let H be hilbert space and let $U\in L(H).$
show that if U is self-adjoint then $$\|u\|= sup_{\|x\|\leq 1} |<Ux,x>|.$$
I showed that $ sup_{\|x\|\leq 1} |<Ux,x>|\leq \|U\|.$
But can you help me to find the other direction please thank you.
Asked
Active
Viewed 213 times
0
calipaw
- 29
- 6
-
2$||Ux||^2=\langle Ux,Ux\rangle =\langle U^{}Ux,x\rangle\leq \sup \langle U^{}Ux,x\rangle \leq ||U^{*}||\sup \langle Ux,x\rangle=||U||\sup \langle Ux,x\rangle$ maybe let helpfull – weymar andres Dec 08 '21 at 16:51
-
yeah since $|U|= sup |Ux|$ so $|U| \leq sup \langle Ux,x \rangle$ – calipaw Dec 08 '21 at 16:58
-
@weymarandres why should $\sup \langle U^Ux,x\rangle \le |U^| \sup \langle Ux,x\rangle$ be true?? – daw Dec 08 '21 at 17:29