# Prove that the Set of Bounded Linear Operators is Banach

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 $$\begin{split} \|T(x)\| &\leq \|T_n(x)-T(x)\|+\|T_n(x)\| \\ &=\lim_{m\rightarrow\infty}\|T_n(x)-T_m(x)\|+\|T_n(x)\| \\ & \leq \limsup_{m\rightarrow\infty}\|T_n(x)-T_m(x)\|+\|T_n(x)\| \\ & \leq \varepsilon \|x\|+\|T_n(x)\| \end{split}$$ 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.
• 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.) – Federico 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. – Ton Dec 28 '13 at 17:04
• 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. – Federico Dec 28 '13 at 18:13

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 $B(V,V')$.

• Assume you know that V' is Banach. I think the same situation applies to show that B(V,V') is Banach as here. Right? – Léo Léopold Hertz 준영 Sep 14 '15 at 13:17
• What happens in proof of 1, if ||x|| is not finite? – Sahiba Arora Dec 17 '15 at 20:30
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