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.
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yes, i see, but i don't understand the answer – Ariel Marcelo Pardo Jul 04 '15 at 05:13
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for example, what is $V^*$? – Ariel Marcelo Pardo Jul 04 '15 at 05:14
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@ArielMarceloPardo $V^*$ denotes the space of continuous linear functions from $V$ into the underlying field (usually real or complex). – user251257 Jul 04 '15 at 05:24
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ok, i see, but how does $\Phi$ work? i don't understand $(\Phi w)(v) = f(v) w$ – Ariel Marcelo Pardo Jul 04 '15 at 05:29
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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.
Theo Bendit
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you can define $1-$norm because $V$ is finite dimensional, so every norm are equivalent, right? – Ariel Marcelo Pardo Jul 04 '15 at 13:43
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Yes, that's exactly why we can assume the norm. In terms of the Cauchiness, I explained it badly. Let me edit. – Theo Bendit Jul 04 '15 at 14:47
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why do you use the unit ball in $V$? is it not a specific case? – Ariel Marcelo Pardo Jul 04 '15 at 15:23
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Well, because it's involved in the definition of the operator norm, the canonical choice of norm on $B(V, W)$, which in this case (because $V$ is finite-dimensional) is equal to $L(V, W)$. For $T \in B(V, W)$, the operator norm is defined by $|T| = \sup_{x \in B_V} |T(x)|.$ – Theo Bendit Jul 04 '15 at 15:53
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1ok, that's why you wote: "Let $x$ be in the unit ball of $V$" – Ariel Marcelo Pardo Jul 04 '15 at 16:10