## 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 $K$.

## 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, $U \ne \mathcal{N}$.

Posted in Descriptive Set Theory, Kechris | 2 Comments

## 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 $\| T \| \le 1$. $\blacksquare$

## 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.

$\blacksquare$

## 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 T$$T$ 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. $\blacksquare$

## 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.

## 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 $A$.