0
$\begingroup$

My book says that $||Ax|| \leq M ||x|| \ \forall x \in X \implies ||A|| \leq M$, because $||A|| = sup_{||x|| \leq 1} \ ||Ax|| \in [0, M]$.

However, I'm unable to see how this follows from the definition. Thank you very much.

$\endgroup$
  • $\begingroup$ Because $\sup_{\|x\| \le 1} \|Ax\| \le M \sup_{\|x\| \le 1} \|x\|$.... $\endgroup$ – user296602 May 10 '16 at 20:01
2
$\begingroup$

For any $x $ with $\|x\|=1$, $$ \|Ax\|\leq M. $$ So $M $ is an upper bound for the set $$\{\|Ax\|:\ \|x\|\leq1\}. $$ And thus $$\sup\{\|Ax\|:\ \|x\|\leq1\}\leq M. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.