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A Convexity Primer

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A Convexity Primer A Convexity Primer Ashwin Trisal & Micah Pedrick September 7, 2019 Credit: Much of this comes from Kadison-Ringrose's text Fundamentals of the Theory of Operator Algebras and Bollob´as' Linear Analysis. This primer was intended for use alongside Chuck Akemann's 201 course homeworks, but should serve in any discussion of basic convexity theory as applied to function spaces. The authors can be contacted at [email protected] and [email protected], respectively. Contents 1 Convex Sets 2 2 Locally Convex Topological Vector Spaces 6 3 Krein-Milman 11 4 Convex Spaces and Convex Functions 12 5 Krein-Milman in Dual Spaces 13 6 The Pettis Integral 15 A Nets 16 B Initial Topologies 17 C Seminorm Topologies 18 1 1 Convex Sets The convexity theory requires real linearity and not complex linearity, even though complex spaces are preferable for spectral theory. Therefore, although we will consider real linear combinations throughout the following section, it is important to bear in mind the results are equally applicable in real and complex vector spaces. Let V be such a vector space. Definition 1.1. A set K ⊆ V is called convex if for every x; y 2 K and t 2 (0; 1), we have tx + (1 − t)y 2 K. That is, a convex set is closed under convex combinations. You may have a lot of intuition about convexity from the finite-dimensional case; this is valuable, but relies heavily on the local compactness of these spaces. As a result, this structure has not been helpful in developing intuition for the setting that we most care about—infinite-dimensional function spaces. Lemma 1.1. Let K ⊆ V be convex. Then for each x1; : : : ; xn 2 K and t1; : : : ; tn 2 [0; 1] P with ti = 1, we have X tixi 2 K: That is, K is closed under arbitrary finite convex combinations. Proof. This follows by induction. It's trivially true for n = 1 (unlike in most algebraic contexts, we do not let the empty sum denote 0 here; we don't know that 0 is in our set, so we leave it undefined). Assume it is true for n = k; then, given x1; : : : ; xk+1 2 K and Pk+1 t1; : : : ; tk+1 2 [0; 1], we show that i=1 tixi 2 K. If tk+1 = 0 or 1, we're done; if not, note that k X ti = 1 − tk+1 i=1 and therefore that k k 1 X X ti t = = 1: 1 − t i 1 − t k+1 i=1 i=1 k+1 Pk ti In particular, xi = x 2 K by the inductive hypothesis. i=1 1−tk+1 Now we see that k+1 k X X ti t x =(1 − t ) x + t x i i k+1 1 − t i k+1 k+1 i=1 i=1 k+1 =(1 − tk+1)x + tk+1xk+1; which is therefore in K by definition. So by induction, K contains all of its finite convex combinations. 2 An infinite convex combination can be made sense of, but requires some analytic structure we currently lack. T Lemma 1.2. If fKαg is a family of convex sets, then K = α Kα is convex. Proof. This is immediate: if for all α we have x; y 2 Kα, then by convexity tx+(1−t)y 2 Kα for t 2 (0; 1). S Lemma 1.3. If fKng is a family of convex sets with K0 ⊆ K1 ⊆ · · · , then K = n Kn is convex. Proof. This is also immediate: suppose x; y 2 K. Then for some N we have x; y 2 KN , and may apply convexity there to conclude tx + (1 − t)y 2 KN ⊆ K. In fact, we didn't need countability above, just a totally ordered indexing set with Kα ⊆ Kβ whenever α ≤ β. Definition 1.2. The convex hull of a set E, denoted co E, is the smallest convex set con- taining E. As with most notions of closures or hulls, it is obtained as the intersection of all convex supersets of E. Equivalently, co E is the set of all convex combinations of elements of E. Definition 1.3. A face of a convex set K is a convex subset F ⊆ K with the property that if for x; y 2 K and t 2 (0; 1), tx + (1 − t)y 2 F , then x; y 2 F . An extreme point of K is a face that has only one point. Faces are inaccessible convex subsets: they cannot be entered via nontrivial convex com- binations of elements of K n F . It is reasonable to imagine proper faces (F ⊂ K) as lying on the boundary of K. Lemma 1.4. If G is a face of F , and F is a face of K, then G is a face of K. Proof. Let x; y 2 K and t 2 (0; 1) be such that tx + (1 − t)y 2 G. Then tx + (1 − t)y 2 F , because G ⊆ F ; F being a face of K then implies that x; y 2 F . But G is a face of F , so also x; y 2 G. Hence, G is a face of K. In particular, this gives us a natural partial order (by inclusion) on the faces of K. Definition 1.4. The dimension of a convex set K is the dimension of the subspace of V generated by fx − y0 j x 2 Kg for any y0 in K. Definition 1.5. A point x 2 K is said to be internal if for every y 2 V there is some c > 0 such that for each 0 ≤ t < c, x + ty 2 K. Note the order of the quantifiers here: in the familiar case V is equipped with a norm, this is weaker than the statement that x is an interior point. In infinite dimensions, this may be strict. 3 1 1 1 Example 1.1. Let V = ` (N) and K = (an) 2 ` (N) j 8k 2 N; jakj ≤ k . This is convex, as 1 jta + (1 − t)b j ≤ t ja j + (1 − t) jb j ≤ k k k k k 1 if jakj ; jbkj ≤ k and t 2 [0; 1]. 1 1 We have (0) internal to K: if (bn) 2 ` (N), then only finitely many terms may exceed k , else the series would not converge. Thus, supk k jbkj < 1, and dividing (bn) by this amount 1 lands in K. However, B((0)) contains the sequences 2 δk for all k 2 N (in particular, for k 1 1 such that 2 > k ), so clearly no -neighborhood of 0 is contained in K. We conclude with more examples of computing the faces of convex sets representative of those we're interested in. Example 1.2. Denote by I the unit interval [0; 1] ⊆ R. Then the faces of I are f0g, f1g, and I itself. First, we show that f0g is a face. Assume that for some x; y 2 I and t 2 (0; 1) we have tx + (1 − t)y = 0. Without loss of generality, suppose x > 0 (we are uninterested in the case x = y = 0). But then y ≥ 0 implies tx + (1 − t)y ≥ tx > 0: a contradiction. Similarly, f1g is a face: without loss of generality, suppose x < 1 in the setting above. Then tx < t implies tx + (1 − t)y < t + (1 − t)y ≤ t + (1 − t) = 1. Finally, any convex set is trivially a face of itself. Conversely, assume that F ⊆ I is a face of I. If F = f0g or F = f1g, we're done. If F contains both 0 and 1, it contains their convex hull; as t1 + (1 − t)0 = t for t 2 [0; 1], said hull is cof0; 1g = I. Otherwise, let c 2 F n f0; 1g ⊆ (0; 1). We can write c = (1 − c)0 + c1 to conclude that 0 and 1 are both in F by the definition of a face. By the argument above, then, F = I. Thus, the faces are exactly as claimed. Example 1.3. Denote by I the unit interval [0; 1] in R, and consider In ⊆ Rn. Then, considering Rn as the space of functions ff : f1; : : : ; ng ! Rg, the faces of In are exactly of the following form: n KA0;A1 = ff 2 R j 8a0 2 A0; f(a0) = 0; 8a1 2 A1; f(a1) = 1; A0 t A1 ⊆ f1; : : : ; ngg : This requires some work. First, that all such sets are faces: it is straightforward to show n that they are convex. For the face property, assume that g1; g2 2 R and t 2 (0; 1) are such that tg1 + (1 − t)g2 2 KA0;A1 . Then for each a0 2 A0, tg1(a0) + (1 − t)g2(a0) = 0, so g1(a0) + g2(a0) = 0 by nonnegativity. Similarly, for each a1 2 A1, g1(a1) = g2(a1) = 1. So g1; g2 2 KA0;A1 , so it is a face. For the other direction, let K be a face of In. Define A0 = fa0 2 f1; : : : ; ng j 8f 2 K; f(a0) = 0g ; and A1 similarly. Then K ⊆ KA0;A1 . Let f 2 KA0;A1 . We show that there is an element of K which agrees with f at every point|that is, that f 2 K. Let x 2 f1; : : : ; ng n (A0 t A1). Then there exists g 2 K such that g(x) 2 (0; 1): x2 = A0, so there is some function with g(x) 6= 0, and x2 = A1, so there is some function with g(x) 6= 1; 4 if these functions have values 1 and 0 respectively, any nontrivial convex combination has g(x) 2 (0; 1).
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