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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This entry was posted in Analysis, Descriptive Set Theory, Kechris, Metric Spaces, Separability, Topology. Bookmark the permalink.

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