L(V,W) is Banach, then W is Banach

Question

Let $V,W$ normed vector spaces, $V$ not empty and with a finite dimension. Prove that $L(V,W)$ is Banach, then $W$ is also Banach.

Answer 1

First, since $V$ is finite-dimensional, the image of all transformations in $L(V, W)$ have finite rank, and are therefore bounded, so we do have the operator norm here. Otherwise I'd insist that we use $B(V, W)$ instead. Additionally, we may assume without loss of generality here that $V$ is $\mathbb{R}^n$ with the $1$-norm, for some $n$, since every two spaces finite-dimensional spaces of the same dimension, regardless of norms, are isomorphic. I will denote the standard basis vectors of $\mathbb{R}^n$ by $e_1, \ldots, e_n$.

Suppose $(w_m)_{m=1}^\infty \in W$ is a Cauchy sequence. Let $T_m : V \rightarrow W$ be the (bounded, linear) transformation, $$T_m(a_1, a_2, \ldots, a_n) = a_1 w_m.$$ I claim that $(T_m)$ is Cauchy. Fix $\varepsilon > 0$. Since $(w_m)$ is Cauchy, there exists an $N$ such that, $$p, q \ge N \implies \|w_p - w_q\| < \varepsilon \implies \|(T_p - T_q)(e_1)\| < \varepsilon.$$ Additionally, $(T_p - T_q)(e_i) = 0$ for $i \neq 1$. Let $x$ be in the unit ball of $V$. Then $x = (x_1, \ldots, x_n) = x_1 e_1 + \ldots + x_n e_n$, where $|x_1| + \ldots + |x_n| \le 1$. This implies $|x_1| \le 1$, hence $(T_p - T_q)(x) = x_1 (w_p - w_q)$, and so $\|(T_p - T_q)(x)\| = |x_1| \|w_p - w_q\| < \varepsilon$. Therefore, $\|T_p - T_q\| \le \varepsilon$, whenever $p, q \ge N$, proving $(T_m)$ is Cauchy.

Therefore $(T_m)$ converges to some $T$. I claim that $T(e_i) = 0$ for $i \neq 1$. We have, $$\|T(e_i)\| = \|(T - T_m)(e_i)\| \le \sup_{v\in B_V} \|(T - T_m)(v)\| = \|T - T_m\| \rightarrow 0,$$ thus $\|T(e_i)\| = 0$, hence $T(e_i) = 0$. Therefore, we may express $T$ in the form $$T(a_1, \ldots, a_n) = a_1 w$$ for some $w$ (in particular, $w = T(e_1)$).

I now claim that $w$ is the limit of $(w_m)$. Fix some $\varepsilon > 0$. Since $T_m \rightarrow T$, there exists some $M$ such that $$m \ge M \implies \|T_m - T\| < \varepsilon \implies \|T_m(e_1) - T(e_1)\| < \varepsilon \implies \|w_m - w\| < \varepsilon,$$ which proves $w_m \rightarrow w$ as required.