Decomposition of the spectrum of a bounded linear operator

1 Definitions

Let $H$ be a complex Hilbert space. If $\lambda \in ℂ$, we also write $\lambda$ to denote $\lambda \cdot {\text{id}}_H\in B(H)$.

For $T\in B(H)$, the spectrum $\sigma (T)$ of $T$ is the set of those $\lambda \in ℂ$ such that the map $T-\lambda$ is not bijective. It happens that it is useful for some purposes to write $\sigma (T)$ as a union of three particular disjoint subsets of itself.

• The point spectrum ${\sigma }_{\text{point}}(T)$ is the set of those $\lambda \in ℂ$ such that $T-\lambda$ is not injective. Equivalently, $\lambda \in {\sigma }_{\text{point}}(T)$ if $\lambda$ is an eigenvalue of $T$.

• The continuous spectrum ${\sigma }_{\text{cont}}(T)$ is the set of those $\lambda \in ℂ$ such that $T-\lambda$ is injective, has dense image, and is not surjective.

• The residual spectrum ${\sigma }_{\text{res}}(T)$ is the set of those $\lambda \in ℂ$ such that $T-\lambda$ is injective and does not have dense image.

It is apparent that the sets ${\sigma }_{\text{point}}(T)$, ${\sigma }_{\text{cont}}(T)$, and ${\sigma }_{\text{res}}(T)$ are disjoint, and that

$$\sigma (T)={\sigma }_{\text{point}}(T)\cup {\sigma }_{\text{cont}}(T)\cup {\sigma }_{\text{res}}(T).$$

If $T\in B(H)$ then $\sigma (T)\ne \mathrm{\varnothing }$, but any of the above three sets may be empty; they merely can’t all be empty for a given operator.

2 Residual spectrum

If $T\in B(H)$ is a normal operator then ${\sigma }_{\text{res}}(T)=\mathrm{\varnothing }$. We prove this. Suppose that $T-\lambda$ is injective. We have to show that $\text{im }(T-\lambda )$ is dense in $H$, and thus that $\lambda \notin {\sigma }_{\text{res}}(T)$. ($\lambda$ might be in ${\sigma }_{\text{cont}}(T)$ or might not be in $\sigma (T)$; we merely want to show that it is not in ${\sigma }_{\text{res}}(T)$.) We have

$$H=\overline{\text{im }(T-\lambda )}\oplus {(\text{im }(T-\lambda ))}^{⟂}.$$

Let $w\in {(\text{im }(T-\lambda ))}^{⟂}$; we have to show that $w=0$. For all $v\in H$,

$$⟨(T-\lambda )v,w⟩=0,$$

so for all $v\in H$ we have $⟨v,{(T-\lambda )}^{*}w⟩=0$ and therefore ${(T-\lambda )}^{*}w=0$, so $w\in \ker {(T-\lambda )}^{*}=\ker (T-\lambda )$. As $T-\lambda$ is injective, $w=0$, completing the proof.

3 Point spectrum

If $A\in B(H)$ is normal then it is straightforward to show that $\ker A=\ker A^{*}$. Also, if $T\in B(H)$ is normal then for any $z\in ℂ$, $T-z$ is normal. Thus, $\ker (T-z)=\{0\}$ if and only if $\ker ({(T-z)}^{*})=\{0\}$. That is, $\lambda \in {\sigma }_{\text{point}}(T)$ if and only if $\overline{\lambda }\in {\sigma }_{\text{point}}(T^{*})$. For $X\subseteq ℂ$ we define $X^{*}=\{\overline{z}:z\in X\}$. We have shown that if $T\in B(H)$ is normal then

$${\sigma }_{\text{point}}{(T)}^{*}={\sigma }_{\text{point}}(T^{*}).$$

4 Continuous spectrum

If $\lambda \in {\sigma }_{\text{cont}}(T)$, then $\text{im }(T-\lambda )$ is dense in $H$. Also,

$${(T-\lambda )}^{-1}:\text{im }(T-\lambda )\to H$$

is a surjective linear map (the inverse of a linear map is itself a linear map) that is not continuous. For $\text{im }(T-\lambda )$ is dense in $H$, so if ${(T-\lambda )}^{-1}$ were continuous then it would have a unique extension to a continuous, hence bounded, map $H\to H$. Using this and the fact that $T-\lambda$ is not surjective will give a contradiction.

5 Approximate point spectrum

Let $\lambda \in ℂ\setminus ({\sigma }_{\text{point}}(T)\cup {\sigma }_{\text{res}}(T))$. I claim that $\lambda \in {\sigma }_{\text{cont}}(T)$ if and only if $\lambda$ is in the approximate point spectrum of $T$, the set of those $\lambda \in ℂ$ such that there is is a sequence $v_n\in H$ with $\parallel v_n\parallel =1$ and $\parallel (T-\lambda )v_n\parallel \to 0$ as $n\to \mathrm{\infty }$. It is a fact that for $T\in B(H)$, $T$ is invertible if and only if $T(H)$ is dense in $H$ and there is some $\alpha >0$ such that $\parallel Tv\parallel \ge \alpha \parallel v\parallel$ for all $v\in H$. If $\lambda \in {\sigma }_{\text{cont}}(T)$, then $T-\lambda$ is not invertible but the image of $T-\lambda$ is dense in $H$, then it must therefore be that $T-\lambda$ is not bounded below. That is, there is no $\alpha >0$ such that for every $w\in H$ we have $\parallel (T-\lambda )w\parallel \ge \alpha \parallel w\parallel$. Then for each $n$ there is some $w\in H$ such that . Let $v_n=\frac{w_n}{\parallel w_n\parallel }$. We have $\parallel v_n\parallel =1$ and , showing that $\lambda \in {\sigma }_{\text{ap}}(T)$.

On the other hand, if $\lambda \in {\sigma }_{\text{ap}}(T)$ then there is a sequence $v_n\in H$ such that $\parallel (T-\lambda )v_n\parallel \to 0$ as $n\to \mathrm{\infty }$. Then for any $\alpha >0$, there is some $v_n$ such that , so $T-\lambda$ is not invertible and hence $\lambda \in \sigma (T)$. Since we assumed that $\lambda \in ℂ\setminus ({\sigma }_{\text{point}}(T)\cup {\sigma }_{\text{res}}(T))$, we then have $\lambda \in {\sigma }_{\text{cont}}(T)$.

Halmos shows in Problem 62 of his Hilbert Space Problem Book that ${\sigma }_{\text{ap}}(T)$ is a closed subset of $ℂ$, and proves in Problem 63 that $\partial \sigma (T)\subseteq {\sigma }_{\text{ap}}(T)$: the boundary of the spectrum of $T$ is contained in the approximate point spectrum of $T$.

6 Normal operators

We showed earlier that if $T\in B(H)$ is normal then ${\sigma }_{\text{point}}{(T)}^{*}={\sigma }_{\text{point}}(T^{*})$, where, for $X\subseteq ℂ$, $X^{*}=\{\overline{z}:z\in X\}$. This is one reason why it can be helpful to know that an operator is normal. Using this we can show something more about normal operators. Let $T\in B(H)$ be normal and suppose that $\lambda ,\mu \in {\sigma }_{\text{point}}(T)$ are distinct. Then there are nonzero $v,w\in H$ with $Tv=\lambda v$ and $Tw=\mu w$. Using $T^{*}w=\overline{\mu }w$ we get

$\lambda ⟨v,w⟩$ $=$ $⟨\lambda v,w⟩$

$=$ $⟨Tv,w⟩$

$=$ $⟨v,T^{*}w⟩$

$=$ $⟨v,\overline{\mu }w⟩$

$=$ $\mu ⟨v,w⟩.$

As $\lambda \ne \mu$, this means that $⟨v,w⟩=0$. In words, if $T\in B(H)$ is normal, then its eigenspaces are mutually orthogonal.

7 Compact operators

Let $K(H)$ be the closure in $B(H)$ of the set of finite-rank operators. We call the elements of $K(H)$ compact operators. The following are equivalent ways to state that an operator is compact.

• $T\in B(H)$ is compact if and only if for every bounded subset $S$ of $H$, the closure of the image $T(S)$ is compact.

• $T\in B(H)$ is compact if and only if for every sequence $v_n\in H$ with $\parallel v_n\parallel =1$, $T(v_n)$ has a convergent subsequence.

• Let $B$ be the closed unit ball in $H$, and let $B$ be a topological space with the weak topology: a net $v_{\alpha }\in B$ converges weakly to $v\in B$ if for all $w\in H$ we have $⟨v_{\alpha },w⟩\to ⟨v,w⟩$. If $H$ is separable, then the weak topology on $B$ is metrizable and thus can be characterized using merely sequences instead of nets. $T\in B(H)$ is compact if and only if the restriction of $T$ to $B$ is continuous $B\to H$, where $B$ has the weak topology and $H$ has the norm topology.

If $K\in K(H)$ and $V$ is a closed subspace of $V$ such that $K(V)\subseteq V$, then the restriction of $K$ to $V$ is an element of $K(V)$.

$B(H)$ is a $C^{*}$-algebra, and $K(H)$ is a $C^{*}$-subalgebra of $B(H)$. If $T\in K(H)$ and $S\in B(H)$, then

$$ST,TS\in K(H).$$

Hence $K(H)$ is an ideal of the $C^{*}$-algebra $B(H)$.

Useful facts about compact operators are proved in Yuri A. Abramovich and Charalambos D. Aliprantis, An Invitation to Operator Theory, p. 272, §7.1.

8 Fredholm alternative

The Fredholm alternative states that if $K\in K(H)$, $\lambda \ne 0$, and $\ker (K-\lambda )=\{0\}$ then ${(K-\lambda )}^{-1}\in B(H)$. Equivalently, if $K\in K(H)$, $\lambda \ne 0$, and $\lambda \notin {\sigma }_{\text{point}}(K)$ then $\lambda \notin \sigma (K)$. Equivalently, if $K\in K(H)$, then

$$\sigma (K)\subseteq {\sigma }_{\text{point}}(K)\cup \{0\}.$$

The above forms are the ones that we want to use. The following is the one that we want to prove, which is equivalent because a nonzero multiple of a compact operator is compact: If $K\in K(H)$ and $\ker ({\text{id}}_H-K)=\{0\}$ then ${({\text{id}}_H-K)}^{-1}\in B(H)$.

We prove two standalone lemmas that we then use to prove the Fredholm alternative.

• Let $K\in K(H)$ let $A={\text{id}}_H-K\in B(H)$, and suppose that $A(H)=H$. Define $K_n=\ker (A^n)$. We have $K_1\subseteq K_2\subseteq \mathrm{\cdots }$. Assume by contradiction that $K_1\ne \{0\}$. Then there is some nonzero $f_1\in K_1$. As $A(H)=H$, there is some $f_2\in H$ with $A{f}_2=f_1$; but $A^2{f}_2=A{f}_1=0$ and $A{f}_2=f_1\ne 0$, so $f_2\in K_2\setminus K_1$. Let $f_{n+1}\in K_{n+1}\setminus K_n$ with $A{f}_{n+1}=f_n$. Therefore $K_1,K_2,\mathrm{\dots }$ are a strictly increasing sequence of subspaces of $H$. Using Gram-Schmidt, there is an orthonormal sequence $e_1,e_2,\mathrm{\dots }$ with $e_n\in K_n$ for all $n$; we caution that we do not necessarily have $A{e}_{n+1}=e_n$. As $A{e}_{n+1}\in K_n$, $⟨A{e}_{n+1},e_{n+1}⟩$, giving

  $${\parallel K{e}_{n+1}\parallel }^2={\parallel e_{n+1}-A{e}_{n+1}\parallel }^2={\parallel e_{n+1}\parallel }^2+{\parallel A{e}_{n+1}\parallel }^2\ge 1{\parallel e_{n+1}\parallel }^2=1.$$

  Each $e_n$ is an element of the closed unit ball $B$, and $e_n\to 0$ weakly (this is the case for any orthonormal sequence in $H$, basis or not, and is proved using Bessel’s inequality). Since $K$ is compact, it is continuous $B\to H$ where $B$ has the weak topology and $H$ has the norm topology; but $e_n\to 0$, $K(0)=0$, and $\parallel K{e}_n\parallel \ge 1$, so $K{e}_n$ does not converge to $0$ in $H$, a contradiction. Therefore $K_1=\{0\}$, that is, $\ker A=\{0\}$.

• Let $K\in K(H)$ and let $A={\text{id}}_H-K\in B(H)$. Suppose by contradiction that $A$ is not bounded below on ${(\ker A)}^{⟂}$. So for every $\alpha >0$ there is some $w\in {(\ker A)}^{⟂}$ such that $\parallel Aw\parallel \ge \alpha \parallel w\parallel$. Then for all $n$ there is some $w_n\in {(\ker A)}^{⟂}$ with . Let $v_n=\frac{w_n}{\parallel w_n\parallel }\in {(\ker A)}^{⟂}$. Then

  As $K$ is compact, there is some subsequence $v_{a(n)}$ such that $K{v}_{a(n)}$ converges to some $v\in H$. $A{v}_n\to Av$ and $A{v}_n\to 0$, so $v\in \ker A$. On the other hand, because $v_n=A{v}_n+K{v}_n\to v$ and ${(\ker A)}^{⟂}$ is closed, we have $v\in {(\ker A)}^{⟂}$, so $v\in \ker A\cap {(\ker A)}^{⟂}=\{0\}$. But as $\parallel v_n\parallel =1$ for each $n$, we have $\parallel v\parallel =1$, a contradiction. Therefore, $A$ is bounded below on ${(\ker A)}^{⟂}$.

If $K\in K(H)$ and $A={\text{id}}_H-K$, then by the second of the two lemmas, we have that $A$ is bounded below on ${(\ker A)}^{⟂}$: there is some $\alpha >0$ such that $\parallel Av\parallel \ge \alpha \parallel v\parallel$ for all $v\in {(\ker A)}^{⟂}$. If

$$w_n\in A(H)=A(\ker A\oplus {(\ker A)}^{⟂})=A({(\ker A)}^{⟂})$$

with $w_n\to w\in H$, then there are $v_n\in {(\ker A)}^{⟂}$ such that $A{v}_n=w_n$. As $w_n$ converges, for all $ϵ>0$ there is some $N$ such that if $n,m\ge N$ then $\parallel w_n-w_m\parallel \le ϵ$, so $\parallel A(v_n-v_m)\parallel =\parallel A{v}_n-A{v}_m\parallel \le ϵ$. But

$$\parallel A(v_n-v_m)\parallel \ge \alpha \parallel v_n-v_m\parallel ,$$

so

$$\parallel v_n-v_m\parallel \le \frac{ϵ}{\alpha },$$

so $v_n$ is a Cauchy sequence and hence converges, say to $v$. $v_n\in {(\ker A)}^{⟂}$, which is closed, so $v\in {(\ker A)}^{⟂}$. Then $w_n=A{v}_n\to Av\in A(H)$. Therefore, if $K\in K(H)$ and $A={\text{id}}_H-K$, then $A(H)$ is closed in $H$.

Let $K\in K(H)$ and $A={\text{id}}_H-K$, and suppose that $\ker A=\{0\}$. By the above paragraph, $A(H)$ is closed in $H$. $K^{*}\in K(H)$ and $A^{*}={\text{id}}_H-K^{*}$, so by the above paragraph we also get that $A^{*}(H)$ is closed in $H$. It is a fact that if $T\in B(H)$ then $\ker T^{*}={(T(H))}^{⟂}$, so using this with $T=A^{*}$ we get

$$\ker A={(A^{*}(H))}^{⟂};$$

taking orthogonal complements and using the fact that the double orthogonal complement of a subspace is its closure and that $A^{*}(H)$ is closed, we obtain

$${(\ker A)}^{⟂}=A^{*}(H).$$

Since $\ker A=\{0\}$, we have $A^{*}(H)=H$. Then we can apply the first of the two lemmas: as $A^{*}={\text{id}}_H-K^{*}$, $K^{*}\in K(H)$, and $A^{*}(H)=H$, we have $\ker A^{*}=\{0\}$. We now apply the second of the two lemmas: $A^{*}$ is bounded below on ${(\ker A^{*})}^{⟂}=H$. Using the fact that if $T\in B(H)$ is bounded below and has dense image then $T^{-1}\in B(H)$, we get ${(A^{*})}^{-1}\in B(H)$ ($A^{*}(H)=H$ so $A^{*}$ certainly has dense image). Taking adjoints commutes with taking inverses, so $A^{-1}\in B(H)$. This completes the proof of the Fredholm alternative.

9 Compact self-adjoint operators

It is a fact that if $T\in B(H)$ is self-adjoint then

$$\parallel T\parallel =\underset{\parallel v\parallel \le 1}{sup}|⟨Tv,v⟩|=\underset{\parallel v\parallel =1}{sup}|⟨Tv,v⟩|.$$

Let $T\in B(H)$ be compact and self-adjoint and $T\ne 0$. Since $T$ is self-adjoint, $⟨Tv,v⟩\in ℝ$, so either $\parallel T\parallel ={sup}_{\parallel v\parallel =1}⟨Tv,v⟩$ or $\parallel T\parallel =-{inf}_{\parallel v\parallel =1}⟨Tv,v⟩$. Say the first is the case. Let $\parallel v_n\parallel =1$ and $⟨T{v}_n,v_n⟩\to \parallel T\parallel$ as $n\to \mathrm{\infty }$. Then, as $T=T^{*}$,

$⟨T{v}_n-\parallel T\parallel v_n,T{v}_n-\parallel T\parallel v_n⟩$ $=$ $⟨T{v}_n,T{v}_n⟩-⟨T{v}_n,\parallel T\parallel v_n⟩-⟨\parallel T\parallel v_n,T{v}_n⟩$

$+⟨\parallel T\parallel v_n,\parallel T\parallel v_n⟩$

$=$ ${\parallel T{v}_n\parallel }^2-2\parallel T\parallel ⟨T{v}_n,v_n⟩+{\parallel T\parallel }^2{\parallel v_n\parallel }^2$

$\le$ ${\parallel T\parallel }^2{\parallel v_n\parallel }^2-2\parallel T\parallel ⟨T{v}_n,v_n⟩+{\parallel T\parallel }^2{\parallel v_n\parallel }^2$

$=$ $2{\parallel T\parallel }^2-2\parallel T\parallel ⟨T{v}_n,v_n⟩.$

Thus $\parallel T{v}_n-\parallel T\parallel v_n\parallel \to 0$ as $n\to \mathrm{\infty }$, that is, $T{v}_n-\parallel T\parallel v_n\to 0$. On the other hand, as $\parallel v_n\parallel =1$ for each $n$, there is some subsequence $v_{a(n)}$ such that $T{v}_{a(n)}$ converges, say to $v$. Together with $T{v}_n-\parallel T\parallel v_n\to 0$ this gives $\parallel T\parallel v_{a(n)}\to v$ as $n\to \mathrm{\infty }$, from which we get $\parallel v\parallel =\frac{1}{\parallel T\parallel }>0$. Thus $v\ne 0$. And

$$(T-\parallel T\parallel )v=(T-\parallel T\parallel )\underset{n\to \mathrm{\infty }}{lim}{v}_{a(n)}=\underset{n\to \mathrm{\infty }}{lim}(T-\parallel T\parallel )v_{a(n)}=0,$$

which means that $\parallel T\parallel \in {\sigma }_{\text{point}}(T)$. Likewise, in the case $\parallel T\parallel =-{inf}_{\parallel v\parallel =1}⟨Tv,v⟩$ we get $-\parallel T\parallel \in {\sigma }_{\text{point}}(T)$.

10 Multiplication operators

Let $(X,\mathrm{\Sigma },\mu )$ be a $\sigma$-finite measure space. $L^2(X)$ is a Hilbert space with inner product

$$⟨f,g⟩=\int f{g}^{*}𝑑\mu ,$$

where $g^{*}(x)=\overline{g(x)}$.

A multiplication operator on $L^2(X)$ is an operator $M_{\varphi }:L^2(X)\to L^2(X)$, $\varphi \in L^{\mathrm{\infty }}(X)$, of the form

$$(M_{\varphi }f)(x)=\varphi (x)f(x).$$

As

$${\parallel M_{\varphi }f\parallel }^2={\int }_X\varphi (x)f(x)\overline{\varphi (x)f(x)}𝑑\mu (x)\le {\parallel \varphi \parallel }_{\mathrm{\infty }}^2{\parallel f\parallel }^2,$$

where ${\parallel \varphi \parallel }_{\mathrm{\infty }}$ is the essential supremum of $\varphi (x)$ for $x\in X$, we have $\parallel M_{\varphi }\parallel \le \parallel \varphi \parallel$ and so $M_{\varphi }\in B(H)$. $L^{\mathrm{\infty }}(X)$ is a $C^{*}$-algebra, and so is $B(L^2(X))$. If $X$ is $\sigma$-finite then I claim that

$$\varphi \mapsto M_{\varphi },L^{\mathrm{\infty }}(X)\to B(L^2(X))$$

is an injective homomorphism of $C^{*}$-algebras. It is straightforward to show that this map is a homomorphism of $C^{*}$-algebras, and this does not use the assumption that $X$ is $\sigma$-finite. For our benefit, we shall show that $\varphi \mapsto M_{\varphi }$ is injective. If $\varphi \ne 0$, then ${\parallel \varphi \parallel }_{\mathrm{\infty }}>0$, so for

$$E=\{x\in X:|\varphi (x)|\ge \frac{1}{2}{\parallel \varphi \parallel }_{\mathrm{\infty }}\}$$

we have . Because $(X,\mu )$ is $\sigma$-finite, there is some subset $F$ of $E$ with . As $f={\varphi }^{*}\cdot {\chi }_F\in L^2(X)$, we have

$M_{\varphi }f$ $=$ ${\displaystyle {\int }_F}\varphi (x)\overline{\varphi (x)}𝑑\mu (x)$

$\ge$ ${\displaystyle {\int }_F}{\displaystyle \frac{1}{4}}{\parallel \varphi \parallel }_{\mathrm{\infty }}^2𝑑\mu (x)$

$=$ ${\displaystyle \frac{1}{4}}{\parallel \varphi \parallel }_{\mathrm{\infty }}^2\cdot \mu (F)$

$>$ $0,$

so $M_{\varphi }\ne 0$. Generally, an injective homomorphism of $C^{*}$-algebras is an isometry, so $\parallel M_{\varphi }\parallel ={\parallel \varphi \parallel }_{\mathrm{\infty }}$.

As $M_{\varphi {\varphi }^{*}}=M_{\varphi }{M}_{{\varphi }^{*}}=M_{\varphi }{M}_{\varphi }^{*}$ and $M_{\varphi {\varphi }^{*}}=M_{{\varphi }^{*}\varphi }$, we have $M_{\varphi }{M}_{\varphi }^{*}=M_{\varphi }^{*}{M}_{\varphi }$, namely, a multiplication operator is a normal operator. Since residual spectrum of a normal operator is empty, the residual spectrum of a multiplication operator is empty.

For $\varphi \in L^{\mathrm{\infty }}(X)$, we define the essential range of $\varphi$ to be the set

Equivalently, the essential range of $\varphi$ is the set of those $z\in ℂ$ such that for all $ϵ>0$,

$$\mu ({\varphi }^{-1}(D_{ϵ}(z)))>0,$$

in words, those $z\in ℂ$ such that the inverse image of every $ϵ$-disc about $z$ has positive measure. Equivalently, the essential range of $\varphi$ is the intersection of all closed subsets $K$ of $ℂ$ such that for almost all $x\in X$, $\varphi (x)\in K$. It is a fact that if $\varphi \in L^{\mathrm{\infty }}(X)$ then the essential range of $\varphi$ is a compact subset of $ℂ$.

Let $\varphi \in L^{\mathrm{\infty }}(X)$. If $\lambda$ is not in the essential range of $\varphi$, then there is some $ϵ>0$ such that $\mu ({\varphi }^{-1}(D_{ϵ}(\lambda )))=0$, which means that for almost all $x\in X$ we have $|\varphi (x)-\lambda |\ge ϵ$. Define $\psi (x)=\frac{1}{\varphi (x)-\lambda }$. For almost all $x\in X$,

$$|\psi (x)|=\frac{1}{|\varphi (x)-\lambda |}\le \frac{1}{ϵ},$$

hence $\psi \in L^{\mathrm{\infty }}(X)$. Then

$$M_{\psi }{M}_{\varphi -\lambda }=M_{\varphi -\lambda }{M}_{\psi }=M_{(\varphi -\lambda )\cdot \psi }=M_1={\text{id}}_{L^2(X)},$$

so $M_{\varphi -\lambda }$ is invertible. But $M_{\varphi -\lambda }=M_{\varphi }-\lambda$, so $M_{\varphi }-\lambda$ is invertible and hence $\lambda \notin \sigma (M_{\varphi })$.

If $\lambda$ is in the essential range of $\varphi$, then for each $n$ we have ; since $\mathrm{\Sigma }$ is $\sigma$-finite, for each $n$ there is a subset $E_n$ of ${\varphi }^{-1}(D_{1/n}(\lambda ))$ with , and so ${\chi }_{E_n}\in L^2(X)$. We have, since for $x\in E_n$,

${\parallel (M_{\varphi }-\lambda ){\chi }_{E_n}\parallel }^2$ $=$ ${\displaystyle {\int }_X}{|(\varphi (x)-\lambda ){\chi }_{E_n}(x)|}^2𝑑\mu (x)$

$=$ ${\displaystyle {\int }_{E_n}}{|\varphi (x)-\lambda |}^2𝑑\mu (x)$

$\le$ ${\displaystyle \frac{1}{n^2}}{\displaystyle {\int }_{E_n}}𝑑\mu (x)$

$=$ ${\displaystyle \frac{1}{n^2}}{\displaystyle {\int }_X}{\chi }_{E_n}𝑑\mu (x)$

$=$ ${\displaystyle \frac{1}{n^2}}{\parallel {\chi }_{E_n}\parallel }^2,$

so for each $n$,

$$\parallel (M_{\varphi }-\lambda ){\chi }_{E_n}\parallel \le \frac{1}{n}\parallel {\chi }_{E_n}\parallel .$$

It follows that $M_{\varphi }-\lambda$ is not invertible, as it is not bounded below. Therefore $\lambda \in \sigma (M_{\varphi })$. Therefore the essential range of $\varphi \in L^{\mathrm{\infty }}(X)$ is equal to the spectrum of $M_{\varphi }\in B(L^2(X))$.

We say that $\varphi \in L^{\mathrm{\infty }}(X)$ is invertible if there is some $\psi \in L^{\mathrm{\infty }}(X)$ such that $\varphi (x)\psi (x)=1$ for almost all $x\in X$. (It would not make sense to demand that $\varphi (x)\psi (x)=1$ for all $x\in X$.) For $\varphi \in L^{\mathrm{\infty }}(X)$ to be invertible, it is necessary and sufficient that there is some $\alpha >0$ such that $|\varphi (x)|\ge \alpha$ for almost all $x\in X$ (lest its inverse not have an essential supremum).

If $\lambda$ is not just an element of the essential range $\varphi$ but is an isolated element of the essential range, then we can say more than just that $\lambda \in \sigma (M_{\varphi })$. In this case, there is some $ϵ>0$ such that the intersection of $D_{ϵ}(\lambda )$ and the essential range of $\varphi$ is equal to the singleton $\{\lambda \}$. Let $E$ be a subset of ${\varphi }^{-1}(D_{ϵ}(\lambda ))$ with . For almost all $x\in X$, $\varphi (x)$ is an element of the essential range of $\varphi$, hence for almost all $x\in E$ we have $\varphi (x)=\lambda$. Therefore, for almost all $x\in X$ we have

$$(M_{\varphi }{\chi }_E)(x)=\varphi (x){\chi }_E(x)=\lambda {\chi }_E(x).$$

Hence $(M_{\varphi }-\lambda ){\chi }_E=0$, and as $\mu (E)>0$ we have ${\chi }_E\ne 0$. Therefore, if $\lambda$ is an isolated element of the essential range of $\varphi$ then $\lambda \in {\sigma }_{\text{point}}(M_{\varphi })$.

11 Functional calculus

Let $T\in B(H)$ be self-adjoint. The spectrum $\sigma (T)$ is a compact subset of $ℝ$, and one checks that the set $C(\sigma (T))$ of continuous functions $\sigma (T)\to ℂ$ is a $C^{*}$-algebra, with norm $\parallel f\parallel ={sup}_{\lambda \in \sigma (T)}|f(\lambda )|$.

Let $ℂ[x]$ be the set of polynomials with complex coefficients. For $T\in B(H)$ self-adjoint and $p(x)={\sum }_{k=0}^n{a}_k{x}^k\in ℂ[x]$, we define

$$p(T)\in B(H)$$

by

$$p(T)=\sum _{k=0}^n{a}_k{T}^k.$$

$T\mapsto p(T)$ is a homomorphism of $C^{*}$-algebras, where

$${\left(\sum _{k=0}^n{a}_k{x}^k\right)}^{*}=\sum _{k=0}^n\overline{a_k}{x}^k.$$

It is a fact that

$$\sigma (p(T))=p(\sigma (T)),$$

where for $M\subseteq ℂ$ we define $p(M)=\{p(z):z\in M\}$. It is also a fact that

$$\parallel p(T)\parallel =\parallel p\parallel =\underset{\lambda \in \sigma (T)}{sup}|p(\lambda )|;$$

this is proved using the result that the norm of a normal operator $T$ is equal to its spectral radius, which is given by the two following expressions that one proves are equal:

$$r(T)=\underset{n\to \mathrm{\infty }}{lim}{\parallel T^n\parallel }^{1/n}=\underset{\lambda \in \sigma (T)}{\max }|\lambda |.$$

The above is used to define $f(T)$ for any continuous function $f:\sigma (T)\to ℂ$. This map $C(\sigma (T))\to B(H)$ is called the continuous functional calculus. It is an isometric homomorphism of $C^{*}$-algebras. The continuous functional calculus can be used to prove things about the spectrum of self-adjoint operators that do not obviously have to do with making sense of continuous functions applied to these operators.

Let $T\in B(H)$ be self-adjoint and let $\lambda \in \sigma (T)$ be an isolated point in $\sigma (T)$. I will show that that $\lambda \in {\sigma }_{\text{point}}(T)$. Since $\lambda$ is isolated in $\sigma (T)$, the function $f:\sigma (T)\to ℂ$ defined by

is continuous. Since $f$ is continuous, $f(T)\in B(H)$, and because $f=f^{*}$, $f(T)$ is self-adjoint. Let $P=f(T)$. As $\parallel P\parallel =\parallel f\parallel =1$, $P\ne 0$. Define $g\in C(\sigma (T))$ by $g(x)=(x-\lambda )f(x)$. Then $g(x)=0$ for all $x\in \sigma (T)$, so $g(T)=0$, i.e.

$$(T-\lambda )P=0.$$

Hence $\text{im }P\subseteq \ker (T-\lambda )$. As $P\ne 0$, there is some $v\in H$ with $Pv\ne 0$. Then $(T-\lambda )Pv=0$, $Pv\ne 0$, so $\lambda \in {\sigma }_{\text{point}}(T)$.

12 Spectral measures

It is a fact that if $f\ge 0$ then $f(T)\ge 0$, where, for a self-adjoint operator $T$, $T\ge 0$ means $⟨Tv,v⟩\ge 0$ for all $v\in H$. For $T\in B(H)$ self-adjoint and $v\in H$, using the continuous functional calculus talked about in the previous section we define $\varphi :C(\sigma (T))\to ℂ$ by

$$\varphi (f)=⟨v,f(T)v⟩.$$

$\varphi$ is a positive linear functional: if $f$ is real-valued and $f(x)\ge 0$ for all $x\in \sigma (T)$, then $\varphi (f)\ge 0$. $\sigma (T)$ is indeed a locally compact Hausdorff space and since $\sigma (T)$ is compact the continuous functions of compact support on $\sigma (T)$ are precisely the continuous functions on $\sigma (T)$, so we satisfy the conditions of the Riesz-Markov theorem. Thus there exists a unique regular Borel measure $\mu$ on the Borel $\sigma$-algebra of $\sigma (T)$ such that, for all $f\in C(\sigma (T))$,

$$⟨v,f(T)v⟩=\varphi (f)={\int }_{\sigma (T)}f(x)𝑑\mu (x).$$

Lebesgue’s decomposition theorem states that

$$\mu ={\mu }_{\text{ac}}+{\mu }_{\text{sing}}+{\mu }_{\text{pp}},$$

where

• ${\mu }_{\text{ac}}$ is absolutely continuous with respect to Lebesuge measure: if $A$ is a measurable subset of $\sigma (T)$ and its Lebesgue measure is $0$, then ${\mu }_{\text{ac}}(A)=0$.

• ${\mu }_{\text{sing}}$ and Lebesgue measure are mutually singular, and if $\lambda \in \sigma (T)$ then ${\mu }_{\text{sing}}(\{\lambda \})=0$.

• There is a countable subset $J$ of $\sigma (T)$ such that

  $${\mu }_{\text{pp}}=\sum _{\lambda \in J}{a}_{\lambda }{\delta }_{\lambda },a_{\lambda }\in ℂ.$$

We define $H_{\text{ac}}$ to be the set of those $v\in H$ such that $\mu$ is equal to the absolutely continuous part of its Lebesgue decomposition, i.e. the other two parts are 0, and we define $H_{\text{sing}}$ and $H_{\text{pp}}$ likewise. (Note that we first took $v\in H$ and then defined $\mu$ using $v$.) One proves that $H_{\text{ac}}$, $H_{\text{sing}}$ and $H_{\text{pp}}$ are closed subspaces of $H$ and that they are invariant under $T$, and defines the absolutely continuous spectrum of $T$ to be the spectrum of the restriction of $T$ to $H_{\text{ac}}$; the singular spectrum of $T$ to be the spectrum of the restriction of $T$ to $H_{\text{sing}}$; and the pure point spectrum of $T$ to be the spectrum of the restriction of $T$ to $H_{\text{pp}}$. It is a fact that

$$\sigma (T)={\sigma }_{\text{ac}}(T)\cup {\sigma }_{\text{sing}}(T)\cup \overline{{\sigma }_{\text{pp}}(T)},$$

but these three sets might not be disjoint.