from book A Short Course on Spectral Theory , William Arveson
Theorem 4.2.4 say
Let $A$ be a bounded operator on $H^{2}$ satisfying $S^{*}AS=A$. There is a unique function $\phi\in L^{\infty}$ such that $A=T_{\phi}$, and one has $||A||=||\phi||_{\infty}$
I'm stuck in the end, he say
uniqueness of $\phi$ follows that $||T_{\phi}||=||\phi||_{\infty}$
i don't know how deduce this using what was seen earlier in this book.
Thanks
He could have written it slightly better.
What the line means is that, since $\|Tg\|=\|g\|_\infty$ for any $g\in L^\infty$, if $T_\phi=T_\psi$ then for any $f\in L^2$ $$ 0=T_\phi f-T_\psi f=P_+\phi f-P_+\psi f=P_+(\phi-\psi)f=T_{\phi-\psi}f. $$ So $T_{\phi-\psi}=0$, and then $$ 0=\|T_{\phi-\psi}\|=\|\phi-\psi\|_\infty. $$ implying that $\psi=\phi$ in $L^\infty$.