Quadratic form of Hilbert space operator is equal when operators are equal

Question

Show that $\omega(A) \leq ||A|| \leq 2 \omega(A)$ and deduce that $\langle A \xi , \xi \rangle = \langle B \xi, \xi \rangle$ when $A = B$.

I already showed the inequalities but I am stuck on showing $\langle A \xi , \xi \rangle = \langle B \xi, \xi \rangle$ when $A = B$.. Note that $A, B$ are Hilbert space operators and $\omega(A) = \sup \{ |\langle A \xi , \xi \rangle| : ||\xi|| = 1 \}$ is the numerical radius. Also note that we consider complex Hilbert spaces.

Answer 1

Obviously when $A=B$ we have that $A\xi=B\xi$ so $\langle A\xi,\xi\rangle=\langle B\xi,\xi\rangle$, so I assume you mean the converse.

Note that if $\langle A\xi,\xi\rangle=\langle B\xi,\xi\rangle$ then $\langle (A-B)\xi,\xi\rangle=0$, so it suffices to show that if $T$ is a bounded operator with $\langle Tx,x\rangle=0$ for all $x$, then $T=0$. But if this is the case then $\omega(T)=0$, and from the inequalities we get that $\|T\|=0$, thus $T=0$.