12. The spectral theorem - Borel calculus for bounded normal ...
12. The spectral theorem - Borel calculus for bounded normal ...
12. The spectral theorem - Borel calculus for bounded normal ...
Create successful ePaper yourself
Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.
<strong>12.</strong> <strong>The</strong> <strong>spectral</strong> <strong>theorem</strong> - <strong>Borel</strong> <strong>calculus</strong> <strong>for</strong><br />
<strong>bounded</strong> <strong>normal</strong> operators<br />
Klaus Thomsen matkt@imf.au.dk<br />
Institut <strong>for</strong> Matematiske Fag<br />
Det Naturvidenskabelige Fakultet<br />
Aarhus Universitet<br />
December 2005<br />
Klaus Thomsen <strong>12.</strong> <strong>The</strong> <strong>spectral</strong> <strong>theorem</strong> - <strong>Borel</strong> <strong>calculus</strong> <strong>for</strong> <strong>bounded</strong> norm
We read Sections <strong>12.</strong>23 and <strong>12.</strong>24 in Rudins book - with some<br />
changes.<br />
Klaus Thomsen <strong>12.</strong> <strong>The</strong> <strong>spectral</strong> <strong>theorem</strong> - <strong>Borel</strong> <strong>calculus</strong> <strong>for</strong> <strong>bounded</strong> norm
<strong>The</strong> <strong>spectral</strong> <strong>theorem</strong> - the statement<br />
<strong>The</strong>orem<br />
Let H be a Hilbert space and T ∈ B(H) a <strong>normal</strong> operator.<br />
<strong>The</strong>re is a unique resolution of the identity E on the spectrum<br />
σ(T) of T such that<br />
<br />
<br />
T = λ dE(λ) = idσ(T) dE .<br />
σ(T)<br />
σ(T)<br />
<strong>The</strong> reader should compare this to the following result from linear<br />
algebra:<br />
Let T ∈ Mn(C) be a <strong>normal</strong> matrix. <strong>The</strong>re are then orthogonal<br />
projections E1,E2,... ,En and complex numbers λ1,λ2,... ,λn such<br />
that 1 = E1 + E2 + · · · + En and<br />
n<br />
T = λiEi.<br />
i=1<br />
Klaus Thomsen <strong>12.</strong> <strong>The</strong> <strong>spectral</strong> <strong>theorem</strong> - <strong>Borel</strong> <strong>calculus</strong> <strong>for</strong> <strong>bounded</strong> norm
On the notion of spectrum<br />
Lemma<br />
Let D be a unital C ∗ -algebra and E ⊆ D a unital C ∗ -subalgebra.<br />
Let x ∈ E. <strong>The</strong>n x is invertible in E if and only if x is invertible in<br />
D.<br />
Proof.<br />
Assume that x is invertible in D. <strong>The</strong>n x ∗ is invertible in D and<br />
x −1 = x ∗ (xx ∗ ) −1 . We must show that x −1 ∈ E.<br />
Set a = xx ∗ and observe that it suffices to prove that a −1 ∈ E. Set<br />
b = a −1 and observe that a,b and 1 generate an abelian<br />
C ∗ -subalgebra A of D.<br />
By <strong>The</strong>orem 11.18 in Rudin’s book there is a compact Hausdorff<br />
space ∆ and a ∗-isomorphism Φ : A → C(∆) such that Φ(1) = 1.<br />
Set f = Φ(a),g = Φ(b) and note that g = 1<br />
f . We claim that<br />
1<br />
f ∈ C ∗ (1,f ). (1)<br />
Klaus Thomsen <strong>12.</strong> <strong>The</strong> <strong>spectral</strong> <strong>theorem</strong> - <strong>Borel</strong> <strong>calculus</strong> <strong>for</strong> <strong>bounded</strong> norm
On the notion of spectrum<br />
Proof.<br />
To prove this, note first that there is a interval [ǫ,N] in the positive<br />
half-axis such that |f (x)| 2 ⊆ [ǫ,N] <strong>for</strong> all x ∈ ∆.<br />
By the classical Weierstrass <strong>theorem</strong> there is a sequence {Pn} of<br />
polynomials such that<br />
<br />
lim<br />
n→∞ sup<br />
t∈[ǫ,N]<br />
<br />
<br />
Pn(t) − 1<br />
<br />
t = 0.<br />
<strong>The</strong>n limn→∞ Pn ◦ |f | 2 = 1<br />
|f | 2 in C(∆). Since<br />
Pn ◦ |f | 2 = Pn ◦ (ff ∗ ) ∈ C ∗ (1,f ) <strong>for</strong> all n, it follows that<br />
1<br />
|f | 2 ∈ C ∗ (1,f ).<br />
Since 1<br />
f<br />
= f 1<br />
|f | 2, we see that (1) holds. It follows from (1) that<br />
b = Φ −1 (g) ∈ A. Since A ⊆ E we conclude that b ∈ E, as<br />
desired.<br />
Klaus Thomsen <strong>12.</strong> <strong>The</strong> <strong>spectral</strong> <strong>theorem</strong> - <strong>Borel</strong> <strong>calculus</strong> <strong>for</strong> <strong>bounded</strong> norm
On the notion of spectrum<br />
Lemma<br />
Let D be a unital C ∗ -algebra and E ⊆ D a unital C ∗ -subalgebra.<br />
Let x ∈ E. <strong>The</strong>n x has the same spectrum in E as it has in D.<br />
Proof.<br />
By Lemma 2, (λ − x) is invertible in E if and only if it is invertible<br />
in D.<br />
Klaus Thomsen <strong>12.</strong> <strong>The</strong> <strong>spectral</strong> <strong>theorem</strong> - <strong>Borel</strong> <strong>calculus</strong> <strong>for</strong> <strong>bounded</strong> norm
Proof of the <strong>spectral</strong> <strong>theorem</strong> - <strong>The</strong>orem <strong>12.</strong>23(existence)<br />
Let A be the C ∗ -subalgebra of B(H) generated by T and 1. This is<br />
an abelian C ∗ -subalgebra of B(H) and <strong>The</strong>orem <strong>12.</strong>22 applies to<br />
give us a resolution of the identity E ′ on the <strong>Borel</strong> σ-algebra of the<br />
Gelfand-spectrum ∆ of A such that<br />
<br />
T = Td E.<br />
∆<br />
On the other hand, <strong>The</strong>orem 11.19 (suitably intepretated) gives us<br />
a homeomorphism κ : ∆ → σ(T) defined such that<br />
κ(ω) = ω(T).<br />
Observe that in <strong>The</strong>orem 11.19 the relevant spectrum σ(T) is<br />
taken relative to A - but by Lemma 3 this is the same as the<br />
spectrum relative to B(H).<br />
Now transfer the resolution E ′ to a resolution on σ(T) via κ. More<br />
precisely, when B ⊆ σ(T) is a <strong>Borel</strong> set, κ −1 (B) is a <strong>Borel</strong> set in ∆<br />
and we set<br />
E (B) = E ′ κ −1 (B) .<br />
Klaus Thomsen <strong>12.</strong> <strong>The</strong> <strong>spectral</strong> <strong>theorem</strong> - <strong>Borel</strong> <strong>calculus</strong> <strong>for</strong> <strong>bounded</strong> norm
Proof of the <strong>spectral</strong> <strong>theorem</strong> - <strong>The</strong>orem <strong>12.</strong>23(existence)<br />
This is a resolution of the identity: E(∅) = 0 and<br />
E (σ(T)) = E ′ (∆) = 1. And E (B) is obviously a self-adjoint<br />
projection because E ′ κ −1 (B) is.<br />
When x ∈ H we see that<br />
Ex(B) = 〈E(B)x,x〉 = E ′ κ −1 (B) x,x = E ′ x ◦ κ−1 (B),<br />
which is a measure because E ′ x is (!).<br />
<strong>The</strong> next step is to show that<br />
<br />
T =<br />
σ(T)<br />
Let x ∈ H and note that<br />
<br />
λ dE x,x =<br />
σ(T)<br />
<br />
σ(T)<br />
= λ dE ′ x ◦ κ −1 <br />
= κ dE ′ x.<br />
σ(T)<br />
λ dE(λ). (2)<br />
∆<br />
λ Ex<br />
(3)<br />
Klaus Thomsen <strong>12.</strong> <strong>The</strong> <strong>spectral</strong> <strong>theorem</strong> - <strong>Borel</strong> <strong>calculus</strong> <strong>for</strong> <strong>bounded</strong> norm
Proof of the <strong>spectral</strong> <strong>theorem</strong> - <strong>The</strong>orem <strong>12.</strong>23(existence)<br />
<strong>The</strong> last equation may not be too well-known so here is an<br />
argument: We claim that in general<br />
<br />
<br />
f dE<br />
σ(T)<br />
′ x ◦ κ −1 =<br />
f ◦ κ dE<br />
∆<br />
′ x<br />
when f ∈ BB (σ(T)).<br />
<strong>The</strong> last identity in (3) follows from (4) by taking f to be the<br />
identity function on σ(T).<br />
To establish (4) note that it suffices to check on simple functions.<br />
Hence it suffices in fact to establish the equality when f = 1B <strong>for</strong><br />
some <strong>Borel</strong> set B ⊆ σ(T). But 1B ◦ κ = 1κ−1 (B) and hence the (4)<br />
reads E ′ <br />
x κ−1 (B) = E ′<br />
x κ−1 (B) in this case - a statement which<br />
is hard to argue against.<br />
(4)<br />
Klaus Thomsen <strong>12.</strong> <strong>The</strong> <strong>spectral</strong> <strong>theorem</strong> - <strong>Borel</strong> <strong>calculus</strong> <strong>for</strong> <strong>bounded</strong> norm
Proof of the <strong>spectral</strong> <strong>theorem</strong> - <strong>The</strong>orem <strong>12.</strong>23(existence)<br />
To complete the calculation started in (3) observe that κ = T<br />
because κ(ω) = ω(T) = T(ω) <strong>for</strong> all ω ∈ ∆.<br />
Hence<br />
<br />
<br />
=<br />
∆<br />
λ dE<br />
σ(T)<br />
= 〈Tx,x〉 .<br />
T dE ′ x =<br />
x,x<br />
<br />
∆<br />
=<br />
T dE<br />
Since x ∈ H was arbitrary, this proves (2).<br />
κ dE<br />
∆<br />
′ x<br />
<br />
x,x<br />
(5)<br />
Klaus Thomsen <strong>12.</strong> <strong>The</strong> <strong>spectral</strong> <strong>theorem</strong> - <strong>Borel</strong> <strong>calculus</strong> <strong>for</strong> <strong>bounded</strong> norm
Proof of the <strong>spectral</strong> <strong>theorem</strong> - <strong>The</strong>orem <strong>12.</strong>23(uniqueness)<br />
To establish the uniqueness part in the <strong>spectral</strong> <strong>theorem</strong> we have to<br />
work a little more than Rudin does because we have not required<br />
the measures arising from a resolution of the identity to be<br />
reagular. <strong>The</strong> next lemma gives us what we need:<br />
Lemma<br />
Let X be a compact metric space, and let µ be a finite <strong>Borel</strong><br />
measure on X. <strong>The</strong>n µ is both inner- and outer regular, i.e<br />
and<br />
<strong>for</strong> every <strong>Borel</strong> set B ⊆ X<br />
µ(B) = sup {µ(K) : K ⊆ B compact}<br />
µ(B) = inf {µ(V) : E ⊆ V open}.<br />
Klaus Thomsen <strong>12.</strong> <strong>The</strong> <strong>spectral</strong> <strong>theorem</strong> - <strong>Borel</strong> <strong>calculus</strong> <strong>for</strong> <strong>bounded</strong> norm
Proof of the <strong>spectral</strong> <strong>theorem</strong> - <strong>The</strong>orem <strong>12.</strong>23(uniqueness)<br />
(proof of lemma)<br />
Let R be the collection of subsets E in X with the following<br />
property:<br />
∀ǫ > 0, ∃F closed and U open such that F ⊂ E ⊂<br />
U and µ (U\F) < ǫ.<br />
We claim that R is a σ-algebra. To see this, observe first that<br />
X ∈ R. Let then A ∈ R.<br />
To see that A c = X \A ∈ R, let ǫ > 0. <strong>The</strong>re is an open set U and<br />
a closed set F such that F ⊆ A ⊆ U such that µ (U\F) < ǫ. Set<br />
F ′ = X \U and U ′ = X \F and note that F ′ is closed, that U ′ is<br />
open and that F ′ ⊆ A c ⊆ U ′ . Since U ′ \F ′ = U\F we find that<br />
µ (U ′ \F ′ ) = µ (U\F) < ǫ.<br />
Klaus Thomsen <strong>12.</strong> <strong>The</strong> <strong>spectral</strong> <strong>theorem</strong> - <strong>Borel</strong> <strong>calculus</strong> <strong>for</strong> <strong>bounded</strong> norm
Proof of the <strong>spectral</strong> <strong>theorem</strong> - <strong>The</strong>orem <strong>12.</strong>23(uniqueness)<br />
(proof of lemma)<br />
This shows that A c ∈ R. Let next A1,A2,A3,... be a sequence of<br />
sets from R. To see that ∞<br />
i=1 Ai ∈ R, let ǫ > 0. For each n there<br />
are Fn closed, Un open such that Fn ⊆ An ⊆ Un and<br />
µ (Un\Fn) < ǫ<br />
3 n.<br />
Set C = <br />
n Fn. Since µ(C) < ∞ we see that 1C is µ-integrable.<br />
Note that 1 S n<br />
k=1 F k increases to 1C pointwise so that (e.g.)<br />
Lebesgue’s <strong>theorem</strong> on monotone convergence implies that<br />
µ(C) = limn→∞ µ ( n<br />
k=1 Fk). (This can also be seen by use of<br />
Proposition 2 on page 255 in Royden: ’Real Analysis’.)<br />
We can there<strong>for</strong>e choose an n so big that µ (C\ n . It<br />
follows then that F = n i=1 Ai,<br />
k=1 Fn) < ǫ<br />
2<br />
k=1 Fn is a closed subset of ∞ U = ∞ k=1 Uk is an open set containing ∞ i=1 Ai and that<br />
µ (U\F) ≤ µ (U\C) + µ (C\F)<br />
<br />
∞<br />
<br />
≤ µ Uk\Fk + ǫ<br />
2 ≤<br />
∞ ǫ ǫ<br />
+ ≤ ǫ.<br />
3k 2<br />
k=1<br />
k=1<br />
Klaus Thomsen <strong>12.</strong> <strong>The</strong> <strong>spectral</strong> <strong>theorem</strong> - <strong>Borel</strong> <strong>calculus</strong> <strong>for</strong> <strong>bounded</strong> norm
Proof of the <strong>spectral</strong> <strong>theorem</strong> - <strong>The</strong>orem <strong>12.</strong>23(uniqueness)<br />
(proof of lemma)<br />
We have now shown that R is a σ-algebra. Let F ⊆ X be a closed<br />
set. Let d be a metric <strong>for</strong> the topology on X. For each n,<br />
<br />
Un = x ∈ X : ∃y ∈ F : d(x,y) < 1<br />
<br />
n<br />
is an open set in X.<br />
We leave the reader to check that U1 ⊇ U2 ⊇ U3 ⊇ ..., and that<br />
∞<br />
n=1 Un = F. When this is established, it follows that<br />
limn→∞ µ (Un) = µ(F), see e.g Proposition 2 on page 255 in<br />
Royden: ’Real Analysis’.<br />
For any given ǫ > 0 we can the find an n such that µ (Un\F) < ǫ,<br />
and we conclude that F ∈ R. Thus R is a σ-algebra containing all<br />
closed sets, and must there<strong>for</strong>e contained all <strong>Borel</strong> sets.<br />
It follows that µ is both inner and outer regular.<br />
Klaus Thomsen <strong>12.</strong> <strong>The</strong> <strong>spectral</strong> <strong>theorem</strong> - <strong>Borel</strong> <strong>calculus</strong> <strong>for</strong> <strong>bounded</strong> norm
Proof of the <strong>spectral</strong> <strong>theorem</strong> - <strong>The</strong>orem <strong>12.</strong>23(uniqueness)<br />
Now to the proof of the uniqueness part in <strong>The</strong>orem <strong>12.</strong>23:<br />
Let E and F be two resolutions of the identity on σ(T) such that<br />
<br />
<br />
λ dE(λ) = T = λ dF(λ). (6)<br />
σ(T)<br />
σ(T)<br />
By <strong>The</strong>orem <strong>12.</strong>21 there are then ∗-homomorphisms<br />
ΨE : L ∞ (E) → B(H) and ΨF : L ∞ (F) → B(H) such that<br />
and<br />
<br />
〈ΨE(f )x,x〉 =<br />
<br />
〈ΨF(g)x,x〉 =<br />
σ(T)<br />
σ(T)<br />
f dEx, f ∈ L ∞ (E),<br />
g dFx, g ∈ L ∞ (F).<br />
Klaus Thomsen <strong>12.</strong> <strong>The</strong> <strong>spectral</strong> <strong>theorem</strong> - <strong>Borel</strong> <strong>calculus</strong> <strong>for</strong> <strong>bounded</strong> norm
Proof of the <strong>spectral</strong> <strong>theorem</strong> - <strong>The</strong>orem <strong>12.</strong>23(uniqueness)<br />
<strong>The</strong> equation (6) means that ΨE and ΨF agree on the identity<br />
function id σ(T) on σ(T). Since ΨE and ΨF are both<br />
∗-homomorphisms it follows that they agree on any polynomial in<br />
the variables z and z.<br />
Since these polynomials are dense in C (σ(T)) by the<br />
Stone-Weierstrass <strong>theorem</strong>, it follows by continuity that ΨE and<br />
ΨF agree on C (σ(T)).<br />
Now the regularity comes in: Let x ∈ H. We need to prove that<br />
Ex(B) = Fx(B) <strong>for</strong> every <strong>Borel</strong> set B ⊆ σ(T).<br />
Let ǫ > 0. Since σ(T) is a compact metric space and both<br />
meausures, Ex and Fx, are finite we conclude from Lemma 4 that<br />
there are compact subsets KE ⊆ B and KF ⊆ B and open sets<br />
B ⊆ UE and B ⊆ UF such that<br />
and<br />
Ex (UE \KE) ≤ ǫ<br />
Fx (UF \KF) ≤ ǫ.<br />
Klaus Thomsen <strong>12.</strong> <strong>The</strong> <strong>spectral</strong> <strong>theorem</strong> - <strong>Borel</strong> <strong>calculus</strong> <strong>for</strong> <strong>bounded</strong> norm
Proof of the <strong>spectral</strong> <strong>theorem</strong> - <strong>The</strong>orem <strong>12.</strong>23(uniqueness)<br />
Set K = KE ∪ KF and U = UE ∩ UF, and note that K ⊆ B ⊆ U.<br />
Since K is compact and U open, Urysohn’s lemma provides us with<br />
a function K ≺ f ≺ U.<br />
Note first that Ex (U\K) ≤ Ex (UE \KE) ≤ ǫ and<br />
Fx (U\K) ≤ Fx (UF \KF) ≤ ǫ. <br />
<br />
<strong>The</strong>n |1B − f | ≤ 2 1U\K and hence Ex (B) − <br />
<br />
σ(T) f dEx<br />
≤<br />
<br />
σ(T) |1B − f | dEx ≤ <br />
σ(T) 2 1U\K dEx = 2Ex (U\K) ≤ 2ǫ.<br />
<strong>The</strong> same estimates with E replaced by F yields also that<br />
<br />
Fx (B) − <br />
<br />
σ(T) f dFx<br />
≤ 2ǫ.<br />
Since <br />
σ(T) f dEx = <br />
σ(T) f dFx we find that<br />
|Fx (B) − Ex(B)| ≤ 4ǫ. Since ǫ > 0 was arbitrary we conclude that<br />
Fx (B) = Ex(B). Since B was an arbitrary <strong>Borel</strong> set and x an<br />
arbitrary vector in H, we conclude that E = F.<br />
Klaus Thomsen <strong>12.</strong> <strong>The</strong> <strong>spectral</strong> <strong>theorem</strong> - <strong>Borel</strong> <strong>calculus</strong> <strong>for</strong> <strong>bounded</strong> norm