Upper bound for operator valued integral in Proposition 7.11 in Brian Hall‘s Quantum theory for mathematicians

Question

I am currentyl working through Brian Hall‘s „Quantum Theory for Mathematicians“ and I am stuck at a seemingly „easy“ steo in the proof of Proposition 7.11. To a bounded measurable function $f$ and projection-valued measure $\mu$ we associate the integral $Q_f(\psi):= \int_X f d\mu_\psi.$ I can see why we have
$$|Q_f(\psi)|\leq \left(\sup_{x\in X} |f(x)|\right)\lVert \psi\rVert^2.$$ Since $Q_f$ is a bounded quadratic form there exists an operator $A_f$ such that $$Q_f(\psi)=\left< \psi, A_f\psi \right>.$$ Hall then clames that it follows that $$||A_f||\leq \sup_{x\in X} |f(x)|$$ but I don‘t see how. By definition we have $$\lVert A_f \rVert=\sup_{\lVert\psi\rVert=\lVert\phi\rVert=1}|\left<\phi,A_f\psi\right>|\geq \sup_{\lVert\psi\rVert=1}|\left< \psi, A_f\psi\right>|=\sup_{\lVert\psi\rVert=1}|Q_f(\psi)|\leq \sup_{x\in X}|f(x)|.$$ So I assume we actually want to show that $$\sup_{\lVert\phi\rVert=\lVert\psi\rVert=1}|\left<\phi, A_f\psi\right>|=\sup_{\lVert\psi\rVert=1}|\left<\psi,A_f\psi\right>|.$$ Do you have a tipp how to show that? Is that a general property for an operator associated to a quadratic form or does it depend on the definition of $Q_f$?.