The space of bounded linear operators into a banach space is complete

Question

This is a common theorem and is proven in many books. I am confused with a particular part of the proof. This image has been taken from Christopher Heils notes.

If $A_n$ is a cauchy sequence in $B(X,Y)$ then we know that $||An-Am||\rightarrow 0$ as $n,m\rightarrow 0$. It then follows that for any $f\in X$ it must be that $||A_nf-A_mf||\rightarrow 0$. But I do not understand how we get $||A_nf-A_mf||\leq||A_n-A_m|| \,||f||$.

Answer 1

Recall the (usual) norm on $B(X, Y)$: $$\|A\| := \sup_{\|x\|_X \le 1} \|Ax\|_Y.$$ Then, for all $x \in X \setminus \{0\}$, we have $$\|A\| \ge \left\|A\left(\frac{x}{\|x\|_X}\right)\right\|_Y = \frac{\|Ax\|_Y}{\|x\|_X} \implies \|A\|\|x\|_X \ge \|Ax\|_Y.$$ (The final inequality also holds trivially for $x = 0$ too.)

So, just apply this principle here! We have, $$\|A_n f - A_m f\| = \|(A_n - A_m)f\| \le \|A_n - A_m\| \|f\|.$$

Answer 2

I realize that I did not properly think about the definition of the operator norm.

Being explicit: We let $\|\cdot\|_Y$ be the norm on $X$, $\|\cdot\|_Y$ be the norm on $Y$, $\|\cdot\|_B$ be the operator norm. Recall that $\|A\|_B=\sup_{x\in X\{0\}}\frac{\|Ax\|_Y}{\|x\|_X}$.

This gives the result $\|A_nx-A_mx\|_Y\leq \|A_n-A_m\|_B \|x\|_X$.