Convex Sets and Minimal Sublinear Functions Amitabh Basu¤ G¶erardCornu¶ejols y Giacomo Zambelliz March 2009, revised January 2010 Abstract We show that, given a closed convex set K containing the origin in its interior, the support function of the set fy 2 K¤ j 9x 2 K such that hx; yi = 1g is the pointwise smallest sublinear function σ such that K = fx j σ(x) · 1g. 1 Introduction The purpose of this note is to prove the following theorem. For K ⊆ Rn, we use the notation K¤ = fy 2 Rn j hx; yi · 1 for all x 2 Kg K^ = fy 2 K¤ j hx; yi = 1 for some x 2 Kg: The set K¤ is called the polar of K and it is convex. The set K^ is contained in the relative boundary of K¤ and it is not convex in general. The support function of a nonempty set T ½ Rn is de¯ned by n σT (x) = suphx; yi for all x 2 R : y2T It is straightforward to show that support functions are sublinear, that is they are convex and positively homogeneous (A function f : Rn ! R is positively homogeneous if f(tx) = n n tf(x) for every x 2 R and t > 0), and σT = σconv(T ) [3]. We will show that, if K ½ R is a closed convex set containing the origin in its interior, then K = fx j σK^ (x) · 1g. The next theorem shows that σK^ is the smallest sublinear function with this property. ¤Carnegie Mellon University, [email protected] yCarnegie Mellon University, [email protected]. Supported by NSF grant CMMI0653419, ONR grant N00014-03-1-0188 and ANR grant BLAN06-1-138894. zUniversit`adi Padova, [email protected] 1 Theorem 1 Let K ½ Rn be a closed convex set containing the origin in its interior. If σ : Rn ! R is a sublinear function such that K = fx 2 Rn j σ(x) · 1g, then n σK^ (x) · σ(x) for all x 2 R . In the remainder we de¯ne ½K = σK^ . Let K ½ Rn be a closed convex set containing the origin in its interior. A standard concept in convex analysis [3, 5] is that of gauge (sometimes called Minkowski function), which is the function γK de¯ned by ¡1 n γK (x) = infft > 0 j t x 2 Kg for all x 2 R : By de¯nition γK is nonnegative. One can readily verify that K = fx j γK (x) · 1g. It is ¤ well known that γK is the support function of K (see [3] Proposition 3.2.4). Given any sublinear function σ such that K = fx j σ(x) · 1g, it follows from positive homogeneity that σ(x) = γK (x) for every x where σ(x) > 0. Hence σ(x) · γK (x) for n all x 2 R . On the other hand we prove in Theorem 1 that the sublinear function ½K n satis¯es ½K (x) · σ(x) for all x 2 R . In words, γK is the largest sublinear function such n that K = fx 2 R j σ(x) · 1g and ½K is the smallest. Note that ½K can take negative values, so in general it is di®erent from the gauge γK . Indeed the recession cone of K, which is the set rec(K) = fx 2 K j tx 2 K for all t > 0g, coincides with fx 2 K j σ(x) · 0g for every sublinear function σ such that K = fx j σ(x) · 1g. In particular ½K (x) can be negative for x 2 rec(K). For example, 2 ¤ ^ let K = fx 2 R j x1 · 1; x2 · 1g. Then K = convf(0; 0); (1; 0); (0; 1)g and K = 2 convf(1; 0); (0; 1)g. Therefore, for every x 2 R , γK (x) = maxf0; x1; x2g and ½K (x) = maxfx1; x2g. In particular, ½K (x) < 0 for every x such that x1 < 0, x2 < 0. Theorem 1 has applications in integer programming. In particular it is used to establish the relationship between minimal inequalities and maximal lattice-free convex sets [1], [2]. By HÄormander'stheorem [4], a sublinear function σ : Rn ! R is the support function n of a unique bounded closed convex set C ½ R , say σ = σC . So the condition K = fx 2 Rn j σ(x) · 1g says K = C¤. Thus Theorem 1 can be restated in its set version. Theorem 2 Let K ½ Rn be a closed convex set containing the origin in its interior. If C ½ Rn is a bounded closed convex set such that K = C¤, then K^ ½ C. When K is bounded, this theorem is trivial (the hypothesis K = C¤ becomes K¤ = C. The conclusion K^ ½ C is obvious because one always has K^ ½ K¤.) So the interesting case of Theorem 1 is when K is unbounded. 2 2 Proof of Theorem 1 We will need Straszewicz's theorem [6] (see [5] Theorem 18.6). Given a closed convex set C, a point x 2 C is extreme if it cannot be written as a proper convex combination of two distinct points in C. A point x 2 C is exposed if there exists a supporting hyperplane H for C such that H \ C = fxg. Clearly exposed points are extreme. We will denote by ext(C) the set of extreme points and exp(C) the set of exposed points of C. Theorem 3 Given a closed convex set C, the set of exposed points of C is a dense subset of the set of extreme points of C. Let K be a closed convex set containing the origin in its interior. Let σ be a sublinear function such that K = fx j σ(x) · 1g. The boundary of K, denoted by bd(K), is the set fx 2 K j σ(x) = 1g. Lemma 4 For every x2 = rec(K), σ(x) = ½K (x) = supy2K¤ hx; yi. Proof. Let x2 = rec(K). Then t = σ(x) > 0. By positive homogeneity, σ(t¡1x) = 1, hence t¡1x 2 bd(K). Since K is closed and convex, there exists a supporting hyperplane for K containing t¡1x. Since 0 2 int(K), this implies that there existsy ¹ 2 K¤ such that ¡1 ^ (t x)¹y = 1. In particulary ¹ 2 K, hence by de¯nition ½K (x) ¸ hx; y¹i = t. Furthermore, for any y 2 K¤, ht¡1x; yi · 1, hence hx; yi · t, which implies t ¸ supy2K¤ hx; yi. Thus ½K (x) ¸ t ¸ sup hx; yi ¸ suphx; yi = ½K (x); y2K¤ y2K^ where the last inequality holds since K^ ½ K¤, hence equality holds throughout. ¤ Corollary 5 K = fx j ½K (x) · 1g. Lemma 6 Given an exposed point y¹ of K¤ di®erent from the origin, there exists x 2 K such that hx; y¹i = 1 and hx; yi < 1 for all y 2 K¤ distinct from y¹. Proof. Ify ¹ 6= 0 is an exposed point of K¤, then there exists a supporting hyperplane H = fy j ha; yi = ¯g such that ha; y¹i = ¯ and ha; yi < ¯ for every y 2 K¤ n fy¹g. Since 0 2 K¤ andy ¹ 6= 0, ¯ > 0. Thus the point x = ¯¡1a 2 K¤¤ = K satis¯es the statement. ¤ The next lemma states that K^ and K^ \ exp(K¤) have the same support function. n Lemma 7 For every x 2 R , ½K (x) = supy2K^ \exp(K¤)hx; yi. 3 ^ Proof. We ¯rst show that ½K (x) = supy2K^ \ext(K¤)hx; yi. Given y 2 K we show that there exists an extreme point y0 of K¤ in K^ such that hx; yi · hx; y0i. Since y 2 K^ , there existsx ¹ 2 K such that hx;¹ yi = 1. The point y is a convex combination of extreme ¤ i 1 k ^ points y1; : : : ; yk of K , and each yi satis¯es hx;¹ y i = 1. Thus y ; : : : ; y 2 K, and hx; yii ¸ hx; yi for at least one i. By Straszewicz's theorem (Theorem 3) the set of exposed points in K¤ is a dense subset of the extreme points of K¤. By Lemma 6, all exposed points of K¤ except ^ ¤ ^ ¤ ^ the origin are in K, hence exp(K ) \ K is dense in ext(K ) \ K. Therefore ½K (x) = supy2K^ \exp(K¤)hx; yi. ¤ n A function σ is subadditive if σ(x1+x2) · σ(x1)+σ(x2) for every x1; x2 2 R . It is easy to show that σ is sublinear if and only if it is subadditive and positively homogeneous. Proof of Theorem 1. By Lemma 4, we only need to show σ(x) ¸ ½K (x) for points x 2 rec(K). By Lemma 7 it is su±cient to show that, for every exposed pointy ¹ of K¤ contained in K^ , σ(x) ¸ hx; y¹i. Lety ¹ be an exposed point of K¤ in K^ . By Lemma 6 there existsx ¹ 2 K such that hx;¹ y¹i = 1 and hx;¹ yi < 1 for all y 2 K¤ distinct fromy ¹. Note thatx ¹ 2 bd(K). We observe that for all ± > 0,x ¹ ¡ ±¡1x2 = rec(K). Indeed, since x 2 rec(K), x¹ + ±¡1x 2 K. Hencex ¹ ¡ ±¡1x2 = int(K) becausex ¹ 2 bd(K). Since 0 2 int(K) and x¹ ¡ ±¡1x2 = int(K), thenx ¹ ¡ ±¡1x2 = rec(K). Thus by Lemma 4 σ(¹x ¡ ±¡1x) = sup hx¹ ¡ ±¡1x; yi: (1) y2K¤ Sincex ¹ 2 bd(K), σ(¹x) = 1. By subadditivity, 1 = σ(¹x) · σ(¹x ¡ ±¡1x) + σ(±¡1x): By positive homogeneity, the latter implies that σ(x) ¸ ± ¡ ±σ(¹x ¡ ±¡1x) for all ± > 0. By (1), σ(x) ¸ inf [±(1 ¡ hx;¹ yi + hx; yi]; y2K¤ hence σ(x) ¸ sup inf [±(1 ¡ hx;¹ yi + hx; yi]: ¤ ±>0 y2K ¤ Let g(±) = infy2K¤ ±(1 ¡ hx;¹ yi + hx; yi.