# Spectral Measures: Lebesgue

Preface

Dominated convergence: $$f_n(\omega)\to f(\omega)\quad(\omega\in\Omega)\implies f_n(E)\to f(E)$$ (This gives a tool for analysis of operators.)

Problem

Given a Borel space $\Omega$ and a Hilbert space $\mathcal{H}$.

Consider a spectral measure: $$E:\mathcal{B}(\Omega)\to\mathcal{B}(\mathcal{H})$$

And its Borel measures: $$\nu_\varphi(A):=\|E(A)\varphi\|^2$$ $$\mu_{\varphi\psi}(A):=\langle E(A)\varphi,\psi\rangle$$

Regard its measurable calculus: $$\mathcal{D}f(E):=\left\{\varphi\in\mathcal{H}:\int_\Omega|f|^2\mathrm{d}\nu_\varphi\right\}:\quad\langle f(E)\varphi,\psi\rangle:=\int_\Omega f\mathrm{d}\mu_{\varphi\psi}$$

Suppose pointwise convervence: $$f_n(\omega)\to f(\omega)$$

Denote their common domain by: $$\mathcal{D}:=\bigcap_{n\in\mathbb{N}}\mathcal{D}f_n(E)$$

Define their limiting operator by: $$\mathcal{D}T:=\{\varphi\in\mathcal{D}:f_n(E)\varphi\to\psi\}:\quad T\varphi:=\psi$$ (That is the one taken in Stone's theorem.)

Operator Domain

Can it happen that this domain shrinks?

Domain Criterion

Does a common domain suffice: $$\mathcal{D}f_n(E)\supseteq\mathcal{D}f(E):\quad T\supseteq f(E)$$ In that case one obtains: $$\mathcal{D}f_n(E)\supseteq\mathcal{D}f(E)\iff |f_n|^2\leq C_n(|f|^2+1)$$ So one may miss a dominant for: $$\|f(E)\varphi-f_n(E)\varphi\|^2=\|(f-f_n)(E)\varphi\|^2=\int|f-f_n|^2\mathrm{d}\nu_\varphi$$

But what is an explicit nonexample?

Dominant Criterion

Assuming a dominant one gets: $$|f_n|^2\leq C(|f|^2+1):\quad T=f(E)$$

Clearly one has: $$\mathcal{D}f(E)\subseteq\mathcal{D}f_n(E)$$

So the above shows: $$\|f(E)\varphi-f_n(E)\varphi\|^2=\ldots=\int_\Omega|f-f_n|^2\mathrm{d}\nu_\varphi\to0$$

How to prove the converse inclusion?

Operator Core

Does one have as core: $$\mathcal{D}f_n(E)\supseteq\mathcal{D}f(E):\quad\overline{T}=f(E)$$ (This solves the failure above!)

• what is "finite measure space"? Finite set? – Yurii Savchuk Dec 10 '14 at 8:12
• what is $\mathcal D$ ? – Yurii Savchuk Dec 10 '14 at 8:13
• @YuriiSavchuk: Oh, it actually it must be only a measurABLE space; for instance a topological space together with its Borel algebra. – C-Star-W-Star Dec 10 '14 at 13:15
• @YuriiSavchuk: The domain of an operator: $\mathcal{D}(T)$ – C-Star-W-Star Dec 10 '14 at 13:18

Ok, I think I finally got it...

(However, if anybody finds bugs, typos, loopholes etc. then please let me know. Thanks!)

Framework

Given the natural numbers $\Omega:=\mathbb{N}$ and the Hilbert space $\mathcal{H}:=\ell^2(\mathbb{N})$.

Choose the canonical basis by: $e_n:=\chi_n$

Consider the spectral measure $E(\{n\}):=P_n$.

Operator Domain

Regard the series of measurable functions $f_n:=\sum_{k=1}^nk\chi_{\{k\}}$.

Denote their sum by: $f:=\sum_{k=1}^\infty k\chi_{\{k\}}$

It approaches pointwise its sum: $f_n\to f$

It has a dominant: $|f_n|^2\leq1(|f|^2+1)$

However, not every element converges: $$\varphi:=\sum_{k=1}^\infty\frac{1}{k}e_k:\quad f_n(E)\varphi=\sum_{k=1}^n1e_k\nrightarrow\psi$$

So the domain may shrink: $\mathcal{D}T\subsetneq\mathcal{D}$

Domain-Criterion

Regard the sequence of measurable functions $f_n:=n\chi_{\{n\}}$.

It vanishes pointwise: $f_n\to0$

Now, it fulfils the domain criterion: $$\mathcal{D}f_n(E)=\mathcal{H}\supseteq\mathcal{H}=\mathcal{D}f(E)$$ However, its measurable calculus does not converge: $$\varphi:=\sum_{k=1}^\infty\frac{1}{k}e_k:\quad f_n(E)\varphi=e_n\nrightarrow0=f(E)\varphi$$

So a common domain is not sufficient: $T\subsetneq f(E)$

Dominant-Criterion

Suppose $\varphi\in\mathcal{D}T$ with $T\varphi=\psi$.

By Fatou applied to the measurable calculus one has: $$0\leq\int|f-f_n|^2\mathrm{d}\nu_\varphi\leq\liminf_m\int|f_m-f_n|^2\mathrm{d}\nu_\varphi=\liminf_m\|f_m(E)\varphi-f_n(E)\varphi\|^2\to0$$ But that implies especially: $\varphi\in\mathcal{D}f(E)$

Moreover, one has then: $f(E)\varphi=\lim_nf_n(E)\varphi=\psi=T\varphi$

Concluding that the operators agree: $f(E)=T$

Operator Core

Suppose $\varphi\in\mathcal{D}f(E)$ with $f(E)\varphi=\psi$.

By Egorov's theorem one finds: $$\nu_\varphi(\Omega_N)\to\nu_\varphi(\Omega):\quad\|f-f_n\|_{\Omega_N}\stackrel{n\to\infty}{\to}0$$

Set the approximation to be: $$\varphi_N:=1_N(E)\varphi:=E(\Omega_N)\varphi$$

So on the hand one has: $\varphi_N\to\varphi$

Also note that: $f_n(E)1_N(E)=(f_n1_N)(E)$

Now by uniform convergence: $$|f_n1_N|\leq R_N(|f1_N|+1)\quad|f1_N|\leq R_N(|f_n1_N|+1)$$ So one has a common domain: $\mathcal{D}(f1_N)(E)=\mathcal{D}(f_n1_N)(E)$

But also one has trivially: $$|f1_N|\leq1(|f|+1)$$ Thus they all include: $\mathcal{D}f(E)\subseteq\mathcal{D}(f1_N)(E)$

Hence one can compute: $$\left\|(f1_N)(E)\varphi-(f_n1_N)(E)\varphi\right\|^2=\int_{\Omega_N}|f-f_n|^2\mathrm{d}\nu_\varphi\leq\|f-f_n\|_{\Omega_N}\|\varphi\|^2\stackrel{n\to\infty}{\to}0$$ That is $\varphi_N\in\mathcal{D}T$ with $T\varphi_N=(f1_N)(E)\varphi$.

But by the preceding dominant-criterion: $$|f1_N|\leq1(|f|+1):\quad(f1_N)(E)\varphi\to f(E)\varphi$$ So as was desired: $T\varphi_N\to f(E)\varphi$

Concluding that it is a core: $\overline{T}=f(E)$