12. The spectral theorem - Borel calculus for bounded normal ...

12. The spectral theorem - Borel calculus for 

bounded normal operators 

Klaus Thomsen matkt@imf.au.dk 

Institut for Matematiske Fag 

Det Naturvidenskabelige Fakultet 

Aarhus Universitet 

December 2005 

Klaus Thomsen 12. The spectral theorem - Borel calculus for bounded norm

We read Sections 12.23 and 12.24 in Rudins book - with some 

changes. 


The spectral theorem - the statement 

Theorem 

Let H be a Hilbert space and T ∈ B(H) a normal operator. 

There is a unique resolution of the identity E on the spectrum 

σ(T) of T such that 

 

 

T = λ dE(λ) = idσ(T) dE . 

σ(T) 

σ(T) 

The reader should compare this to the following result from linear 

algebra: 

Let T ∈ Mn(C) be a normal matrix. There are then orthogonal 

projections E1,E2,... ,En and complex numbers λ1,λ2,... ,λn such 

that 1 = E1 + E2 + · · · + En and 

n 

T = λiEi. 

i=1 


On the notion of spectrum 

Lemma 

Let D be a unital C ∗ -algebra and E ⊆ D a unital C ∗ -subalgebra. 

Let x ∈ E. Then x is invertible in E if and only if x is invertible in 

D. 

Proof. 

Assume that x is invertible in D. Then x ∗ is invertible in D and 

x −1 = x ∗ (xx ∗ ) −1 . We must show that x −1 ∈ E. 

Set a = xx ∗ and observe that it suffices to prove that a −1 ∈ E. Set 

b = a −1 and observe that a,b and 1 generate an abelian 

C ∗ -subalgebra A of D. 

By Theorem 11.18 in Rudin’s book there is a compact Hausdorff 

space ∆ and a ∗-isomorphism Φ : A → C(∆) such that Φ(1) = 1. 

Set f = Φ(a),g = Φ(b) and note that g = 1 

f . We claim that 

1 

f ∈ C ∗ (1,f ). (1) 



Proof. 

To prove this, note first that there is a interval [ǫ,N] in the positive 

half-axis such that |f (x)| 2 ⊆ [ǫ,N] for all x ∈ ∆. 

By the classical Weierstrass theorem there is a sequence {Pn} of 

polynomials such that 

 

lim 

n→∞ sup 

t∈[ǫ,N] 

 

 

Pn(t) − 1 

 

t = 0. 

Then limn→∞ Pn ◦ |f | 2 = 1 

|f | 2 in C(∆). Since 

Pn ◦ |f | 2 = Pn ◦ (ff ∗ ) ∈ C ∗ (1,f ) for all n, it follows that 

1 

|f | 2 ∈ C ∗ (1,f ). 

Since 1 

f 

= f 1 

|f | 2, we see that (1) holds. It follows from (1) that 

b = Φ −1 (g) ∈ A. Since A ⊆ E we conclude that b ∈ E, as 

desired. 



Lemma 

Let D be a unital C ∗ -algebra and E ⊆ D a unital C ∗ -subalgebra. 

Let x ∈ E. Then x has the same spectrum in E as it has in D. 

Proof. 

By Lemma 2, (λ − x) is invertible in E if and only if it is invertible 

in D. 


Proof of the spectral theorem - Theorem 12.23(existence) 

Let A be the C ∗ -subalgebra of B(H) generated by T and 1. This is 

an abelian C ∗ -subalgebra of B(H) and Theorem 12.22 applies to 

give us a resolution of the identity E ′ on the Borel σ-algebra of the 

Gelfand-spectrum ∆ of A such that 

 

T = Td E. 

∆ 

On the other hand, Theorem 11.19 (suitably intepretated) gives us 

a homeomorphism κ : ∆ → σ(T) defined such that 

κ(ω) = ω(T). 

Observe that in Theorem 11.19 the relevant spectrum σ(T) is 

taken relative to A - but by Lemma 3 this is the same as the 

spectrum relative to B(H). 

Now transfer the resolution E ′ to a resolution on σ(T) via κ. More 

precisely, when B ⊆ σ(T) is a Borel set, κ −1 (B) is a Borel set in ∆ 

and we set 

E (B) = E ′ κ −1 (B) . 



This is a resolution of the identity: E(∅) = 0 and 

E (σ(T)) = E ′ (∆) = 1. And E (B) is obviously a self-adjoint 

projection because E ′ κ −1 (B) is. 

When x ∈ H we see that 

Ex(B) = 〈E(B)x,x〉 = E ′ κ −1 (B) x,x = E ′ x ◦ κ−1 (B), 

which is a measure because E ′ x is (!). 

The next step is to show that 

 

T = 

σ(T) 

Let x ∈ H and note that 

 

λ dE x,x = 

σ(T) 

 

σ(T) 

= λ dE ′ x ◦ κ −1 

= κ dE ′ x. 

σ(T) 

λ dE(λ). (2) 

∆ 

λ Ex 

(3) 



The last equation may not be too well-known so here is an 

argument: We claim that in general 

 

 

f dE 

σ(T) 

′ x ◦ κ −1 = 

f ◦ κ dE 

∆ 

′ x 

when f ∈ BB (σ(T)). 

The last identity in (3) follows from (4) by taking f to be the 

identity function on σ(T). 

To establish (4) note that it suffices to check on simple functions. 

Hence it suffices in fact to establish the equality when f = 1B for 

some Borel set B ⊆ σ(T). But 1B ◦ κ = 1κ−1 (B) and hence the (4) 

reads E ′ 

x κ−1 (B) = E ′ 

x κ−1 (B) in this case - a statement which 

is hard to argue against. 

(4) 



To complete the calculation started in (3) observe that κ = T 

because κ(ω) = ω(T) = T(ω) for all ω ∈ ∆. 

Hence 

 

 

= 

∆ 

λ dE 

σ(T) 

= 〈Tx,x〉 . 

T dE ′ x = 

x,x 

 

∆ 

= 

T dE 

Since x ∈ H was arbitrary, this proves (2). 

κ dE 

∆ 

′ x 

 

x,x 

(5) 


Proof of the spectral theorem - Theorem 12.23(uniqueness) 

To establish the uniqueness part in the spectral theorem we have to 

work a little more than Rudin does because we have not required 

the measures arising from a resolution of the identity to be 

reagular. The next lemma gives us what we need: 

Lemma 

Let X be a compact metric space, and let µ be a finite Borel 

measure on X. Then µ is both inner- and outer regular, i.e 

and 

for every Borel set B ⊆ X 

µ(B) = sup {µ(K) : K ⊆ B compact} 

µ(B) = inf {µ(V) : E ⊆ V open}. 



(proof of lemma) 

Let R be the collection of subsets E in X with the following 

property: 

∀ǫ > 0, ∃F closed and U open such that F ⊂ E ⊂ 

U and µ (U\F) < ǫ. 

We claim that R is a σ-algebra. To see this, observe first that 

X ∈ R. Let then A ∈ R. 

To see that A c = X \A ∈ R, let ǫ > 0. There is an open set U and 

a closed set F such that F ⊆ A ⊆ U such that µ (U\F) < ǫ. Set 

F ′ = X \U and U ′ = X \F and note that F ′ is closed, that U ′ is 

open and that F ′ ⊆ A c ⊆ U ′ . Since U ′ \F ′ = U\F we find that 

µ (U ′ \F ′ ) = µ (U\F) < ǫ. 




This shows that A c ∈ R. Let next A1,A2,A3,... be a sequence of 

sets from R. To see that ∞ 

i=1 Ai ∈ R, let ǫ > 0. For each n there 

are Fn closed, Un open such that Fn ⊆ An ⊆ Un and 

µ (Un\Fn) < ǫ 

3 n. 

Set C = 

n Fn. Since µ(C) < ∞ we see that 1C is µ-integrable. 

Note that 1 S n 

k=1 F k increases to 1C pointwise so that (e.g.) 

Lebesgue’s theorem on monotone convergence implies that 

µ(C) = limn→∞ µ ( n 

k=1 Fk). (This can also be seen by use of 

Proposition 2 on page 255 in Royden: ’Real Analysis’.) 

We can therefore choose an n so big that µ (C\ n . It 

follows then that F = n i=1 Ai, 

k=1 Fn) < ǫ 

2 

k=1 Fn is a closed subset of ∞ U = ∞ k=1 Uk is an open set containing ∞ i=1 Ai and that 

µ (U\F) ≤ µ (U\C) + µ (C\F) 

 

∞ 

 

≤ µ Uk\Fk + ǫ 

2 ≤ 

∞ ǫ ǫ 

+ ≤ ǫ. 

3k 2 

k=1 

k=1 




We have now shown that R is a σ-algebra. Let F ⊆ X be a closed 

set. Let d be a metric for the topology on X. For each n, 

 

Un = x ∈ X : ∃y ∈ F : d(x,y) < 1 

 

n 

is an open set in X. 

We leave the reader to check that U1 ⊇ U2 ⊇ U3 ⊇ ..., and that 

∞ 

n=1 Un = F. When this is established, it follows that 

limn→∞ µ (Un) = µ(F), see e.g Proposition 2 on page 255 in 

Royden: ’Real Analysis’. 

For any given ǫ > 0 we can the find an n such that µ (Un\F) < ǫ, 

and we conclude that F ∈ R. Thus R is a σ-algebra containing all 

closed sets, and must therefore contained all Borel sets. 

It follows that µ is both inner and outer regular. 



Now to the proof of the uniqueness part in Theorem 12.23: 

Let E and F be two resolutions of the identity on σ(T) such that 

 

 

λ dE(λ) = T = λ dF(λ). (6) 

σ(T) 

σ(T) 

By Theorem 12.21 there are then ∗-homomorphisms 

ΨE : L ∞ (E) → B(H) and ΨF : L ∞ (F) → B(H) such that 

and 

 

〈ΨE(f )x,x〉 = 

 

〈ΨF(g)x,x〉 = 

σ(T) 

σ(T) 

f dEx, f ∈ L ∞ (E), 

g dFx, g ∈ L ∞ (F). 



The equation (6) means that ΨE and ΨF agree on the identity 

function id σ(T) on σ(T). Since ΨE and ΨF are both 

∗-homomorphisms it follows that they agree on any polynomial in 

the variables z and z. 

Since these polynomials are dense in C (σ(T)) by the 

Stone-Weierstrass theorem, it follows by continuity that ΨE and 

ΨF agree on C (σ(T)). 

Now the regularity comes in: Let x ∈ H. We need to prove that 

Ex(B) = Fx(B) for every Borel set B ⊆ σ(T). 

Let ǫ > 0. Since σ(T) is a compact metric space and both 

meausures, Ex and Fx, are finite we conclude from Lemma 4 that 

there are compact subsets KE ⊆ B and KF ⊆ B and open sets 

B ⊆ UE and B ⊆ UF such that 

and 

Ex (UE \KE) ≤ ǫ 

Fx (UF \KF) ≤ ǫ. 



Set K = KE ∪ KF and U = UE ∩ UF, and note that K ⊆ B ⊆ U. 

Since K is compact and U open, Urysohn’s lemma provides us with 

a function K ≺ f ≺ U. 

Note first that Ex (U\K) ≤ Ex (UE \KE) ≤ ǫ and 

Fx (U\K) ≤ Fx (UF \KF) ≤ ǫ. 

 

Then |1B − f | ≤ 2 1U\K and hence Ex (B) − 

 

σ(T) f dEx 

≤ 

 

σ(T) |1B − f | dEx ≤ 

σ(T) 2 1U\K dEx = 2Ex (U\K) ≤ 2ǫ. 

The same estimates with E replaced by F yields also that 

 

Fx (B) − 

 

σ(T) f dFx 

≤ 2ǫ. 

Since 

σ(T) f dEx = 

σ(T) f dFx we find that 

|Fx (B) − Ex(B)| ≤ 4ǫ. Since ǫ > 0 was arbitrary we conclude that 

Fx (B) = Ex(B). Since B was an arbitrary Borel set and x an 

arbitrary vector in H, we conclude that E = F.

12. The spectral theorem - Borel calculus for bounded normal ...

Create successful ePaper yourself

Delete template?

Save as template?