# Generating a projection valued measure from a resolution of the identity

Let $$H$$ be a complex Hilbert space. Suppose $$P : \mathbb{R} \to \mathcal{L}(H)$$ is a resolution of the identity, meaning:

1. Each $$P(\lambda)$$ is an orthogonal projection.
2. $$P(\lambda_1) \le P(\lambda_2)$$ for $$\lambda_1 \le \lambda_2$$.
3. $$P(\lambda_n)$$ converges strongly to $$P(\lambda)$$ whenever $$\lambda_n \searrow \lambda$$.
4. $$P(\lambda_n)$$ converges strongly to $$0$$ ($$\text{Id}$$) as $$\lambda_n \searrow -\infty$$ ($$\lambda_n \nearrow +\infty$$).

For each $$\psi \in H$$, by 1., 2. and 3., we get an increasing right continuous function $$\lambda \mapsto \langle \psi, P(\lambda) \psi \rangle,$$ so there is a unique Borel measure $$\mu_\psi$$ obeying $$\mu_\psi(\lambda_1, \lambda_2] = \langle \psi, P(\lambda_1, \lambda_2] \psi \rangle$$ where we define $$P(\lambda_1, \lambda_2] := P(\lambda_2) - P(\lambda_1)$$. By 4., we have $$\mu_\psi(\mathbb{R}) = \lim_{n \to \infty}\mu_\psi(-n,n] = \lim_{n \to \infty}( \langle \psi, P(-n) \psi \rangle - \langle \psi, P(n) \psi \rangle) = \|\psi\|^2 < \infty.$$

Next, for each pair $$(\varphi, \psi) \in H \times H$$, define the complex measure $$\mu_{\varphi, \psi} := \frac{1}{4}\Big( \mu_{\varphi + \psi} - \mu_{\varphi - \psi} +i(\mu_{\varphi - i\psi} - \mu_{\varphi + i\psi})\Big).$$

A brief calculation shows that, for each $$(\varphi, \psi) \in H \times H$$ and each interval of the form $$(\lambda_1, \lambda_2]$$, one has $$\mu_{\varphi, \psi}(\lambda_1, \lambda_2]= \langle \varphi, P(\lambda_1, \lambda_2]\psi \rangle.$$ The left side is sesquilinear in $$(\varphi, \psi)$$ because the right side is. Since the half-open intervals generate the Borel $$\sigma$$-algebra on $$\mathbb{R}$$, this sesquilinearity of the left side continues to hold with $$(\lambda_1, \lambda_2]$$ replaced by an arbitrary Borel set $$\Omega$$. Moreover, using Cauchy Schwarz for sesquilinear forms $$|\mu_{\varphi, \psi}(\Omega) | \le \mu_{\varphi, \varphi}(\Omega)^{1/2} \mu_{\psi, \psi}(\Omega)^{1/2} \le \|\varphi\|\|\psi\|.$$ Thus, by the Lax-Milgram lemma, there is a unique bounded operator $$P(\Omega)$$ on $$H$$ with $$\mu_{\varphi, \psi}(\Omega) = \langle \varphi, P(\Omega)\psi \rangle$$ (and incidentally $$\|P(\Omega) \|_{H \to H} \le 1$$). Moreover, $$P(\Omega)$$ is self adjoint because $$\langle P(\Omega) \psi, \psi \rangle = \overline{\langle \psi,P(\Omega) \psi\rangle} = \overline{\mu_{\psi,\psi}(\Omega)} = \mu_{\psi,\psi}(\Omega) = \langle \psi,P(\Omega) \psi\rangle.$$

I would now like to show that each $$P^2(\Omega) = P(\Omega)$$ (i.e., each $$P(\Omega)$$ is a projection), but I've gotten stuck.

It would suffice to show that for any $$\psi$$, $$\Omega_1$$ and $$\Omega_2$$, $$\langle \psi, P(\Omega_2)P(\Omega_1)\psi \rangle = \langle \psi, P(\Omega_2 \cap\Omega_1)\psi \rangle.$$

Using additivity of measures it's easy to see that $$P(\mathbb{R} \setminus \Omega) = \text{Id} - P(\Omega)$$ as well as $$P(\Omega_2 \cup \Omega_1) + P(\Omega_2 \cap \Omega_1) = P(\Omega_2) + P( \Omega_1)$$, though I'm not sure these identities are useful.

We can prove that each $$P^2(\Omega) = P(\Omega)$$ as follows. First, one shows (without too much effort), using the properties of $$\lambda \mapsto P(\lambda)$$, that $$P((a_1, b_1] \cap (a_2, b_2]) = P((a_1, b_1]) P( (a_2, b_2])$$
Then for fixed half-open interval $$(a_1, b_2]$$, it follows that the measures $$\Omega \mapsto \mu_{P(a_1, b_2]\psi, \psi} (\Omega), \, \mu_{\psi} ((a_1, b_2]\cap \Omega),$$
agree on the half-open intervals, and hence are identical. Subsequently, for any fixed Borel $$\Omega_2$$, the measures
$$\Omega \mapsto \mu_{\psi, P(\Omega_2)\psi} (\Omega), \, \mu_{\psi} (\Omega\cap \Omega_2),$$