If space of bounded operators L(V,W) is Banach, V nonzero, then W is Banach (note direction of implication)

Solution 1:

By Hahn-Banach, there exists a nonzero bounded linear functional $f$ on $V$. Then there exists $v_0 \in V$ with $f(v_0) \ne 0$; by rescaling we can get $f(v_0)=1$. For each $w \in W$, let $T_w \in L(V,W)$ be the operator defined by $T_w v = f(v) w$, which is a bounded operator because $f$ is a bounded functional.

Suppose $\{w_n\}$ is a Cauchy sequence in $W$. Then note that $\|T_{w_n} - T_{w_m}\|_{L(V,W)} \le \|f\|_{V^*} \|w_n - w_m\|_{W}$. Hence $\{T_{w_n}\}$ is Cauchy in $L(V,W)$ so by assumption it converges to some $T \in L(V,W)$. In particular $w_n = T_{w_n} v_0 \to T v_0$ so $w_n$ converges.