Compute spectral/projection-valued measures explicitly?

Question

Spectral/projection-valued measures have very handy applications theoretically, but I got stuck when asked to compute explicitly certain projection-valued measures. Let's focus on the following:

Let $ N: {\mathcal{L}^{2}}([0,1]) \to {\mathcal{L}^{2}}([0,1]) $ be the normal operator defined by $$ \forall f \in {\mathcal{L}^{2}}([0,1]),\forall t \in [0,1]: \quad [N(f)](t) \stackrel{\text{def}}{=} t \cdot f(t). $$ What is the projection-valued measure corresponding to $ N $?

As $ \sigma(N) = [0,1] $, we need a projection-valued measure $ P $ supported on $ [0,1] $. Theoretically, it should be defined just by $$ P(E) = {\chi_{E}}(N), $$ where $ \chi_{E} $ is the characteristic function of $ E \subseteq [0,1] $ and $ {\chi_{E}}(N) $ is obtained via the Borel functional calculus of $ N $.

However, to find $ {\chi_{E}}(N) $, we need to find a sequence of polynomial functions converging to $ \chi_{E} $, which is not an easy job.

I somehow feel that the projection-valued measure is given simply by $ P(E) = {\chi_{E}}(N) $, but I am not sure.

Can someone give a hint on how to compute the spectral/projection-valued measure explicitly?

Thanks!

Answer 1

Here is a solution that illustrates how one gets his/her hands dirty with spectral measures. We fix some notation first.

• Let $ \mathcal{C}([0,1]) $ denote the space of all continuous functions from $ [0,1] $ to $ \mathbb{C} $.

• Let $ {\mathcal{B}_{b}}([0,1]) $ denote the space of all bounded Borel-measurable functions from $ [0,1] $ to $ \mathbb{C} $.

---

The following lemma will serve as our workhorse.

Lemma Let $ \mu $ be a finite positive Borel measure on $ [0,1] $. Then every $ \varphi \in {\mathcal{B}_{b}}([0,1]) $ is the $ \mu $-a.e. pointwise limit of a bounded sequence $ (\psi_{n})_{n \in \mathbb{N}} $ in $ \mathcal{C}([0,1]) $.

Proof: Using Lusin’s Theorem (cf. Theorem 2.24 of Rudin’s Real and Complex Analysis, Third Edition), we can inductively create a sequence $ (E_{n})_{n \in \mathbb{N}} $ of Borel subsets of $ [0,1] $ such that for all $ n \in \mathbb{N} $,

(1) $ \mu(E_{n}) < \dfrac{1}{2^{n}} $ and

(2) there exists a $ \psi_{n} \in \mathcal{C}([0,1]) $ satisfying $$ \| \psi_{n} \|_{\infty} \leq \| \varphi \|_{\infty} \quad \text{and} \quad \varphi|_{[0,1] \setminus E_{n}} = {\psi_{n}}|_{[0,1] \setminus E_{n}}. $$

We now show that $ (\psi_{n})_{n \in \mathbb{N}} $ meets the necessary requirements. By (2), the boundedness of $ (\psi_{n})_{n \in \mathbb{N}} $ is ensured. From (1), we obtain $ \displaystyle \sum_{n=1}^{\infty} \mu(E_{n}) < \infty $; it follows that almost every $ x \in [0,1] $ lies in $ [0,1] \setminus E_{n} $ for all $ n \in \mathbb{N} $ sufficiently large. Therefore, $ \displaystyle \lim_{n \to \infty} {\psi_{n}}(x) = \varphi(x) $ for almost every $ x \in [0,1] $. This completes the proof. $ \quad \spadesuit $

Let $ P $ be the spectral decomposition of $ N $, which is a projection-valued measure defined on the Borel subsets of $ \sigma(N) = [0,1] $. To each $ \varphi \in {\mathcal{B}_{b}}([0,1]) $, there corresponds a bounded linear operator $ T(\varphi) $ that is uniquely determined by the condition $$ \forall f,g \in {L^{2}}([0,1]): \quad \langle [T(\varphi)](f),g \rangle = \int_{[0,1]} \varphi ~ d{P_{f,g}}, $$ where $ P_{f,g} $ is the complex Borel measure on $ [0,1] $ defined by $$ {P_{f,g}}(E) \stackrel{\text{def}}{=} \langle [P(E)](f),g \rangle $$ for all Borel subsets $ E $ of $ [0,1] $. We then have the following result (click here for a proof).

Theorem The mapping $ \varphi \longmapsto T(\varphi) $ is a unital *-homomorphism from $ {\mathcal{B}_{b}}([0,1]) $ to $ B({L^{2}}([0,1])) $.

Note that $ T(\varphi) $ is commonly denoted by $ \varphi(N) $.

Let $ E $ be a Borel-measurable subset of $ [0,1] $. Then $ \chi_{E} \in {\mathcal{B}_{b}}([0,1]) $ and \begin{align} \forall f,g \in {L^{2}}([0,1]): \quad \langle [T(\chi_{E})](f),g \rangle &= \int_{[0,1]} \chi_{E} ~ d{P_{f,g}} \\ &= {P_{f,g}}(E) \\ &= \langle [P(E)](f),g \rangle. \end{align} We thus obtain $ {\chi_{E}}(N) = T(\chi_{E}) = P(E) $.

Fix $ f \in {L^{2}}([0,1]) $. Notice that $ P_{f,f} $ is a finite positive Borel measure on $ [0,1] $. By the lemma, there exists a bounded sequence $ (\psi_{n})_{n \in \mathbb{N}} $ in $ \mathcal{C}([0,1]) $ that converges pointwise $ P_{f,f} $-a.e. to $ \chi_{E} $. Using the theorem, we get \begin{align} \forall n \in \mathbb{N}: \quad \| [T(\chi_{E})](f) - [T(\psi_{n})](f) \|^{2} &= \| [T(\chi_{E}) - T(\psi_{n})](f) \|^{2} \\ &= \| [T(\chi_{E} - \psi_{n})](f) \|^{2} \\ &= \langle [T(\chi_{E} - \psi_{n})](f),[T(\chi_{E} - \psi_{n})](f) \rangle \\ &= \left\langle [T(\chi_{E} - \psi_{n})]^{*} [T(\chi_{E} - \psi_{n})](f),f \right\rangle \\ &= \left\langle \left[ T \left( \overline{\chi_{E} - \psi_{n}} \right) \right][T(\chi_{E} - \psi_{n})](f),f \right\rangle \\ &= \left\langle \left[ T \left( \overline{\chi_{E} - \psi_{n}} (\chi_{E} - \psi_{n}) \right) \right](f),f \right\rangle \\ &= \left\langle \left[ T \left( |\chi_{E} - \psi_{n}|^{2} \right) \right](f),f \right\rangle \\ &= \int_{[0,1]} |\chi_{E} - \psi_{n}|^{2} ~ d{P_{f,f}}. \end{align} The Dominated Convergence Theorem then yields $$ 0 = \lim_{n \to \infty} \int_{[0,1]} |\chi_{E} - \psi_{n}|^{2} ~ d{P_{f,f}} = \lim_{n \to \infty} \| [T(\chi_{E})](f) - [T(\psi_{n})](f) \|^{2}. $$ Therefore, \begin{align} [T(\chi_{E})](f) &= \lim_{n \to \infty} [T(\psi_{n})](f) \\ &= \lim_{n \to \infty} \psi_{n} \cdot f \quad (\text{By the Continuous Functional Calculus.}) \\ &= \chi_{E} \cdot f. \quad (\text{By the Dominated Convergence Theorem again.}) \end{align}

Conclusion: For all $ f \in {L^{2}}([0,1]) $, we have $ [{\chi_{E}}(N)](f) = [P(E)](f) = \chi_{E} \cdot f $.