Let $V,W$ be normed vector spaces, and $L(V,W)$ be the space of bounded linear operators. Usually I would only see the statement "If $W$ is Banach, then $L(V,W)$ is Banach.". But Wikipedia writes that there is a converse: "If $L(V,W)$ is Banach, and if $V$ is non-trivial, then $W$ is Banach". This is pretty interesting since I never seen a converse before. I was wondering if anyone has a nice proof.
I tried reversing the proof for the usual direction, but the inequalities can't be reversed. I also tried to start with a Cauchy sequence in $W$, and construct linear operators (using Hahn-Banach to control the operator norms), but alas I cannot say much about the distance between these operators (much less say that it is Cauchy)
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.