Functional Analysis HW #1 Sangchul Lee October 9, 2015 1 Solutions Solution of #1.1. Suppose that X <. The only possible Hausdorff topology on X is the discrete topology, and thus every function on X is continuous. Thus C(X) C X and hence dim C(X) = dim C X = X <. Suppose that X =. Given any n Z +, choose a set of n distinct points X n = {x 1,, x n } X. We claim that Claim 1.1. The mapping α : C(X) C(X n ) given by α( f ) = f Xn is a surjective linear map. Assuming this, we know that dim C(X) C(X n ) = n for any n. Therefore dim C(X) =. To finish the proof we are required to show Claim 1.1. It is straightforward that α is a well-defined linear map, thus it suffices to prove that α is surjective. Since X is compact Hausdorff, it is a normal space. Then by the Urysohn s lemma, for each i, j {1,, n} with i j we can choose f i j C(X) such that f i j (x i ) = 1 and f i j (x j ) = 0. Now define g i = j:j i f j. Then g i C(X) and g i (x j ) = δ i j. This proves that α is surjective and hence the claim. Solution of #1.2. Suppose that X is finite-dimensional with dim X = n. Choose a basis {x 1,, x n } of X and let α : C n X be the vector-space isomorphism defined by α(u 1,, u n ) = u 1 x 1 + + u n x n. Then we claim that Claim 1.2. There exists a positive constant C such that u C α(u) u C n. 1
Assuming this, the proof is straightforward. Let ϕ : X C be any linear functional on X. Then writing x = α(u) for some u = (u 1,, u n ) we have ϕ(x) u 1 ϕ(x 1 ) + + u n ϕ(x n ) C( ϕ(x 1 ) + + ϕ(x n ) ) x. Therefore ϕ < and ϕ is continuous. So it suffices to show Claim 1.2. Notice that for any u, v C n, we have α(u) α(v) = α(u v) u 1 v 1 x 1 + u n v n x n ( x 1 + + x n ) u v and consequently α is continuous. Moreover, since the set K = {u C n : u = 1} is compact and α never vanishes on K, the map u α(u) attains a positive minimum δ = min u K α(u) > 0. Then with C = δ 1 we have and therefore the claim is proved. u C u α(u/ u ) = C α(u) Assume that X is infinite-dimensional. Choose a linearly independent set {x 1, x 2, } X and let ϕ be a linear functional on X satisfying ϕ(x n ) = n x n. (The existence of such ϕ is guaranteed by the Zorn s lemma.) Then ϕ = and hence ϕ is not continuous. Sketch of Solution of #1.3. Construction of such X is a very straightforward and routine work, so we only give a sketch of idea: Let S be the set of Cauchy sequences in M and define an equivalence relation on S by ( f n ) (g n ) iff f n g n 0. Then the quotient space X = S/ gives rise to a Banach space with the norm defined as follows: for any equivalence class [( f n )] X, [( f n )] = lim n f n. It is routine to check that this is a well-defined norm on X with respect to which X is complete. Then M is isometrically embedded into X by the mapping f ( f, f, f, ). Now if Y is any Banach space in which M is dense, then we can define an isomorphism Y X as follows: whenever f Y is written as f = lim n f n with f n M, f [( f n )]. Indeed, this definition is a well-defined isometric isomorphism. 2
Solution of #1.5. c 0 (Z + ) is separable since the following set D = { f c 0 (Z + ) : f (k) Q for all k and f (k) = 0 for all sufficiently large k} is a countable dense subset of c 0 (Z + ). Indeed, for any f c 0 (Z + ) and ε > 0, Choose N such that f (k) < ε for k > N. Choose g(1),, g(n) Q such that f (k) g(k) < ε for 1 k N. Extending g onto Z + by letting g(k) = 0 for k > N. we find that g D and f g < ε. l 1 (Z + ) is separable with a dense subset D as above. Indeed, for any f l 1 (Z + ) and ε > 0, Choose N such that k >N f (k) < ε/2. Choose g(1),, g(n) Q such that f (k) g(k) < ε/2n for 1 k N. Extending g onto Z + by letting g(k) = 0 for k > N. we find that g D and f g < ε. l (Z + ) is inseparable. Indeed, for each U Z + define the indicator function 1 U by 1 U (k) = 1, k U 0, k U. Then the uncountable family {1 U : U Z + } satisfies the following condition: for any U V, 1 U 1 V = 1. In particular, there is an uncountable family of mutually disjoint open balls of radius 1/2 and thus no countable subset of l (Z + ) can be dense. l (Z + ) is also inseparable by the following claim: Claim 1.3. If X is a Banach space and X is separable, then so is X. The following simple tweak of Hahn-Banach theorem is quite useful for our proof. Lemma 1.4. Let M be a closed proper subspace of the Banach space X and x 0 X \ M. Then there exists ϕ X such that ϕ(y) = 0 for y M, ϕ(x 0 ) = dist(x 0, M), and ϕ 1. We defer the proof of this lemma to the next section. Assuming this, the claim is easy to verify. Indeed, let {ϕ 1, ϕ 2, } be a dense subset of X. (For technicality, we also assume that ϕ n 0 for all n.) For each n, choose x n X such that ϕ n (x n ) x n 1 2 ϕ n Then we claim that the set D of all possible Q-linear combinations of {x 1, x 2, } is dense in X. Since D is countable, this will prove that X is separable as well. 3
To this end, we assume otherwise. Then the the closure M of D in X is not equal to X. Notice also that M is in fact a closed subspace of X. Using this, choose any x 0 X \ M and pick ϕ X as in Lemma 1.4. Clearly ϕ is not identically zero. Also we have ϕ ϕ n ϕ(x n) ϕ n (x n ) x n = ϕ n (x n ) x n 1 2 ϕ n. Since {ϕ 1, ϕ 2, } is dense in X, we can choose a subsequence (ϕ n j ) such that ϕ ϕ n j 0. Then the above inequality shows that ϕ = lim j ϕ n j = 0, a contradiction. This completes the proof. Sketch of Solution of #1.13. Since (X ) 1 is compact Hausdorff by Banach-Alaoglu theorem, a simple application of Urysohn metrization theorem shows that (X ) 1 is metrizable if and only if it is second-countable. Assume that (X ) 1 is second-countable. Then X = n (X ) n is also second-countable and hence separable. This implies that X is separable, by Claim 1.3. Assume that X is separable. Choose a dense subset D = { f 1, f 2, } of X and a countable base (U k ) of the usual topology on C. Then we consider the set V n,k = {ϕ (X ) 1 : ϕ( f n ) U k }. Clearly {V n,k } n,k 1 is a countable family of open sets. Now aassume that (ψ α ) α A is a net in (X ) 1 such that lim α A ψ α ( f n ) = ψ( f n ) for each n. We remark that this information is completely determined by the collection {V n,k } n,k 1. Also, by Proposition 1.21, this implies that (ψ α ) w*-converges to ψ. Thus {V n,k } n,k 1 determines the topology of (X ) 1 and hence (X ) 1 is second countable. Solution of #1.14. Let M be the closure of N, which is a proper closed subspace of X. Utilize Lemma 1.4 to choose ϕ X such that ϕ n (y) = 0 for y M, ϕ n (x) = 1, and ϕ n 1/d. Notice also that, for any y N, we have ϕ ϕ(x y) x y = 1 x y. Taking supremum over y N, it follows that ϕ 1 d. Therefore ϕ = 1 d and the construction is done. Solution of #1.15. We know from linear algebra that the mapping f ˆ f is linear. So it suffices to prove that this mapping preserves norm. To this end, let f X. Then for any ϕ X with ϕ = 1, ˆ f (ϕ) = ϕ( f ) ϕ f = f. 4
Taking supremum over ϕ = 1, we find that ˆ f f. Also, by Corollary 1.27, there exists ϕ with ϕ = 1 such that ϕ( f ) = f. Thus we have ˆ f ˆ f (ϕ) = ϕ( f ) = f. Therefore the inequality is saturated and hence ˆ f = f. Solution of #1.16. If X is finite-dimensional, then we know from linear algebra that dim X = dim X. Also, from the previous exercise we know that X is naturally embedded into X. Therefore X and X is isometrically isomorphic and hence reflexive. Inseparability of c 0 (Z + ). We know that c 0 (Z + ) l (Z + ), which is much larger than c 0 (Z + ). Thus c 0 (Z + ) is not reflexive. Inseparability of l 1 (Z + ). We know that l 1 (Z + ) is separable. If it is reflexive, then l 1 (Z + ) l 1 (Z + ) must be also separable. But by Claim 1.3, this implies that l 1 (Z + ) l (Z + ) is separable, which is a contradiction. Therefore l 1 (Z + ) is not reflexive. Inseparability of C([0, 1]). Notice that for each x [0, 1], the Dirac-delta δ x : f f (x) is a bounded linear functional on C([0, 1]): δ x ( f ) = f (x) f. Moreover, for x y we can find f C([0, 1]) such that f = 1, f (x) = 1 and f (y) = 0. This implies that δ x δ y δ x ( f ) δ y ( f ) f = 1. Thus C([0, 1]) is inseparable and the same is true for C([0, 1]). On the other hand, C([0, 1]) is separable by a simple application of Stone-Weierstrass theorem. Therefore C([0, 1]) is not reflexive. Inseparability of L 1 ([0, 1]). We know that L 1 ([0, 1]) is separable. On the other hand, it is easy to modify the argument for inseparability of l (Z + ) to show that L ([0, 1]) L 1 ([0, 1]) is inseparable. Thus L 1 ([0, 1]) cannot be reflexive. 2 Proofs Definition 2.1. p is called a complex sublinear functional on X if the followings are satisfied: p(x + y) p(x) + p(y) x, y X and p(λx) = λ p(x) λ C, x X, (2.1) Lemma 2.2. Let M be a closed proper subspace of the Banach space X and x 0 X \ M. Then there exists ϕ X such that ϕ(y) = 0 for y M, ϕ(x 0 ) = dist(x 0, M), and ϕ 1. 5
Proof. It is easy to check that dist(x, M) = inf y M x y is a complex sublinear functional on X. Now let d = dist(x 0, M) and consider the linear functional ϕ 0 : Cx 0 C given by ϕ 0 (λx 0 ) = λd. Since ϕ 0 (λx 0 ) = λ d = dist(λx 0, M), we can follow the proof of Theorem 1.26 mutandis mutatis to construct a linear functional ϕ that extends ϕ 0. Indeed, consider Re ϕ 0 as a R-linear functional, extend this to a R-linear functional ψ on X (X considered as a R-normed space) dominated by the sublinear functional dist(, M), and define φ(x) = ψ(x) iψ(ix). Then we notice that ϕ(x 0 ) = ϕ 0 (x 0 ) = d. To show that ϕ is bounded and lies in the unit ball (X ) 1, write ϕ(x) = re iθ and notice that ϕ(x) = e iθ ϕ(x) = ϕ(e iθ x) = ψ(e iθ x) dist(e iθ x, M) = dist(x, M) x. The inequality above also shows that ϕ(y) dist(y, M) = 0 for y M. This completes the proof. 6