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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!

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You can't explicitly compute this projection since you don't have explicit description of Borel sets. –  no identity Dec 13 '12 at 18:58
    
@Hui Yu: It is not true that every bounded Borel-measurable function on a complete metric space is the pointwise limit of a sequence of polynomial functions. This is because the pointwise limit of a sequence of continuous functions on a metric space, if such a limit exists, must be continuous on a dense set of points. You can prove this using the Baire Category Theorem. –  Haskell Curry Jan 7 '13 at 10:23
    
@Hui Yu: Hence, $ \chi_{[0,1] \cap \mathbb{Q}} $ cannot be the pointwise limit of a sequence of polynomial functions. Instead of pointwise convergence, one can ask for pointwise convergence almost everywhere. This is possible by Lusin's Theorem. –  Haskell Curry Jan 7 '13 at 10:27
    
@Norbert: Strange. The Borel subsets are staring at me right in my face. Are they not just the usual Borel subsets of $ [0,1] $? –  Haskell Curry Jan 7 '13 at 10:32
    
@HaskellCurry Borel sets are of very complicated structure. You can use intersection (finite or countable), complementation, union (finite or countable) and other operations to construct Borel sets. But even after applying them countable amount of times you will not get all the Borel sets. You must make this procedure up to $\omega_1$ oridinal to cover the whole variety of Borel sets. This is what I meant when I said that structure of Borel sets is complicated. –  no identity Jan 9 '13 at 17:20

1 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 $.

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