Questions

I have been unable to solve the following problems from Kechris.  Any hints, advice, or solutions would be appreciated.

1. Exercise 4.9: Let X, Y be separable Banach spaces.  The weak topology on L(X,Y) is the one generated by the functions (from L(X,Y) into the scalar field) T \mapsto \langle Tx, y^* \rangle ; x \in X, y^* \in Y^*.  Show that if Y is reflexive, L_1(X,Y) = \{ T \in L(X,Y) : \| T \| \le 1 \} with the weak topology is compact metrizable.  Find a compatible metric.

2. Exercise 4.10:

A topological vector space is a vector space X (over \mathbb{K}, where \mathbb{K} = \mathbb{R} \text{ or } \mathbb{C}), equipped with a topology in which addition and scalar multiplication are continuous.  So Banach spaces and their duals with the weak*-topology are topological vector spaces. A point x in a convex set K is extreme (in K) if x = \lambda y + (1 - \lambda)z, with 0 < \lambda < 1, y,z \in K, implies y = z = x.  Denote by \partial_e K the extreme boundary of K, i.e., the set of extreme points of K.

Show that if K is a compact metrizable (in the relative topology) convex subset of a topological vector space, the the set \partial_e K is G_\delta in K.

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On trees: Compactness and finite splitting

Terms and notation are defined here.

(Kechris exercise 4.11)
(a) Let T be a pruned tree on A.  It follows that [T] is compact iff T is finite splitting.
(b) In particular, if K \subset \mathcal{N} = \mathbb{N}^\mathbb{N} is compact, there is x \in \mathcal{N} such that for all y \in K, y(n) \le x(n) for every n.  Conclude that \mathcal{N} is not a countable union of compact sets.

Proof.

(a) “\Rightarrow”. Suppose T is not finite splitting.  So there exists s \in T, of length n - 1, such that infinitely many a \in A satisfy s \land a \in T.  For each a here, there exists x_a \in [T] such that x_a|n = s \land a \in T. Pick sequence (x_{a_i})_{i = 0}^\infty.  Then, d(x_{a_i}, x_{a_j}) = 2^{-n - 1}, for any i, j \in \mathbb{N}.  Hence, there is no convergent subsequence.  Hence, [T] is not compact.

\Leftarrow”.  Suppose T is finite splitting.  Let (x_n) be a sequence in [T].  If s \in [T], k \in \mathbb{N}, let P(s, k) be the following statement: B_{s}^k := \{ t \in (x_n) : t|k = s | k \} is an infinite
set. Then, it is clear that there exists y_1 \in (x_n) such that P(y_1, 1) holds, since there are only finitely many sequences of length 1 in T (since T is finite splitting).

Now, assume inductively that there exists y_k \in (x_n) such that P(y_k, k) holds.  Then, again since T is finite splitting, there are only finitely many choices for y_k(k + 1).  Hence, since B_{y_k}^k is infinite, there must exist y_{k + 1} \in B_{y_k}^k such that P(y_{k + 1}, k + 1)
holds.

Hence we have inductively defined a sequence (y_k) \subset (x_n) such that y_k \to s for some s \in [T].  That is, (x_n) contains a convergent subsequence, so [T] is compact.

(b) Let K \subset \mathcal{N} be compact.  Then K is closed, and hence corresponds to a pruned tree T, such that K = [T].  So, for each n, \{ y(n): y \in K \} := A_n is finite (by the above).  So define x \in \mathcal{N} by x(n) = \text{ max}A_n.  This x satisfies the above property.

Now, suppose U = \bigcup_{i = 0}^\infty K_i, where each K_i is compact.  Then for each K_i we have x_i such that y \in K_i implies y(n) \le x_i(n) for all n.  Now define z: \mathbb{N} \to \mathbb{N} by z(n) = x_n(n) + 1.  Then, z \notin K_i for any i.  Therefore, U \ne \mathcal{N}.

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Linear extension of “linear” map on dense subset of Banach space

Let X, Y be Banach spaces over the field \mathbb{F} (= \mathbb{R} \text{ or } \mathbb{C}).  Let Q be dense in \mathbb{F}, D be dense in X, such that D is closed under Q-linear combinations.  Let f \in Y^D, such that for all x, y \in D, p, q \in Q, f(px + qy) = pf(x) + qf(y) and for all x \in D, \| f(x) \| \le \| x \|. Then, there exists T \in L_1(X,Y) such that T|_D = f.
Proof.
(a)

First, we define T:X \to Y.  Let x \in X. There exists (d_n) \subset D, such that d_n \to x.  Define T(x) := \lim_{n \to \infty} f( d_n ). We need to check that T is well-defined.  Let e_n \to x, e_n \in D.  Now, let \epsilon > 0.  Then, since (d_n) is Cauchy, there exists N such that n, m \ge N implies \| d_n - d_m \| < \epsilon.  Therefore, \| f(d_n - d_m) \| \le \| d_n - d_m \| < \epsilon, so (f(d_n)) converges in Y.  Now, there exists N such that n, m \ge N implies \| d_n - x \| < \epsilon, \| e_m - x \| < \epsilon. Therefore, \| e_m - d_n \| < 2 \epsilon, by the triangle inequality.  So \|f(e_m) - f(d_n) \| < 2 \epsilon. Hence
\| \lim_{m \to \infty} f( e_m ) - \lim_{n \to \infty} f( d_n ) \| \le 2 \epsilon.
Since \epsilon, x were arbitrary we see that T is well-defined on X.
(b) 

Now, we check that T is linear.
Let x, y \in X.  Then, x = \lim_{n \to \infty} d_n, y = \lim_{n \to \infty} e_n,  where \{ d_n, e_n : n \in \mathbb{N} \} \subset D. Then, T( x + y ) = \lim_{n \to \infty} f( d_n + e_n ) = \lim_{n \to \infty} f( d_n ) + f( e_n ) = T(x) + T(y), so T is additive. The demonstration that T(rx) = rT(x), for r \in \mathbb{F} is quite similar.
(c)

Now, we need to show that T is bounded.  \| T(x) \| = \| \lim_{n \to \infty} f( d_n ) \| = \lim_{n \to \infty} \| f( d_n ) \| \le \lim_{n \to \infty} \| d_n \| = \| x \|. Thus T is bounded with \| T \| \le 1. \blacksquare

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König’s Lemma (p20 (A))

(Exercise 4.12)
Let T be a tree on A.  If T is finite splitting, then [T] \neq \emptyset iff T is infinite.  Show that this fails if T is not finite splitting.

Proof.

\Rightarrow”. Suppose T is finite.  Then, obviously, [T] = \emptyset.
\Leftarrow”. Suppose T is infinite.  Since T is finite splitting, for every n \in \mathbb{N}, the number of strings in T of length n is finite. (see Fact here)

Now, there is no terminal string of maximal length. For if there were, T would be finite.  That is, for each n, there exists terminal string s^n, where \text{length}(s^n) = m \ge n.  Let C := \{ s^n : n \in \mathbb{N} \}. Now, there must be an infinite subset B_1 \subset C such that r, s \in B_1 implies r|1 = s|1, for otherwise the fact that there are only finitely many sequences of length 1 would be contradicted.  Define t^1 := r|1.

Now, suppose that for k \in \mathbb{N}, there exists B_k \subset B_{k - 1} such that B_k is infinite, r, s \in B_k implies s|k = r|k := t^k, and t^k \supset t^{k - 1}.  Now, there must exist infinite B_{k + 1} \subset B_k such that r, s \in B_{k + 1} implies r|(k + 1) = s|(k + 1) := t^{k + 1}, for otherwise the fact that there are only finitely many sequences of length k + 1 would be contradicted. Also, t^{k + 1} \supset t^k.

So, we have inductively constructed a sequence (t^n) \subset T with t^n \subset t^{n + 1} for all n.  Thus, there exists x \in A^{\mathbb{N}} such that x|n = t^n \in T; that is [T] \neq \emptyset.

Finally, define T := \{ \emptyset \} \cup \{ (n) : n \in \mathbb{N} \}.  Then T is infinite and [T] = \emptyset.  But T is not finite splitting.

\blacksquare

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Trees

Definitions and notation for trees.

A^{< \mathbb{N}} = \bigcup_n A^n.  If x \in A^n \cup A^\mathbb{N}, and k \in \mathbb{N}, x|k = (x_0, \ldots, x_{k - 1}) \in A^k.  If s \in A^n, t \in A^m, the concatenation of s and t is defined by s \land t = (s_0, \ldots, s_{n - 1}, t_0, \ldots, t_{m -1}) \in A^{n + m}.  If s \in A^n, t \in A^m, then s is an initial segment of t and t is an extension of s (written s \subset t), if s = t | k, for some k \le m.

A tree on a set A is a subset T \subset A^{< \mathbb{N}} closed under initial segments. The body of T, [T] = \{ x \in A^{\mathbb{N}} : \forall n(x | n \in T) \}.  An element s of T is called terminal if s \in T and s \land a \notin T for all a \in A.

Let T is a tree on A. T is finite splitting if for every s \in T there only finitely many a \in A such that s \land a \in TT is pruned if every s \in T has a proper extension t \supsetneqq s, t \in T.

Fact.   If T is finite splitting, for every n \in \mathbb{N}, the number of strings in T of length n is finite.
Proof.
Assume that for some n, the number of strings in T of length n is infinite.  Then, there must be a minimal number k \le n, which satisfies the property that infinitely many of the strings differ at position k, but agree at all previous positions.  Then T is not finite splitting. \blacksquare

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A complete, compatible metric for the strong topology (p15 (A))

Let X, Y be separable Banach spaces over \mathbb{F} (= \mathbb{R} \text{ or } \mathbb{C}).
Show that the following is a complete, compatible metric for the strong topology \tau_s on L_1(X,Y) = \{ T: X \to Y : T \text{ is linear }, \| T \| \le 1 \}:
d(S,T) = \sum_{n = 0}^\infty 2^{-n-1} \| (S - T)( x_n ) \|,
where (x_n) is a dense sequence in the unit ball of X.

Proof.
It is not difficult to check that distance function given above is a metric.  Denote the topology induced by the metric by \tau_d.

Lemma 1. Let T \in L_1(X,Y).  Given N, \epsilon, there is a ball B around T in \tau_d, such that
S \in B, n < N implies \| (S - T)(x_n) \| < \epsilon.
Proof of Lemma 1.
Let \epsilon > 0, N \in \mathbb{N}.  If \eta is a positive real number, then
d(S,T) = \sum_{n=0}^{\infty} 2^{-n-1} \|(S-T)(x_n)\| < \eta
implies
\sum_{n = 0}^{N} 2^{-n-1} \|(S-T)(x_n)\| < \eta.
Hence,
2^{N + 1} \left( \sum_{n = 0}^N 2^{-n-1} \| (S - T)(x_n) \| \right) < 2^{N+1} \eta,
which is the same as
\sum_{n = 0}^N 2^{N - n} \|(S-T)(x_n)\| < 2^{N + 1} \eta.
Hence, n < N implies \|(S - T)(x_n) \| < 2^{N + 1} \eta.
So if we let \eta = \frac{\epsilon}{2^{N+1}}, we have found a ball with the above property.

Lemma 2. \tau_d \subset \tau_s.

Proof of Lemma 2.

Let V_{y_1, \ldots, y_m; \epsilon; T} = \{ S \in L_1(X,Y) : \| (S - T)(y_1) \| < \epsilon, \ldots, \| (S - T)(y_m) \| < \epsilon \} be a basis element for \tau_s.
Let i \in \{1, \ldots, m \}.  Let \hat{y}_i := \frac{y_i}{\| y_i \|}. Let \eta > 0. Since (x_n) is dense in the unit ball of X,
there exists x_K such that \| \hat{y}_i - x_K \| < \eta / 4.
Now, for any S \in L_1(X,Y),
\|(S-T)(x_K) - (S-T)(\hat{y}_i)\| = \| (S-T)(x_K - \hat{y}_i) \| \le \|S - T \|\|x_K - \hat{y}_i \| < \eta / 2.
Therefore, | \| (S-T)(x_K) \| - \| (S - T)(\hat{y}_i) \| | < \eta / 2.
By Lemma 1, there is a ball B around T such that S \in B implies \|(S - T)(x_K) \| < \eta / 2.
Then, for S \in B, \|(S-T)(\hat{y}_i)\| < \eta, which yields \|(S-T)(y_i)\| < \eta \| y_i \|.
Therefore, if we choose \eta = \epsilon / \| y_i \|, we have B_i such that S \in B_i implies \|(S-T)(y_i)\| < \epsilon.

Let B = B_1 \cap \ldots \cap B_m.  Then B \subset V.

Lemma 3. \tau_s \subset \tau_d.

Let B_T(\epsilon) be an arbitrary ball of \tau_d.  Choose N such that
\sum_{n=N}^\infty 2^{-n-1}(2) < \epsilon / 2.
Let S \in V_{x_0, \ldots, x_{N-1}; \epsilon / (2N);T}.  Then n < N \Rightarrow \|(S-T)(x_n)\| < \epsilon / (2N).
So

\sum_{n=0}^\infty 2^{-n-1} \|(S - T)(x_n) \| = \\ \sum_{n=0}^{N - 1} 2^{-(n + 1)} \|(S - T)(x_n) \| + \sum_{n = N}^\infty 2^{-(n + 1)} \|(S - T)(x_n) \| < N \cdot \frac{\epsilon}{2N} + \frac{\epsilon}{2} = \epsilon

That is, d(S,T) < \epsilon, so S \in B_T(\epsilon).

By Lemmas 2 and 3, \tau_d = \tau_s.

Finally, we check that d is a complete metric. Let (S_n) be a Cauchy sequence in d. That is, for all \epsilon > 0 there exists
N such that m,n \ge N implies d(S_n,S_m) < \epsilon. That is,
\sum_{k=0}^\infty 2^{-k-1} \|(S_n - S_m)(x_k)\| < \epsilon.
So, in particular, 2^{-k-1} \|(S_n - S_m)(x_k)\| < \epsilon if n,m \ge N.  So (S_n(x_k))_{n=0}^\infty is a Cauchy sequence in Y.
So it converges to some y \in Y, which we will name S(x_k).   This defines S \in Y^D.  Now, consider the set \hat{D} := \{ \sum_{i = 1}^m \alpha_i x_{n_i} : \alpha_i \in \mathbb{F}, m \in \mathbb{N} \}.  It is closed under \mathbb{F}-linear combinations, and we easily extend S to \hat{D}, by defining S(d) = \lim_{n \to \infty} S_n(d), for d \in \hat{D}; it is easily verified S satisfies S(\alpha_1 d_1 + \alpha_2 d_2) = \alpha_1 S( d_1 ) + \alpha_2 S( d_2 ), for \alpha_1, \alpha_2 \in \mathbb{F}, d_1, d_2 \in \hat{D}.  Furthermore, it is clear that \hat{D} is dense in X.  Then, by this result, S extends to a linear map in L_1(X, Y), which we will also call S.

Then,
\lim_{n \to \infty} \sum_{k = 0}^\infty 2^{-k-1} \|(S_n - S_m)(x_k)\| = \sum_{k=0}^\infty \lim_{n \to \infty} 2^{-k-1} \| (S_n - S_m)(x_k) \|
= \sum_{k = 0}^\infty 2^{-k-1} \|(S - S_m)(x_k) \| = d(S , S_m),

by the Dominated Convergence Theorem.  If m > N, we see that d(S, S_m) < \epsilon.
Thus (S_m) \to S, and d is a complete metric.

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Any subspace of a separable metric space is separable

Let X be a separable, metric space, A \subset X, and let Q \subset X be countable and dense.  For each q \in Q, n \in \mathbb{N}, choose a_n^q \in A \cap B_{1/n}(q) if such intersection is nonempty.  Then, A^* = \bigcup_{n,q} \{ a_n^q \} is a countable subset of A.

Let a \in A, 0 < \epsilon < 1. Then choose q \in B_{\epsilon / 4}(a) \cap Q.  Now, choose N such that \epsilon / 4 \le 1 / N \le \epsilon / 2.  Then, B_{1/N}(q) \cap A \neq \emptyset, so there exists a^* \in A^*, such that d(a^*, q) < 1/N.

Now d(a^*,a) \le d(a^*,q) + d(q,a) < \epsilon / 2 + \epsilon / 4 < \epsilon.  So a^* \in B_\epsilon(a) \cap A.  Hence, the closure of A^* in A is equal to A.

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