# Lecture 14: Spectral Theorem for Bounded Operators
$\newcommand{\C}{\mathbb{C}}$
###### tags: `224a`
The goal of this lecture is to prove the following important theorem of von Neumann.
**Theorem.(Spectral Theorem for Bounded Operators.** If $A=A^*\in L(H)$, there is a finite measure space $(X,\mu)$, a function $g\in L^\infty(X,\mu)$ and a unitary $U:H\rightarrow L^2(X,\mu)$ such that
$$ (UAU^{-1})(f) = M_g f.$$
## Continuous Functional Calculus
Let $C(X)$ denote the Banach space of continuous functions on a compact set $X$ with the sup norm.
**Theorem. (Continuous Functional Calculus)** If $A=A^*$ there is a linear map $\phi=\phi_A:C(\sigma(A))\rightarrow L(H)$ with the following properties:
1. $*$-homomorphism: $\phi(fg)=\phi(f)\phi(g), \phi(1)=I, \phi(\overline{f})=\phi(f)^*$.
4. Spectral Mapping: $\sigma(\phi_A(f))=f(\sigma(A))$
2. Isometry: $\|\phi(f)\|=\|f\|_{\infty}$.
5. Eigenvector Mapping: If $A\psi=\lambda\psi$ then $\phi_A(f)\psi=f(\lambda)\psi$.
3. Positivity: If $f\ge 0$ then $\phi(f)\ge 0$.
With the normalization that $\phi_A(x)=A$, the above is unique (in fact, assuming only (1) and continuity). We denote it by $\phi_A(f)=f(A)$.
To prove the existence of such a mapping, we first show that it exists for the dense subspace $P\subset C(\sigma(A))$ of polynomial functions.
**Lemma 1.(Spectral Mapping for Polynomials)** If $p\in \C[x]$, $p(\sigma(A))=\sigma(p(A))$.
*Proof.* Reed and Simon VII.1$\square$
**Lemma 2.(Isometry for Polynomials)** If $p\in \C[x]\subset C(\sigma(A))$, $$\|p(A)\| = \|p\|_\infty.$$
The BLT theorem now implies that $\phi_A(\cdot)$ has a unique extension from $P$ to $C(\sigma(A))$. Note that self-adjointness was used in two places: (1) the polynomial functions are dense in $C(\sigma(A))$ (2) $r(A)=\|A\|$ for self-adjoint matrices. The theorem is simply not true without self-adjointness; consider the $2\times 2$ Jordan block with $\sigma(J)=\{0\}$ but $\|J\|=1$.
## Consequences and Spectral Measures
**Theorem.** If $A$ is self-adjoint $\lambda$ is an isolated point in $\sigma(A)$, then $\lambda\in \sigma_p(A)$.
*Proof.* Since $\lambda$ is isolated, there is an $f\in C(\sigma(A))$ such that $f(\lambda)=1$ and $f(x)=0$ on $\sigma\setminus\{\lambda\}$. We also have $f^2=f$ and $\overline{f}=f$, so by the functional calculus, $f(A)$ must be a nonzero orthogonal projection $P$. Since $(x-\lambda)f(x)\equiv 0$, we have $(A-\lambda)P=0$, whence $\lambda\in\sigma_p(A)$.$\square$
A more substantial punch line is that the calculus yields a canonical way to associate a probability measure with a state. Given self-adjoint $A$ and any unit vector $\psi\in H$, consider the bounded linear map
$$ \ell_\psi(f)= (\psi, f(A) \psi).$$ Notice that $\ell_\psi$ is *positive* because $(\psi, f(A)\psi)\ge 0$ whenever $f\ge 0$ by the functional calculus.
We now appeal to the fact that positive linear functionals on continuous functions always come from positive measures.
**Riesz-Markov Theorem.** If $\ell:C(X)\rightarrow \C$ is a positive, continuous linear functional, there is a unique finite positive Borel measure $(X,\mu)$ such that
$$ \ell(f)=\int_X f(x)d\mu(x).$$
Applying this to $\ell_\psi,$ we obtain a canonical mapping $\psi\mapsto \mu_\psi$. By pluggin in $f=1$ we see that $\mu(X)=(\psi,\psi)=1$, so it is a probability measure.
## Cyclic Vectors
**Defn.** A vector $\psi\in H$ is called *cyclic* if the closure of the span of $\{p(A)\psi:p\in \C[x]\}$ is all of $H$.
**Thm.** If $A=A^*\in L(H)$ has a cyclic vector $\psi$, then there is an isometry $U:L^2(\sigma(A),\mu_\psi)$ such that
$$ (U^{-1}AU)(f) = M_\lambda f.$$
*Proof.* Reed and Simon VII.2 Lemma 1.$\square$
The full spectral theorem follows because for any $A\in L(H)$, $H$ can be decomposed into (possibly infinitely many) orthogonal invariant subspaces $H_n$ such that $A:H_n\rightarrow H_n$ has a cyclic vector for every $n$.

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