How to prove $n(T)=\sup\{|\langle Tx,x \rangle |, \|x\|=1\}$ is a norm on $B(H)$ and $n(T)\lt\|T\|\lt2n(T)$ where $T\in B(H)$?

Asked Jun 19 '16 at 05:27

Active Jun 19 '16 at 08:05

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Let $H$ be a Hilbert space over $\mathbb C$. If $T\in B(H)$, how to prove that

$$n(T)=\sup\{|\langle Tx,x \rangle |, \|x\|=1\}$$ is a norm on $B(H)$ and $$n(T)\lt||T||\lt2n(T)\ \textrm{?}$$ I couldn't show that $\|T\|\lt2n(T)$.

Thanks for any hint

edited Jun 19 '16 at 08:05

Math1000

asked Jun 19 '16 at 05:27

user330305

All I can say off the top of my head is that we must somehow use the fact that $H$ is not $2$-dimensional, for in $2$ dimensions a rotation $T$ thru $90^o$ has $|T|=1$ and $n(T)=0.$ – DanielWainfleet Jun 19 '16 at 07:49
@user254665 : That won't happen if $H$ is complex. – Disintegrating By Parts Jun 19 '16 at 20:16
I am pretty sure those inequalities should be $\leqslant$ instead of $<$. See https://math.stackexchange.com/a/2281150/161825 – Jonas Dahlbæk Jun 20 '17 at 11:38

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