## 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.