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:
- 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 $T\in V'$. - 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.
$$
T_n(\lambda x+\mu y)=\lim_{n\rightarrow\infty}T_n(\lambda x+\mu y)=\lim_{n\rightarrow\infty}(\lambda T_n(x)+\mu T_n(y))=\lambda T(x)+\mu T(y)
$$ - 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.
Best Answer
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 $B(V,V')$.