Final Exam Practice Problems Math 428, Spring 2017

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1 Final xam Practice Problems Math 428, Spring 2017 Name: Directions: Throughout, (X,M,µ) is a measure space, unless stated otherwise. Since this is not to be turned in, I highly recommend that you work on these problems independently. P1. If f : X [0, ] is a nonnegative measurable function and f dµ = 0, prove that X f = 0 a.e. Solution 1. Let = {x : f (x) > 0}, so = {x : f (x) = 0} and X = is a disjoint union. There are many ways to show that µ() = 0. Since f (x) > 0 for all x, there exists s S + such that 0 < s(x) < f (x) for all x. Suppose µ() 0. Then, s dµ µ()min{α i } where s(x) = α i χ Ai (x) and α i > 0 for all i. So, 0 < s dµ < f dµ. Since f = 0 on, we have f dµ = f dµ > 0 which is a contradiction. Thus, µ() = 0 so f = 0 almost everywhere. X P2. Suppose f L 1 (µ) and M. Given ε > 0, show there exists simple s such that f s dµ < ε.

2 Solution 2. Since f L 1 (µ), f + dµ < and f dµ <. There exist sequences {s + n} and {s n} of nonnegative simple functions such that s + n f + and s n f, where 0 s + n f + and 0 s n f for all n. Let t n = s + n s n. Then, {t n } is a sequence of simple functions (possibly negative) such that t n f, since t n f s + n f + + f s n. Furthermore, the sequence {t n } is dominated by f, since s + n s n s + n + s n f. By the dominated convergence theorem (DCT), we have lim t n f dµ = 0. P3. If f is measurable and {f n } is a sequence of measurable functions, we say f n f in measure if the following statement holds; for all ε > 0 there exists K(ε) such that µ({x : f n (x) f (x) > ε}) < ε for all n > K(ε). Prove that f n f in measure if and only if for all ε > 0. Solution 3. First, assume that lim µ({x : f n(x) f (x) > ε}) = 0 n lim µ({x : f n(x) f (x) > ε}) = 0 n for all ε > 0. Given ε > 0 we have: for all η > 0 there exists K(η) such that µ({x : f n (x) f (x) > ε}) < η for all n > K(η). Since this holds for all η > 0, we can simply take η = ε, hence f n f in measure. On the other hand, assume that for all ε > 0 there exists K(ε) such that µ({x : f n (x) f (x) > ε}) < ε

3 for all n > K(ε). We need to show that, given ε > 0, we have: for all η > 0 there exists N(η) such that µ({x : f n (x) f (x) > ε}) < η for all n > N(η). If ε η, we are done (take N(η) = K(ε)). Suppose η < ε. By assumption, µ({x : f n (x) f (x) > η}) < η for all n > K(η). Since η < ε, if f n (x) f (x) > ε then f n (x) f (x) > η, hence for all n > K(η) (with N(η) = K(η)). µ({x : f n (x) f (x) > ε}) < η P4. Suppose f 0 a.e. and F,,F M. Prove that f dµ f dµ. F Solution 4. Let N = {x : f (x) < 0}, so µ(n) = 0. For any measurable set A X, (A N) (A N ) is disjoint, so f dµ = f dµ + f dµ = f dµ A A N A N A N since µ(a N) = 0. Without loss of generality, we assume,f are subsets of N. Since F, we have F = (F\), hence since f 0 on F\. f dµ = F f dµ + f dµ F\ f dµ, P5. If f L 1 (µ) and M, prove that for all ε > 0 there exists δ(ε) > 0 such that f dµ < ε whenever µ() < δ(ε). Solution 5. You should prove the following claim. Claim 0.1. If f : X [0, ] is such that f dµ <, then the set {x : f (x) = } has µ-measure 0.

4 A measurable function g : X R is essentially bounded if there exists M < such that g(x) M a.e. In this case, the essential supremum of g, g, is defined by g = inf{m : g(x) M a.e. }. If f L 1 (µ) and g is essentially bounded (i.e. g L (µ)), we have f g dµ g f dµ. Now, since f L 1 (µ), if M f dµ f µ(). Hence, if µ() < ε f we have f dµ < ε. Similarly, it could be argued directly that except possibly on a set of µ-measure 0. sup f (x) <, x P6. Let X be a set. A collection τ of subsets of X is a topology in X if (O1) (O2) (O3) X τ, τ O 1 τ and O 2 τ = O 1 O 2 τ O α τ for all α O α τ. α The elements of τ are called open sets. Property (O1) states that both X and the empty set are open sets; (O2) states that the intersection of two open sets is open; (O3) states that an arbitrary union(need not be countable) of open sets will be open. A set X along with a topology τ in X is a topological space. A topological space X is called a Hausdorff space if given distinct points x,y X there are disjoint open sets O 1,O 2 such that x O 1 and y O 2. Note that any metric on X defines a topology in X and any metric space is a Hausdorff space. Prove the following statements, keeping in mind that there is no metric or norm, so we must use the topology. a) Let X be a topological space. Suppose K X is compact and C X is closed. If C K then C is compact.

5 b) Suppose X is a Hausdorff space, K X is compact and y K. Then there are open sets O,V such that y V, K O and O V =. c) If K X is compact, then K is closed. d) If K X is compact and C X is closed, then C K is compact. Solution 6. Solution 6a) Since C is closed, C = X\C is an open set. Let {O λ } λ Λ be an open cover of C. Then, {C,{O λ } λ Λ } is an open cover of K. Since K is compact, there exists a finite subcover of K, which is a finite subcover of C. Thus, C is compact. Solution 6b) For each x K there exist open sets O x,v x such that O x V x = and y V x, x O x. Since K is compact, there exists a finite subcover O x1,...,o xn (since O x clearly covers K). Then, defining we have O K and y V. O := n O xi and V := i=1 Solution 6c) By definition of closure, K K where K is the smallest closed set containing K. Assume, for contradiction, that K K and let y K\K. Then, there exist open sets O,V such that y V, K O, and O V =. But, V is a closed set and K V. Moreover, K V K K, which is a contradiction since V K is a closed set smaller than K. Hence, K K = K = K, so K is closed. Solution 6d) Since K is closed, C K is a closed subset of K, which is compact, so C K is compact. n i=1 V xi x K

6 P7. Let A be a Banach space. Suppose an operation of multiplication (a map from A A A) is defined which satisfies (M1) (M2) (M3) (M4) x(yz) = (xy)z, x(y + z) = xy + yz, (x + y)z = xz + yz, α(xy) =(αx)y = x(αy), where x,y,z A and α C (we assume the field is complex, unless stated otherwise). If the multiplication on A satisfies (M1)-(M4) and xy x y, (0.1) then A is called a Banach algebra. If A is a Banach algebra and there exists e A such that ex = x = xe for all x A, then e is called the identity (or unit) and A is called a Banach algebra with identity. Note that not all Banach algebras have an identity. If A is a Banach algebra for which xy = yx for all x,y A, then A is called a commutative Banach algebra (or abelian). Prove the following statements. Unless otherwise stated, you may assume A has an identity. a) Suppose A is a Banach algebra. Show that the operation of multiplication is a continuous operation. b) Let be a Banach space. Show that A = B() is a Banach algebra. Is it commutative? Is there an identity e? c) Let X be a compact metric space (in general, a compact Hausdorff space works here). Consider the space C(X) of continuous, complex-valued funtions defined on X with the uniform norm f = sup f (x). x X Show that A = C(X) is a commutative Banach algebra with multiplication of functions defined pointwise.

7 d) Let A be a Banach algebra with identity. An element x A is called invertible if there exists an element x 1 A such that xx 1 = e = x 1 x. If x A is such that x < 1, then e + x is invertible and (e + x) 1 = ( 1) n x n. Prove this, then determine an upper bound for (e + x) 1. e) Let A be a Banach algebra with identity. The resolvent R λ (x) of x A is defined by R λ (x) = (λe x) 1 for all λ C for which the inverse exists. Prove the first resolvent formula R λ (x) R µ (x) = (µ λ)r λ (x)r µ (x). Solution 7. Solution 7a) Consider the multiplication map A A A defined by (x,y) xy for x,y A. Suppose x n x and y n y in A (convergence in norm). It follows that x n x and y n y, hence x n y n xy = x n y n x n y + x n y xy x n (y n y) + y(x n x) x n y n y + y x n x 0 since x n x 0 and y n y 0. Thus, x n x and y n y implies x n y n xy, so multiplication is continuous. Solution 7b) There s not much to show here, just a recollection of facts from previous material. Since is complete, B() is complete. If A,B B(), the product AB is defined by composition; if x then (AB)x = A(Bx). Since A,B are both linear, it is clear that AB is linear. The properties (M1)-(M4) have been established, but are straightforward to re-verify. The inequality AB A B has already been established, where A = sup Ax, x =1 which establishes (0.1) and the boundedness of AB. So, A,B B() implies AB B(). Thus, B() is a Banach algebra. It has an identity, which is the identity operator I B(), but is noncommutative, in general, since AB BA.

8 Solution 7c) Let ρ be a metric on X (or use a norm x y X = ρ(x,y)). If f,g C(X), define f g by (f g)(x) = f (x)g(x) for all x X. Products of continuous functions are continuous, so f g C(X). Then, f g = sup x X f (x)g(x) sup x X f (x) sup g(x) = f g x X which establishes (0.1). The pointwise product of functions is clearly commutative, so C(X) is commutative. The identity on C(X) is just the constant function e(x) = 1 for all x X. If C(X) is complete, then C(X) is a Banach algebra. Proof of completeness. Suppose {f n } C(X) is a Cauchy sequence. Since f n f m 0 as n,m and f n (x) f m (x) f n f m for all x X, the sequence {f n (x)} C is Cauchy for each x X. Since C is complete, {f n (x)} converges for each x X. Define f (x) = lim n f n (x) for each x X. We need to show that f is a continuous function and that f n f in C(X); that is, f n f 0 as n. Let M(ε) be such that f n f m < ε for all n,m > M(ε). Then, for all k 0. If n > M(ε), then f n (x) f (x) f n (x) f n+k (x) + f n+k (x) f (x) f n (x) f (x) < ε + f n+k (x) f (x) for all x X, k 0. Letting k, the second term goes to 0 for all x X, hence f n f. Let n 0 > M( ε /3). ach f n is continuous, in particular, f n0 is continuous. Let δ(ε,z) = δ n0 ( ε /3,z), where δ n0 works for f n0. That is, if ρ(x,z) < δ n0 ( ε /3,z), then f n0 (x) f n0 (z) < ε /3. Assume ρ(x,z) < δ(ε,z). Then, f (x) f (z) f (x) f n0 (x) + f n0 (x) f n0 (z) + f n0 (z) f (z) < ε 3 + ε 3 + ε 3 = ε. Solution 7d) Sicne x < 1, we have x n x n for any n N. Let k s k = ( 1) n x n. Then (e + x)s k = e + ( 1) k x k+1,

9 hence (e + x)s k e = x k+1 x k+1 0 as k since x < 1. Thus, (e + x) 1 = ( 1) n x n. Then, (e + x) 1 = ( 1) n x n x n 1 = 1 x. Solution 7e) Assuming R λ (x) and R µ (x) exist for λ,µ C, we have (µ λ)e = ((µe x) (λe x)) = (µe x) ( e (µe x) 1 (λe x) ) = (µe x) ( (λe x) 1 (µe x) 1) (λe x) = (µe x) ( R λ (x) R µ (x) ) (λe x). Multiplying by the respective inverses, we have (µ λ)r µ (x)r λ (x) = R λ (x) R µ (x).