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Suppose that $X$ is a compact metric space, $\mathcal{H}$ is a Hilbert space, and that $A: \mathcal{B}(X) \rightarrow \mathcal{B}(\mathcal{H})$ is a Positive Operator Valued Measure (POVM), where $\mathcal{B}(X)$ denotes the Borel $\sigma$-algebra of subsets of $X$. That is,

  • $A(\Delta)$ is a positive operator in $\mathcal{B}(\mathcal{H})$ for all $\Delta \in \mathcal{B}(X)$;
  • $A(\emptyset) = 0$ and $A(X) = \text{id}_{\mathcal{H}}$ (the identity operator on $\mathcal{H}$);

  • If $\{ \Delta_n \}_{n=1}^{\infty}$ is a sequence of pairwise disjoint sets in $\mathcal{B}(X)$, and if $g,h \in \mathcal{H}$, then

$$ \left \langle A\left( \bigcup_{n=1}^{\infty} \Delta_n \right)g , h \right \rangle = \sum_{n=1}^{\infty} \langle A(\Delta_n)g, h \rangle.$$

If $g,h \in \mathcal{H}$, consider the complex measure $A_{g,h}(\cdot) := \langle A(\cdot)g, h \rangle$. I am trying to show that the total variation norm of this measure is less than or equal to $||g||\cdot||h||$. This is true if $A$ is a projection valued measure, and I am pretty sure it is also true for POVM's, but I am not sure how to prove it for the POVM case. Any help or references would be appreciated! Thanks!

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  • $\begingroup$ Are you sure you want to denote the Borel algebra and the space of continuous operators with the same symbol $\mathcal B$? $\endgroup$ – Joonas Ilmavirta Oct 4 '14 at 17:08
  • $\begingroup$ Hi Mike, this is where I started. Suppose that $\Delta_1, …, \Delta_n$ is a pairwise disjoint collection of Borel subsets. We then want to show that $\sum_{j=0}^{N-1} |\langle A(\Delta_j) g, h \rangle | \leq ||g||||h||$. The fact that $||A(\Delta)|| \leq 1$ is important but I am not sure that $\sum_{j=0}^{N-1} |\langle A(\Delta_j) g, h \rangle | \leq ||g||||h||$ follows immediately from this fact. If it is obvious and I am not seeing it, I apologize in advance. The proof for the PVM case is done in Conway's Function Analysis in Ch. 9, and does require some work. Thanks for your help! $\endgroup$ – trubee Oct 4 '14 at 17:33
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Let me denote the positive measure $\langle g, A(\cdot) g\rangle$ by $\mu_g$. By Cauchy-Schwarz for the sesquilinear form $\langle g, A(M) h\rangle$, $$ \sum \left|\langle g, A(M_j)h\rangle \right| \le \sum \left( \mu_g(M_j)\mu_h(M_j)\right)^{1/2} \le \left( \sum \mu_g(M_j)\sum \mu_h(M_j) \right)^{1/2} \le \left(\mu_g(X)\mu_h(X)\right)^{1/2} =\|g\|\, \|h\| . $$

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