Is the following a correct way to show that operator norms are consistent?

$$ \|AB\|=\max_{Bx \ne 0}\frac{\|ABx\|_\alpha }{\|x\|_\alpha} =\max_{ Bx\ne 0}\frac{\|ABx\|_\alpha}{\|Bx\|_\alpha} \frac{\|Bx\|_\alpha}{\|x\|_\alpha}\le \max_{y \ne 0} \frac{\|Ay\|_\alpha}{\|y\|_\alpha} \max_{x \ne 0} \frac{\|Bx\|_\alpha}{\|x\|_\alpha} = \|A\| \|B\| $$ It shows that: $$ \|AB\|\le \|A\| \|B\| $$

  • $\begingroup$ Yep. Also see here. If you are satisfied with this, I will convert this to an answer. $\endgroup$ – Ahaan S. Rungta Nov 28 '13 at 17:17
  • $\begingroup$ possible duplicate of Proof of matrix norm property: submultiplicativity $\endgroup$ – copper.hat Nov 28 '13 at 17:26
  • $\begingroup$ @AhaanRungta Tom Lyche is the lecturer of my course. I found the same proof in the slide you attached so I think this must be correct. $\endgroup$ – dresden Nov 28 '13 at 17:31
  • $\begingroup$ @copper.hat: The other question was intended for matrix norm which I think is not the same with operator norm $\endgroup$ – dresden Nov 28 '13 at 17:32
  • $\begingroup$ @dresden_p Nice! I'll convert to an answer. $\endgroup$ – Ahaan S. Rungta Nov 28 '13 at 17:32

Yes, that is correct, as demonstrated here.


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