# Showing that if $A\in\mathcal{B}(H)$ and $v$ is an eigenvector of $A$ with $Av=cv$ then $f(A)v=f(c)v$ for a bounded measurable $f$ on $\sigma(A)$

This is the exercise 6 of chapter 8 of Hall's book: Quantum Theory for Mathematicians.

Suppose $$A\in\mathcal{B}(H)$$ is self-adjoint and $$v$$ is an eigenvector for $$A$$ with $$Av = cv$$ for some $$c\in\mathbb{R}$$. Then for any bounded and measurable function $$f$$ on the spectrum $$\sigma(A)$$ we have $$f(A)v = f(c)v$$

where $$H$$ is a Hilbert space over $$\mathbb{C}$$ and $$f(A)$$ is an operator defined as

$$f(A) = \int_{\sigma(A)}f(c)d\mu^A(c)$$

with $$\mu^A$$ being such a unique projection-valued measure on the Borel $$\sigma$$-algebra in $$\sigma(A)$$ with values in projections on $$H$$ such that

$$\int_{\sigma(A)}cd\mu^A(c) = A$$

A given hint is to use the prior exercise 5:

Suppose $$A\in \mathcal{B}(H)$$ is self-adjoint operator and $$V$$ is a closed subspace of $$H$$ that is invariant under $$A$$. Then a.) the spectrum of the restriction to $$V$$ of $$A$$ is contained in the spectrum of $$A$$ b.) if $$f$$ is a bounded measurable function on $$\sigma(A)$$ then $$V$$ is invariant under $$f(A)$$ and $$f(A)\mid_V = f(A\mid_V)$$.

Thoughts/problem: My problem is that I don't really know how to apply $$f(A)$$ to $$v$$ or any element $$w\in H$$ for that matter. Connection to exercise 5 is probably by the span of $$v$$. Namely, define $$S:=\{\alpha v\mid \alpha\in\mathbb{C}\}$$. It is not hard to show that the span of $$v$$, $$S$$, is a closed subspace of $$H$$ invariant under $$A$$. Then if we can determine what $$f(A)w$$ for $$w\in H$$ is in terms of $$w$$, the result may follow quite easily. But I am lost even on what $$\chi_E(A)\psi := \int_{\sigma(A)}\chi_E(c)d\mu^A(c) = \mu^A(E)\psi$$ should be for a measurable $$E\subset\sigma(A)$$, let alone how to move the limit outside the integral when transition from simple functions to measurable functions.

• The quantity $f(c)$ is not well defined. May 25 at 12:18
• @RyszardSzwarc How so? May 25 at 12:43
• The function $f$ is measurable. If it was continuuos or monotonic left continuous then the point values would be well defined. May 25 at 15:28
• The evaluation is well defined! Here, $f$ is an actual function and not just an element of $L^\infty.$ May 30 at 23:56

Another approach:

Let $$V$$ be the span of the eigenvector $$v.$$ To apply the other exercise, note that $$\sigma(A\vert_V)=\{c\},$$ then: $$f(A)\vert_V=f(A\vert_V)=\int_{\sigma(A\vert_V)}f(\lambda)d\mu^{A\vert_V}(\lambda)=f(c)\int_{\sigma(A\vert_V)}1d\mu^{A\vert_V}(\lambda)=f(c)\mu^{A\vert_V}(\mathbb{R})=f(c)\operatorname{id}_V.$$

The first equality is the other exercise; the second is the spectral theorem for $$A\vert_V;$$ the third follows since the spectrum consists only of the single point $$c,$$ so $$f$$ is constant with value $$f(c)$$ on $$\sigma(A\vert_V)$$; the last equality is a property of the projection-valued measure $$\mu^{A\vert_V}.$$

Now apply both sides of $$f(A)\vert_V=f(c)\operatorname{id}_V$$ to $$v$$ to conclude $$f(A)v=f(c)v.$$

Based on my limited understanding, here is my answer,

I presume or think that: Let $$V_{E_j} = <\{v:Av = qv, q \in E_{j}\}>.$$ Since $$f(A \restriction_V ) = f(A)\restriction_V \implies \int_{E_{j}^c} f(c) d\mu_A(c) w = 0$$ for all $$w \in V_{E_j}.$$

Now, $$f(A) = \int f(c) d\mu_A(c)$$ and let $$\int_{E_{j}^c} f(c) d\mu_A(c) + f(u_j) \mu_A(E_j) \rightarrow \int f(c) d\mu_A(c)$$ as $$j \rightarrow \infty.$$

Now by continuity of measure and for now assuming continuity of $$f$$, $$f(u_j) \mu_A(E_{j}) v \rightarrow f(c) \mu_A(c) v.$$

as $$j \rightarrow \infty.$$

where $$E_j = (c-\epsilon_j,c+\epsilon_j)$$, $$\epsilon_j \rightarrow 0$$, $$u_j = c \in E_j$$ for all $$j.$$

For $$e \notin E_j$$ for all $$j \geq J$$ with $$Av_e = e v_e$$, since $$\int_{E_{j}^c} f(c) d\mu_A(c)v_e = \int f(c) d\mu_A(c)v_e = f(A) v_e$$, we have that, $$f(u_j) \mu_A(E_{j}) v_e \rightarrow 0.$$

But $$f(u_j) \mu_A(E_{j}) v_e \rightarrow f(c) \mu_A(c) v_e.$$ Hence $$f(c) \mu_A(c) v_e = 0$$

By choosing $$f(x) = 1$$ for all $$x$$, we have that $$\mu_A(c) v_q = 0$$ for all $$q \neq c$$ and $$Av_q = qv_q.$$

Now by choosing $$f(x) = x$$ for all $$x$$, we have that, $$c\mu_A(c)v = Av = cv \implies$$ $$\mu_A(c)v = v.$$ Here in the above i assumed $$c \neq 0$$ because otherwise $$\mu_A(c)$$ may not be well defined by the integral definition $$\int c d\mu_A = A.$$

Now if $$f$$ is measurable and not continuous, we now choose $$u_j \rightarrow f(c)$$ and $$E_j = f^{-1}((f(c)-\epsilon_j,f(c)+\epsilon_j))$$, $$\epsilon_j \rightarrow 0.$$ Now by continuity of measure, by same argument as above with above choice of $$E_j$$, $$u_j \mu_A(E_{j}) v \rightarrow f(c)\mu_A(f^{-1}(f(c)))v = f(c) \sum_{\{c_i: f(c_i) = f(c)\}} \mu_A(c_i) v = f(A)v.$$

But then $$\mu_A(c_i)v = 0$$ for all $$c_i \neq c.$$ Hence, we have

$$u_j \mu_A(E_{j}) v \rightarrow f(c)\mu_A(f^{-1}(f(c)))v = f(c) \sum_{\{c_i: f(c_i) = f(c)\}} \mu_A(c_i) v = f(c) \mu_A(c)v = f(c)v= f(A)v.$$

Now if $$\{c_i: f(c_i) = f(c)\}$$ is not a discrete set, you need to use continuity of $$f(x) = 1$$ to conclude:

$$f(c) \mu_A(f^{-1}(f(c)))v = f(c) \int_{\{c_i: f(c_i) = f(c)\}} d\mu_A v = f(c) \mu_A(c)v = f(c)v = f(A)v.$$

• Sorry for the late response. I have few questions: 1.) Why does $f(u_j)\mu_A(E_j)\to 0,j\to\infty$?, 2.) What is $u_j$? May 29 at 13:18
• $u_j = c$ (written in the answer) and $f(u_j) \mu_A(E_j) \rightarrow f(x) \mu_A(c)$ by continuity of measure. May 30 at 2:19
• Ah, so you have taken $\left(u_j\right)_{j}$ to be a constant sequence with $u_j = c$ for all $j$? May 30 at 7:01
• yes. $u_j$ is constant for all $j$. May 30 at 10:46
• Similar question: math.stackexchange.com/questions/4709333/… May 30 at 13:19