# Proving that $B(X,Y)$ is a Banach Space if $Y$ is.

Let $$B(X,Y)$$ be the family of all bounded maps from $$X$$ to $$Y,$$ normed linear maps. Then, $$B(X,Y)$$ is a Banach Space if $$Y$$ is.

Remark: I've seen this question before $$Y$$ is a Banach space if $$B(X,Y)$$ is a Banach space, but it is the converse of my question statement.

MY TRIAL

Let $$T_n\in B(X,Y),\;\forall\;n\in \Bbb{N}$$ s.t. $$T_n\to T,\;\text{as}\;n\to\infty.$$ So, $$T_n\in B(X,Y),\;\forall\;n\in \Bbb{N}$$ implies for each $$x\in X,\;T_{n}(x)\in Y.$$ Since $$Y$$ is complete, $$T_n(x)\to T(x)\in Y,\;\text{as}\;n\to\infty,\;\forall\;x\in X.$$ i.e., $$T:X\to Y.$$

Also, $$T_n\in B(X,Y),\;\forall\;n\in \Bbb{N}$$ implies there exists $$K\geq 0,$$ s.t. $$\forall\;n\in \Bbb{N},\;\forall\;x\in X,$$ \begin{align} \Vert T_n(x)\Vert \leq K \Vert x\Vert. \end{align} As $$n\to\infty,$$ \begin{align} \lim\limits_{n\to \infty}\Vert T_n(x)\Vert= \Vert \lim\limits_{n\to \infty}T_n(x)\Vert= \Vert T(x)\Vert\leq K \Vert x\Vert, \end{align} which implies $$T\in B(X,Y)$$ and hence, $$B(X,Y)$$ is a Banach space.

Please, kindly check if I'm right or wrong. If it turns out that I'm wrong, kindly provide an alternative proof. Regards!

What you want to show is that given any Cauchy sequence $$\{T_n\}_{n=1}^{\infty}$$ in $$B(X,Y)$$ there exists a mapping $$T\in B(X,Y)$$ such that $$\|T_n-T\|\rightarrow 0$$.

You seem to show that if $$\|T_n-T\|\rightarrow 0$$ then $$T\in B(X,Y)$$.

You mention completeness of $$Y$$ to motivate that $$T_n(x)\rightarrow T(x)$$ as $$n\rightarrow \infty$$ and that is indeed correct, to motivate it rigorously we have that for a fix $$x\in X$$ $$\|T_n(x)-T_m(x)\|\leq \|T_n-T_m\|\|x\|\rightarrow 0$$ as $$n,m\rightarrow \infty$$. Now completeness of $$Y$$ implies that $$T_n(x)$$ converges point wise and so we define $$T:X\rightarrow Y$$ by

$$T(x) = \lim_{n\rightarrow \infty}T_n(x).$$

After that you want to show that $$\|T-T_n\|\rightarrow 0$$. This follows from the fact that

$$\|(T-T_n)(x)\| = \lim_{m\rightarrow \infty}\|(T_m-T_n)(x)\|\leq \lim_{m\rightarrow \infty}\|T_m-T_n\|\|x\|$$

and therefore picking $$n$$ large enough so that $$\|T_m-T_n\|<\varepsilon$$ if $$m>n$$ this shows that

$$\frac{\|(T-T_n)(x)\|}{\|x\|}<\varepsilon$$

Credits to Olof Rubin. So, I post the full proof for future readers.

Let $$\{T_n\}_{n=1}^{\infty}\in B(X,Y)$$, be a Cauchy sequence and $$\epsilon>0$$ be given. Then, there exists $$N$$ s.t. forall $$m\geq n\geq N,$$ $$\|T_n-T_m\|<\epsilon.$$ Since $$\|T_n-T_m\|=\sup\limits_{\|x\|\leq 1}\|T_n(x)-T_m(x)\|,\;\;\forall\;m,n\in \Bbb{N},$$ we have that $$\|T_n(x)-T_m(x)\|\leq\|T_n-T_m\|<\epsilon,\;\;\forall\;m\geq n\geq N,\;\text{for each}\;x\in X.$$

This implies that $$T_n(x)\to T(x),\;\text{as}\;n\to\infty$$, pointwise and since $$Y$$ is complete, $$T(x)\in Y$$

Fix $$n\geq N$$ and allow $$m\to\infty.$$ Then, for each $$x\in X,$$ $$\|T_n(x)-T(x)\|\leq\epsilon.$$ Taking $$\sup$$ over $$\|x\|\leq 1,$$ we have

$$\|T_n-T\|=\sup\limits_{\|x\|\leq 1}\|T_n(x)-T(x)\|\leq\epsilon,\;\;\forall n\geq N.$$ Hence, $$T\in B(X,Y)$$ and we're done!