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)

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    Hint: it's enough to consider operators of rank 1. – Nate Eldredge Dec 10 '13 at 23:53
  • What's the meaning of "non-trivial" here? Nonzero? – Martin Argerami Dec 11 '13 at 0:06
  • @MartinArgerami: Yes, nonzero – suncup224 Dec 11 '13 at 3:39
  • @NateEldredge: So I tried the following approach. Let $w1,w2,...$ be a Cauchy sequence in W. Fix $v\in V$, where $v$ has norm 1. For $w_i$, we construct operator $T_i$ as follows: we send $v$ to $w_i$, and we will extend this, using Hahn-Banach, to a map $T_i$ from $V$ to $span(v)$ (which is isomorphic to $R$), such that norm of $T_i$ remains at most norm of $w_i$. But I still cannot bound norm of $(T_i-T_j)$, to show that $T_n$ gives a Cauchy sequence. – suncup224 Dec 11 '13 at 3:43
up vote 5 down vote accepted

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.

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    The existence of $f$ can be proved by taking a vector $v_0$ with norm 1 and setting $f(tv_0)=t$, so that $f$ is defined with norm 1 on the span of $v_0$, and Hahn-Banach extends this to $f\in V^\ast$ with the same norm, right? – MickG Oct 25 '15 at 11:46
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    That $\Phi$ be an isometry is because $(\Phi w)(v_0)=w$ so the norm is at least 1, and yet at most 1 because $\|(\Phi w)(v)\|=|f(v)|\|w\|$ and $|f(v)|\leq1$ always, right? – MickG Oct 25 '15 at 11:50
  • Not quite. Replace 1 by $\|w\|$ when concerning the norm of $\Phi(w)$, and $|f(v)|\leq1$ on the unit ball, where $\|\Phi(w)\|$ is the sup over that ball, so that is enough. – MickG Oct 26 '15 at 13:47

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