Show that if $(X,\|\cdot\|_X)$ is a normed space and $(Y,\|\cdot\|_Y)$ is a Banach space then $(B(X,Y),\|\cdot\|_{op})$ is a Banach space.
$B(X,Y) = \{T\colon X\to Y : T$ is bounded and linear$\}$
What I did :
Let $\|x\|_X \le 1$ and $\|T\|_{op} = \sup\{\|T(x)\|_Y\ : \|x\|_X \le 1\}$ and $\sum_{n=1}^\infty \|T_n\|_{op}$ is convergent i.e. $\sum_{n=1}^\infty T_n$ is absolute convergent series in $B(X,Y)$
We have $\|T_n(x)\|_Y \le \|T_n\|_{op}$ from definition
Since $\sum_{n=1}^\infty \|T_n\|_{op}$ is convergent $\sum_{n=1}^\infty \|T_n(x)\|_Y$ is convergent too.
Since $(Y,\|\cdot \|_Y)$ is a Banach space $\sum_{n=1}^\infty T_n(x)$ is convergent.
Let $S_k= \sum_{n=1}^k T_n(x)$
$\exists y \in Y$ such that $S_k \to y$
I've stuck in that point. How must I pursue? I think there are some mistakes or unnecessary things. Could you please fix me? How can I show $\sum_{n=1}^\infty T_n$ is convergent in $B(X,Y)$?
Thanks
We have established that $\sum_{n=1}^\infty T_nx$ converges for all $x \in X$. Define a linear map $T : X \to Y$ with $$Tx = \sum_{n=1}^\infty T_nx$$
For $N \in \mathbb{N}$ and $x \in X$ we have
$$\left\|Tx - \sum_{n=1}^N T_nx\right\| = \left\|\sum_{n=N+1}^\infty T_nx\right\| \le \sum_{n=N+1}^\infty \|T_nx\| \le \left(\sum_{n=N+1}^\infty \|T_n\|\right)\|x\|$$ taking the supremum over $\|x\| = 1$ we get
$$\left\|T - \sum_{n=1}^N T_n\right\| \le \sum_{n=N+1}^\infty \|T_n\| \xrightarrow{N\to\infty} 0$$
so $\sum_{n=1}^\infty T_n = T$ with respect to the operator norm.
Furthermore, let $N \in \mathbb{N}$ such that $\left\|T - \sum_{n=1}^N T_n\right\| \le 1$. We have
$$\|T\| \le \left\|T - \sum_{n=1}^N T_n\right\| + \left\|\sum_{n=1}^N T_n\right\| \le 1 + \sum_{n=1}^N \|T_n\| < +\infty$$
so $T$ is bounded.
We conclude that $\sum_{n=1}^\infty T_n$ converges to an element of $B(X,Y)$ so $B(X,Y)$ is a Banach space.