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Banach Algebra Notes
Dwight Holland
February 16, 2015
(Corrected February 25, 2015)
Contents
1 Nets
2
2 Banach algebras
3
3 Gelfand Transform
6
4 Spectral Theory
11
Abstract
These notes are for a series of lectures given in functional analysis during Winter term of 2015. The
two influences of the presentation here are Banach Algebra Techniques in Operator Theory (2e) by Ronald
G. Douglas and Lecture Notes on the Spectral Theorem by Dana P. Williams.
The notes start with a presentation of the two facts about nets that are required for the subject
matter at hand. We then move to (unital) Banach algebras (over C), the Gelfand Transform and Spectral
theory. The only facts that are proven for Banach algebras and the Gelfand transform are those that seem
necessary to eventually arrive at the Gelfand-Naimark theorem. This is then used as the primary tool to
define the continuous functional calculus, which is used to prove a number of spectral theory results.
1
1
Nets
We start first with some topological notions useful in Functional Analysis.
Definition 1. Let A be a partially ordered set under ≤. We call A a directed set if for all α, β in A, there
exists γ in A such that α ≤ γ and β ≤ γ.
Definition 2. Let X be a set. For a directed set A, a function x : A → X is called a net. We write xα for
x(α), as we do with sequences.
Definition 3. Let A be a directed set and let X be a topological space. A sequence {xα : α ∈ A} converges
to a point x in X if for all neighborhoods U of x, there exists αU in A so that xα is in U for all α ≥ αU .
We write xα → x and limα∈A xα = x for {xα } converges to x.
Lemma 4. Let X and Y be topological spaces. Then
(a) a set C ⊆ X is closed if and only if whenever {xα } is a net in C with limit x in X then we have that x is
in C, and
(b) a function f : X → Y is continuous if and only if xα → x implies f (xα ) → f (x) for all nets.
In the proof of both parts, we consider the directed set Tx , the set of neighborhoods of x partially ordered
under reverse inclusion. In particular, if U and V are neighborhoods of x, then so is U ∩ V with U ≤ U ∩ V
and V ≤ U ∩ V .
Proof of (a). Suppose that C is closed and {xα } is a net in C converging to a point x in X. If x is not in C,
then there exists a neighborhood U ⊆ X \ C of x. However, as xα → x, we must have that there is αU such
that xα ∈ U for α ≥ αU , a contradiction to {xα } being contained in C.
We now suppose that every convergent net in C converges to a point in C. Take a point x in C. For every
neighborhood U of x, take a point xU in C ∩ U , which is non-empty by definition of C. We then have that
{xU } is a net with respect to the directed set Tx . We further have that xU → x. To see this, note that for
every neighborhood U of x, we have xV is in U for V ≥ U . By hypothesis, we then have that x is in C, and
so C ⊆ C.
The proof for (b) is similar, and recorded here for completeness.
Proof of (b). Suppose that f is continuous and that xα → x. For any neighborhood U of f (x), there exists
an αU such that xα is in f −1 (U ) for all α ≥ αU . Thus, f (xα ) is in U for all α ≥ αU . By definition, it follows
that f (xα ) → f (x).
For the reverse direction, suppose that xα → x implies f (xα ) → f (x). We will then show that f (A) ⊆ f (A)
for all subsets A of X, which is known to be equivalent to f being continuous. To show this, take a point x in
A. If for each U in Tx we take xU to be a point in U , we then have a net. Further, this net converges to x by
the same reasoning as in (a). Thus, f (xU ) → f (x), and so (a) gives that f (x) is in f (A).
As we saw in the previous lemma, nets are sufficient to fully characterize a topological space. In the
following example, we show that sequences are not enough. That is, there are situations in which discussing
behavior sequences is not sufficient to prove topological properties such as continuity.
2
Example 5. Let I = [0, 1] be the closed interval with the standard topology, X = I I be the set of all
functions from I to I with the product topology, and let Y be the subset of all integrable functions with
the subspace topology. If fn → f in the subspace topology, then we can conclude that fn converges to f
pointwise. In particular, X has the basis
B=


U ⊆X:U =

Y
Uy , Uy is open and all but finitely many Uy = I


.

y∈I
For each x in [0, 1] and r > 0, we then have a neighborhood U (x; r) =
Q
y∈I
Uα with Ux = B(x; r) and
Uy = [0, 1] for x 6= y. If fn → f , then fn is eventually in U (x; r), and so |f (x) − fn (x)| < r for sufficiently
large n.
Let T : Y → R be defined by T f =
R1
0
f . As fn → f implies pointwise convergence, and as all fn are
dominated by 1, we can apply Lebesgue Dominated Convergence to say T fn → T f . However, T is not
continuous on Y with this topology. This can be seen by considering the set
A = {f ∈ Y : f (x) 6= 0 for finitely many x} .
It is clear that T f = 0 for all f in A, and so T (A) = {0}. Impressively enough however, the function g that
is identically 1 is in A. To see this, consider a basis neighborhood U around g. Define fU by fU (y) = 0 if
Uy = I, and fU (y) = 1 otherwise. Then fU is in A as only finitely many Uy are not I, and fU is in U as
g(y) − fU (y) = 0 when Uy is not I. As T g = 1, we then have that T (A) 6⊆ T (A). It can be further seen that
no sequence of elements in A converge to g as well.
2
Banach algebras
With these topological notions in hand, we are now ready to start with the theory of Banach algebras.
Definition 6. Let B be a Banach space over C. We call B a Banach algebra if it is also a ring. That is,
there exists an operation from B × B to B, (x, y) 7→ xy, such that for all x, y and z in B and α in C we have
(i) (xy)z = x(yz) (the operation is associative),
(ii) (αx + y)z = αxz + yz and z(αx + y) = αzx + zy (the operation is bilinear), and
(iii) kxyk ≤ kxk kyk (the norm is submultiplicative).
For the purposes of this discussion, we also assume that B is unital. That is, there exists an element 1 in B
such that 1x = x = x1 and k1k = 1. We will typically write λ for λ1.
Example 7. If X is a Banach space, then L(X), the set of continuous linear maps from X to itself is a
non-commutative Banach algebra under composition.
Example 8. If K is a compact Hausdorff space, the set of complex-valued continuous maps C(K) is a
commutative Banach algebra under the uniform norm.
Example 9. If X is an infinite dimensional Banach space, then the set of all compact operators from X to
itself is a non-commutative Banach algebra without identity.
Example 10. L1 (Rn ) is a commutative Banach algebra without identity under convolution.
3
We will now show a few elementary properties of Banach algebras, most of which are generalizations of
things from first term.
Proposition 11. Let B be a Banach algebra. Then multiplication is continuous from B × B to B.
Proof. For x, x0 , y and y 0 in B, we have
kxy − x0 y 0 k = kxy − xy 0 + xy 0 − x0 y 0 k = kx(y − y 0 ) + (x − x0 )y 0 k ≤ kxk ky − y 0 k + kx − x0 k ky 0 k
We then have if xα × yα → x × y, kxy − xα yα k → 0.
Proposition 12. Let B be a Banach algebra. If kxk < 1, then 1 − x is invertible with
(1 − x)−1 ≤
Proof. Let η = kxk and let yn =
Pn
k=0
1
.
1 − kxk
xk . For ε > 0, take N ≥ 0 such that η N +1 < ε(1 − η). We then have
for N ≤ m < n
kyn − ym k ≤
n
X
k
kxk ≤
k=m+1
Thus, {yn } is a Cauchy sequence, and y = limn→∞ yn =
gives
(1 − x)y = (1 − x)
∞
X
xk =
k=0
P∞
η m+1
< ε.
1−η
k=0
∞
X
xk exists. The continuity of multiplication
xk −
k=0
∞
X
xk+1 = 1,
k=0
with similar for y(1 − x). Finally, we also have
kyk ≤
∞
X
k
kxk =
k=0
1
,
1 − kxk
which establishes our norm bounds on the inverse.
Corollary 13. Let B be a Banach algebra. If k1 − xk < 1, then x is invertible with
−1 x ≤
1
.
1 − k1 − xk
Proof. Apply the previous proposition to 1 − (1 − x).
Proposition 14. For a Banach algebra B, the set of all invertible elements G is an open set.
−1
Proof. Suppose that x is invertible. We will show that the set U = B(x, x−1 ) is contained in G . For
y in U , we have 1 − x−1 y ≤ x−1 kx − yk < 1, and so x−1 y is invertible. However, we then have that
x(x−1 y) = y is invertible, and so we are done.
Proposition 15. If B is a Banach algebra, then the map on G defined by x 7→ x−1 is continuous. That is,
G is a topological group.
Proof. Take x and y in G such that kx − yk < 1/(2 x−1 ). We then have that 1 − x−1 y < 1/2 as
1 − x−1 y = x−1 (x − y) ≤ x−1 kx − yk < 1 ,
2
4
and so the previous theorem gives a bound on the inverse of x−1 y. In particular, this gives that
−1 −1 −1 −1 −1 −1 y ≤ y x x = (x y) x ≤
−1 x −1 x .
≤
2
1 − k1 − x−1 yk
It thus follows that
−1
x − y −1 = x−1 (x − y)y −1 ≤ x−1 kx − yk y −1 ≤ 2 x−1 2 kx − yk .
≤ 2 x−1 2 kx − xα k → 0, and so we have continuity.
If we take xα → x, this gives that x−1 − x−1
α
Definition 16. For a Banach algebra B and an element x of B, we define the spectrum of x to be the set
σB (x) = {λ ∈ C : x − λ is not invertible} .
Further, we define the resolvent set of x to be ρB (x) = C \ σB (x), and the spectral radius of x to be
rB (x) = sup {|λ| : λ ∈ σB (x)}. In general, we will write σ(x), ρ(x) and r(x) if we are speaking of a single
Banach algebra with no expectation of confusion.
Lemma 17. Let x be an invertible element in a Banach algebra B. Then λ is in σ(x) if and only if λ−1 is
in σ(x−1 ).
Proof. Note that as x is invertible, λ 6= 0. We can then prove this lemma by noting that
x−1 − λ−1 = −λ−1 x−1 (x − λ).
As −(λx)−1 is invertible, we have that x−1 − λ−1 is not invertible if and only if x − λ is not invertible.
Proposition 18. If B is a Banach algebra and x is in B, then σ(x) is compact with r(x) ≤ kxk.
Proof. Fix x in B and define f (λ) = x − λ. Then f is a continuous map from C to B, which gives that
ρ(x) = f −1 (G ) is an open set. As σ(x) is the complement of an open set, it must be closed. Further, take λ in
C with |λ| > kxk. We then have that λ−1 x < 1, and so 1 − λ−1 x is invertible. Thus, −λ(1 − λ−1 x) = x − λ
is invertible. As σ(x) is closed and bounded, it must be compact.
More difficult is the following:
Proposition 19. If x is an element of a Banach algebra B, then σ(x) is not empty.
Proof. Take λ and µ in ρ(x). We then have
(x − λ)−1 − (x − µ)−1
(x − µ) − (x − λ)
= (x − λ)−1
(x − µ)−1 = (x − λ)−1 (x − µ)−1 .
λ−µ
λ−µ
By continuity of inversion, we then have
(x − λ)−1 − (x − µ)−1
= lim (x − λ)−1 (x − µ)−1 = (x − λ)−2 .
µ→λ
µ→λ
λ−µ
lim
5
If we take ` in the dual of B, we then have that f (λ) = `((x − λ)−1 ) is a holomorphic function with derivative
f 0 (λ) = `((x − λ)−2 ). If σ(x) is empty, it then follows that f is entire. For λ such that |λ| > kxk, we have
1
1
(x − λ)−1 = 1 (1 − λ−1 x)−1 ≤ 1
=
,
|λ|
|λ| 1 − |λ|−1 kxk
|λ| − kxk
and so lim|λ|→∞ |f (λ)| = 0. As f was entire, we can conclude that f is identically zero. Because ` was chosen
arbitrarily, we conclude that (x − λ)−1 is zero. This is impossible as zero cannot be an inverse. We thus
conclude that σ(x) is non-empty.
Definition 20. A division algebra is an algebra in which every non-zero element is invertible.
Corollary 21 (Gelfand-Mazur). If B is a Banach algebra which is a division algebra, then there is a unique
isometric isomorphism of B onto C.
Proof. Let x be a non-zero element of B. As σ(x) is non-empty, there exists λx in σ(x) such that x − λx is
not invertible. However, this implies that x − λx = 0 as B is a division algebra, and so x = λx . If µ 6= λx ,
then λx − µ is invertible, and so σ(x) = {λx }. We then have the map ψ : B → C given by ψ(x) = λx is well
defined, and can be shown to be an isometric isomorphism. If ψ 0 were any other isometric isomorphism, we
would have that ψ 0 (x) would have to be in σ(x), and so ψ 0 (x) = ψ(x).
3
Gelfand Transform
Definition 22. Let B be a Banach algebra. A complex linear function ϕ is called multiplicative if
- ϕ(xy) = ϕ(x)ϕ(y), and
- ϕ(1) = 1. Equivalently, ϕ is not identically zero.
The set of all multiplicative linear functionals is denoted by MB or M is there is no confusion. We do not
assume that ϕ is continuous.
Example 23. Let K be a compact Hausdorff space with x in K fixed. Then the linear functional ϕx defined
on C(K) by ϕx (f ) = f (x) is multiplicative. That is, ϕx is in MC(K) .
Proposition 24. Let B be a Banach algebra. For ϕ in M , we have that ϕ is continuous and kϕk = 1.
Proof. Suppose that x is invertible. Then 1 = ϕ(xx−1 ) = ϕ(x)ϕ(x−1 ), and so ϕ(x) 6= 0. It follows that if
M = ker ϕ, no element of M is invertible, and so k1 + xk ≥ 1 for all x in M . Further, we have for all x in B
that ϕ(x − ϕ(x)1) = 0, and so every element in B can be represented in the form x + λ, with x in M and λ
in C. We then have
kϕk = sup
y6=0
|ϕ(y)|
|ϕ(x + λ)|
|λ|
1
= sup
= sup
= sup
= 1,
kyk
kx
+
λk
kx
+
λk
kx
+
1k
x∈M
x∈M
x∈M
λ6=0
λ6=0
and so ϕ is bounded.
Lemma 25. If B is a Banach algebra, M is compact in the weak-∗ topology.
6
Proof. It suffices to show that M is weak-∗ closed, as it is a subset of the closed unit ball in B 0 , which we
know to be compact in the weak-∗ topology. Let {ϕα } be a net of multiplicative linear functions with limit ϕ.
For x, y in B, we then have
ϕ(xy) = lim ϕα (xy) = lim ϕα (x)ϕα (y) = ϕ(x)ϕ(y)
α∈A
α∈A
and similarly
ϕ(1) = lim ϕα (1) = 1.
α∈A
Thus, ϕ is multiplicative, and it follows that M is closed.
Before I feel comfortable defining the Gelfand transform, we need the following ”structure” theorem.
Theorem 26 (Banach). Let X be a Banach space, and let K be the norm closed unit ball in X 0 equipped
with the weak-∗ topology. It then follows that X is isometrically isomorphic to a closed subspace of C(K).
Proof. For x in X, define a map x 7→ x̂ by x̂(`) = `(x). That x̂ is continuous follows from the fact that K has
the weak-∗ topology. It is a linear map as
\
αx
+ y(`) = `(αx + y) = α`(x) + `(y) = αx̂(`) + ŷ(`).
We further have
kx̂k∞ = sup |x̂(`)| = sup |`(x)| ≤ sup k`k kxk ≤ kxk .
`∈K
`∈K
`∈K
As there exists a linear functional ` such that `(x) = kxk with k`k = 1 by Hahn-Banach, we have kf (x)k∞ = kxk.
We now have that X is isometrically isomorphic to its image under this map.
That is, we can identify every element of a Banach space X with a continuous function on the closed unit
ball in X 0 , which allows us to make the following definition.
Definition 27. Let B be a Banach algebra with M non-empty and equipped with the weak-∗ topology. We
define the Gelfand Transform Γ : B → C(M ) by Γ(x) = x̂|M .
Proposition 28. If B is a Banach Algebra and Γ is the Gelfand transform on B, then
(1) Γ is an algebra homomorphism, and
(2) kΓxk∞ ≤ kxk for x in B.
Proof. Take x and y in B. That Γ is linear follows from the proof that x 7→ x̂ is itself linear. To show that Γ
preserves the multiplicative structure, note that M is multiplicative and so we have
Γ(xy)(ϕ) = ϕ(xy) = ϕ(x)ϕ(y) = (Γx(ϕ)) (Γy(ϕ)) = [Γ(x)Γ(y)](ϕ)
for all ϕ in M .
For the second statement, we have
kΓxk∞ = kx̂|M k∞ ≤ kx̂k∞ = kxk ,
which finishes the proof.
7
An implicit assumption of the previous theorem is that M is non-empty. It can be shown that if B is
commutative, then it is guaranteed that M is non-empty. We also have that M is large enough that it can
determine which elements of B are invertible. However, to show this result we need quotient algebras.
Definition 29. Let B be a Banach algebra and M be a closed two-sided ideal. We define the Quotient
algebra B/M to be the Banach space B/M with multiplication operation [x][y] = [xy], where
[x] = x + M = {x + m : m ∈ M } .
Justification of Definition. For x, y in B, we want to show that [x][y] is well defined. That is, suppose that
[x] = [x0 ] and [y] = [y 0 ]. We need to verify that [xy] = [x0 y 0 ]. To this end, we have that x − x0 = m1 and
y − y 0 = m2 , m1 and m2 in M . It follows that
xy − x0 y 0 = (x0 + m1 )(y 0 + m2 ) − x0 y 0 = x0 m2 + y 0 m1 + m1 m2 ∈ M ,
and so [xy] − [x0 y 0 ] = [xy − x0 y 0 ] = [0]. Because our multiplication is well defined, we are justified in saying
that B/M is an algebra. We now want to show that the norm k[x]k = inf m∈M kx + mk is sub-multiplicative
and k[1]k = 1. The latter is straight forward, for we have
1 = k1 + 0k ≥ inf k1 + mk ≥ 1.
m∈M
The last inequality follows from the fact that if k1 + mk < 1, then m is invertible. We would then have that
M = B. To show that our norm is sub-multiplicative, we simply compute. By doing so we have
k[x][y]k = k[xy]k
= inf kxy + mk
m∈M
≤
≤
inf
k(x + m1 )(y + m2 )k
inf
kx + m1 k ky + m2 k
m1 ,m2 ∈M
m1 ,m2 ∈M
= inf kx + m1 k inf ky + m2 k
m1 ∈M
m2 ∈M
= k[x]k k[y]k .
We thus have that B/M is in fact a Banach algebra, with the map x 7→ [x] a contractive homomorphism.
Proposition 30. If B is a commutative Banach algebra, then the set M of multiplicative linear functions on
B is in one-to-one correspondence with the set of maximal two-sided ideals in B.
Proof. Let ϕ be a non-trivial multiplicative linear functional with kernel M . The kernel is a proper two-sided
ideal, as ϕ(xy) = ϕ(x)ϕ(y) = 0 if either x is in M or y is in M . If x is not in M , we then have that
1 = (1 − y) + y, where y = x/ϕ(x). We have that ϕ(1 − y) = 0, and so 1 − y is in M . It follows that the linear
span of x with M contains the identity, and so any ideal containing both must in fact be all of B. Thus, M
is maximal.
Now suppose that M is a maximal (proper) two-sided ideal in B. Since each x in M is not invertible, we
have that k1 − xk ≥ 1 for all x in M . This gives that M does not contain 1. By continuity of multiplication,
we can see that M is also a two-sided proper ideal. We thus have M ⊆ M ⊂ B, and so M = M . As
8
we now know that M is closed, we can consider the quotient algebra B/M . As M is maximal and B is
commutative, it is possible to show that B/M is a division ring. By the Gelfand-Mazur theorem, there exists
an isometric isomorphism ψ : B/M → C. If we let π : B → B/M denote the natural homomorphism, then
the composition ψ ◦ π = ϕ is a non-zero multiplicative linear functional with kernel M .
We want to show that the correspondence between multiplicative linear functionals and the kernel of
said multiplicative linear functionals is one to one. To this end, suppose that ϕ1 and ϕ2 are such that
ker ϕ1 = ker ϕ2 = M . We want to show that ϕ1 (x) − ϕ2 (x) = 0 for all x in B. To see this, note that x − ϕ1 (x)
is in the kernel of ϕ1 , and thus in the kernel of ϕ2 . Similarly, x − ϕ2 (x) is in the kernel of ϕ2 , and so in the
kernel of ϕ1 . This gives that
ϕ1 (x) − ϕ2 (x) = (x − ϕ2 (x)) − (x − ϕ1 (x))
is in M , as the difference of elements in M . However, ϕ1 (x) − ϕ2 (x) is also a multiple of 1 in B. We can
thus conclude that ϕ1 (x) − ϕ2 (x) = 0, for otherwise M would contain 1. That is, ker ϕ1 = ker ϕ2 implies
ϕ1 = ϕ2 .
Because of the previous theorem, we will call MB the maximal ideal space for B.
Proposition 31. If B is a commutative Banach algebra and x is in B, then x is invertible if and only if Γx
is invertible in C(M ).
Proof. If x is invertible in B, then Γx has inverse Γ(x−1 ). Suppose that x is not invertible. We can then
generate a proper two-sided ideal containing x by letting M0 = {yx : y ∈ B}. We then have that M0 is
contained in some maximal ideal M , which is the kernel of some multiplicative linear functional ϕ. Thus,
0 = ϕ(x) = Γx(ϕ).
Corollary 32. If B is a commutative Banach algebra and x is in B, then σ(x) = range Γx and r(f ) = kΓxk∞ .
The second statement neatly follows from the first part and the definition of r(x).
For a multiplicative linear functional ϕ, we have ϕ(λ) = λϕ(1) = λ for all complex numbers λ. This gives
that Γx = λ if and only if Γ(x − λ) is not invertible. That is, Γx = λ if and only if x − λ is not invertible.
Definition 33. Let B be a Banach algebra, x be an element of B and f : C → C be entire with series
P∞
P∞
expansion f (z) = n=0 an z n . We then define f (x) by f (x) = n=0 an xn .
The previous definition makes sense since the series converges absolutely for all z, and so the series must
converge as an element of a Banach space.
Corollary 34 (Spectral Mapping Theorem). Let B be a Banach algebra. If x is in B and f is an entire
function, then f (σ(x)) = σ(f (x)).
Proof. Let B0 be the closed subalgebra of B generated by 1, x and elements of the form (x−λ)−1 for λ in ρ(x).
We then have that B0 is a commutative Banach algebra with σB (x) = σB0 (x) and σB (f (x)) = σB0 (f (x)).
That is, we can reduce to the case that B is commutative.
P∞
Let f (z) = n=0 an z n be the series expansion of f . We then have by linearity and continuity of Γ that
Γf (x) = Γ
∞
X
!
an xn
=
∞
X
n=0
n=0
9
an (Γx)n = f (Γx).
Together with the previous corollary, we then have the statement
σ(f (x)) = range Γ(f (x)) = range f (Γx) = f (range Γx) = f (σ(x)),
which concludes the proof.
1/n
Theorem 35. If B is a Banach algebra and x is in B, then r(x) = limn→∞ kxn k
.
Proof. Because z n is entire and the previous corollary, we have σ(xn ) = σ(x)n = {λn : λ ∈ σ(x)}. It thus
1/n
follows that r(x)n = r(xn ) ≤ kxn k. This establishes that r(x) ≤ lim inf n→∞ kxn k
Now consider the function
G(λ) = −
.
∞
1 X xn
.
λ n=0 λn
We have G(λ) = (x−λ)−1 for |λ| > kxk. We have previously argued that for ϕ in B 0 , the function ϕ((x−λ)−1 )
P∞
is holomorphic for |λ| > r(x). This gives that − n=0 ϕ(λ−n−1 xn ) is holomorphic for λ > r(x), as it agrees
with ϕ((x − λ)−1 ) on an open set. In particular, this gives that limn→∞ ϕ(λ−n−1 xn ) = 0 for all ϕ in B 0 .
As weakly bounded implies norm bounded by the uniform boundedness principle, we then have the existence
of a number Mλ > 0 such that λ−n−1 xn ≤ Mλ . Thus,
lim sup kxn k
1/n
≤ lim sup |λn |
n→∞
1/n
(λMλ )1/n = |λ| .
n→∞
As this holds for all |λ| > r(x), we have that the lim sup is bounded by r(x). Thus,
1/n
r(x) ≤ lim inf kxn k
n→∞
≤ lim sup kxn k
1/n
≤ r(x).
n→∞
This gives that the limit exists and is equal to r(x).
Theorem 36 (Šilov). If B is a Banach algebra and B0 is a closed subalgebra of B, then the boundary of
σB0 (x) is contained in the boundary σB (x) for all x in B0 .
Proof. If x − λ has no inverse in B, then it cannot have an inverse in B0 . Thus, σB (x) ⊆ σB0 (x). In light of
this, we only need to argue that the boundary of σB0 (x) is contained in σB (x).
Suppose that λ0 is in the boundary of σB0 (x). We then know there exists a sequence {λn } in ρσ0 (x)
−1
such that λn → λ0 . Recall that we have shown that B(x − λn ; (x − λn )−1 ) is contained in the set of all
invertible elements. Because x − λ0 is not invertible (in B0 ), we must have that
|λn − λ0 | = k(x − λ0 ) − (x − λn )k ≥
1
.
k(x − λn )−1 k
Thus,
1
= ∞.
|λ0 − λn |
If x − λ0 was invertible in B, then we would have that (x − λ)−1 is bounded in a neighborhood of λ0 . That
would imply that (x − λn )−1 is eventually bounded. This contradiction gives that λ0 is in σB (x).
lim (x − λn )−1 ≥ lim
n→∞
n→∞
Corollary 37. If B is a Banach algebra with closed subalgebra B0 , then σB0 (x) is obtained by adding
bounded components of C \ σB (x) to σB (x) for elements x in B0 .
10
Proof. Note that if a set A has empty boundary in a connected space U , then either A = U or A = ∅. This
is because A having empty boundary implies that both A and Ac are open, which is only possible if A is
everything or empty.
This fact implies the theorem. If V is a bounded component of C \ σB (x), then the boundary of σB0 (x)
has empty intersection with V , as the boundary is contained in σB (x). Thus, either V ∩ σB0 (x) = ∅ or
V ⊆ σB0 (x). If V∞ is the unbounded component of C \ σB (x), then V∞ ∩ σB0 (x) = ∅ as σB0 (x) cannot
contain an unbounded set. Further, every element of σB0 (x) not in σB (x) must be contained in some
connected component of C \ σB (x). This completes the proof.
Example 38. Suppose that B is a Banach algebra and B0 is a closed sub-algebra. Suppose further that x
in B0 is such that σB (x) = {z : |z| = 1}. Then we have either σB0 (x) = {z : |z| ≤ 1} or σB0 (x) = σB (x).
4
Spectral Theory
Definition 39. If B is a Banach algebra, then an involution on U is a mapping x 7→ x∗ which satisfies
(i) x∗∗ = x for x in B,
(ii) (αx + βy)∗ = αx∗ + βy ∗ for x, y in B and α, β in C, and
(iii) (xy)∗ = y ∗ x∗ for x and y in B.
Lemma 40. Let B be a Banach algebra with involution ∗. Then 1∗ = 1 and x is invertible implies that x∗ is
invertible with inverse (x−1 )∗ .
Proof. The first fact follows from the computation
1∗ = 11∗ = 1∗∗ 1∗ = (11∗ )∗ = (1∗ )∗ = 1∗∗ = 1.
For the second, we have
x∗ (x−1 )∗ = (x−1 x)∗ = 1∗ = 1,
with similar for (x−1 )∗ x∗ .
2
Definition 41. Let U be a Banach algebra with involution ∗. If kx∗ xk = kxk , then we call U a C ∗ -algebra.
In general, I will let U denote a C ∗ -algebra, and use B for Banach algebras.
Example 42. Let K be a compact Hausdorff space. Then C(K) is a Banach algebra with involution
f ∗ (x) = f (x). Further, C(K) is a C ∗ -algebra under this involution.
Example 43. Let H be a Hilbert space. Then the space L(H) of bounded linear operators is a C ∗ -algebra
with the involution of T being defined as the adjoint of T .
Example 44. Let B = `1 (Z) be a Banach algebra under convolution (xy)i =
P
k∈Z xk yi−k .
∗
If we define an
involution (x )i = x−i , then B is a Banach algebra with involution that is not a C -algebra. In particular,
P
(x∗ x)i = k∈Z xk xk−i , and so x defined by x0 = 1, x1 = x2 = −1 and xi = 0 otherwise has the property that
∗
2
kx∗ xk = 5 while kxk = 9.
Lemma 45. Let U be a C ∗ -algebra. Then kxk = kx∗ k for all x in U . Thus, ∗ is an isometry on U .
11
2
Proof. We have kxk = kx∗ xk ≤ kx∗ k kxk, and so kxk ≤ kx∗ k. As x = x∗∗ , we then must have that
kxk = kx∗ k.
Definition 46. Let U be a Banach algebra with involution ∗. We call x self-adjoint if x∗ = x, normal if
x∗ x = xx∗ and unitary if xx∗ = x∗ x = 1.
Theorem 47. If x is a self-adjoint element of a C ∗ -algebra U , then x has real spectrum.
Proof. Define the element y by y = exp(ix). As x is self-adjoint, we can see that y ∗ = exp(−ix). Moreover,
by multiplying the series for exp out, we get that
yy ∗ = exp(ix) exp(−ix) = exp(ix − ix) = 1 = exp(−ix + ix) = exp(−ix) exp(ix) = y ∗ y,
2
and so y is unitary. Moreover, we have 1 = k1k = ky ∗ yk = kyk , and so kyk = ky ∗ k = y −1 = 1. We earlier
showed that λ in σ(y) implies that λ−1 is in σ(y −1 ). This gives that σ(y) ⊆ {z : |z| = 1}, as |λ| ≤ 1 and
−1 λ ≤ 1.
To finish the proof, note that the spectral mapping theorem gives that σ(y) = σ(exp(−ix)) = exp(−iσ(x)),
and so σ(x) must consist of only real numbers.
Theorem 48 (Spectral Permanence). If U is a C ∗ -algebra and U0 is a closed self-adjoint subalgebra, then
σU (x) = σU0 (x) for all x in U0 .
Proof. If x − λ is invertible in U0 , then it is invertible in U . It only remains to show that if x − λ is invertible
in U , then its inverse is in U0 . Thus, suppose that x − λ is invertible in U and let y = (x − λ)∗ (x − λ). It then
follows that y is a self-adjoint element of U which is invertible. Since σU (y) is real by the previous theorem,
the only connected component of C \ σU (y) is unbounded and so σU (y) = σU0 (y). Thus, y is invertible in
U0 . We then have that
−1
(x − λ)−1 = (x − λ)−1 (x∗ − λ)−1 (x∗ − λ) = [(x − λ)∗ (x − λ)] (x − λ)∗ = y −1 (x − λ)∗
is an element of U0 , as it is the product of elements in U0 .
Definition 49. Let B1 , B2 be Banach algebras with involutions ∗ and ?. We call a map T : U1 → U2
a ∗-homomorphism if it is an algebraic homomorphism and T (x∗ ) = T (x)? . Similarly, we call T a
∗-isomorphism if it is an invertible ∗-homomorphism.
Recall that the Stone-Weierstrass theorem says that if A is a closed, self adjoint subspace of C(K) that
separates points and contains the constant functions, then A = C(K). We use this to prove the following
theorem.
Theorem 50 (Gelfand-Naimark; Abstract Spectral Theorem). If U is a commutative C ∗ -algebra and M is
the maximal ideal space of U , then the Gelfand map is a isometric ∗-isomorphism of U onto C(M ).
Proof. We only need to show that Γx = Γ(x∗ ) and kΓ(x)k∞ = kxk. Along with giving that Γ is injective and
a ∗-homomorphism, we will have surjective by use of the Stone-Weierstrass theorem.
For x in U , let u = 12 (x + x∗ ) and v =
1
∗
2i (x − x ),
so that x = u + iv with u and v self-adjoint. As u and v
are self-adjoint, we know that σ(u) and σ(v) are contained in R. As range (Γu) = σ(u) and range (Γv) = σ(v),
we have that Γ(u) and Γ(v) are both real valued. Thus,
Γx = Γ(u + iv) = Γu + iΓv = Γu − iΓv = Γ(u − iv) = Γ(x∗ ).
12
Hence, Γ is a ∗-homomorphism.
By induction, we have that
k
k 1/2
2
kxk = kx∗ xk = (x∗ x)2 ,
k
k 1/2
1/n
2
. However, we also have that limn→∞ kxn k
= r(x) = kΓxk∞ . Thus,
and so kxk = limk→∞ (x∗ x)2 2
2
kxk = kΓ(x∗ x)k∞ = ΓxΓx∞ = |Γx| ∞
2
= kΓxk∞ ,
and so Γ is an isometry. In particular, this gives that the range of Γ is closed. As Γλ = λ, we have
that the range of Γ contains the constant functions. To see that it separates points, take ϕ1 and ϕ2 in
M with ϕ1 6= ϕ2 . We must then have that there exists an x such that ϕ1 (x) 6= ϕ2 (x). This gives that
Γx(ϕ1 ) = ϕ1 (x) 6= ϕ2 (x) = Γx(ϕ2 ), and so Γx separates ϕ1 and ϕ2 . By the Stone-Weierstrass function, it
follows that Γ is surjection, and so Γ is a isometric ∗-isomorphism onto C(M ).
Theorem 51 (Spectral Theorem). If H is a Hilbert space and T is a bounded normal operator on H, then
the C ∗ -algebra CT generated by T is commutative. Moreover, the maximal ideal space of CT is homeomorphic
to σ(T ), and hence the Gelfand map is a isometric ∗-isomorphism of CT onto C(σ(T )).
Proof. As T is normal, T commutes with T ∗ , and so the collection of all polynomials of T and T ∗ is a
commutative sub-algebra of L(H). Further, it is necessarily contained in the C ∗ -algebra generated by T , and
so its closure must be the C ∗ -algebra generated by T . By using sequences, we can see that the closure is
commutative, and so CT is commutative. We thus have a non-trivial maximal ideal space M for CT . Further,
by the Gelfand-Naimark theorem, we have that Γ is a isometric ∗-isomorphism from CT to C(M ), and we only
need to show that M is homeomorphic to σ(T ). In particular, a homeomorphism from M to σ(T ) induces an
isometric isomorphism from C(M ) to C(σ(T )).
To show that the maximal ideal space M of CT is homeomorphic to σ(T ), we will construct a homeomorphism. Define Ψ : M → σ(T ) by Ψ(ϕ) = Γ(T )(ϕ). As the range of Γ(T ) is σ(T ), we have that Ψ is surjective.
To see that it is injective, suppose that Ψ(ϕ1 ) = Ψ(ϕ2 ). We then have Γ(T )(ϕ1 ) = Γ(T )(ϕ2 ), and
Γ(T ∗ )(ϕ1 ) = Γ(T )(ϕ1 ) = Γ(T )(ϕ2 ) = Γ(T ∗ )(ϕ2 ).
More explicitly, we have ϕ1 (T ) = ϕ2 (T ) and ϕ1 (T ∗ ) = ϕ2 (T ∗ ), and so ϕ1 and ϕ2 agree on all polynomials in
T and T ∗ . As the set of all such polynomials is dense in CT by construction, we have that ϕ1 = ϕ2 . Thus, Ψ
is injective. It now remains to show that Ψ is continuous, as a continuous map from a compact space to a
Hausdorff space is necessarily a closed map. As the inverse of Ψ exists, that implies that Ψ−1 is continuous as
well.
Let {ϕα }α∈A be a net in M such that limα∈A ϕα = ϕ. We then have
lim Ψ(ϕα ) = lim Γ(T )(ϕα ) = lim ϕα (T ) = ϕ(T ) = Ψ(ϕ).
α∈A
α∈A
α∈A
Thus, Ψ is continuous, and σ(T ) is homeomorphic to M .
Definition 52. Let T be a bounded normal operator on a Hilbert space H. We define the continuous
functional calculus f (T ) for a continuous function f on σ(T ) by f (T ) = Φ(f ), where Φ : C(σ(T )) → CT is
the inverse of the Gelfand map.
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Lemma 53. Let T be a bounded normal operator on a Hilbert space H, and let λ be an element of σ(T ).
Then there exists a sequence of unit vectors {xn } such that (T − λ)xn converges to 0.
Proof. Let f (z) = z − λ, and fn be a continuous function on σ(T ) such that kfn k∞ = 1, fn (λ) = 1 and fn = 0
outside of B(λ; 1/n). We then have that f (T ) = T − λ and kf fn k∞ ≤ 1/n → 0. As Γ is an isometry, this
gives that kf (T )fn (T )kL(H) → 0 and that kfn (T )kL(H) = 1. As kfn (T )kL(H) = 1, there exists yn in H such
that kfn (T )yn kH = 1 and kyn kH ≤ 2. Let xn = fn (T )yn . We then have
lim k(T − λ)xn kH ≤ lim kf (T )fn (T )kL(H) kyn kH ≤ lim 2 kf (T )fn (T )kL(H) = 0,
n→∞
n→∞
n→∞
which finishes the proof.
Corollary 54. For a bounded normal operator T on a Hilbert space H, the following are equivalent:
(1) T is a positive operator on H, i.e. hT x, xi ≥ 0 for all x in H,
(2) σ(T ) ⊆ [0, ∞), and
(3) There is a bounded operator S on H such that T = S ∗ S.
Proof. Take λ in σ(T ) and choose xn as in the previous lemma. Then h(T − λ)xn , xn i → 0 implies that
2
hT xn , xn i → λ kxn kH = λ. However, as hT xn , xn i ≥ 0, we must have that λ ≥ 0, and so σ(T ) ⊆ [0, ∞).
√
If σ(T ) ⊆ [0, ∞), then f (x) = x is continuous on σ(T ). Let S = f (T ). Then S ∗ = Φ(f )∗ = Φ(f ) =
Φ(f ) = S, and S 2 = Φ(f )2 = Φ(f 2 ) = Φ(idσ(T ) ) = T .
Finally, suppose that there exists a bounded operator S on H such that T = S ∗ S. Then for all x in H, we
have
2
hT x, xi = hS ∗ Sx, xi = hSx, Sxi = kSxkH ≥ 0,
which gives the final equivalence.
Theorem 55. Suppose that K is a normal compact operator on a Hilbert space H. Then
(1) Every nonzero λ in σ(K) is an eigenvalue,
(2) the eigenspace Eλ is finite dimensional, and
(3) σ(K) is countable with 0 the only possible accumulation point.
Proof. Let {xn } be a sequence of norm 1 vectors such that (K − λ)xn → 0. As K is compact, we can assume
that Kxn → x. (In particular, a subsequence will converge to some element x.) We then have
0 = lim (K − λ)xn = x − lim λxn ,
n→∞
n→∞
and so λxn → x. This gives that |λ| = limn→∞ kλxn k = kxk, and so x is not zero. Further, it gives that
Kx = limn→∞ λKxn = λx. Thus, x is an eigenvector.
To see that Eλ is finite dimensional, suppose that it is infinite dimensional. We then have an infinite
orthonormal system {xn } (not necessarily complete) such that Kxn = λxn . It follows that {Kxn } cannot
2
have compact closure as kK(xn − xm )k = 2λ.
Recall that if λ1 and λ2 are distinct eigenvalues, then their eigenvectors are orthogonal for a normal
operator. If σ(K) \ B(0; ε) was infinite, we would again have an infinite set of vectors whose image under
14
K could not have a convergent subsequence as the distance between Kxn and Kxm would be at least
√
2ε.
It follows that there are finitely many eigenvalues outside a ball of radius 1/n around 0, and so there are
countably many eigenvalues with 0 the only possible accumulation point.
Theorem 56 (Spectral Theorem for Compact Operators). Let K be a normal compact operator on a Hilbert
space H, and let {λn : n ∈ I} be the nonzero eigenvalues of K. Let Pn : H → Eλn be the orthogonal projection
operator. We then have that
K=
X
λ n Pn
n∈I
where the sum converges in operator norm.
Proof. Let f be the identity function on σ(K), so that f (K) = K. Further, let fn be the indicator function for
P
λn . Then fn is continuous on σ(K) as λn is isolated for all n. We can show that f = n∈I λn fn in norm
by
P
noting that either I is finite, or if it is infinite then λn → 0. We can conclude that (f − n≤N λn fn )(λ) < ε,
P
and so we have convergence in k·k norm. If we let Pn = fn (T ), then we have that T −
λn Pn =0
∞
n∈I
L(H)
as Γ is an isometry. Thus, we only need to verify that Pn is an orthogonal projection.
Recall that Pn is an orthogonal projection if and only if it is self-adjoint and idempotent, i.e. Pn2 = Pn .
Note that fn is real valued with fn2 = fn . We then have
Pn2 = Φ(fn )2 = Φ(fn2 ) = Φ(fn ) = Pn and Pn∗ = Φ(fn )∗ = Φ(fn ) = Φ(fn ) = Pn .
This finishes the result.
15