Let $E$ be regular bounded $B(H)-$valued measure on a compact, $T_2$ space $X$. Then the linear map $\phi_E:C(X)\to B(H)$ is bounded

Question

Let $X$ be a compact, $T_2$ space, $\mathcal{B}$ be borel $\sigma-$algebra on $X$ and $H$ be a hilbert space. A $\mathcal{B}(H)-$valued measure on $X$ is a map $E:\mathcal{B}\to\mathcal{B}(H)$ such that if it satisfies the following condition- if $\{B_i\}$ is a countable collection of disjoint Borel sets with union $B$ then $$\langle E(B)x,y\rangle=\sum\langle E(B_i)x,y\rangle\ \forall x,y\in H$$

The measure is bounded if $\lVert E\rVert:=\text{sup}\{\lVert E(B)\rVert: B\in\mathcal{B}\}$ is finite. The measure is regular if the following complex measure is regular for all $x,y\in H$ $$\mu_{x,y}(B)=\langle E(B)x,y\rangle$$

Given such regular bounded measure $E$ on $X$, we define a linear map $\phi_E:C(X)\to\mathcal{B}(H)$ by $$\langle \phi_E(f)x,y\rangle=\int f\ d\mu_{x,y}$$

I want to prove $\phi_E$ is bounded. Let $x,y\in H$ be two unit vectors. Then $$|\langle \phi_E(f)x,y\rangle| =\left|\int f\ d\mu_{x,y}\right| \le\int |f|\ d|\mu_{x,y}| \le \lVert f\rVert_\infty |\mu_{x,y}|(X).$$

I want to have a bound on this $|\mu_{x,y}|(X)$. Let $\{B_i\}$ be a partition of $X$. Then $$\sum_i|\mu_{x,y}(B_i)|=\sum_i|\langle E(B_i)x,y\rangle|\le \lVert x\rVert\lVert y\rVert\sum_i\lVert E(B_i)\rVert=\sum_i\lVert E(B_i)\rVert.$$

I want to show the right hand term is bounded. I know that $\sum_i E(B_i)$ converges weakly to $E(X)$ and $\lVert E(B)\rVert\le \lVert E\rVert\ \forall B$. I need to use these facts to prove our claim.

Can anyone help me complete the proof? Thanks for your help in advance.

Answer 1

$\def\abajo{\\[0.3cm]}$ $\def\Re{\operatorname{Re}}$

The estimate you are trying to use is too crude. One needs a more subtle idea, which is the same idea as in the scalar case (following Rudin's ideas here). The key is the following lemma (proof at the bottom):

Lemma. Let $z_1,\ldots,z_n\in\mathbb C$. Then there exists $S\subset\{1,\ldots,n\}$ and such that $$\Big|\sum_{k\in S}z_k\Big|\geq\frac1\pi\,\sum_{k=1}^n|z_k|.$$

We have $$ |\mu_{x,y}|(X)=\sup\Big\{\sum_j|\mu_{x,y}(B_j)|:\ \{B_j\}\ \text{ partition of }X\Big\} $$ Fix such a partition $\{B_j\}_{j=1}^n$. By the Lemma there exists $S\subset\{1,\ldots,n\}$ such that $$ \sum_j|\mu_{x,y}(B_j)|\leq\pi\Big|\sum_{j\in S}\mu_{x,y}(B_j)\Big|. $$ Let $$X_S=\bigcup_{j\in S}B_j. $$ Then \begin{align} \sum_j|\mu_{x,y}(B_j)| &\leq\pi\Big|\sum_{j\in S}\mu_{x,y}(B_j)\Big|=\pi\,\Big|\sum_{j\in S}\langle E(B_j)x,y\rangle\Big|\\[0.3cm] &=\pi|\langle E(X_S)x,y\rangle|\leq\pi\|E(X_S)\|\,\|x\|\,\|y\|\\[0.3cm] &\leq \pi\|E\|\,\|x\|\,\|y\|. \end{align} As this works for any partition, $$ |\mu_{x,y}|(X)\leq \pi\|E\|\,\|x\|\,\|y\|. $$

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Proof of the Lemma. Write the polar form $z_k=|z_k|\,e^{i\gamma_k}$. Define sets $$s(\theta)=\{k:\ \cos(\gamma_k-\theta)>0\}.$$ These sets are for sure nonempty when $\theta$ is close enough to some $\gamma_k$, namely when $|\gamma_k-\theta|<\tfrac\pi2$. We have \begin{align*} \Big|\sum_{k\in S(\theta)}z_k\Big|&=\Big|\sum_{k\in S(\theta)}e^{-i\gamma_k}|z_k|\Big| =\Big|\sum_{k\in S(\theta)}|z_k|\,e^{i(\gamma_k-\theta)}\Big|\abajo &\geq\Re\sum_{k\in S(\theta)}|z_k|\,e^{i(\gamma_k-\theta)} =\sum_{k\in S(\theta)}|z_k|\,\cos(\gamma_k-\theta)\abajo &=\sum_{k=1}^n|z_k|\,\cos^+(\gamma_k-\theta). \end{align*} The use of the positive part of the cosine allows us to sum over all indices $k$. As this last sum is a continuous expression on $\theta$, we can choose $\theta_0$ such that it is maximum. As such, it is greater than its average; so \begin{align*} \Big|\sum_{k\in S(\theta_0)}z_k\Big|&\geq\frac1{2\pi}\int_{0}^{2\pi}\sum_{k=1}^n|z_k|\cos^+(\gamma_k-\theta)\,d\theta =\sum_{k=1}^n|z_k|\frac1{2\pi}\int_0^{2\pi}\cos^+(\gamma_k-\theta)\,d\theta\abajo &=\sum_{k=1}^n|z_k|\frac1{2\pi}\int_0^{2\pi}\cos^+\theta\,d\theta =\sum_{k=1}^n|z_k|\frac1{2\pi}\int_{-\pi/2}^{\pi/2}\cos\theta\,d\theta\abajo &=\frac1\pi\,\sum_{k=1}^n|z_k|. \end{align*}