Does anyone know a simple proof of the following theorem stating that a positive Borel operator measure $P$ on $\mathbb{R}$ can be written as $V^{\star}EV$ for a Borel spectral measure $E$?

Theorem (Operator Measure Dilation): Let $P$ be a function from the Borel subsets of $\mathbb{R}$ into the bounded linear operators $\mathcal{L}(\mathcal{H})$ on a complex Hilbert space $\mathcal{H}$ such that $\mu_{x}(S)=(P(S)x,x)$ is a positive Borel measure for each $x\in\mathcal{H}$, with $P(\mathbb{R})=I$. Then there is complex Hilbert space $\mathcal{K}$, a spectral measure $E$ from the Borel subsets of $\mathbb{R}$ into $\mathcal{L}(\mathcal{K})$, and an isometry $V : \mathcal{H}\rightarrow\mathcal{K}$ such that $P(S)=V^{\star}E(S)V$.


1 Answer 1


It is convinient to use the following notation $\mu_{x, y}(S) := \left<P(S)x, y\right>$ and shortly $\mu_{x}$ for $\mu_{x,x}$.

Let $B(\mathbb{R})$ denote the abelian C*-algebra of bounded Borel functions on $\mathbb{R}$.

Define a map $\Phi_P \colon B(\mathbb{R}) \rightarrow \mathcal{L}(\mathcal{H})$ by $$\left<\Phi_P(f)x, y\right> = \int_{\mathbb{R}} f \ \mathrm{d}\mu_{x,y}.$$ Note that $\Phi_P$ is well-defined, since $$\left<\Phi_P(f)x, x\right> = \int_{\mathbb{R}} f \mathrm{d} \mu_x \leq \|f\|_{\infty}\|x\|,$$ (the assumption that $P(\mathbb{R})=I$ was used) and moreover $\Phi_P$ is a bounded linear map itself.

Positivity of $\mu_x$ implies that $\Phi_P$ is a positive map, furthermore it is an easy exercise to show that $\Phi_P$ is a completely positive map since $B(\mathbb{R})$ is an abelian C* algebra.

Now let us apply the Stinespring factorisation theorem:

Therefore there exists a Hilbert space $\mathcal{K}$ and a unital *-homomorphism $$\pi \colon B(\mathbb{R}) \rightarrow \mathcal{B}(\mathcal{K})$$ such that $$ \Phi_P(f) = V^*\pi(f)V,$$ where $V \in \mathcal{B}(\mathcal{H}, \mathcal{K})$. Moreover, $V$ is an isometry and $$ I = \Phi_P(1) = V^* \pi(1)V = V^*V.$$

Using the fact that $\pi$ is a unital *-homomorphism, we obtain that there is a unique spectral measure $E \colon B(\mathbb{R}) \rightarrow \mathcal{B}(\mathcal{K})$ such that $$ \pi(f) = \int_{\mathbb{R}} f \ \mathrm{d}E.$$

Finally, let $S$ be an arbitrary Borel subset of $\mathbb{R}$. Then

$$ \left<P(S)x, y \right> = \int_{\mathbb{R}} \mathbf{1}_S \ \mathrm{d}\mu_{x,y} = \left< \Phi_P(\mathbf{1}_S)x, y\right> $$ $$= \left< V^*\pi(\mathbf{1}_S)Vx, y\right>= \left< V^*\int_{\mathbb{R}} \mathbf{1}_S \ \mathrm{d}EVx, y\right> =\left< V^*E(S)Vx, y\right>$$ for all $x, y \in \mathcal{H}$.

  • $\begingroup$ I was really hoping for a proof that didn't use quite so much from Operator Algebras. But this may be about as simple as it gets. Thank you for this well-done answer! $\endgroup$ Oct 24, 2014 at 21:14
  • $\begingroup$ T.A.E. I really doubt that you can avoid using Stinespring here (or some version of it). This is a very nice proof. $\endgroup$ Oct 24, 2014 at 21:16
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    $\begingroup$ @mgn, I don't think it's quite right as you really need indicators of sets which are not necessarily clopen and if $\Omega$ is connected you, you don't have any such sets which are non-trivial. I guess you should still consider the C*-algebra of bounded Borel functions on $\Omega$. $\endgroup$ Oct 25, 2014 at 11:05
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    $\begingroup$ Yes, you are correct, you take the same C*-algebra (bounded Borel functions). Continuous functions are not a good choice due to the lack of indicators. I have just found a good notes on projective measures and dilating them, here is the link: math.tamu.edu/~larson/hlll.pdf (on page 11 and 12 you can find a version of the theorem with a compact $\Omega$, and the sketch seems the same as the proof in the answer). $\endgroup$
    – m_gnacik
    Oct 25, 2014 at 12:03
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    $\begingroup$ Thank you mgn and @TomekKania for the clarification. Seeing how general this approach is, I doubt that a simpler proof exists. Concerning just the case on $\mathbb{R}$: to think that one could simultaneously dilate all the positive operators $P(S)$ to $V^{\star}E(S)V$ using a fixed $V$ and a spectral measure $E$ is a result that I would have guessed to be wrong had I not known better. That's why I was cautious in assuming that the fully general case can be handled with the same elegant approach. The arguments of $C^{\star}$ algebras are exceedingly abstract and effective. $\endgroup$ Oct 25, 2014 at 17:06

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