Definition 4.3. An element x of Rd is a singular point of a convex function f : Rd → R±∞ if there are elements y and z of Rd with lim f(x)f(xy 1/r)r = ∞ r→∞ and lim f(x)f(xz 1/r)r = −∞.Thesetofsingular points of f is denoted by r→∞ M(f).

Definition 4.4. Let f : Rd → R±∞ be convex. For elements x and y of Rd,the directed derivative of f at x towards y is f (x; y)= if x ∈ M(f) then −∞ else lim f(x)f(xy 1/r)r. (4.2) r→∞ Cf. [1, 13], [5, (3,8)], [7, (11.4)].

Example. Let X be the epigraph of the exponential function exp: R → (R, ≤). Consider x =(0, −1),y =(−1, 0), and z =(1, 0). Note HX (x)=0.Then lim HX (x)HX (xy 1/r)r = ∞,while lim HX (x)HX (xz 1/r)r = −∞.Thusx r→∞ r→∞ ±∞ is a singular point of HX : R → R .

Proposition 4.5. Let f : Rd → R±∞ be convex. Then ∀x, y ∈ Rd,f(x; y) ≤ f(y). (4.3)

Proof. Note that (4.3) is immediate if f(x) ∈{±∞},iff(y) ∈{±∞},orifx is singular. Otherwise, it follows from f(x)f(xy 1/r)r ≤ f(x)f(x)f(y)1/r r = f(y) for r>1. 

Proposition 4.6. Let f : Rd → R±∞ be convex. Then for each x in Rd,the function Rd → R±∞ ; y → f (x; y) is convex. Vol. 49, 2003 Support functions of general convex sets 311

Proof. For y,z in Rd and p in I◦,theinequality

f (x; yzp) ≤ f (x; y)f (x; z)p (4.4) must be proved. It is trivial if x is singular, if ∞∈{f (x; y),f(x; z)}⊂R∞,orif f(x)isnotfinite. Otherwise, it follows from

f(x)f(xyzp 1/r)r = f(x)f(xxpyzp 1/r)r = f(x)f(xy 1/r xz 1/r p)r ≤ f(x)f(xy 1/r)r f(x)f(xy 1/r)r p , in which idempotence is used for the first equality, the entropic law for the second, and the Monotonicity Lemma is used for the inequality. 

Lemma 4.7. Let f : Rd → R±∞ be positively homogeneous and convex. Then for each x in Rd,onehas f (x;0)∈{0, −∞}.

Proof. The result is clear if x is singular or if f(x)isnotfinite. Otherwise, one has

f (x;0)= lim f(x)f(x01/r)r = lim f(x)f(x[1 − 1 ])r r→∞ x→∞ r = lim {f(x)[1 − r]+f(x)[1 − 1 ]r} =0.  x→∞ r

Theorem 4.8. Let f : Rd → R±∞ be positively homogeneous and convex. Then for each x in Rd,thefunction

Rd → R±∞ ; y → f (x; y)(4.5) is also positively homogeneous and convex.

Proof. In view of Proposition 4.6 and Lemma 4.7, it remains to be shown that

f (x;0yq)=f (x;0)f (x; y)q (4.6) for y in Rd and q>0. By Proposition 4.6 and Lemma 4.1, the equality (4.6) is immediate if f (x;0)=−∞.Otherwise,Lemma 4.7 shows that (4.6) reduces to

f (x;0yq)=0f (x; y)q (4.7) with f(x) finite. By the positive homogeneity of f,onehas   f(x0yq 1/r)=0f xy(0q 1/r)/(1q 1/r) (1q 1/r). (4.8)

Case (a): f (x; y)=∞.Inthiscasef(xy 1/r)=∞ for r>1. By (4.8), one then has f(x0yq 1/r)=∞ for r>1. Thus both sides of (4.7) are ∞. 312 D.-H. ChoiandJ.D.H. Smith Algebra univers.

Case (b): f (x; y) finite. In this case (4.8) yields f (x;0yq)= lim f(x)f(x0yq 1/r)r r→∞   = lim f(x)0f xy (0q 1/r)/(1q 1/r) (1q 1/r) r r→∞     = lim q f(x)f xy(0q 1/r)/(1q 1/r) (1q 1/r)/(0q 1/r) r→∞ = q lim f(x)f(xy 1/s)s = qf(x; y), s→∞ verifying(4.7).  Proposition 4.9. Let f : Rd → R±∞ be convex and positively homogeneous, with f(0) = 0.Thenfor each y in Rd, f (0; y)=f(y). (4.9) Proof. If f(0) = −∞,onehas f(y)=−∞ by Lemma 4.1, whence f (0; y)=−∞ by Proposition4.5.Otherwise,fory in Rd,onehas lim 0f(y/r)r = lim f(y)=f(y). r→∞ r→∞ By Lemma 4.1, there is no y in Rd with f(y)=−∞.Thus0isnotasingular point of f,and(4.8)follows. 

5. for support functions

This section describes the construction, inductive over the dimension d,ofthe ± d codomain modal Dd for the support functions of convex subsets of R .Alongwith ± the series of codomain modals Dd for d ≥ 1, there will be three other series of ± ± ± inductively defined modals Ed (for d ≥ 0), Fd (for d ≥ 1), and Gd (for d ≥ 0). ± ◦ R0 ◦ The induction basis is a modal (G0 , +,I )identifiedwith the modal ( T,+,I ) ± {∅ { }} ∅ ≤ { } as in (3.3). Thus G0 = , 0 with join semilattice order + 0 and with {∅} ◦ ± ◦ as a sink for the I -reduct (G0 ,I ), in accordance with (3.2). The induction ± Rd−1 ◦ step begins with a modal Gd−1 identified with the modal ( T,+,I )asin(3.3). Note that the modal D of (2.4) is cancellative,inthesense that xyp = xzp ⇒ y = z (5.1) for p in I◦ and x, y, z in R.Moreover,(D, +) is a chain, and also Rd−1T has ∅ as a zero element. The ordinal product ± ◦ ± Ed = D Gd−1 (5.2) ◦ × ± ◦ is the Cartesian product (D, I ) (Gd−1,I )equippedwiththelexicographic order (d, g) ≤ (d ,g ) ⇔ d

± ◦ Using the modal (Ed , +,I )of(5.2), define a partial orderonthedisjoint union ± ± Dd = {(−∞, ∅)}∪ Ed ∪{(∞, ∅)} (5.4) as the ordinal sumoftheuniands in (5.4), i.e. with (−∞, ∅)

6. The G±-conditions

This section presents the so-called G±-conditions on a function d ± f : R → Dd ; x → (Hf (x),Cf (x)), 314 D.-H. ChoiandJ.D.H. Smith Algebra univers.

± ± qualifying it for membership in the subset Gd of Fd .Somedefinitions are required for the formulation of the G±-conditions. Definition 6.1. Fix a function H : Rd → R±∞. d d (a) Suppose H(x)isfinite for somenon-zero x in R .Define πx : {z ∈ R | (z|x)=H(x)}→Rd−1 to be the restriction of the projection from Rd along xR onto the orthogonal complement of xR. d d−1 (b) If x is zero or H(x)isnotfinite, define πx : R → R to be the zero map. Definition 6.2. Let H : Rd → R±∞ be positively homogeneous and convex. An element x of Rd is a linearity point of H if H(x)H(−x)1/2 =0. (6.1) Define L(H)={x ∈ Rd | H(x)H(−x)1/2 =0} as the set of linearity points of H. Proposition 6.3. Let H : Rd → R±∞ be positively homogeneous and convex. (a) H(0) = 0 ⇒ 0 ∈ L(H). (b) x ∈ L(H) ⇒ H(x) ∈ R and H(−x)=−H(x). (c) L(H) is a subspace of the vector space Rd. Proof. (a) is immediate by the idempotence of 1/2. (b): For x in L(H), one has 0 = H(0) = H(x(−x)1/2) ≤ H(x)H(−x)1/2 =0, { − }⊂R 1 1 − so that H(x),H( x) by (2.7) and then 0 = 2 H(x)+ 2 H( x). (c): Given (b), (c) follows by the traditional argument [1, §13].  d ± For a function f : R → Dd with positively homogeneous and convex extended d real function part Hf ,andforx in R ,define the supercrust d d Kf (x)={z ∈ R |∀y ∈ R , (z|y) ≤ Hf (x; y)}. (6.2) ◦ Define Kf (x) to be the relative interior of Kf (x). (Equivalently, one may define ◦ ◦ ◦ (Kf (x) ,I )asthesmallest non-empty sink of the barycentric algebra (Kf (x),I ) [4, 386].) ± Definition 6.4. The set Gd is the set of elements d ± f : R → Dd ; x → (Hf (x),Cf (x)) satisfying the following G±-conditions: ± d ±∞ (G C1) Hf : R → R is positvely homogeneous; ± d ±∞ (G C2) Hf : R → R is convex; ± d (G C3) ∀x ∈ R ,Cf (x) ⊆ πx(Kf (x)); ± ◦ (G C4) ∀x ∈ L(Hf ),Cf (x) ⊇ πx(Kf (x) ); (G±C5) For all nonzero x, y in Rd, −1 d −1 d πx (Cf (x)) ∩{z ∈ R | (z|y)=Hf (y)} = πy (Cf (y)) ∩{z ∈ R | (z|x)=Hf (x)}. Vol. 49, 2003 Support functions of general convex sets 315

7. Support functions of general convex sets

d d Let X be a convexsubset of R .Inthissection, the support function hX : R → ± ± Dd of X will be defined, and shown to be an element of Gd .Theextended real d ± function part of hX : R → Dd is just the extended real valued support function d ±∞ HX : R → R of Definition 3.1. By Propositions 3.3 and 3.4, this extended real function part of hX is positively homogeneous and convex, so that hX already satisfies the first two conditions of Definition 7.3. For a vector x in Rd,thecrust of X (in the x-direction)istheintersection of X with its supporting hyperplane in the x-direction (as in Definition 3.1). The crust shadow of X (in the x-direction)isdefined as the image   d CX (x)=πx X ∩{z ∈ R | (z|x)=HX (x)} (7.1) d ±∞ of the crust of X under the “projection” πx obtained from HX : R → R using Definition 7.1.

Definition 7.1. The support function of X is the function d ± hX : R → Dd ; x → (HX (x),CX (x)) (7.2) ± in Fd .Thesupercrusts KhX (x)ofhX are denoted KX (x). Proposition 7.2. For each vector x in Rd, d KX (x)=X ∩{z ∈ R | (z|x)=HX (x)}. (7.3)

Proof. The result is immediate if X = ∅.Otherwise,ifx ∈ M(HX )orHX (x)=∞, then the KX(x)isasubset of the right hand side of (7.3). Otherwise, one has HX (x; x)=HX (x)andHX (x; −x)=−HX (x), so that  d KX (x)= {z ∈ R | (z|y) ≤ HX (x; y)} y∈Rd     d d ⊆ {z ∈ R | (z|y) ≤ HX (y)} ∩ {z ∈ R | (z|y) ≤ HX (x; y)} y∈Rd y∈{±x} d = X ∩{z ∈ R | (z|y)=HX (x)} using Proposition 4.5 and Proposition 3.2. Conversely, it must be shown that KX(x)containstheright hand side of (7.3) for X = ∅.Ifthisrighthand side is empty, the containment is trivial. Otherwise, HX (x) ∈ R. Consider an element t of the right hand side. By Proposition 3.2, one has d (t|y)=(t|x)(t|xy 1/r)r ≤ HX (x)HX (xy 1/r)r for all y in R and r in (1, ∞). d In particular, there is no point z in R with lim HX (x)HX (xz 1/r)r = −∞,so r→∞ 316 D.-H. ChoiandJ.D.H. Smith Algebra univers.

that x is not a singular point of HX .By(4.1), one then has (t|y) ≤ HX (x; y)for d all y in R ,sothatt lies in the supercrust KX (x).  d ± d Theorem 7.3. The support function hX : R → Dd of a convex subset X of R satisfies the G±-conditions. Proof. The conditions (G±C1) and (G±C2) have already been verified. Moreover, (G±C3) is immediate from (7.1) and (7.3). To verify (G±C4), consider a linearity d point x of HX .By(3.8), one has X ⊆{z ∈ R | (z|x)=HX (x)}.By(7.3),

x ∈ L(HX ) ⇒ X = KX (x). (7.4) ◦ ◦ ◦ By [2, Theorem 3.4(d)], one has X =(X) .ThusCX (x)=πx(X) ⊇ πx(X )= ◦ ◦ ± ± πx((X) )=πx(KX (x) ), verifying (G C4). Finally, (G C5) holds, since each side of the equality reduces to d d X ∩{z ∈ R | (z|x)=HX (x)}∩{z ∈ R | (z|y)=HX (y)}. 

8. Characterization of support functions

By Theorem 7.3, there is a well-defined function d ± η : R T → Gd ; X → hX . (8.1) d ± ± For a function f : R → Dd ; x → (Hf (x),Cf (x)) in Gd ,define the convex subset d −1 Pf (x)={z ∈ R | (z|x)

PhX (x). By (6.2), (4.8), and (3.8), note KX(0) = X. d d ◦ Proposition 8.1. For X in R T and each x in R ,onehas KX (0) ⊆ PX (x). ◦ Proof. For empty X,onehas KX (0) = ∅ = PX (x)foreachx in X.Nowsuppose ◦ d that X is non-empty. Note KX (0) ⊆ R = PX (0), since CX (0) = {0}.For ± ◦ ◦ 0 = x ∈ L(HX), (7.4) and (G C4) yield πx(KX (0) )=πx(KX (x) ) ⊆ Cf (x), ◦ −1 whence KX (x) ⊆ πx (Cf (x)) by (8.2). If x is not a linearity point of HX ,then ◦ ◦ d KX(0) =(X) ⊆{z ∈ R | (z|x)

d Proposition 8.2. For X in R T and f = hX ,onehas X = Xf .

d Proof. For X = ∅,(7.1) and (8.2) yield Pf (x)=∅ for all x in R ,whence X = Xf . Now suppose that X is non-empty. By (3.8), (7.1), and (8.2), one certainly has X ⊆ Xf .Itremainstobe shown that Xf ⊆ X.By[2,Theorem3.1], [2, Theorem ◦ ◦ ◦ 3.4(d)], and Proposition 8.1, one has ∅ ⊂ X =(X) = KX(0) ⊆ Xf .Letu be an element of Xf .Itmustbeshownthatu ∈ X. d−1 By the positive homogeneity of HX ,onehas Pf (ax)=Pf (x)forx in S and d−1 d−1 −1 a>0. If ∃x ∈ S , (u|x) ≥ HX (x), then ∃x ∈ S ,u ∈ πx (CX (x)) ⊆ X, whence u ∈ X.Otherwise,

d−1 ∀x ∈ S , (u|x)

d−1 In this case L(HX )={0},fory ∈ L(HX ) ∩ S implies  d {z ∈ R | (z|x)

Define

d ±∞ g : R → R ; x → if HX (x)=∞ then ∞ else HX (x) − (u|x)(8.6) and V = {z ∈ Rd |∀x ∈ Sd−1, (z|x) ≤ 0g(x)1/2}. (8.7) Note that V is a closed, convex subset of Rd.By(8.5), one has 0 ∈ V .Now ◦ L(g) ⊆ L(HX )={0},sotherelative interior V of V coincides with its interior. ◦ By (8.5), onehasO ∈ V .Thusforsome positive r,thereisanopen ball Br(0) = d ◦ {z ∈ R | (z|z)

± d d ± Proposition 8.3. The maps ξ : Gd → R T and η : R T → Gd are mutually inverse.

Proof. By Proposition 8.2, ηξ =1RdT .Conversely, consider f : x → (Hf (x),Cf (x)) ± in Gd .IfHf (0) = −∞,eachsupercrust Kf (x)isempty,whence each crust shadow ± Cf (x)isemptyby(G C3). By (8.2) and (8.3), it follows that Xf = ∅,andthen fξη = h∅ = f,asrequired. The remainder of the proof that ξη =1G± follows the d lines of [5, Lemma 5.10].  318 D.-H. ChoiandJ.D.H. Smith Algebra univers.

Proposition 8.4. For 0 = x ∈ Rd,forp ∈ I◦,andforA, B ∈ RdT ,

A ∩{z | (z|x)=HA(x)} B ∩{z | (z|x)=HB(x)} p =

ABp ∩{z | (z|x)=HABp(x)}. (8.8)

Proof. If both A and B are non-empty, the proof proceeds as for [5, Lemma 5.12]. Otherwise, both sides of (8.8) are empty. 

Proposition 8.5. The map d ◦ ± ◦ ϕ:(R T,+,I ) → (Fd , +,I );X → hX (8.9) is a modal homomorphism.

Proof. For A, B in RdT and x in Rd,itmustbeshownthat

(HA+B(x),CA+B (x)) = (HA(x),CA(x)) + (HB(x),CB (x)) (8.10) and (HABp(x),CABp(x)) = (HA(x),CA(x))(HB (x),CB (x))p (8.11) for p in I◦.IfA and B are both non-empty, the proof proceeds as for [5, Lemma 5.13]. Otherwise, suppose without loss of generality that A is empty. Then the left hand side of (8.10) is (HB(x),CB (x)), while the right hand side is (−∞, ∅)+ (HB(x),CB (x)) = (HB(x),CB (x)). The left sideof(8.11) is (−∞, ∅), while the right hand side is (−∞, ∅)(HB(x),CB (x))p =(−∞, ∅), as required. 

± ± ◦ Corollary 8.6. The set Gd is a submodal of (Fd , +,I ). d ± ± d ± Proof. Note that η : R T → Gd is the corestriction to Gd of ϕ: R T → Fd .By ± Proposition 8.3, the set Gd is the image of the modal isomorphism ϕ. 

Combining Propositions 8.3 and 8.5 with Corollary 8.6, one achieves the com- pletion of the inductive constructions of Section 5.

d ◦ d ± ◦ Theorem 8.7. The modals (R T,+,I ) of convex subsets of R and (Gd , +,I ) of support functions are isomorphic.

References

[1] T. Bonnesen and W. Fenchel, Theorie der Konvexen K¨orper,Chelsea, New York 1948. [2] A. Brønsted, An Introduction to Convex Polytopes, Springer, New York 1983. [3] H. Minkowski, Gesammelte Abhandlungen 2. Band Teubner, Leipzig 1911. [4] A. B.Romanowska and J. D. H. Smith, Modal Theory – An Algebraic Approach to Order, Geometry and Convexity,Heldermann Verlag, Berlin 1985. [5] A. B.Romanowska and J. D. H. Smith, Support functions and ordinal products, Geom. Dedicata 30 (1989), 281–296. Vol. 49, 2003 Support functions of general convex sets 319

[6] E. Slatinsk´y, Die arithmetische Operation der Summe,ArchivumMath.20 (1984), 9–20. [7] J. D. H.Smith,Modes and modals,Discuss. Math. Algebra Stochastic Methods 19 (1999), 9–40.

Dug-Hwan Choi School of Electrical and Computer Engineering, Sungkyunkwan University, 300 ChunChun-Dong, JangAn-Gu, Suwon, Kyunggi-Do 440-746, Korea Jonathan D. H. Smith Iowa State University, Ames, Iowa 50011, USA

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