Show that the following metric is complete on $L_1(X,Y)$

Question

I am working through the exercises of Kechris' Classical Descriptive Set Theory for personal edification. I've solved Exercise 3.5, but the proof was two pages. Did I miss a more straightforward method (either a "ground-up" approach, or one where the proof follows straightforwardly from a larger result)?

(3.5) Show that the following is a complete compatible metric for the strong topology on $L_1(X,Y)$: $$d(S,T) = \sum_{n=0}^\infty 2^{-n-1} \|(S-T)(x_n)\|$$ where $\{x_n\}$ is a dense sequence in the unit ball of $X$.

Above, $X$ and $Y$ are Banach spaces and $L_1(X,Y)$ is defined as the space of bounded linear operators $X\to Y$ with norm $\leq 1$, that is, the unit ball in $L(X,Y)$. The strong topology is defined as the topology on $L(X,Y)$ generated by sets of the form $$V_{y_1,\ldots,y_m;\varepsilon;T}:=\{S\in L(X,Y):\|Sy_1-Ty_1\|<\varepsilon,\ldots,\|Sy_m-Ty_m\|<\varepsilon\}$$ for $y_1,\ldots,y_m\in X$, $\varepsilon >0$, $T\in L(X,Y)$.

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Here is a summary of my proof:

1. Let $\{S_k\}_{k\in \mathbb{N}}$ be a Cauchy sequence in the metric $d$.
2. Show that $S_k$ converges pointwise on $\{x_n\}$ to a function $S:\{x_n\}\to Y$.
3. Show that $S$ is Lipschitz continuous with constant $1$.
4. Extend $S$ by continuity to the unit ball in $X$, then linearly to all of $X$, and show that the resulting function is in $L_1(X,Y)$.
5. Show that $S_k \to S$ in the $d$ metric (hence $d$ is complete).
6. Given a basis set in the strong topology, show that a $d$-open ball can fit inside it.
7. Given a $d$-open ball, show that a strong topology basis set can fit inside it.

(I can also post the proof as an answer here if it's useful).

Answer 1

Let's start with a $d$-Cauchy $(S_k)_{k=1}^\infty$ and, as you have already done, show that $\lim_k S_kx_n$ exists for all $n=0,1,2,\ldots$.

Let $(M,d)$, $(N,\rho)$ be metric spaces with $(N,\rho)$ complete and let $(f_n)_{n=1}^\infty$ be $1$-Lipschitz functions which converge pointwise on a dense subset $A$ of $M$. Fix $\epsilon>0$, $x\in M$, and $y\in A$ such that $d(x,y)<\epsilon$. Then \begin{align*} \underset{k,l}{\lim\sup}\rho(f_k(x),f_l(x)) & \leqslant \underset{k,l}{\lim\sup}\text{ }\rho(f_k(x),f_k(y))+\rho(f_k(y),f_l(y))+\rho(f_l(y),f_l(x)) \\ & = \underset{k,l}{\lim\sup}\text{ }\rho(f_k(x),f_k(y))+\rho(f_l(y),f_l(x)) \leqslant 2\epsilon.\end{align*} First inequality is triangle inequality, second is because the functions converge at $y\in A$, and the last is because $f_k,f_l$ are $1$-Lipschitz and $d(x,y)<\epsilon$. This shows that $(f_k(x))_{k=1}^\infty$ is Cauchy, and therefore convergent in $N$ by completeness. Therefore $(f_k)_{k=1}^\infty$ is pointwise convergent on all of $M$.

Apply this with $f_k=S_k$, $A=\text{span}\{x_n:n\in\mathbb{N}\}$, $M=X$, and $N=Y$. We note that $\lim_k S_k\sum_{x\in F}a_xx=\sum_{x\in F}a_x\lim_k S_kx$ for any linear combination $\sum_{x\in F}a_xx$ of members $F$ of $\{x_n:n\in\mathbb{N}\}$, so that pointwise convergence on $\{x_n:n\in \mathbb{N}\}$ and linearity give convergence on the entire span.

Then define $Sx=\lim_k S_kx$. Note that the pointwise limit of $1$-Lipschitz functions is $1$-Lipschitz (true on general metric spaces) and the limit of linear operators is linear. For the first fact, $\|Sx-Sy\|=\lim_k \|S_kx-S_ky\|\leqslant \|x-y\|$. For the second, $$S(ax+by)=\lim_k S_k(ax+by)=a\lim_k S_kx+b\lim_k S_ky=aSx+bSy.$$ So $S\in L_1(X,Y)$.

Then I would do 6 and 7 (or its equivalent with nets) without using completeness of $d$. Then we get that $\lim_k d(S,S_k)=0$ for free, because this is the same as saying $S_k\to S$ pointwise, which we already know it does by the preceding paragraphs