Let $B(V,W)$ be the space of bounded linear maps from $V$ to $W$. Then it is complete with respect to the operator norm. Can you tell me if my proof is correct? Thanks.
It's easy to verify that the operator defines a norm. Let $T_n$ be Cauchy in $B(V,W)$ with respect to $\|\cdot\|$. Let $\varepsilon > 0$. We want to show that $T_n \to T$ for some $T \in B(V,W)$.
We have that $T_n v$ is a Cauchy sequence in $W$ since $\|T_nv - T_mv\|_W \leq \|T_n-T_m\| \|v\|_V < \varepsilon$ for $n,m$ large enough since $T_n$ is Cauchy with respect to $\|\cdot\|$ by assumption. Since $W$ is complete, the pointwise limit $Tv$ of $T_nv$ is in $W$ for all $v$. Now we need to show that $T$ is linear, bounded and $T_n \to T$ in the operator norm.
(i) $T(\alpha v + \beta w) = \lim_{n \to \infty} T_n(\alpha v + \beta w) = \alpha \lim_{n \to \infty} T_nv + \beta \lim_{n \to \infty} T_n w = \alpha T v + \beta T w $
(ii) $\|T\|=\sup_{\|v\|\leq 1}\|Tv\| = \sup_{\|v\| \leq 1} \|Tv + T_n v - T_n v\| \leq \sup_{\|v\| \leq 1} \|Tv - T_n v\| + \| T_n v\|$
(iii) $\|(T-T_n)v \|_W < \varepsilon $ for all $v$ if $n$ large enough, hence for $n$ large enough, $\|T-T_n\| < \varepsilon $