# Math 209B Homework 2

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1 Math 29B Homework 2 Edward Burkard Note: All vector spaces are over the field F = R or C 4.6. Two Compactness Theorems. 4. Point Set Topology Exercise 6 The product of countably many sequentally compact spaces is sequentially compact. (Use the diagonal trick as in the proof of Theorem 4.44.) Exercise 64 Let (X, ρ) be a metric space. A function f C(X) is called Hölder continuous of exponent α (α > ) if the quantity f(x) f(y) N α (f) = sup x y ρ(x, y) α is finite. If X is compact, {f C(X) f u and N α (f) } is compact in C(X). Let F = {f C(X) f u and N α (f) }. Let f F. Then we have that N α (f) implies that f(x) f(y) ρ(x, y) α. Let ɛ > and let x X. Then to show equicontinuity, let U = B (x), hence: ɛ α f(x) f(y) (ɛ α ) α = ɛ which implies, since x was arbitrary, F is equicontinuous on X. Recall that f u = sup x X f(x). So since every function in F satisfies f u it is clear that F is uniformly bounded. So by the Arzelà-Ascoli Theorem I we have that, since X is compact and Hausdorff (since it is a metric space), the closure of F in C(X) is compact. So to prove the claim it remains to show that F is closed. Let {f n } be a sequence in F converging to some function f. Observe: f u = lim f n u = sup lim f n (x) x X = sup lim f n (x) x X = lim sup f n (x) x X = lim = N α (f) = sup x y lim f n (x) lim f n (y) ρ α (x, y) lim(f n (x) f n (y)) = sup x y ρ α (x, y) = sup lim (f n(x) f n (y)) x y ρ α (x, y) (f n (x) f n (y)) = lim sup x y ρ α (x, y) = lim = Hence F is closed, and therefore F is compact in C(X).

2 The Stone-Weierstrass Theorem. Exercise 68 Let X and Y be compact Hausdorff spaces. The algebra generated by functions of the form f(x, y) = g(x)h(y), where g C(X) and h C(Y ), is dense in C(X Y ). Since both X and Y are compact and Hausdorff, by Tychonoff s theorem and Proposition 4. we have that X Y is compact and Hausdorff. Now consider the algebra described above. Call it A. Clearly A contains the constant functions as we can let g and h both be constants and hence we have an f(x, y) = r for all r R. Also clearly A separates points (this can be done by letting g(x) =, h(x) = 2, k(y) = 3 and given any two points (a, b) and (c, d) plugging them into f (x, y) = g(x)k(y) = 3 and f 2 (x, y) = h(x)k(y) = 6 respectively). Since A is closed by definition, we can apply the Stone-Weierstrass Theorem to it to conclude that A is dense in C(X Y ). 5.. Normed Vector Spaces. Exercise 3 Complete the proof of Proposition Elements of Functional Analysis We need to verify that T := lim T n is linear and T n T. First: Hence T is linear. Now check the norm of T : T (ax + by) = lim T n (ax + by) = lim(t n (ax) + T n (by)) = lim(at n (x) + bt n (y)) = lim(at n (x)) + lim(bt n (y)) = a lim T n (x) + b lim T n (y) = at (x) + bt (y) T = sup{ T x x = } = sup{ lim n T nx x = } = sup{ lim n T nx x = } = lim n sup{ T nx x = } = lim n T n Now fix x X and let ɛ >, then since {T n } is Cauchy there is an N N such that for all m, n N: Notice: T n (x) T m (x) ɛ T n (x) T (x) = T n (x) lim m T m(x) = lim m (T n(x) T m (x)) = lim m T n(x) T m (x) < ɛ Letting ɛ we see that T n (x) T (x). So since x was arbitrary we have T n T. Exercise 6 Suppose that X is a finite dimensional vector space. Let e,..., e n be a basis for X, and define n a je j = n a j. is a norm on X. The map (a,..., a n ) n a je j is continuous from F n with the usual Euclidean topology to X with the topology defined by. c. {x X x = } is compact in the topology defined by. d. All norms on X are equivalent. (Compare any norm to.)

3 3 a j e j + b j e j = = n λ a j e j = a j + b j a j + b j a j e j + b j e j λa j = λ a j = λ a j e j Now if n a je j =, then we have n a j =, which is a sum of nonnegative numbers, hence a j = for all j, thus n a je j =. So we conclude that is a norm. Define the map η : F n X by η(a,..., a n ) = a j e j. Using the norm (a,..., a n ) = a j on F n suggested by Folland on page 53 we can see: (a,..., a n ) = a j = η(a,..., a n ) = a j e j which makes it obvious that η is continuous. c. Let K = {(a i,..., a j ) (a,..., a j ) = }. Then K is clearly compact in F n. Now since we have η(k) = {x X x = } (by the definition of η) we can see that η(k) is compact as it is the continuous image of a compact set. Thus we have our conclusion. d. Let be any norm on X. Define M = max( e,..., e n ). Then we have a j e j a j e j = a j e j M a j = M a j e j. Thus we have x M x for all x X. As this is true, we can conclude that the map sending one norm to the other, in this case use ζ : (X, ) (X, ), is continuous (this is really just a dialation of the space). Recall the set K from the previous part. Since K is compact and ζ is continuous, ζ(k) is compact. By the extreme value theorem ζ assumes a minimum value as measured by, so we may as well say assumes a minimum value on K, call it m. Since is a norm, m and since x / K, m >. Given any element x X, we can construct the element x which is in K and hence x x m which implies m x x and so: m x x M x. Since this holds for any x X we have that is equivalent to. Moreover, since was arbitrary, every norm on X is equivalent to. Exercise 7 Let X be a Banach space. If T L(X, X ) and I T < where I is the identity operator, then T is invertible; in fact, the series (I T )n converges in L(X, X ) to T. If T L(X, X ) is invertible and S T < T, then S is invertible. Thus the set of invertible operators is open in L(X, X ).

4 4 Since X is a Banach space we have that it is complete, and hence by Proposition 5.4 L(X, X ) is complete. Let T L(X, X ) be such that I T <. Consider the sum (I T )n which is in L(X, X ) since it is an absolutely convergent sum (since I T < and hence I T n is a convergent geometric series) by Theorem 5.. Notice that showing I = T (I T )n proves that T exists and in fact T = (I T )n. A simple use of induction shows that T N (I T )n = I (I T ) N+. Now notice: N I (T (I T ) n ) = I (I (I T )N+ ) = (I T ) N+ I T N + which gives that I (T N (I T )n ) as N since I T <, hence I T (I T ) n = and so or equivalently T (I T ) n = I T = (I T ) n. Thus T is invertible. Let T L(X, X ) be invertible and let S L(X, X ) be such that S T < T (an equivalent statement: T S T < ). So: I ST = T (T S) T S T <. By the proof in part a, we have that ST is invertible, hence: I = (ST )(ST ) = S(T (ST ) ) and thus S = T (ST ). This proves that for any S in L(X, X ) sufficiently close to T, S is invertible, hence the set of invertible operators is open in L(X, X ). Exercise If < α, let Λ α ([, ]) be the space of Hölder continuous functions of exponent α on [, ]. That is, f Λ α ([, ]) iff f Λα <, where f(x) f(y) f Λα = f() + sup x,y [,], x y x y α. Λα is a norm that makes Λ α ([, ]) into a Banach space. Let λ α ([, ]) be the set of all f Λ α ([, ]) such that f(x) f(y) x y α as x y, for all y [, ]. If α <, λ α ([, ]) is an infinite-dimensional closed subspace of Λ α ([, ]). If α =, λ α ([, ]) contains only constant functions. Exercise 2 Let X be a normed vector space and M a proper closed subspace of X. x + M = inf{ x + y y M} is a norm on X /M. For any ɛ > there exists x X such that x = and x + M ɛ. c. The projection map π(x) = x + M from X to X /M has norm. d. If X is complete, so is X /M. (Use Theorem 5.) e. The topology defined by the quotient norm is the quotient topology as defined in Exercise 28.

5 5 Let x + M, y + M X /M. Then: (x + y) + M = inf{ (x + y) + z z M} { = inf (x + 2 ) ( z + y + ) } 2 z z M {( since is a norm on X inf x + ) ( 2 z + y + ) } 2 z z M let w = z 2 = inf{ x + w w M} + inf{ y + z w M} = x + M + y + M Thus since x and y were arbitrary, satisfies the triangle inequality for all x + M, y + M X /M. Now let x + M X /M and let c R. Then: c(x + M) = inf{ c(x + y) y M} since is a norm on X = inf{ c x + y y M} = c inf{ x + y y M} = c x + M Now suppose that x + M =. Then clearly x M, and hence x + M = in X /M, because if it were not, then it would be impossible for x + M to be as we are taking the infimum over vectors in M and x / M. Thus it is a norm on X /M. As M is a proper closed subset, we can find an x X /M such that x + M = a >. Define y = ax, then y + M =. Now by the definition of infimum we have that there is some m M with y + m + ɛ 2. Thus there is a element Υ B ɛ(y + m) with Υ =. So we have y + M Υ + M Υ y + M = Υ (y + m) + M Υ (y + m) ɛ Yielding the inequality: or equivalently c. Observe: d. e. y + M Υ + M = Υ + M ɛ π(x) = x + M = ɛ Υ + M. { x if x / M if x M Hence π(x) x which means π. However part b tells us π and hence π = Linear Functionals. Exercise 7 A linear functional f on a normed vector space X is bounded iff f ({}) is closed. (Use Exercise 2) (= ) Assume that f is bounded. Then it is continuous. So since {} is closed in R and f is continuous, f ({}) is closed as desired. ( =) Suppose that f were not bounded. Then n > there is an x n X with x n = and f(x n ) > 2 n. ( ) Define y n = xn f(x. Then y x n) n = n f(x = n) f(x x n) 2 and f(y n n ) = f n f(x n) Now fix x X and consider the sequence a n = x f(x)y n. Then a n x as n since y n as n. However, f(a n ) = f(x) f(x)f(y n ) = f(x) f(x) =. Thus we have a n f ({}). This would imply that f ({}) is dense in X. So there are two choices, either f : X {} or f ({}) is not closed. Since the first choice is obviously not true, it must be that f ({}) is not closed. Hence by contrapositive we have our desired conclusion. = f(x n) f(x n) =.

6 6 Exercise 8 Let X be a normed vector space. If M is a closed subspace and x X /M then M + Cx is closed. (Use Theorem 5.8) Every finite-dimensional subspace of X is closed. Exercise 9 Let X be an infinite-dimensional normed vector space. There is a sequence {x j } in X such that x j = for all j and x j x k 2 x j inductively, using Exercised 2b and 8.) X is not locally compact. for j k. (Construct Exercise 25 If X is a Banach space and X is separable, then X is separable. (Let {f n } be a countable dense subset of X. For each n choose x n X with x n = and f n (x n ) 2 f n. Then the linear combinations of {x n } are dense in X.) Note: Separability of X does not imply separability of X The Baire Category Theorem and Its Consequences. Exercise 27 There exist meager subsets of R whose complements have Lebesgue measure zero. Exercise 29 Let Y = L (µ) where µ is counting measure on N, and let X = {f SCY {n f(n) < }, equipped with the L norm. X is a proper dense subspace of Y ; hence X is not complete. Define T : X Y by T f(n) = nf(n). Then T is closed but not bounded. c. Let S = T. Then S : Y X is bounded and surjective, but not open. c. Exercise 3 Let Y = C([, ]) and X = C ([, ]), both equipped with the uniform norm. X is not complete. The map ( d dx) : X Y is closed (see Exercise 9) but not bounded. Consider the function f(x) = x 2 defined on [, ]. By lemma 4.47, ɛ > there is a polynomial P (x) defined on R such that P () = and x P (x) u < ɛ for x [, ]. If we let Q(x) = P ( ) x 2 we can rephrase the lemma to say ɛ > there is a polynomial Q(x) such that Q ( 2) = and x 2 Q(x) u < ɛ for x [, ]. Since these Q s are polynomials they are clearly in C ([, ]), and thus we may create a sequence of them as follows: Define Q n (x) to be one of the polynomials such that x 2 Q n (x) u < n for all n N. This creates a convergent sequence {Q n } in C ([, ]) which converges to x 2 on [, ], however x 2 / C ([, ]). Thus C ([, ]) is not complete. Exercise 32 Let and 2 be norms on the vector space X such that 2. If X is complete with respect to both norms, then the norms are equivalent.

7 5.4. Topological Vector Spaces. Exercise 45 The space C (R) of all infinitely differentiable functions on R has a Fréchet space topology with respect to which f n f iff f (k) n f (k) uniformly on compact sets for all k. Exercise 47 Suppose that X and Y are Banach spaces. If {T n } L(X, Y ) and T n T weakly (or strongly), then sup N T n <. Every weakly convergent sequence in X, and every weak -convergent sequence in X, is bounded (with respect to the norm). 7

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