$B(H)$-valued measure: show that a sesquilinear form is bounded Consider the following fragment from Paulsen's book "Completely bounded maps and operator algebras":


To obtain the bounded operator $\phi_E(f)\in B(H)$, we must show that the sesquilinear form
$$(x,y) \mapsto \int_X fd\mu_{x,y}$$
is bounded. However, I don't manage to show this. Concretely, I would have to show the following: There exists $M \ge 0$ such that
$$|\int_X fd \mu_{x,y}| \le M\|x\|\|y\|$$
for all $x,y \in H$. Of course, we have
$$|\int_X f d\mu_{x,y}| \le \|f\| |\mu_{x,y}|(X) $$
where $|\mu_{x,y}|(X)$ denotes the total variation of the measure $\mu_{x,y}$ so it suffices to show that there exists $M \ge 0$ such that
$$|\mu_{x,y}|(X) \le M\|x\|\|y\|$$
for all $x,y \in H$, or in other words, that
$$\sup_{x,y} |\mu_{x,y}|(X)< \infty$$ where $x,y$ range over the unit ball of $H$.
I found an argument on p68 of Murphy's book "C*-algebras and operator theory" but the proof there assumes that $E$ has image in the set of orthogonal projections in $H$, which is not the case in my context here. Yet, it illustrates that some care has to be taken to prove this.
 A: First we can write $E=E_1+i E_2$ with $E_1(B)=\frac 1 2(E(B)+E(B)^\ast)$ and $E_2(B)=\frac 1{2i}(E(B)-E(B)^\ast)$. Thus we may assume that $E(B)$ is symmetric for all $B\in\mathcal B$. In particular, $\mu_{x,x}(B)$ is real-valued for all $x\in\mathcal H$ and $B\in\mathcal B$.
I will first show that $|\mu_{x,x}|(X)\leq C\|x\|^2$ with $C=2\sup\{\|E(B)\|:B\in\mathcal B\}$: If $(B_j)$ is a disjoint sequence of Borel sets such that $\bigcup_j B_j=X$, let $J_+=\{j\in\mathbb N\mid \mu_{x,x}(B_j)\geq 0\}$ and $J_-=\mathbb N\setminus J_+$. Then
\begin{align*}
\sum_{j=1}^\infty |\mu_{x,x}(B_j)|&=\sum_{j\in J_+}\langle E(B_j) x,x\rangle-\sum_{j\in J_-}\langle E(B_j)x,x\rangle\\
&=\left\langle E\left(\bigcup_{j\in J_+}B_j\right) x,x\right\rangle-\left\langle E\left(\bigcup_{j\in J_-}B_j\right) x,x\right\rangle\\
&\leq 2\sup\{\|E(B)\|:B\in\mathcal B\}\|x\|^2\\
&=C\|x\|^2.
\end{align*}
By the polarization identity,
$$
\mu_{x,y}=\frac 1 4\sum_{k=0}^3 i^k\mu_{x+i^k y},
$$
where I used the shorthand $\mu_z$ for $\mu_{z,z}$. Thus
$$
|\mu_{x,y}|(X)\leq \frac 14\sum_{k=0}^3 |\mu_{x+i^k y}|(X)\leq \frac C 4\sum_{k=0}^3\|x+i^k y\|^2\leq 2C(\|x\|^2+\|y\|^2).
$$
The right side is clearly bounded for $x$ and $y$ in the unit ball.
