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Let $B(V,V')$ be the vector space formed by set of linear operators $T:V\rightarrow V'$. where $V,V'$ are normed vector spaces. Equip $B(V,V')$ with the norm $$ \|T\|=\sup\frac{\|T(x)\|}{\|x\|} $$ where $x\in V$.
Show that if $V'$ is Banach then $(B(V,V'),\|\cdot\|)$ is Banach.

(!!Beware that I am bad with quantifiers!!)

Proof:

  1. Find potential limit.
    Let $\{T_n\}_{n\geq1}$ be a Cauchy in $B(V,V')$, then $\forall\varepsilon>0\ \exists N\geq 1 $ s.t. $n,m\geq N\Rightarrow$ $$ \sup\|T_n(x)-T_m(x)\|<\varepsilon \|x\| , $$ then $$ \|T_n(x)-T_m(x)\|\leq\sup\|T_n(x)-T_m(x)\|<\varepsilon \|x\|, $$ and so $\{T_n\}_{n\geq1}$ is Cauchy in $V'$ which is Banach, and so has a limit $T\in V'$.
  2. Show that the limit is in the space.
    W.t.s. that $T$ is bounded and linear \begin{equation} \begin{split} \|T(x)\| &\leq \|T_n(x)-T(x)\|+\|T_n(x)\| \\ &=\lim_{m\rightarrow\infty}\|T_n(x)-T_m(x)\|+\|T_n(x)\| \\ & \leq \limsup_{m\rightarrow\infty}\|T_n(x)-T_m(x)\|+\|T_n(x)\| \\ & \leq \varepsilon \|x\|+\|T_n(x)\| \end{split} \end{equation} Linearity.
    $$ T_n(\lambda x+\mu y)=\lim_{n\rightarrow\infty}T_n(\lambda x+\mu y)=\lim_{n\rightarrow\infty}(\lambda T_n(x)+\mu T_n(y))=\lambda T(x)+\mu T(y) $$
  3. Show that the Cauchy sequence converges in norm.
    W.t.s. $\lim_{n\rightarrow\infty}\sup\|T_n(x)-T(x)\|=0$.
    For all $n,m\geq N$ $$ \|T_n(x)-T_m(x)\|\leq\sup\|T_n(x)-T_m(x)\|<\varepsilon \|x\|. $$ Take the limit in $m\rightarrow\infty$ $$ \|T_n(x)-T(x)\|<\varepsilon \|x\|. $$ Take the supremum $$ \sup\|T_n(x)-T(x)\|<\varepsilon \|x\|. $$ but $\varepsilon$ is arbitrary and $\|x\|$ finite.
Matthew Conroy
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Rgkpdx
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    What is your question? You have proved the easy part. (it seems to be correct.). The converse is true also: if $B(V,V')$ with operator norm is Banach, then $V'$ is Banach. This is the hardest part. (Hint: you need the Hahn-Banach theorem.) – user91126 Dec 28 '13 at 15:48
  • I just wanted to see if I structured the proof properly using the quantifiers in the correct way besides the eventual "corerctness" of the proof. Thanks I'll look into the opposite implication. – Rgkpdx Dec 28 '13 at 17:04
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    Quantifiers look good ;) there is some technical issue. For example, in $\sup|T_n(x)-T_m(x)|<\varepsilon |x|$ you don't have to take the $\sup$, since you're working with $x$ fixed in that context. – user91126 Dec 28 '13 at 18:13

1 Answers1

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Your proof of (1) is incorrect. You start with a Cauchy sequence $\{T_n\} \subset B(V,V')$, and prove that it has a limit in $V$?!

What you want to do (and I suspect this is what you are trying to do) is as follows :

(1) For each $x\in V$, you check that $\{T_n(x)\} \subset V'$ is a Cauchy sequence. Since $V'$ is a Banach space, this sequence as a limit, which you can denote by $\alpha_x$.

(2) Now check that the elements $\{\alpha_x : x\in V\}$ satisfy the properties of a linear transformation. ie for any $x,y \in V$ and $c\in \mathbb{K}$, $$ \alpha_{x+y} = \alpha_x + \alpha_y $$ $$ \alpha_{cx} = c\alpha_x $$

(3) So define $T : V \to V'$ by $T(x) = \alpha_x$, and check that $T$ defines a bounded linear map.

Now conclude that $T_n \to T$ in the operator norm defined on $B(V,V')$.

Prahlad Vaidyanathan
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