$\newcommand{\N}{\mathbf N}\renewcommand{\leq}{\leqslant}\renewcommand{\geq}{\geqslant} \newcommand{\eps}{\varepsilon}$I was looking through the functional analysis notes of TWK (on his webpage https://www.dpmms.cam.ac.uk/~twk/) and was trying to answer the following question: If $X$ is a complete normed space and $Y$ is a normed space, is $\mathrm B(X,Y)$ complete? ($\mathrm B(X,Y)$ is the normed space of bounded linear operators from $X$ to $Y$, with as norm the operator norm). It seems to me that this is not the case, and I have thought of (what I think is) a counterexample. I'm not completely sure about the last step I made and therefore would like to ask you to verify that it is correct, show me how to correct it, or tell me that my counterexample is wrong and why.
Claim If $X$ is a complete normed space and $Y$ a normed space, $\mathrm B(X,Y)$ is not necessarily complete:
Proof Consider $X=\ell^\infty(\N),Y=c_{00}(\N)$ ($c_{00}(\N)$ is the set of sequences of compact support), with the norm on $Y$ the $\ell^1$ norm. Let, for $n\in\N_{\geq 1}$, $T_n:X\to Y$ be given by $$(x_1,x_2,x_3,\ldots,x_n,x_{n+1},\ldots)\mapsto\left(\frac{x_1}{1^2},\frac{x_2}{2^2},\frac{x_3}{3^2},\ldots,\frac{x_n}{n^2},0,0,\ldots\right)$$ For every $n\in \N_{\geq 1}$, $T_n$ is linear. Furthermore each $T_n$ is bounded: Let $x=(x_n)\in X$. Then $\|x\|=\sup|x_n|<\infty$. So $$\|T_nx\|=\sum\limits_{k=1}^n\frac{|x_k|}{k^2}\leq \sum\limits_{k=1}^n\frac{\sup|x_n|}{k^2}=\sum\limits_{k=1}^n\frac{1}{k^2}\|x\|\leq\frac{\pi^2}{6}\|x\|$$ We conclude that for all $n\in\N_{\geq 1}, T_n\in \mathrm B(X,Y)$, because $$\|T_n\|\leq \frac{\pi^2}{6}$$ Furthermore $(T_n)$ is a Cauchy sequence in $\mathrm B(X,Y)$: Let $\eps>0$. $\sum_{k=1}^\infty\frac{1}{k^2}$ converges, so there is an $N\in\N$ such that $\forall m,n\geq N$ we have $\sum_{k=m}^n \frac{1}{k^2}=\left|\sum_{k=m}^n \frac{1}{k^2}\right|<\frac{\eps}{2}$. Now let $x=(x_n)\in X$. Then $\|x\|=\sup |x_n|<\infty$ so for $m,n\geq N$ (wlog we assume $m<n$), we have $$\|(T_n-T_m)x\|=\sum\limits_{k=m+1}^n\frac{|x_k|}{k^2}\leq\sum\limits_{k=m+1}^n \frac{\sup|x_n|}{k^2}<\frac{\eps}{2}\|x\|$$ And from this it follows that for $m,n\geq N$ $$\|T_n-T_m\|\leq \frac{\eps}{2}<\eps$$ Finally, $(T_n)$ does not converge: This is the step that I'm not completely sure about. My idea is as follows
We have $\mathrm B(X,\ell^1(\N))\supset \mathrm B(X,Y)$. $\mathrm B(X,\ell^1(\N))$ is complete (because $\ell^1(\N)$ is complete) and $(T_n)$ is also a Cauchy sequence in $\mathrm B(X,\ell^1(\N))$ and so it converges to a $T\in \mathrm B(X,\ell^1(\N))$. In fact we can give $T$ explicitly, for $x=(x_n)\in\ell^\infty(\N)$ we have $(Tx)_n=\frac{x_n}{n^2}$ for each $n\in\N_{\geq 1}$. Now it is clear that $T$ does not map $\ell^\infty(\N)$ into $c_{00}(\N)$ (consider the sequence with all entries equal to $1$ and what it gets mapped to) and so $T\not\in \mathrm B(X,Y)$. If $(T_n)$ would converge in $\mathrm B(X,Y)$ then it would converge to the same limit in $\mathrm B(X,\ell^1(\N))$ (because we consider $\mathrm B(X,Y)$ as a metric subspace of $\mathrm B(X,\ell^1(\N))$) so by uniqueness of limits in $\mathrm B(X,\ell^1(\N))$ we conclude that $(T_n)$ does not converge in $\mathrm B(X,Y)$.
This shows that $X$ being a complete normed space does not necessarily imply that $\mathrm B(X,Y)$ is complete.
