Show by induction that for a bounded linear operator $A,\, ||A^n||\le ||A||^n$

Question

Of course the case for $n=1$ is trivial, so I can suppose that for some arbitrary $n\in \mathbb{N}$, $||A^n||\le||A||^n$

Then $||A^{n+1}||=||A\cdot A^n||$. I'd like to proceed by somehow getting this norm into the form $||A||\cdot||A^n||$ either through equality or inequality, but I don't think that's a result which directly follows.

For what it's worth I have proven this result non-inductively using properties of bounded linear operators, but I'm specifically looking for an inductive method.

Any help appreciated.

score 1 · Accepted Answer · answered Oct 04 '20 at 14:28

1

This folows from the fact that $\|A.B\|\leqslant\|A\|.\|B\|$. In particular,$$\|A.A^n\|\leqslant\|A\|.\|A^n\|\leqslant\|A\|.\|A\|^n=\|A\|^{n+1}.$$

answered Oct 04 '20 at 14:28

José Carlos Santos

427,504

Yeah that's the result I was hoping for, but I've been unable to locate it directly in my notes. Do you have a quick proof or something you could link me showing that $||A\cdot B||\le ||A||\cdot ||B||$ – mike123abc Oct 04 '20 at 14:39
You will find it here. – José Carlos Santos Oct 04 '20 at 14:43
Thank you, appreciated – mike123abc Oct 04 '20 at 18:33

Show by induction that for a bounded linear operator $A,\, ||A^n||\le ||A||^n$

1 Answers1