Mathematics 442C
Suggested solutions to exercise sheet 5
1. Let A and B be Banach algebras, and let A0 be a dense subalgebra of A. If θ0 : A0 → B
is a continuous homomorphism, show that there is a unique continuous homomorphism
θ : A → B such that θ|A0 = θ0 .
Solution. We first prove something slightly different, which you may have seen before:
if θ0 : A0 → B is a continuous linear map between a vector space A0 which is dense in
a Banach space A, and a Banach space B, then there is a unique continuous linear map
θ : A → B such that θ|A0 = θ0 . For uniqueness, recall that any two continuous mappings
A → B which agree on a dense subset of A must be equal (in fact, this is true for any
topological spaces A and B). For existence, if a ∈ A, let θ(a) = limn→∞ θ0 (an ) where an
is any sequence in A0 which converges to a. To check that this all makes sense, observe
that
• since A0 is dense in A, such a sequence (an ) exists for every a ∈ A;
• since θ0 is continuous, it maps convergent sequences in A to Cauchy sequences in B,
which are convergent since B is complete. So the limit limn→∞ θ0 (an ) above exists;
• if an → a and bn → a as n → ∞ where an , bn ∈ A0 , then an − bn → 0 ∈ A0 ,
so θ0 (an − bn ) = θ0 (an ) − θ0 (bn ) → 0, so limn→∞ θ0 (an ) = limn→∞ θ0 (bn ). So the
definition of θ(a) above does not depend on the choice of convergent sequence (an );
• if a ∈ A0 then by choosing the constant sequence an = a we see that θ(a) =
limn→∞ θ0 (a) = θ0 (a), so θ|A0 = θ0 ;
• if a, b ∈ A and λ ∈ C and an , bn ∈ A0 with an → a and bn → b as n → ∞,
then an + λbn → a + λb, so θ(a + λb) = limn→∞ θ0 (an + λbn ) = limn→∞ θ0 (an ) +
λ limn→∞ θ0 (bn ) = θ(a) + λθ(b). So θ is linear; and
• θ0 is a continuous linear map, hence is bounded, and if A0 3 an → a ∈ A then
kθ(a)k = k limn→∞ θ0 (an )k = limn→∞ kθ0 (an )k ≤ kθ0 k limn→∞ kan k = kθ0 k kak, so
kθk = kθ0 k. Hence θ is a bounded linear map, so is continuous.
Now consider the situation in the original question where A0 is a dense subalgebra
of A and θ0 is a continuous homomorphism. We wish to show that the continuous linear
map θ constructed above is actually a homomorphism. If a, b ∈ A then there are sequences
(an )n≥1 and (bn )n≥1 in A0 such that an → a and bn → b as n → ∞. Since multiplication
is jointly continuous in a Banach algebra, we also have an bn → ab as n → ∞. Since θ is
continuous and θ0 is a homomorphism, we have
θ(ab) = lim θ0 (an bn ) by the definition of θ
n→∞
= lim θ0 (an )θ0 (bn ) since θ0 is a homomorphism
n→∞
= lim θ0 (an ) lim θ0 (bm ) since multiplication is jointly continuous in B
n→∞
m→∞
= θ(a)θ(b).
So θ is a homomorphism.
2. Let a be a normal element of a unital C*-algebra. Prove the following assertions.
(a) a is hermitian if and only σ(a) ⊆ R.
(b) a is unitary if and only if σ(a) ⊆ T.
(c) a is a projection if and only if σ(a) ⊆ {0, 1}.
If a is not normal, then these assertions may all fail. Find an example to demonstrate
this.
Solution. (a) If a is hermitian then σ(a) ⊆ R by Corollary 4.1.10.
Conversely, suppose that a is normal and σ(a) ⊆ R. Let z : σ(a) → C, t →
7 t.
∗
∗
Observe that z ∈ C(σ(a)) and z = z since z (t) = z(t) = t = t for t ∈ σ(a). The
continuous functional calculus at a is a ∗-homomorphism with z(a) = a, so
∗
a∗ = z(a) = (z ∗ )(a) = z(a) = a.
So a is hermitian.
(b) If a is unitary then σ(a) ⊆ T by Corollary 4.1.10.
Conversely, suppose that a is normal and σ(a) ⊆ T. Let z : σ(a) → C, λ 7→ λ.
Observe that z ∈ C(σ(a)) and zz ∗ = z ∗ z = 1, since (z ∗ z)(λ) = λλ = |λ|2 = 1 = 1(λ)
for λ ∈ σ(a). The continuous functional calculus at a is a unital ∗-homomorphism with
z(a) = a, so
a∗ a = z(a)∗ z(a) = z ∗ (a)z(a) = (zz ∗ )(a) = 1(a) = 1
and similarly, aa∗ = 1. So a is unitary.
(c) If a is a projection then a2 = a, so a2 − a = 0. If λ ∈ σ(a) then λ2 − λ ∈
σ(a2 − a) = {0} by the polynomial spectral mapping theorem. So λ(λ − 1) = 0, so
λ ∈ {0, 1}.
Conversely, suppose that a is normal and σ(a) ⊆ {0, 1}. We know that a is hermitian,
by (a). Let z : σ(a) → C, t 7→ t. Observe that z ∈ C(σ(a)) and z 2 = z since z 2 (t) =
z(t)2 = t2 = t = z(t) for t ∈ σ(a). The continuous functional calculus at a is a ∗homomorphism with z(a) = a, so
a2 = z(a2 ) = z(a)2 = (z 2 )(a) = z(a) = a.
So a∗ = a = a2 , so a is a projection.
To see that the assertions may fail if a is not normal, let A = M2 (C) and let a = 10 11 .
Then σ(a) = {1} = R ∩ T ∩ {0, 1}, but a is not normal. So a is neither hermitian, unitary
nor a projection.
3. Let A be a unital C*-algebra. For a ∈ A, recall that ea was defined in Exercise 2.9 as
ea = lim pn (a) where pn (x) =
n→∞
n
X
xk
k=0
k!
Show that if h is a hermitian element of A, then eih is unitary.
2
.
∗
Solution. Since h = h∗ , we have (ih)k = (−ih)k . Hence pn (ih)∗ = pn (−ih). Since
the adjoint is continuous, this gives
∗
(eih )∗ = lim pn (ih) = lim pn (ih)∗ = lim pn (−ih) = e−ih .
n→∞
n→∞
n→∞
Since ih clearly commutes with −ih, Exercise 2.9(f) gives
(eih )∗ eih = e−ih eih = e−ih+ih = e0 = 1,
and similarly,
eih (eih )∗ = eih e−ih = eih−ih = e0 = 1.
So eih is unitary.
Another way to do this question is to show that if a is normal, then ea = exp(a)
where exp ∈ C(σ(a)) is the restriction of the exponential function C → C to σ(a), and
then use the spectral mapping theorem and the fact that exp(iR) = T and Exercise 2.
4. Prove Lemma 4.3.5:
Let A be a unital C*-algebra. If a is a normal element of A then C ∗ (1, a) is a unital
abelian C*-algebra.
Solution. By the definition of C ∗ (1, a), it is a C*-subalgebra containing 1. Hence it is
a unital C*-algebra, with unit 1.
By Lemma 4.3.4 and using the normality of a (aa∗ = a∗ a) we have that C ∗ (1, a) is
the closure of the linear span of
S = {1, an (a∗ )m : n, m ≥ 1}.
If x, y ∈ S then xy = yx, since a is normal and 1 commutes with any element of A. Hence
any two elements of the linear span of S commute. Taking limits and using the continuity
of the product in A, it follows that any two elements of the closure of the linear span of S
commute. Hence C ∗ (1, a) is abelian.
5. Prove Lemma 4.3.8:
Suppose that X and Y are compact Hausdorff topological spaces and ψ : X → Y is
a homeomorphism. The map ψ t : C(Y ) → C(X), f 7→ f ◦ ψ is a well-defined isometric
unital ∗-isomorphism.
Solution. Since ψ is continuous, if f ∈ C(Y ) then f ◦ ψ = ψ t (f ) is a continuous map
X → C. So ψ t is well-defined. Moreover, for x ∈ X we have ψ t (1C(Y ) )(x) = 1C(Y ) (ψ(x)) =
1 = 1C(X) (x), so ψ t (1C(Y ) ) = 1C(X) and ψ t is unital.
For f, g ∈ C(Y ) and λ ∈ C we have:
ψ t (f + λg)(x) = (f + λg)(ψ(x)) = f (ψ(x)) + λg(ψ(x)) = (ψ t (f ) + λψ t (g))(x),
ψ t (f g)(x) = (f g)(ψ(x)) = f (ψ(x)) · g(ψ(x)) = (ψ t (f ) · ψ t (g))(x),
ψ t (f ∗ )(x) = f ∗ (ψ(x)) = f (ψ(x)) = ψ t (f )(x) = (ψ t (f )∗ )(x).
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So ψ t is a ∗-homomorphism. Moreover, since ψ is surjective, we have
kψ t (f )k = sup |f (ψ(x))| = sup |f (y)| = kf k
x∈X
y∈Y
so ψ t is an isometry and in particular, ψ t is injective. For h ∈ C(X) we have h =
h ◦ ψ −1 ◦ ψ = ψ t (h ◦ ψ −1 ). Since h ◦ ψ −1 ∈ C(X), this shows that ψ t is surjective.
6. Let A be a unital C*-algebra.
(a) Show that if a, b ∈ A+ with ab = ba then ab ∈ A+ .
(b) Give an example with A = M2 (C) to show that there may be a, b ∈ A+ such that
ab 6∈ A+ .
Solution. (a) Let B = C ∗ ({1, a, b}), a unital abelian C*-subalgebra of A. Since the
Gelfand representation γ : B → C(Ω(B)) for B is a unital ∗-isomorphism, it is spectrumpreserving and sends hermitian elements to hermitian elements. So γ(B + ) = C(Ω(B))+ .
Since a, b ∈ A+ we have a, b ∈ B + , so γ(a), γ(b) ∈ C(Ω(B))+ . The (pointwise) product
of two positive-valued continuous functions is positive-valued, so γ(a)γ(b) = γ(ab) ∈
C(Ω(B))+ . Since γ −1 is a ∗-isomorphism, it also preserves positivity, so ab = γ −1 (γ(ab)) ∈
B + ⊆ A+ .
1 i
1 0
(b) For example, consider a, b ∈ M2 (C) given by a = 21 −i
1 and b = 0 0 . Then
it is easy to check that a and b are hermitian with a2 = a and b2 = b, so they are both
∗
1 0
1 i
projections. In particular, they are positive. However, ab = ( −i
0 ) 6= (ab) = ( 0 0 ), so ab
is not hermitian. So ab is not positive.
7. Prove Lemma 4.4.9:
If A is a unital C*-algebra and a ∈ A+ then a ≤ kak1.
Solution. We have r(a) ≤ kak and a is hermitian, so σ(a) ⊆ [−kak, kak]. Hence
σ(kak1 − a) = kak − σ(a) ⊆ kak − [−kak, kak] = [0, 2kak] ⊆ R+ . Since kak1 − a is
hermitian, this shows that it is positive. Hence kak1 − a ≥ 0, so a ≤ kak1.
8. Let A be a unital C*-algebra. Show that S(A) is a convex subset of A∗ .
Solution. We have
S(A) = {τ ∈ A∗ : kτ k = τ (1A ) = 1} = {τ ∈ A∗ : kτ k ≤ 1, τ (1A ) = 1}.
Let τ1 , τ2 ∈ S(A) and let t ∈ [0, 1], and let τ = tτ1 + (1 − t)τ2 . Since A∗ is a vector space
we have τ ∈ A∗ . Moreover, τ (1A ) = tτ1 (1A ) + (1 − t)τ2 (1A ) = t + 1 − t = 1, and by the
triangle inequality, kτ k ≤ tkτ1 k + (1 − t)kτ2 k = 1. So τ ∈ S(A). Hence S(A) is convex.
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9. Let A be a unital abelian Banach algebra. Let a be a normal element of A and let
f ∈ C(σ(a)) and g ∈ C(σ(f (a))).
(a) Check that f (a), (g ◦ f )(a) and g(f (a)) are well-defined elements of A. Explain why
it is not obvious that (g ◦ f )(a) = g(f (a)).
(b) Show that (g ◦ f )(a) = g(f (a)) if g is of the following special forms:
(i). g(λ) = λk for some integer k ≥ 0;
(ii). g(λ) = λk λj for some integers k, j ≥ 0;
(iii). g is a trigonometric polynomial, meaning that
g(λ) =
n
X
αk,j λk λj ,
λ ∈ σ(f (a))
k,j=0
for some αk,j ∈ C.
(c) Use the Stone-Weierstrass theorem to deduce that (g ◦ f )(a) = g(f (a)) for any
g ∈ C(σ(f (a))). [Hint: approximate g.]
Solution. [Whoops, I didn’t actually want A to be abelian in this question. So I won’t
assume that below].
(a) Since f ∈ C(σ(a)) and a is normal, f (a) = θa (f ) is well-defined by the continuous
functional calculus.
The domain of g is σ(f (a)) = f (σ(a)) by the spectral mapping theorem. So the
domain of g is the range of f , and g ◦ f is a well-defined function. It is the composition
of two continuous functions, so g ◦ f is continuous; so g ◦ f ∈ C(σ(a)). Since a is normal,
(g ◦ f )(a) = θa (g ◦ f ) is well-defined by the continuous functional calculus.
The element f (a) is in the abelian C*-algebra C ∗ (1, a), and any element of an abelian
C*-algebra is normal (since its adjoint is in the algebra, so commutes with it). Hence f (a)
is normal, and g ∈ C(σ(f (a))). So g(f (a)) = θf (a) (g) is well-defined by the continuous
functional calculus.
The way in which (g ◦ f )(a) = θa (g ◦ f ) and g(f (a)) = θf (a) (g) are defined, using the
functional calculus based at two different elements (a and f (a)) of A makes it unclear
why they might be equal.
(b) (i) If g(λ) = λk then g ◦ f = f k (the pointwise kth power of f ) so, using the
homomorphism property of the functional calculus (and the fact that it is unital, for
k = 0) gives
(g ◦ f )(a) = (f k )(a) = f (a)k = g(f (a)).
(ii) We have g = g1 g2 = g1 g2∗ where g1 , g2 are of the form considered in (i). So
g ◦ f = (g1 g2 ∗ ) ◦ f = (g1 ◦ f )(g2∗ ◦ f ) = (g1 ◦ f )(g2 ◦ f )∗ . Since the functional calculus is a
∗-homomorphism and using (i) we have
(g ◦ f )(a) = (g1 ◦ f )(a) (g2 ◦ f )(a)∗ = g1 (f (a)) g2 (f (a))∗
= g1 (f (a)) (g2∗ )(f (a)) = (g1 g2∗ )(f (a)) = g(f (a)).
5
(iii) This is immediate from (ii), since the functional calculi at a and at f (a) are
both linear.
(c) By the Stone-Weierstrass theorem, the set of trigonometric polynomials is dense
in C(σ(f (a)); indeed, it is a unital ∗-subalgebra containing the function λ 7→ λ, so it
separates the points of σ(f (a)).
Let g ∈ C(σ(f (a))). If ε > 0 then, by the previous paragraph, there is a trigonometric
polynomial h ∈ C(σ(f (a)) with kg − hk < ε. Then kg ◦ f − h ◦ f k = k(g − h) ◦ f k < ε in
the norm of C(σ(a)) [Why?]. By (c) we have h(f (a)) = (h ◦ f )(a). Now the continuous
functional calculus is an isometry, so
kg(f (a)) − h(f (a))k = k(g − h)(f (a))k = kθf (a) (g − h)k = kg − hk < ε
and
k(g ◦ f )(a) − (h ◦ f )(a)k = k((g − h) ◦ f )(a)k = kθa ((g − h) ◦ f )k = k(g − h) ◦ f k < ε.
By the triangle inequality, kg(f (a)) − (g ◦ f )(a)k < 2ε and ε was an arbitrary positive
number. Hence g(f (a)) = (g ◦ f )(a).
10. Let I be an index set, and for each i ∈ I, let Hi be a Hilbert space and let Ti ∈ B(Hi )
with supi∈I kTi k < ∞. Show that the operator
M M
T =
Ti ∈ B
Hi
i∈I
i∈I
satisfies kT k = supi∈I kTi k.
[All norms in the statement of this question are operator norms.]
L
Solution. Let H = i∈I Hi (the Hilbertian direct sum). Let M = supi∈I kTi k. Then
M = sup{kTi hi k : i ∈ I, hi ∈ Hi , khi k ≤ 1}
so if kT k < M then there is i ∈ I and a vector hi ∈ Hi with khi k ≤ 1 such that
kT k < kTi hi k. Let x = (xj )j∈I be given by xj = 0 if j 6= i, and xi = hi . Clearly, x ∈ H
and kxk = khi k ≤ 1, and if y = T x then yj = 0 if j 6= i and yi = Ti hi . Hence
X
kTi hi k2 > kT k2 ≥ kT xk2 = kyk2 =
kyj k2 = kTi hi k2 ,
j∈I
which is a contradiction. So kT k ≥ M .
Conversely, let h ∈ H with khk ≤ 1. We have
X
X
X
kT hk2 =
kTi hi k2 ≤
kTi k2 khi k2 ≤ sup kTi k2
khi k2 = M 2 khk2 .
i∈I
i∈I
i∈I
So kT k ≤ M . By the previous paragraph, we have equality.
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i∈I
11. (a) Let H be a Hilbert space. A unit vector is a vector x ∈ H with kxk = 1. Given a
unit vector x, let ωx : B(H) → C, T 7→ hT x, xi. Show that ωx is a state on B(H)
for every unit vector x.
(b) Let A be a unital C*-algebra and let τ ∈ S(A), and consider the Hilbert space H
and the ∗-homomorphism π : A → B(H) constructed in Lemma 4.5.9. Find a unit
vector x ∈ H such that
τ (a) = hπ(a)x, xi for a ∈ A.
(c) The maps ωx considered in (a) are called vector states. If A is a unital C*-algebra,
consider the Hilbert space H and the isometric ∗-isomorphism π : A → B(H) constructed in the GNS Theorem 4.5.10. We call π the universal representation of A.
Show that “every state is a vector state in the universal representation”; or, more
precisely, show that
S(A) = {ωy ◦ π : y is a unit vector in H}.
[Warning: H and π are different in parts (b) and (c)!]
Solution. (a) Let ω = ωx where x is a unit vector. We have
ω(S + λT ) = h(S + λT )x, xi = hSx, xi + λhT x, xi = ω(S) + λω(T )
so ω is a linear functional; and by the Cauchy-Schwarz inequality,
|ω(T )| = |hT x, xi| ≤ kT xk kxk ≤ kT k
so ω ∈ B(H)∗ with kωk ≤ 1; and
ω(I) = hIx, xi = hx, xi = kxk2 = 1.
So ω ∈ B(H)∗ with kωk = ω(I) = 1, so ω is a state on B(H).
(b) Using the notation of the proof of Lemma 4.5.9, let x = [1A ]. Then x ∈ H and
kxk = h[1A ], [1A ]i = (1A , 1A ) = τ (1∗A 1A ) = τ (1A ) = 1. So x is a unit vector, and
2
hπ(a)x, xi = hπ(a)[1A ], [1A ]i = h[a1A ], [1A ]i = (a, 1A ) = τ (a).
(c) If y is a unit vector in H then τ = ωy ◦ π : A → C satisfies
kτ k = kωy ◦ πk ≤ kωy k kπk = 1
and
τ (1A ) = ωy (π(1A )) = ωy (I) = 1
so τ ∈ S(A).
Conversely, let τ ∈ S(A), and let Hτ and πτ be the corresponding Hilbert space and
∗-homomorphism fromLLemma 4.5.9. By (b), τ (a) = hπτ (a)x, xi for some unit vector
x ∈ Hτ . Let y ∈ H = σ∈S(A) Hσ be the vector with yσ = 0 if σ 6= τ and yτ = x. Then y
is a unit vector, and
X
(ωy ◦ π)(a) = hπ(a)y, yi =
hπσ (a)yσ , yσ i = hπτ (a)x, xi = τ (a).
σ∈S(A)
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