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Notes for Functional Analysis Notes for Functional Analysis Wang Zuoqin (typed by Xiyu Zhai) November 6, 2015 1 Lecture 18 1.1 The convex hull Let X be any vector space, and E ⊂ X a subset. Definition 1.1. The convex hull of E is the set co(E) = fx = t1x1 + ··· + tnxn 2 X j x1; ··· ; xn 2 E; ti ≥ 0; t1 + ··· + tn = 1g (1) (Such an element x is called a convex combination of x1; ··· ; xn.) From the definition it is easy to see that • E ⊂ co(E), co(E) is convex. • If A ⊂ X is convex and E ⊂ A, then co(E) ⊂ A. Since any intersection of convex sets is still convex, we immediately get the following equivalent description of the convex hull (which is also widely used as definition, e.g., PSET 3-1): Proposition 1.2. (Alternate definition) The convex hull of a set E is \ co(E) = A: E⊂A;A convex Now let X be a topological vector space. Definition 1.3. We call the closure co(E) the closed convex hull of E. Obviously co(E) is a closed convex set, and it is the smallest closed convex set containing E. It is also the intersection of all closed convex sets that contains E. Example 1.1. In PSET 2-2, Problem 4(2) (c.f. the arguments on page 37 of Rudin's book) we have seen that if X = Lp([0; 1]) for 0 < p < 1, and E is any set containing at least one interior point, then co(E) = X. 1 Example 1.2. For any x; y 2 X, co(x; y) is the line segment connecting x and y. n n Example 1.3. Let X = R and E = S(0; 1) = fx 2 R : kxk = 1g. Then n co(E) = B(0; 1) = fx 2 R : kxk ≤ 1g: The following proposition is very useful. It says that if a sequence xn converges weakly to x, then one can find a new sequence yn, each is a convex combination of xn's, so that yn converges to x in the original topology! ! Proposition 1.4. If xn −! x in a locally convex topological vector space X, then x 2 co(fxng): ! ! Proof. xn −! x implies x 2 co(fxng) . But in a locally convex topological vector space, ! co(fxng) = co(fxng) since co(fxng) is convex. We have profound interest in compact convex sets. Proposition 1.5. If A1; ··· ;An are compact convex subsets in a topological vector space X, then co(A1 [···[ An) is compact. Proof. Let S be the simplex n S = f(t1; ··· ; tn) 2 R : ti ≥ 0; t1 + ··· + tn = 1g: Consider the map f : S × A1 × · · · × An ! X; ((t1; ··· ; tn); x1; ··· ; xn) 7! t1x1 + ··· + tnxn: Then f is continuous. Since S × A1 × · · · × An is compact, the image of K = f(S × A1 × · · · × An) is compact in X. In what follows we will prove K = co(A1 [···[ An). S S • By definition, K ⊂ co(A1 ··· An). • K is convex since we have 0 0 0 0 t(t1x1 + ··· + tnxn) + (1 − t)(t1x1 + ··· + tnxn) 0 0 0 0 =(tt1x1 + (1 − t)t1x1) + ··· + (ttnxn + (1 − t)tnxn) 0 00 0 00 =(tt1 + (1 − t)t1)x1 + ··· + (ttn + (1 − t)tn)xn 2K; where 0 0 ( ttixi+(1−t)tixi 0 0 2 Ai; ti 6= 0 or t 6= 0 00 (tt1+(1−t)t ) i xi = i : any element in Ai; otherwise 2 n S • Ai ⊂ K since Ai ⊂ K for all i. i=1 So co(A1 [···[ An) = K is compact. In particular, Corollary 1.6. co(fx1; ··· ; xng) is compact. In general, even if K is compact, it is possible that co(K) is not closed or compact. Definition 1.7. A set E in a topological vector space X is called totally bounded if for any open neighborhood U of 0, one can find x1; ··· ; xn 2 X such that n [ E ⊂ (xi + U): i=1 Remark. If X is metrizable and the metric is translation invariant, then E is totally bound- ed iff for any " > 0, there exists x1; ··· ; xn such that n [ E ⊂ B(xk;"): k=1 In other words, for any " one can find an "-net fx1; ··· ; xng of E. In particular, any compact set in X is totally bounded. Proposition 1.8. If X is a locally convex topological vector space, E ⊂ X is totally bounded, then co(E) is totally bounded. Proof. Let U be any neighborhood of 0 in X. Choose a convex neighborhood V of 0 such that V + V ⊂ U. Since E is totally bounded, one can find x1; ··· ; xn such that n [ E ⊂ (xk + V ): k=1 It follows E ⊂ co(fx1; ··· ; xng) + V: Note that the right hand side is convex, so co(E) ⊂ co(fx1; ··· ; xng) + V: Since co(fx1; ··· ; xng) is compact, and fy + V j y 2 co(fx1; ··· ; xng)g 3 is an open covering of co(fx1; ··· ; xng), one can find y1; ··· ; ym 2 co(fx1; ··· ; xng) such that m [ co(fx1; ··· ; xng) ⊂ (yi + V ): k=1 It follows that m m [ [ co(E) ⊂ (yi + V ) + V ⊂ (yi + U): k=1 k=1 So co(E) is totally bounded. As a consequence, we have Corollary 1.9. If X is Frech´et,and K ⊂ X is compact, then co(K) is compact. Proof. Since K compact, it is totally bounded. So co(K) is totally bounded. Now use the following fact whose proof is left as an exercise: Fact: The closure of a totally bounded set in a complete metric space is compact. 1.2 Extreme points First suppose X is a vector space, and E ⊂ X a subset. Definition 1.10. A non-empty subset S in E is called an extreme set of E if for any x 2 S, x = tx1 + (1 − t)x2 for some 0 < t < 1; x1; x2 2 E =) x1 2 S; x2 2 S: In other words, any point in S cannot be written as a convex combination of two points in E that are not both in S. Definition 1.11. A point x is called an extreme point of E if fxg is an extreme set of E. So a point is an extreme point of E if and only if it cannot be written as a convex combination of other elements in E. We will denote the set of all extreme points of E by Ex(E), i.e. Ex(E) = the set of extreme points of E: Example 1.4. E is an extreme set of E. n Example 1.5. Let X = R and E = B(0; 1) = fx 2 X : kxk ≤ 1g: Then the unit sphere S(0; 1) is an extreme set. 4 Example 1.6. Let X = C([0; 1]) and K = B(0; 1) = ff 2 X : kfk1 ≤ 1g: Then the extreme set of K contains only two \points": Ex(K) = fχ[0;1]; −χ[0;1]g: In fact, if f 2 C([0; 1]) and jf(t0)j < 1 for some t0 2 [0; 1], then one can always write f as a linear combination of two other elements in K, whose values at t0 are f(t0) ± " for some small ". Example 1.7. Let 1 1 X = c0 = f(a1; a2; ··· ) 2 l : lim an = 0g ⊂ l n!1 and let K = B(0; 1) ⊂ X. Then Ex(K) = ;: The proof is similar to the previous example and is left as an exercise. The following theorem is very useful in many applications. Theorem 1.12 (Krein-Milman). Let X be a topological vector space such that X∗ separates points, and K is any nonempty compact convex subset of X. Then K = co(Ex(K)): Before we prove the Krein-Milman theorem, we first prove Proposition 1.13. Let X be a topological vector space such that X∗ separates points, K ⊂ X is compact and non-empty. Then any compact extreme set of K contains at least one extreme point of K. (In particular, since K is a compact extreme set of K, Ex(K) 6= ;.) Proof. We shall use Zorn's lemma. Let K0 be any compact extreme set of K and let P = fall compact extreme set of K that is contained in K0g: We define a partial order relation on P by A B () A ⊂ B: For any totally ordered subset Q of P, we let \ SQ = S: S2Q Then 5 • SQ is compact and SQ ⊂ K0, since it is a closed subset of the compact set K0. • SQ 6= ;. This is a consequence of the so-called finite intersection property, but we will give a direct proof here: Suppose SQ = ;, then [ K0 = K0 n SQ = K0 n S: S2Q But K0 is compact, and each K0 n S is open in K0. So there exists S1; ··· ;SN 2 Q so that N N [ \ K0 = K0 n Sk = K0 n Sk: k=1 k=1 It follows that N \ Sk = ;; k=1 which is impossible since Sk's are in Q and Q is a totally ordered set (with respect to set inclusion), so that the intersection S1 \···\ SN must equals one of the Sk's. • SQ is an extreme set of K: Suppose x 2 SQ, and x = tx1 + (1 − t)x2 for some 0 < t < 1 and x1; x2 2 K. Then for any S 2 Q, one has x 2 S. Since S is an extremal set, we must have x1; x2 2 S for all S. It follows that x1; x2 2 SQ. It follows that SQ is a lower bound of Q.
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