Suppose $X$ and $Y$ are normed space. Associate to each $\Lambda \in \mathcal{B}(X,Y)$ the number $$\lVert \Lambda \rVert = \sup \left\{ \lVert \Lambda x \rVert : x \in X, \lVert x \rVert \leq 1 \right\}.$$ This definition of $\lVert \Lambda \rVert$ makes $\mathcal{B}(X,Y)$ into a normed space. If $Y$ is a Banach space, so is $\mathcal{B}(X,Y)$.
The proof that $\mathcal{B}(X,Y)$ is fine and simple enough. I don't get a subtle step for proving the completeness.
Assume now that $Y$ is complete and that $\left\{ \Lambda_n \right\}$ is a Cauchy sequence in $\mathcal{B}(X,Y)$. Since $$ \lVert \Lambda_n x - \Lambda_m x \rVert \leq \lVert \Lambda_n - \Lambda_m \rVert \lVert x \rVert $$ and since it is assumed that $\lVert \Lambda_n - \Lambda_m \rVert \to 0$ as $n$ and $m$ tend to $\infty$, $\left\{ \Lambda_n x \right\}$ is a Cauchy sequence in $Y$ for every $x \in X$. Hence $$ \Lambda x = \lim_{n\to \infty} \Lambda_n x \;\;\;\; (4) $$ exists. Is is clear that $\Lambda : X \to Y$ is linear. If $\epsilon > 0$, the right side of (4) does not exceed $\epsilon \lVert x \rVert$, provided that $m$ and $n$ are sufficiently large.
And here is the very bit I don't get
It follows that $$ \lVert \Lambda x - \Lambda_m x \rVert \leq \epsilon \lVert x \rVert $$ for all large $m$.
Could you please explain this bit?
Also to me (4) is actually what we want to prove, to me it is obvious there's a $y \in Y$ such that
$$ y = \lim_{n \to \infty} \Lambda_n x $$
but we should prove that there's $\Lambda \in \mathcal{B}(X,Y)$ such that
$$ y = \Lambda x $$
but the author seems to skip this, why can he skip it?
Thank you
Note simply that for all $n,m$ and $x$, $$\|\Lambda x-\Lambda_m x\|\leq\|\Lambda x-\Lambda_nx\|+\|\Lambda_n x-\Lambda_m x\|\quad (*)$$ Let $\epsilon>0$. By (4), there exists $N_0$ such that for all $n>N_0$, the middle term is smaller than $\epsilon\| x\|/2$. Since $\{\Lambda_n\}$ is Cauchy, there exists $N_1$ such that if $n,m > N_1$, then $\|\Lambda_n-\Lambda_n\|<\epsilon/2$. Now let $m>N_1$. Choose $n>\max\{N_0,N_1\}$. Then by (*): $$\|\Lambda x-\Lambda_m x\|\leq\epsilon\|x\|/2+\|\Lambda_nx-\Lambda_mx\|<\epsilon\|x\|/2+\epsilon\|x\|/2=\epsilon\|x\|$$
Added after answer was accepted
Actually the reason is much simpler. By (4) $\Lambda_nx$ tends to $\Lambda x$, so given $\epsilon>0$ there is $N$ such that if $n>N$, we have $\|\Lambda_nx-\Lambda x\|<\epsilon\|x\|$, where $\epsilon\|x\|$ "plays the role" of $\epsilon$. Now just replace $n$ by $m$.