Chapter 1 Convexity: the Brunn–Minkowski theory 1.1 Basic facts on convex bodies n 2 2 1=2 We work in the Euclidean space (R ; j · j), where jxj = x1 + ··· + xn . We denote by h · ; · i the corresponding inner product. We say that a subset K ⊂ Rn is convex if for every x, y 2 K and λ 2 [0; 1], we have λx + (1 − λ)y 2 K. We say that K ⊂ Rn is a convex body if K is convex, compact, with non-empty interior. It is convenient to define a distance on the set of convex bodies in Rn. First, given n K ⊂ R and " > 0, we denote by K" the "-enlargement of K, defined as n K" = fx 2 R : 9y 2 K; jx − yj 6 "g: In other words, K" is the union of closed balls of radius " with centers in K. The Hausdorff distance between two non-empty compact subsets K, L ⊂ Rn is then defined as δ(K; L) = inff" > 0 : K ⊂ L" and L ⊂ K"g: We check (check!) that δ is a proper distance on the space of non-empty compact subsets of Rn. Some basic but important examples of convex bodies in Rn are n n 1. The unit Euclidean ball, defined as B2 = fx 2 R : jxj 6 1g. n n 2. The (hyper)cube B1 = [−1; 1] . n n 3. The (hyper)octahedron B1 = fx 2 R : jx1j + ··· + jxnj 6 1g. 1 n These examples are unit balls for the `p norm on R for p = 2; 1; 1. The `p norm k · kp n is defined for 1 6 p < 1 and x 2 R by n !1=p X p kxkp = jxij i=1 and for p = 1 by kxk1 = limp!1 kxkp = maxfjxij : 1 6 i 6 ng. More generally, the following proposition characterizes symmetric convex bodies as the unit balls for some norm. Proposition 1. Let K ⊂ Rn. The following are equivalent 1. K is a convex body which is symmetric (i.e. satisfies K = −K), 2. there is a norm on Rn for which K is the closed unit ball. To prove Proposition 1 (check!), we may recover the norm from K by the formula x kxk = infft > 0 : 2 Kg: K t A basic geometric fact about convex bodies is given by the Hahn–Banach separation theorem. We give two versions. Theorem 2. Let K, L be two convex bodies in Rn such that K \ L = ;. Then there exist u 2 Rn and α 2 R such that maxhx; ui < α < minhy; ui: x2K y2L Here is the geometric meaning of Theorem 2: the hyperplane H = fh · ; ui = αg sepa- rates K from L, in the sense that each convex body lies in a separate connected component of Rn n H, which is an open half-plane. Theorem 3. Let K be a convex body in Rn and x 2 @K. Then there exists u 2 Rn, u 6= 0, such that maxhy; ui = hx; ui: y2K The hyperplane H = fh · ; ui = hx; ui is said to be a support hyperplane for K at the boundary point x. One can give a geometric proof of Theorem 2 (check!) as follows: choose a couple of points (x; y) 2 K × L which minimizes jx − yj, and take as a separating hyperplane the set of points equidistant from x and y. We can then obtain Theorem 3 as n a corollary by separating K from fxkg, where (xk) is a sequence in R n K converging to x (check!). 2 1.2 The Brunn–Minkowski inequality Given sets K, L in Rn and a nonzero real number λ, we may define λK = fλx : x 2 Kg; K + L = fx + y : x 2 K; y 2 Lg; which we call the Minkowski sum of K and L. We denote by vol( · ) the Lebesgue measure, n or volume, defined on Borel subsets of R . We may write voln instead of vol if we want to precise the dimension. The volume is n-homogeneous, i.e. satisfies vol(λA) = jλjn vol(A), for λ 2 R. The behaviour of the volume with respect to Minkowski addition is governed by the Brunn–Minkowski inequality. Theorem 4 (Brunn–Minkowski inequality). Let K, L be compact subsets of Rn, and λ 2 (0; 1). Then λ 1−λ vol(λK + (1 − λ)L) > vol(K) vol(L) : (1.1) In other words, the function log vol is concave with respect to Minkowski addition. Before proving the Brunn–Minkowski inequality, we point that there is an equivalent form: for every nonempty compact sets A, B in Rn, we have 1=n 1=n 1=n vol(A + B) > vol(A) + vol(B) : (1.2) We check the equivalence between (1.1) and (1.2) by taking advantage of the homo- geneity of the volume. To show (1.2) from (1.1), consider the numbers a = vol(A)1=n and b = vol(B)1=n. The case when a = 0 (and, similarly, b = 0) is easy: it suffices to notice that A + B contains a translate of A (check!). If ab > 0, we may write a A b B A + B = (a + b) + ; a + b a a + b b n and conclude from (1.1) that vol(A+B) > (a+b) , as needed. For the converse implication, we write 1=n 1=n 1=n vol(λK + (1 − λ)L) > vol(λK) + vol((1 − λ)L) = λ vol(K)1=n + (1 − λ) vol(L)1=n λ 1−λ h 1=ni h 1=ni > vol(K) vol(L) ; where the last step is the arithmetic mean-geometric mean (AM-GM) inequality (check!). We present the proof of a functional version of the Brunn–Minkowski inequality. 3 Theorem 5 (Prékopa–Leindler inequality). Let λ 2 (0; 1). Assume that f, g, h : Rn ! [0; 1] are measurable functions such that, n λ 1−λ for every x; y 2 R ; h(λx + (1 − λ)y) > f(x) g(y) : (1.3) Then, λ 1−λ h > f g : ˆRn ˆRn ˆRn Before proving Theorem 5, we notice that it immediately implies Theorem 4 by choosing f = 1K , g = 1L and h = 1λK+(1−λ)L. Proof of Theorem 5. The proof is by induction on the dimension n. We first consider the base case n = 1. By monotone convergence, we may reduce to the case where f, g are bounded, and by homogeneity to the case when kfk1 = kgk1 = 1 (check!). We also use the following formula which relates integrals with measures of level sets (check!): whenever φ : X ! Rn is a measurable function defined on a measure space (X; µ), then 1 φ = µ(fφ > tg) dt: (1.4) ˆX ˆ0 Another information we need is that the Brunn–Minkowski inequality holds in dimension 1: for nonempty measurable sets A, B in R such that A + B is measurable, we have vol(A + B) > vol(A) + vol(B). To prove this, reduce to the case when sup A < +1 and inf B > −∞, and show that A + B contains disjoint translates of A and B (check!). The proof goes as follows: for 0 6 a < 1, we have fh > ag ⊃ λff > ag + (1 − λ)fg > ag; which by the one-dimensional Brunn–Minkowski implies vol(fh > ag) > λ vol(ff > ag) + (1 − λ) vol(fg > ag): We then integrate this inequality when a ranges over [0; 1), and use (1.4) 3 times to obtain h > λ f + (1 − λ) g ˆR ˆR ˆR λ 1−λ > f g ˆRn ˆRn by the AM-GM inequality. We now explain the induction step, assuming the result in dimension n. We decompose Rn+1 as Rn × R. Let f; g; h : Rn+1 ! R satisfying (1.3). For y 2 R, we define 3 n functions on R by the formulas fy(t) = f(t; y), gy(t) = g(t; y), hy(t) = h(t; y). Whenever 4 real numbers y; y1; y2 are such that y = λy1 + (1 − λ)y2, we have hy(λs1 + (1 − λ)s2) > λ 1−λ n fy1 (s1) gy2 (s2) for s1; s2 2 R . In other words, the functions fy1 ; gy2 ; hy satisfy the hypothesis (1.3). By the induction step, it follows that λ 1−λ hy > fy1 gy2 : ˆRn ˆRn ˆRn If we define functions F; G; H on R by F (y) = Rn fy, G(y) = Rn gy and H(y) = Rn hy, this means that the functions F , G, H also satisfy´ (1.3). By´ using the case n =´ 1, and Fubini theorem, it follows that λ 1−λ h > f g : ˆRn+1 ˆRn+1 ˆRn+1 A remarkable corollary of the Brunn–Minkowski theorem is the isoperimetric inequality. One may define the surface area of a subset K ⊂ Rn by vol(K ) − vol(K) a(K) = lim sup " : (1.5) "!0 " This is a simple way to define the (n − 1)-dimensional measure of @K. Theorem 6 (Isoperimetric inequality). Let K ⊂ Rn be a compact set with vol(K) > 0, and B a Euclidean ball with radius chosen so that vol(K) = vol(B). Then, for every " > 0, we have vol(K") > vol(B"), and therefore a(K) > a(B). 1=n n vol(K) n Proof. We may take B = rB2 , for r = n . We have then B" = (r + ")B2 . Note vol(B2 ) n that K" = K + "B2 . By (1.2), we have 1=n 1=n n 1=n vol(K") > vol(K) + vol("B2 ) n 1=n = (r + ") vol(B2 ) = vol(B") as needed.