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

Question

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!

Answer 1

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

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

Answer 2

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 $$ \|T_n(x)-T_m(x)\|\leq\|T_n-T_m\|\cdot \|x\|<\epsilon\|x\|,\;\;\forall\;m\geq n\geq N,\;\text{for each}\;x\in X.$$

This implies that $T_n(x)\to T(x)\in Y\;\text{as}\;n\to\infty$ for each $x\in X$ given that $Y$ is complete.

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!