Uniqueness spectral decomposition unbounded self-adjoint operator

Question

Let $A$ be a self-adjoint (unbounded) operator in a Hilbert space $H$. In Rudin's book "Functional analysis", theorem 13.30, the following is proven:

There exists a unique spectral resolution $E$ of the identity defined on the Borel sets of $\mathbb{R}$ such that $$A= \int_\mathbb{R} \lambda dE(\lambda).$$

The idea of the proof is as follows: One considers the Cayley-transform $U\in B(H)$ of $A$, and then one considers the spectral resolution $E'$ with $$U = \int_{\sigma(U)}\lambda dE'(\lambda).$$ Then we can define the homeomorphism $$g: \Omega := S^1\setminus \{1\} \to \mathbb{R}: \lambda \mapsto \frac{i(1+\lambda)}{1-\lambda}.$$ Then one checks that $$E(\omega):= E'(g^{-1}(\omega)\cap \sigma(U))$$ where $\omega\subseteq \mathbb{R}$ is Borel, works. This settles the existence.

However, I don't understand how the proof in Rudin's book shows uniqueness (if someone asks for it, I can supply the proof here, but I didn't want to make the post needlessly long). Apparently, the idea is to use uniqueness of the spectral decomposition for the unitary $U$ and then conclude the desired uniqueness from this. Can someone please fill in the details? Thanks in advance for your help.

Answer 1

The point is that you have $g(U)=A$. Then $$ A=\int_{\mathbb R} g(\lambda)\,dE'(\lambda)=\int_{S^1\setminus\{1\}}\lambda\,dE'\circ g^{-1}. $$ More generally, $$ h(A)=\int_{\mathbb R} h\circ g(\lambda)\,dE'(\lambda)=\int_{S^1\setminus\{1\}}h(\lambda)\,dE'\circ g^{-1}. $$ Thus $E=E'\circ g^{-1}$.

Answer 2

This is a bit subtle. Recall the following result:

If $T\in B(H)$ is a normal operator, there exists a unique resolution of the identity $E$ on the Borel subsets of $\sigma(T)$ which satisfies $$T = \int_{\sigma(T)} z dE(z).$$

As a consequence of this, let us prove the following:

If $U\in B(H)$ is a unitary operator such that $1-U$ is injective, then there exists a unique spectral resolution of the identity $E'$ on the Borel subsets of $\Omega:=S^1\setminus \{1\}$ such that $$U = \int_{\Omega} z dE'(z).$$

Proof: Uniqueness: Let $E''$ be another spectral resolution of the identity on the Borel subsets of $\Omega$ with $U = \int_\Omega z dE''(z)$. Then for $f\in C_0(\Omega)$ and $x,y\in H$, the Stone-Weierstrass theorem implies that \begin{align*}\int_\Omega f(z)dE'_{x,y}(z)=\langle f(U)x,y\rangle = \int_\Omega f(z) dE''_{x,y}(z) \end{align*} so that by the Riesz representation theorem $E'_{x,y}= E_{x,y}''$. Hence, $E'= E''$.

Existence: Given a Borel subset $A\subseteq \Omega$, define $E'(A) = E(A \cap \sigma(U))$ where $E$ is the spectral resolution associated with $U$.

Then $E'$ works. Note that $E'(\Omega \setminus \sigma(U)) = 0$.

We distinguish two cases:

(1) $1 \notin \sigma(U)$ so that $\sigma(U) \cap \Omega = \sigma(U)$.

$$\int_\Omega z dE'(z) = \int_{\sigma(U)} z dE(z) = U.$$

(2) $1\in \sigma(U)$. Then using that $E(\{1\})= 0$ (here we use the assumption that $1-U$ is injective, so that $1$ is no eigenvalue of $U$), we get

$$\int_\Omega z dE'(z) = \int_{\sigma(U)\setminus \{1\}} z dE(z) = \int_{\sigma(U)} z dE(z) = U.$$

This ends the proof.

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Having established this, it is now easy to prove the uniqueness you are after. What Rudin is really using is the uniqueness of the spectral measure on $\Omega$, instead of on the spectrum $\sigma(U).$

Indeed, recall first that if $A$ is a self-adjoint unbounded operator on $H$, then its Cayley transform $U$ is a (bounded) unitary operator such that $1-U$ is injective. Thus, we have $$U = \int_{\Omega} z dE'(z)$$ for a unique resolution of the identity $E'$ on the Borel subsets of $\Omega$.

Assume now that $A = \int_\mathbb{R} t dE(t)$ where $E$ is a resolution of the identity on the Borel subsets of $\mathbb{R}$. Consider the homeomorphisms $$f: \Omega \to \mathbb{R}: z \mapsto \frac{i(1+z)}{1-z}, \quad g: \mathbb{R}\to \Omega: t \mapsto \frac{t-i}{t+i}.$$ Consider the resolution of the identity $F$ defined on Borel subsets of $\Omega$ by $$F(A) := E(f(A)).$$ Then we have $$\int_\Omega z dF(z) = \int_\Omega g(z) dE(z)= U= \int_\Omega z dE'(z)$$ so that $F= E'$. Then $$E(A) = F(g(A)) = E'(g(A))$$ so $E$ is uniquely determined.