# Spectral theorem for unbounded self-adjoint operators, questions about the proof

I want to understand the proof of the Spectral theorem for unbounded self-adjoint operators. First the theorem:
Let H be a separable complex Hilbert space, $A:D(A)\subseteq H\to H$ a densily defined self-adjoint operator. Then there exists a $\sigma$-finite measure space $(\Omega ,\Sigma ,\mu)$, a measurable function $f:\Omega \to\mathbb{R}$ and a unitary operator $V:H\to L^2(\Omega)$, such that:
$$1. \text{for}\; x\in H: x\in D(A) \Leftrightarrow f\cdot V(x)\in L^2(\Omega)$$ and $$2. VAV^{-1}\varphi =M_f\varphi\;\; \forall \varphi \in D(M_f)$$ where $M_f:D(M_f)=\{u\in L^2(\Omega); f\cdot u\in L^2(\Omega)\}\subseteq L^2(\Omega)\to L^2(\Omega),\; u\mapsto f\cdot u$ is the multiplication operator.

We had 2 theorems before we prove this theorem and which will be needed to prove the Spectral theorem as stated above I think :
Theorem 1: Let H be a separable complex Hilbert space, $A\in L(H)$ self-adjoint. Then there exists a $\sigma$-finite measure space $(\Omega ,\Sigma ,\mu)$, a bounded measurable function $f:\Omega \to\mathbb{R}$ and a unitary operator $V:H\to L^2(\Omega)$, such that:
$(VAV^{-1})\varphi =f\cdot \varphi \;\; \mu$-almost everywhere.

Theorem 2 and definition: Let H be a Hilbert space and $A:D(A)\subseteq H\to H$ a densily defined self-adjoint operator. Then the operator $U=(A+iId)(A-iId)^{-1}$ is well-defined, linear and continuous, unitary and 1 is not in the point spectrum of U. It is $A=i(U+Id)(U-Id)^{-1}$ and we say: U is the Cayley transform of A.

Now the proof of the spectral theorem:
A self-adjoint and therefore the operator $U=(A+iId)(A-iId)^{-1}$ is well-defined, linear and continuous, unitary and 1 is not in the point spectrum of U and it is $A=i(U+Id)(U-Id)^{-1}$ (theorem 2). Then there is a $\sigma$-finite measure space $(\Omega ,\Sigma ,\mu)$ and $g\in L^{\infty}(\Omega)$, g is not necessary realvalued, and a unitary operator $V:H\to L^2(\Omega)$, such that:
$VUV^{-1}=M_g$, respectively $U=V^{-1}M_gV$.
Here I have a first question: Is this the application of theorem 1 on U? But U is unitary and not self-adjoint. And g is only bounded almoust everywhere and not as stated in theorem 1, bounded on $\Omega$. What is going on here?
To continue the proof: Now it is $A=i(U+Id)(U-Id)^{-1}=V^{-1}(i(M_g+Id)(M_g-Id)^{-1})$ and let $M_f$ be $M_f=i(M_g+Id)(M_g-Id)^{-1}$ with $f(\cdot)=i\frac{g(\cdot)+1}{g(\cdot)-1}$. f is clearly measurable but why is f well defined?
we said: It is $V(U-Id)V^{-1}=M_g-Id$ and $U-Id$ is injective and therefore $M_g-Id$ is injective. So $\{x\in\Omega; g(x)=1\}$ is a null set.
Question: Could you explain me, why $\{x\in\Omega; g(x)=1\}$ is a null set? I don't understand this.

And could you explain me, why the first point, $\text{for}\; x\in H: x\in D(A) \Leftrightarrow f\cdot V(x)\in L^2(\Omega)$ is true? We only say it is "clear".
Sorry, a lot of questions. I really appreciate your help. Regards
Update: Ok theorem 1 has to be true for normal operators if it is true what i read in lecture notes. So the only question is, why $$1. \text{for}\; x\in H: x\in D(A) \Leftrightarrow f\cdot V(x)\in L^2(\Omega)$$ in the Spectral theorem is true

• Your theorem 1 looks very much like the statement you want to prove (except for point 1). Are you sure there is no such thing as a spectral theorem for unitary operators which is used in the proof instead of the theorem you cited? To answer your second question: $\{x \in \Omega; g(x) = 1\}$ is a null set because otherwise the indicator function on that set would be non-zero in $L^2(\Omega)$ and thus the kernel of $M_g - \mathop{Id}$ non-trivial. – Three.OneFour Feb 24 '15 at 14:53
• thank you. We only proved a special case of theorem 1 (if A has a cyclic vector) and we had no other theorem which could help. And i'm sure that we don't have a spectral theorem for unitary operators which is used in the proof=(. Is the indicator function on that set in the kernel of $M_g-Id$? – NewBunny Feb 24 '15 at 15:18
• oh wait, i have something to change in theorem 1! This is the version for bounded, linear, self-adjoint operators. Thank you! But i'm not sure, why we could apply theorem 1 for an unitary oeprator here. – NewBunny Feb 24 '15 at 15:23
• Yes, the indicator function is in the kernel of $M_g - \mathop{Id}$: Let $Z = \{x \in \Omega; g(x) = 1\}$ be that set and $\mathbb 1_Z$ the indicator function. Then $(M_g - \mathop{Id})\mathbb 1_Z = g \mathbb 1_Z - \mathbb 1_Z = \mathbb 1_Z - \mathbb 1_Z = 0$. – Three.OneFour Feb 24 '15 at 15:25
• Ok thank you, I understand. Then there are only both of mine other questions – NewBunny Feb 24 '15 at 15:31

## 1 Answer

If the set of $x$ for which $g(x)=1$ is not a set of $\mu$ measure $0$, then $M_{g}$ has an eigenvalue of $1$. But an eigenvalue of $1$ is impossible for $(A+iI)(A-iI)^{-1}$ because $(A+iI)(A-iI)^{-1}x=x$ implies $$x+2i(A-iI)^{-1}x=x \implies (A-iI)^{-1}x=0 \implies x = 0.$$ If $fVx \in L^{2}$ for some $x$, then the following holds for some $y$ $$i\frac{g+1}{g-1}Vx = Vy \in L^{2} \\ i(g+1)Vx = (g-1)Vy \\ i(U+I)x = (U-I)y$$ Now, if you're careful, you can show that $x$ is in the range of $(A-iI)^{-1}$, which is the same as the domain of $A$. To prove this, use the following in the above and solve for $x=(A-iI)^{-1}z$: $$U = (A+iI)(A-iI)^{-1}= I-2i(A-iI)^{-1}.$$ The steps are basically reversible back up to $fVx \in L^{2}$.

• thank you! if you have i(g+1)Vx = (g-1)Vy, what happens to get in the next step i(U+I)x = (U-I)y ? – NewBunny Feb 25 '15 at 15:11
• @NewBunny Your definition of $V$ and $g$ are $V^{-1}M_{g}V=U$, or $gVx = VUx$. – Disintegrating By Parts Feb 25 '15 at 15:51