NOTES ON VECTOR-VALUED INTEGRATION MATH 581, SPRING 2017

Transcription

1 NOTES ON VECTOR-VALUED INTEGRATION MATH 58, SPRING 207 Throughout, X will denote a Banach space. Definition 0.. Let ϕ(s) : X be a continuous function from a compact Jordan region R n to a Banach space X and ρ(s) : R be a real valued weight function. Suppose that there exists x X such that for any f X we have (f ϕ)(s)ρ(s) ds = f(x). We then say that the integral of ϕ over is x, and denote ϕ(s) dµ(s) = x where dµ(s) = ρ(s)ds. Note that if x exists, then the integral is uniquely defined, for if y X is any other vector satisfying the definition, we would have f(x) = f(y) for all f X and hence x = y by the consequences of the Hahn-Banach theorem. Also, this definition should perhaps be called the weak integral of ϕ: there is a notion of a strong integral based on approximating ϕ by simple or step functions, but we shall not need it in this class. This definition also makes sense for a compact metric space equipped a Borel measure µ and ϕ(s) a continuous function (or even a weakly measurable function ϕ such that f ϕ is measurable for any f X ). Definition 0.2. We say that a subset K of a Banach space is convex if whenever 0 t and x, y K. tx + ( t)y K Lemma 0.3. Suppose K R n is compact, convex, and a R n \ K. Then there exists v R n such that a v > x v for all x K. Proof. A standard compactness argument shows there exists a point b K such that d(a, b) = d(a, K). Then set v = a b so that b v < a v as 0 < v 2 = a b 2 = (a b) v = a v b v. If x K and t [0, ], then since b is the closest point to a and K is convex, a b 2 a (tx + ( t)b) 2 = a b 2 2t(a b) (x b) + t 2 x b 2. Subtracting a b 2 from both sides gives 0 2t(a b) (x b) + t 2 x b 2 and hence we may divide by t and take the limit as t 0 + to get that (a b) (x b) 0. Hence v x v b 0 and v x v b < v a.

2 2 NOTES ON VECTOR-VALUED INTEGRATION Definition 0.4. If E X, then we define the convex hull of E, denoted by co(e), to be the smallest convex set that contains E in that if D is convex and contains E then D co(e). Equivalently, we may characterize co(e) as () the intersection of all convex sets containing E (2) { n t jx j : x j E, t j 0, n t j = and n N} (Given vectors x,..., x n a sum of the form n t jx j with t j 0 and n t j = is said to be a convex combination.) Exercise 0.5. Prove that the 3 characterizations of the convex hull here are indeed logically equivalent. It is not too hard to see that if x,..., x n is a finite collection of points in X, then co(x,..., x n ) is a closed, bounded subset contained in a finite dimensional subspace. Hence this convex hull is compact. More generally, we have the following: Lemma 0.6. If X is a Banach space and K X is totally bounded, then co(k) is totally bounded and hence co(k) is compact. Proof. Given ε > 0, let x,..., x n K be such that K n B(x i, ε/2). Therefore K n B(x i, ε/2) {B(x, ε/2) : x co(x,..., x n )}. Let S denote the set on the right. We will show that S is convex and has an ε-net. Since co(k) is the intersection of all convex sets containing K this will show that co(k) S and since S has an ε-net, co(k) does as well. To see that S is convex, suppose y j co(x,..., x n ), z j < ε/2, j =,..., m is a finite collection of points. Then if m t j =, we have tj (y j + z j ) = t j y j + t j z j co(x,..., x n ) + B(0, ε/2) = S as the first vector is in the convex hull and t j z j < t j ε/2 = ε/2. To see that S has an ε-net, observe that co(x,..., x n ) span(x,..., x n ) is closed and bounded and therefore compact. Hence there exists y,..., y m such that co(x,..., x n ) m B(y j, ε/2). Hence if z S, there exists x co(x,..., x n ) such that x z < ε/2. In turn, there exists y j satisfying x y j < ε/2. The triangle inequality now shows that z y j < ε and hence S m B(y j, ε). Lemma 0.7. If S R n is compact then co(s) is compact (that is, co(s) = co(s)). Proof. Let T R n+ be the compact simplex n+ T = {(t,..., t n+ ) : t i =, t i 0}. The key argument in the proof is that if S is compact, then x co(s) if and only if there exists a collection of n + vectors x,..., x n+ S such that (0.) x = t x + + t n+ x n+ for some t S (this is sometimes called Carathéodory s theorem, note that the direction is asserted above). If this holds, then co(s) must be the image of T S n+ under the continuous mapping (t, x,..., x n+ ) t x + + t n+ x n+

3 NOTES ON VECTOR-VALUED INTEGRATION 3 (note that S n+ refers the product of n+ copies of S). Since T S n+ is the finite product of compact sets, it is compact itself, showing that co(s) is the continuous image of a compact set. We now show that if k > n and x = k+ t i x i is a convex combination of some k + vectors x i R n, then x can be rewritten as a convex combination of k of these vectors. If this can be shown then (0.) will follow. The claim is trivial if t i = 0 for some i, so assume that t i > 0 for all i k +. Consider the linear map from R k+ to R n+ determined by (a,..., a k+ ) ( k+ k+ a i x i, note that since x i R n, k+ a i x i R n (it may help to write out the matrix of the linear map in the standard basis). Since k > n, the null space of this map has positive dimension, so there exists (a,..., a k+ ) 0 such that k+ a i x i = 0, k+ a i = 0. Let { t j γ = min a j } : a j 0. Since there exists at least one nonzero a j and t j > 0 for all j, γ is well defined and positive. Choosing λ to be ±γ, we obtain a number satisfying λa i t i for all i and λa j = t j for some j. Set c i = t i λa i for all i, so that c i 0 and we have a i ) k+ k+ k+ c i x i = t i x i λ a i x i = x + 0 = x, k+ k+ k+ c i = t i λ a i = k+ t i =. However, since λa j = t j, we have c j = 0, showing that indeed that x is a convex combination of k vectors. Theorem 0.8. Let R n be a compact Jordan region and ρ(s) : R be a nonnegative continuous weight function satisfying ρ(s) ds = dµ(s) =. If ϕ(s) : X is continuous, then x = ϕ(s)dµ(s) exists and x co(ϕ()). Proof. The first observation is that we may regard X as a real vector space and restrict our attention to real linear functionals. For if f X is a complex linear functional, then we can split it into its real and imaginary parts, both of which will be real linear functionals (cp. our proof of the complex Hahn-Banach theorem). Let F = {f,..., f n } be a finite collection of real linear functionals in X. Now define the set K F = {x co(ϕ()) : f j (x) = f j ϕ(s) dµ(s) for all f j F }. Since ϕ() is compact (as the continuous image of a compact set), we know that by Lemma 0.6, co(ϕ()) is also compact.

4 4 NOTES ON VECTOR-VALUED INTEGRATION We now recall a property of compact sets. We say that a collection of subsets {E α } α A has the finite intersection property if the intersection of any finite subcollection is nonempty n i= E α i. If K is a compact set, and E α K are a collection of closed sets with the finite intersection property, then α A E α. This can be shown by assuming that the infinite intersection is empty and then forming an open cover of K out of the open sets K \ E α. The existence of a finite subcover contradicts the finite intersection property. We will show that all the K F defined above are nonempty. If this holds, then they are closed and have the finite intersection property. Since c j := (f j ϕ)(s) dµ(s) is a scalar quantity, K F = co(ϕ()) ( n i= f j (c j ) ) is an intersection of closed sets, hence K F is closed for any F. Now if F, F 2 are two finite sets of linear functionals, then it can be seen that { } K F K F2 = x co(ϕ()) : f(x) = f ϕ dµ(s) for all f F F 2 = K F F 2. This argument can be repeated for any finite subcollection of the K F s and hence if K F for any F, then the finite intersection property holds. Now given any finite subset F = {f,..., f n } X, regard F = (f,..., f n ) as a map X R n. Define S R n to be the compact subset S = F(ϕ()) and m = (m,..., m n ), where m j = f j ϕ(s) dµ(s). We now claim that m co(s) = co(s) (with the set identity a consequence of the compactness of S from Lemma 0.7). If this holds, we will have that m = l ( l ) t k F(x k ) = F t k x k k= for some x,..., x l ϕ() and t k s satisfying 0 t k, k t k =. Thus x = l k= t kx k co(ϕ()) satisfies m j = f j (x) for all j n. Hence x K F showing that indeed K F. If a / co(s), by Lemma 0.3, there exists v R n such that a v > y v for any y S co(s). Thus since ((f ϕ)(t),, (f n ϕ)(t)) S n a j v j > j= k= n v j f j (ϕ(s)) j= for any s. Since dµ(s) =, integrating both sides of the inequality gives n n n n a j v j = a j v j dµ(s) > v j f j (ϕ(s)) dµ(s) = v j m j j= j= j= j= Hence a m, and a / co(s) is chosen arbitrarily, we must have m co(s). Theorem 0.9. Suppose R n is compact, ρ(s) 0 is continuous and scalarvalued, and ϕ(s) : X is also continuous. Then ϕ(s) dµ(s) ϕ(s) dµ(s)

5 NOTES ON VECTOR-VALUED INTEGRATION 5 Proof. Let x = ϕ dµ. By the consequences of the Hahn-Banach theorem, there exists a linear functional f X such that f(x) = x and f =. Hence for each s we have f(ϕ(s)) ϕ(s). However, this particular f satisfies ϕ(s) dµ(s) = f(x) = f(ϕ(s)) dµ(s) f(ϕ(s)) dµ(s) ϕ(s) dµ(s).