# Collection of all bounded linear operators from $X$ into $Y$ is a normed linear space and is a Banach space if $Y$ is a Banach space

Let $X$ and $Y$ be normed linear spaces, and let $B(X,Y)$ be the collection of all bounded linear operators from $X$ into $Y$ with the operator norm. Show that $B(X,Y)$ is a normed linear space, and $B(X,Y)$ is a Banach space if $Y$ is a Banach space. The vector operations in $B(X,Y)$ are to be defined pointwise: $(A+B)(x) = Ax + Bx$ and $(\alpha A)(x)=\alpha (Ax)$.

Here is what I got so far

Let $\{B_k\}^\infty_{k=1}$ be a Cauchy sequence in $B(X,Y)$. Then for each $x\in X$, $\|B_nx-B_mx\|\leq\|B_n-B_m\|\|x\|\to 0$ as $m,n\to\infty$. So $\{B_k\}^\infty_{k=1}$ is Cauchy in $Y$. As $Y$ is complete there exists $Bx\in Y$ such that $lim_{k\to\infty}\|B_kx-Bx\|=0$.

Using the above, let $x_1,x_2\in X$ and $\alpha,\beta\in \mathbb{R}$. Then:

$B(\alpha x_1+\beta x_2)=lim_{k\to\infty}B_k(\alpha x_1+\beta x_2)=lim_{k\to\infty}(\alpha B_k(x_1)+\beta B_k(x_2))=\alpha B(x_1)+\beta B(x_2)$

...?

Not really sure how to go from there, to showing that $B(X,Y)$ is a normed linear space, or a Banach space. Any suggestions? Thanks!

Expanding on your thoughts above, assume that $Y$ is a Banach space, then we would like to show that $B(X,Y)$ is a Banach space. Given a Cauchy sequence $\{ B_k \}_{k=1}^{\infty}$ in $B(X,Y)$, then for each $x \in X$, $\{ B_k x \}_{k=1}^{\infty}$ is indeed a Cauchy sequence in $Y$. Thus, there exists a limit $y \in Y$ such that $\| B_k x - y \|_Y \to 0$. However, it is not clear that $y =Bx$ for some $B \in B(X,Y)$.
To check this, define a function $B \colon X \to Y$ by $Bx = y$, where $y$ is obtained from $x$ in the manner above. We must check that this function $B$ is a bounded linear operator. For linearity, take $\alpha,\beta \in \mathbb{R}$ and $x_1,x_2 \in X$, then \begin{align} & \| B(\alpha x_1 + \beta x_2) - \alpha B(x_1) - \beta B(x_2) \|\\ &\leq \| B(\alpha x_1 + \beta x_2) - B_k(\alpha x_1 + \beta x_2) \| + \|\alpha B_k(x_1) - \alpha B(x_1) \| + \| \beta B_k(x_2) - \beta B(x_2) \|, \end{align} and all of these terms go to zero as $k \to \infty$. Therefore, $B(\alpha x_1 + \beta x_2) = \alpha B(x_1) + \beta B(x_2)$ i.e. the function $B$ is linear.
Now, recall that a linear operator on a normed linear space is bounded iff it is continuous. Given $x_1, x_2 \in X$, then $$\| B(x_1) - B(x_2) \| \leq \| B(x_1) - B_k(x_1) \| + \| B_k(x_1) - B_k(x_2) \| + \| B_k(x_2) - B(x_2) \|.$$ But, $\| B(x_1) - B_k(x_1) \| \to 0$ and $\| B(x_2) - B_k(x_2) \| \to 0$ as $k \to \infty$, so it remains only to consider the middle term. For each $k$, $B_k$ is bounded and since a Cauchy sequence is bounded, there exists $M > 0$ such that $\| B_k \| \leq M$ for all $k$. It follows that $$\| B_k(x_1) - B_k(x_2) \| \leq \| B_k \| \| x_1 - x_2 \| \leq M \| x_1 - x_2 \|.$$ Therefore, replacing the above inequality into our original inequality and then taking the limit as $k \to \infty$, we find that $$\| B(x_1) - B(x_2) \| \leq M \| x_1 - x_2 \|,$$ that is, $B$ is a bounded operator. We conclude that $B \in B(X,Y)$ and it is a limit of the Cauchy sequence $\{ B_k \}_{k=1}^{\infty}$. Therefore, $B(X,Y)$ is complete in the operator norm, and is thus itself a Banach space.