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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.

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

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

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  • $\begingroup$ 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. $\endgroup$ – T. Stark Jul 28 at 13:11
  • $\begingroup$ In the first part, you have some $\leq $ instead of $\geq $ by definition of the sup. $\endgroup$ – Paul Jul 28 at 13:12
  • $\begingroup$ 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. $\endgroup$ – T. Stark Jul 28 at 13:23
  • $\begingroup$ @Paul Whoops!$\hspace{0pt}$ $\endgroup$ – Theo Bendit Jul 28 at 14:03
  • $\begingroup$ @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. $\endgroup$ – Theo Bendit Jul 28 at 14:07
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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 $\|A_nx-A_mx\|_Y\leq \|A_n-A_m\|_B \|x\|_X$.

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