I am trying to prove that the set of continuous (i.e. bounded) linear maps $B(V,V)$ is complete with respect to the operator norm, where $V$ is a complete normed vector space.
I have got this far:
Let $(l_n)_n$ be a Cauchy sequence, where $l_n \in B(V,V)$. Define $f_{ni}= l_n e_i$ where $(e_i)_i$ is a standard basis of $V$.
Then $\forall \epsilon > 0$, $\exists N$ such that $n,m > N \implies \|l_n - l_m\|_{op} < \varepsilon$.
So, using implicit summation, $\sup_{\|v\|=1} \|l_n v - l_m v\| = \sup_{\|v\|=1} |v_i|\|f_{ni} - f_{mi}\| \le \sum_i \| f_{ni} - f_{mi}\|< \varepsilon$.
Hence $\|f_{ni}-f_{mi}\|<\varepsilon$ so $(f_{ni})_n$ is a Cauchy sequence for each $i$.
$V$ is complete so $(f_{ni})_n$ converges. Define $f_i = \lim_{n \to \infty} f_{ni}$.
Now define a linear map $l$, which sends $e_i \mapsto f_i$. It remains to show that $l_i \to l$ under the operator norm and that $l \in B(V,V)$ i.e. that $l$ is bounded. The latter is simple to do once convergence has been shown.
Let $\varepsilon > 0$. I need to show that there exists $N$ such that if $n > N$ then $\sup_{\| v \| = 1} \| (l_n - l) v \| < \varepsilon$
We have $$\| (l_n - l) v \| = \| (l_n - l_m) v \| + \| (l_m - l) v \|$$
For all $v$ with $\| v \| = 1$, the first term on the RHS can be made arbitrarily small because $(l_n)_n$ is a Cauchy sequence by assumption. I just need to show that for a fixed $v$ with $\| v \| = 1$ that $(l_m v)_m \to lv$, which boils down to showing that $ \sum_i \| f_{mi} - f_i\|< \varepsilon$ for $m$ sufficiently large. I am not sure how to prove this though... any hints would be appreciated.
Let $(f_n)_n $ be a Cauchy sequence in $B(V, V) $. So $||f_n-f_m||<\epsilon $ for $n, m>N$ for some $N$. So $$||(f_n-f_m) (x) ||< \epsilon ||x|| $$ for all $x \in V$. So from this you can show that for each $x$ in $V$ that $(f_n(x)) _n $ is a Cauchy sequence in $V$. Now you can use that $V$ complete.