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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_nin B(X,Y),;forall;nin Bbb{N} $ s.t. $T_nto T,;text{as};ntoinfty. $ So, $T_nin B(X,Y),;forall;nin Bbb{N} $ implies for each $xin X,;T_{n}(x)in Y.$ Since $Y$ is complete, $T_n(x)to T(x)in Y,;text{as};ntoinfty,;forall;xin X. $ i.e., $T:Xto Y. $

Also, $T_nin B(X,Y),;forall;nin Bbb{N} $ implies there exists $Kgeq 0,$ s.t. $forall;nin Bbb{N},;forall;xin X, $
begin{align} Vert T_n(x)Vert leq K Vert xVert. end{align}
As $ntoinfty,$
begin{align} limlimits_{nto infty}Vert T_n(x)Vert= Vert limlimits_{nto infty}T_n(x)Vert= Vert T(x)Vertleq K Vert xVert, end{align}
which implies $Tin B(X,Y)$ 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 $Tin B(X,Y)$ such that $|T_n-T|rightarrow 0$.

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

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

$$T(x) = lim_{nrightarrow infty}T_n(x).$$

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

$$|(T-T_n)(x)| = lim_{mrightarrow infty}|(T_m-T_n)(x)|leq lim_{mrightarrow infty}|T_m-T_n||x|$$

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 $mgeq ngeq N,$
$$ |T_n-T_m|<epsilon.$$
Since $$ |T_n-T_m|=suplimits_{|x|leq 1}|T_n(x)-T_m(x)|,;;forall;m,nin Bbb{N},$$
we have that
$$ |T_n(x)-T_m(x)|leq|T_n-T_m|<epsilon,;;forall;mgeq ngeq N,;text{for each};xin X.$$

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

Fix $ngeq N$ and allow $mtoinfty.$ Then, for each $xin X,$
$$ |T_n(x)-T(x)|leqepsilon.$$
Taking $sup$ over $|x|leq 1,$ we have

$$ |T_n-T|=suplimits_{|x|leq 1}|T_n(x)-T(x)|leqepsilon,;;forall ngeq N.$$
Hence, $Tin B(X,Y)$ and we're done!