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

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 $$||A_nf-A_mf||\leq||A_n-A_m|| \,||f||$$.

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, $$\|A_n f - A_m f\| = \|(A_n - A_m)f\| \le \|A_n - A_m\| \|f\|.$$

• I realized my mistake soon after asking the question. Your answer was posted while I was typing my answer. I will accept your answer when the time limit allows it. – T. Stark Jul 28 at 13:11
• In the first part, you have some $\leq$ instead of $\geq$ by definition of the sup. – Paul Jul 28 at 13:12
• That definition Theo provided is correct. Could you also confirm that this is correct: $\| A_nf-A_mf\|$ is a cauchy sequence because $\|f\|$ is a finite constant and $\|A_n-A_m\|$ is a cauchy sequence. – T. Stark Jul 28 at 13:23
• @Paul Whoops!$\hspace{0pt}$ – Theo Bendit Jul 28 at 14:03
• @T.Stark Sort of, not exactly. It's more because $T \mapsto Tf$ is a Lipschitz map, with Lipschitz map, in that $\|Tf - Sf\| \le K\|T - S\|$, where $K = \|f\|$. As such, it is uniformly continuous, hence Cauchy continuous, and $A_n$ is a Cauchy sequence that maps to Cauchy sequence $A_n f$. Note that $\|A_n - A_m\|$ is not even really a sequence. – Theo Bendit Jul 28 at 14:07

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 $$\|A_nx-A_mx\|_Y\leq \|A_n-A_m\|_B \|x\|_X$$.