Spectral measure and measurable functional calculus

Question

I am learning about measurable functional calculus at the moment. I have learned that given a operator $T \in B(H)$ that is self adjoint, there exists a measurable functional calculus $\overline{\Phi}:B_b(\sigma(T)) \rightarrow B(H)$, i.e. a map from the bounded Borel functions into the set of bounded linear operators.

The next step is to introduce spectral measures and with that the integration with respect to a spectral measure.

Now let $T$ be a linear bounded self-adjoint operator and consider $E_A:=\chi_{A \cap \sigma(T)}(T)$ for $A \in \mathcal {B}(\mathbb{R})$. Where $\sigma(T)$ denotes the spectrum of $T$, $\chi$ the characteristic function,$ \mathcal{B}(\mathbb{R})$ the Borel sigma algebra on the reals and $\chi_{A \cap \sigma(T)}(T)$ in the sense of the measurable functional calculus, i.e. $\chi_{A \cap \sigma(T)}(T)=\Phi(\chi_{A \cap \sigma(T)})$. Note that the way we defined the map $E$, we get a (compactly supported) spectral measure.

On the other hand, given a compactly supported spectral measure $E$ we can define a operator $T$ as follows $T:=\int_{\sigma(T)} \lambda dE$. Then T is linear bounded and self-adjoint. Further if we use the above method to gain a spectral measure $\hat{E }$ from $T$, then $E$ and $\hat{E}$ coincide. Further integration with respect to $E$ gives the uniquely determined functional calculus of $T$.

Summarized we get:

Suppose $E$ is a (compactly supported) spectral measure, define the operator $T$ as $T:=\int_{\sigma(T)} \lambda dE$. Then $E_{\sigma(T)}=I$, $T$ is linear bounded and self-adjoint and its measurable functional calculus is given by $\Psi: B_b(\sigma(T)) \rightarrow B(H), f \mapsto \int_{\sigma(T)} f dE$.

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I would like to prove the above statement (what comes after Summarized.....)

Approach: Part 1: (T is linear, bounded, self-adjoint) Suppose $E$ is a compactly supported spectral measure, where $K$ is compact. The identity function $id_K$ is bounded Borel (measurable),thus $T$ is bounded. By the linearity of the integral, $T$ is linear. And since $id_K$ is real valued $T$ is self-adjoint.

Part 2: ($E_{\sigma(T)}=I$) Next would be to show that $E_{\sigma(T)}=I$. While thinking about it I noticed that I do not really see the reason why we need to show this to begin with. If we wanted to show that the spectral measure we obtain from $T$ concides with the spectral measure we used to define $T$, I would have thought of showing that both spectral measures yield the same result for any Borel set $A$. (I am not even sure if the condition $E_{\sigma(T)}=I$ is here to show that the spectral measure is unique.)

So my first question is: Why do we need to show that $E_{\sigma(T)}=I$?

Part 3: ($\Psi$) If I did understand it correctly, we have to show that integrating a function $f$ w.r.t the spectral measure $E$ which is used to define $T$ gives the same measurable functional calculus. Again, I am a little bit confused on this part. I already learned that there exists a measurable functional calculus $f \mapsto f(T)$. And I have also seen the proof of this statement. But I guess this is different since now we have spectral integrals. How would one approach Part 3? The only thing I could think of is to show that \Psi is an involutive algebra homomorphism, i.e. $\Psi(fg)=\Psi(f) \Psi(g)$, linear ,continuous (i.e. $\lVert \Psi(T) \rVert \leq \lVert f \rVert_{\infty}$) and $\Psi(\overline{f})=\Psi(f)^*$. And do to so I guess one would need to use the definition of integration wrt spectral measures.

Thanks in advance for any help!

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Edit 1: Regarding Part 3: On Linearity: Let $f$ be a bounded Borel function. We have to show that $f \mapsto \int_{\sigma(T)} f dE$ is linear. For that suppose $f,g \in B_b (\sigma(T)) and \mu \in \mathbb{C}$. Then obviously, $\Psi(\mu f+g)=\int_{\sigma(T)} \mu f +g dE=\mu \int_{\sigma(T)} f dE + \int_{\sigma(T)} g dE$, where we just use the linearity of the integral.

On Multiplicativity: I couldn't find a "direct" way, so I will try it with approximation. Since step functions are dense in bounded Borel functions we can approximate. Suppose $f_n$ and $g_n$ are step functions such that $f_n \rightarrow f$ and $g_n \rightarrow g$ uniformly. Denote $f_n=\sum_{i=1}^{\infty} \alpha_i \chi_{A_i} and g_n=\sum_{i=1}^{\infty} \beta_i \chi_{_i}.$ $\Psi(f_n g_n)=\int_{\sigma(T)} (\sum_{i=1}^{\infty} \alpha_i \chi_{A_i})(\sum_{i=1}^{\infty} \beta_i \chi_{B_i}) dE=\sum_{j=1}^n ( \beta_j \chi_{B_j} \sum_{i=1}^n \alpha_i \chi_{A_i}$)

Then since $\chi_{A_i} \chi{B_i}=\chi_{A_i \cap B_i}$ we further get

$=\sum_j \sum \beta_j \alpha_i \int_{\sigma(T)} \chi_{B_j \cap A_i} dE =\sum_j \sum \beta_j \alpha_i E_{B_j \cap A_i} =\sum_j \sum \beta_j \alpha_i E_{B_j} E{ A_i}$

Then we can just rearange the sums to their "original" and get $(\sum_j \beta_j E_{B_j})(\sum_i \alpha_i E_{A_i})=(\sum_j \beta_j \int_{\sigma(T)} \chi_{B_j} dE)(\sum_i \alpha_i \int_{\sigma(T)} \chi_{A_i}) dE$

Thus for step functions $f_n, g_n$ we have $\Psi(f_n g_n)=\Psi (f_n) \Psi(g_n)$.

Then we can use that $\lVert \int f dE \rVert \leq \lVert f \rVert_{\infty}$ to see that $\Psi(f_n g_n)-\Psi(f) \Psi(g) \rightarrow 0 $. Thus we get multiplicativity property.

On $Psi(\overline{f})=\Psi(f)^*$: ?

Answer 1

To the first question: The condition $E(\sigma(T))=I$ is important for the following reason: To get that there exists a unique spectral measure $E$ with $T = \int_{\sigma (T)} \lambda dE(\lambda)$ we must add the condition that $E(\sigma(T)) = I$. The following example shows that $E(\sigma(T))=I$ is a necessary condition for uniqueness:

Example: Consider $T \in B(\mathbb{C}^2)$ given by multiplication with the matrix $$ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}. $$ Then $T$ is self-adjoint and $\sigma (T) = \{1,0\}$. Let $(e_1, e_2)$ be the standard basis of $\mathbb{C}^2$. Define the spectral measure $E$ on $\mathbb{R}$ by $$E(A)x:= \begin{cases} e_1 \langle e_1, x \rangle, \ \text{if } 1 \in A,\\ 0 , \ \text{else}. \end{cases} + \begin{cases} e_2 \langle e_2, x \rangle, \ \text{if } 1000000 \in A,\\ 0 , \ \text{else}. \end{cases} $$ Then for all $x \in \mathbb{C}^2$: $$ \int_{\sigma (T)} \lambda dE(\lambda)x = e_1 \langle e_1, x \rangle = Tx. $$ But at the same time if we define the spectral measure $\tilde{E}$ on $\mathbb{R}$ by $$ \tilde{E}(A) x:= \begin{cases} e_1 \langle e_1, x\rangle, \ \text{if} \ 1 \in A, \\ 0, \ \text{else}. \end{cases} + \begin{cases} e_2 \langle e_2, x\rangle, \ \text{if} \ 0 \in A, \\ 0, \ \text{else}. \end{cases} $$ Then also $\int_{\sigma (T)} \lambda d\tilde{E}(\lambda ) =T$. However $E \neq \tilde{E}$. Only $\tilde{E}$ is truly the spectral measure of $T$. With $E$ we cant even get $\ker T$ and $E$ creates a different functional calculus than $\tilde{E}$. The problem here is that $E(\sigma(T)) \neq I$.

To the second question: Let $\Phi$ be the (original) functional calculus associated to $T$. You already know that $\Phi(s) = \int_{\sigma (T)} s dE$ for all bounded simple functions $s$ by the definition of $E$ and $\int_{\sigma (T)} s dE$ as well as the fact that $\Phi$ is linear. To conclude you only need to show/know that $\forall x,y \in H$: $$\lim_{n \to \infty}\langle \Phi (s_n) x, y\rangle = \langle \Phi (f) x,y \rangle,$$ where $s_n$ is an uniformly bounded sequence of simple functions converging pointwise to an arbitrary bounded measurable function $f$. Because then $\forall x \in H$: $$ \langle \Phi (f) x,x\rangle = \lim_{n \to \infty}\langle \Phi (s_n) x, x\rangle \\ = \lim_{n \to \infty}\int_{\sigma (T)} s_n dE_{x,x} (\lambda) = \int_{\sigma (T)} f dE_{x,x} (\lambda) = \langle \int_{\sigma (T)} f dE x,x\rangle, $$ because of the dominated convergence theorem (for the third equality) and so $\Phi (f) = \int_{\sigma (T)} f dE$.