Mathematics 442C
Suggested solutions to exercise sheet 2
1. (a) Observe that a(1 − ba) = (1 − ab)a. Thus
(1+b(1−ab)−1 a)(1−ba) = 1−ba+b(1−ab)−1 (1−ab)a = 1−ba+ba = 1.
Similarly, (1 − ba)b = b(1 − ab), so
(1 − ba)(1 − b(1 − ab)−1 a) = 1 − ba + b(1 − ab)(1 − ab)−1 a = 1.
(b) Let λ ∈ σ(ab) \ {0}. Then λ − ab = λ(1 − λ−1 ab) 6∈ Inv A, so 1 − λ−1 ab 6∈
Inv A. By (a), 1 − λ−1 ba 6∈ Inv A, so λ − ba = λ(1 − λ−1 ba) 6∈ Inv A. So
λ ∈ σ(ba) \ {0}. Hence σ(ab) \ {0} ⊆ σ(ba) \ {0}. The reverse inclusion
follows by symmetry.
2. Since a and a−1 are invertible, 0 is not in σ(a) or σ(a−1 ). If λ ∈ C with λ 6= 0
then λ−1 − a−1 = λ−1 (a − λ)a−1 by (⋆). Since λ−1 (a − λ) and a−1 commute,
by Remark 1.2.5(iii) we have
λ ∈ σ(a) ⇐⇒ a − λ 6∈ Inv A ⇐⇒ λ−1 − a−1 6∈ Inv A ⇐⇒ λ−1 ∈ σ(a−1 ).
3. For |λ| < 1/r(a) we have r(λa) = |λ|r(a) < 1 by Theorem 1.3.8, so 1 − λa is
invertible and S(λ) is well-defined. For λ, µ in the domain of S, we have
S(µ) − S(λ)
1
=
(1 − µa)−1 − (1 − λa)−1
µ−λ
µ−λ
1
(1 − µa)−1 (1 − λa) − (1 − µa) (1 − λa)−1
=
µ−λ
= (1 − µa)−1 a(1 − λa)−1 .
by (⋆)
The map µ 7→ 1−µa is continuous, inversion is continuous Inv A → Inv A and
multiplication is continuous on A. Hence S(µ)−S(λ)
→ (1 − λa)−1 a(1 − λa)−1
µ−λ
as µ → λ. So S is holomorphic.
4. If a ∈ S ′ then ab = ba for all b ∈ S, so ab = ba for all b ∈ T , so a ∈ T ′ . Hence
S ′ ⊆ T ′.
If a ∈ S then ab = ba for all b ∈ S ′ . Hence a ∈ S ′′ . So S ⊆ S ′′ .
Since S ⊆ S ′′ we have S ′ ⊇ (S ′′ )′ = S ′′′ . Moreover, S ′ ⊆ (S ′ )′′ = S ′′′ . So
we have equality.
5. We know that A/I is a Banach algebra. Since (a + I)(1 + I) = a + I =
(1 + I)(a + I), the element 1 + I is an identity element for A/I, and
k1 + Ik = inf k1 − bk ≤ k1 − 0k = 1.
b∈I
If k1 + Ik < 1 then there is b ∈ I with k1 − bk < 1. By Theorem 1.2.7,
b = 1 − (1 − b) ∈ Inv A. So b ∈ I ∩ Inv A; since A is a proper ideal, this
contradicts Lemma 1.5.5. So k1 + Ik = 1, and 1 + I is the unit for A/I.
Here’s an even more elementary way to prove that k1 + Ik ≥ 1: since I
is a proper ideal, 1 6∈ I, so 1 + I 6= 0, hence k1 + Ik 6= 0. Moreover,
k1 + Ik = k(1 + I)2 k ≤ k1 + Ik2 ,
and cancelling k1 + Ik gives k1 + Ik ≥ 1.
6. If θ : M2 (C) → C is a homomorphism then θ(ab − ba) = θ(ab) − θ(ba) =
θ(a)θ(b) − θ(b)θ(a) = 0 for all a, b ∈ M2 (C).
Let for i, j = 1, 2, let eij ∈ M2 (C) be the matrix with a 1 in the (i, j)
position and 0s elsewhere. Then eij ekℓ = δjk eiℓ , so
θ(e12 ) = θ(e12 e22 − e22 e12 ) = 0
Hence θ(e11 ) = θ(e12 e21 ) = θ(e12 )θ(e21 ) = 0 and θ(e22 ) = θ(e21 e12 ) =
θ(e21 )θ(e12 ) = 0, and θ(e21 ) = θ(e21 e11 ) = θ(e21 )θ(e11 ) = 0. Since M2 (C) =
span{e11 , e12 , e21 , e22 } and θ sends each of these generators to 0, we have
θ = 0.
7. (a) If fn ∈ K and fn → f ∈ C(D) then fn converges uniformly to f , so fn
converges pointwise to f . If z ∈ T then fn (z) = 0 for all n ≥ 1. Hence
f (z) = limn→∞ fn (z) = 0, so f ∈ K. Hence K is closed in C(D). If
h ∈ C(T) and f, g ∈ K and λ ∈ C then for z ∈ T we have (f + λg)(z) =
f (z) + λg(z) = 0 and (hf )(z) = (f h)(z) = f (z)h(z) = 0 · h(z) = 0, so
f + λg and hf = f h are in K. Hence K is an ideal in C(T).
(b) The mapping θ is well-defined, since if f +K = g +K then f −g ∈ K, so
(f −g)|T = 0, so f |T = g|T . It is easy to check that θ is linear. Moreover,
θ(f g) = (f g)|T = (f |T ) · (g|T ) = θ(f )θ(g), so θ is a homomorphism.
Let f ∈ C(D). If g ∈ K then
kf |T k = sup |f (z) − g(z)| ≤ sup |f (z) − g(z)| = kf − gk.
z∈T
z∈D
Hence kf |T k ≤ inf g∈K kf − gk = kf + Kk.
Let ε > 0. Since f is a continuous function on a compact set, it is
uniformly continuous. Hence there is δ > 0 such that |f (z) − f (w)| ≤ ε
whenever z, w ∈ D with |z − w| ≤ δ. Let s be a continuous function
[0, 1] → [0, 1] with s(1) = 0 and s|[0,1−δ] = 1, and let g(z) = s(|z|)f (z)
for z ∈ D. Observe that g is continuous, and if 1 − δ ≤ |z| ≤ 1 then
z
z
| ≤ δ, so |f (z) − f ( |z|
)| ≤ ε. Hence
|z − |z|
kf + Kk ≤ kf − gk = sup |f (z) − g(z)| =
=
sup (1 − s(|z|))|f (z)| ≤
1−δ≤|z|≤1
≤
sup
1−δ≤|z|≤1
|f (z) − g(z)|
sup
1−δ≤|z|≤1
z∈D
sup
|f (z)|
1−δ≤|z|≤1
z
z
)| + |f (z) − f ( |z|
)| ≤ kf |T k + ε.
|f ( |z|
Since ε > 0 was arbitrary, this shows that kf + Kk ≤ kf |T k = kθ(f )k.
The reverse inequality has already been established, so we have equality.
Hence θ is isometric. In particular, θ is injective.
If g ∈ C(T), consider
(
z
)
if z 6= 0,
|z|g( |z|
f : D → C, z 7→
0
if z = 0.
Then f ∈ C(D) and θ(f + K) = f |T = g. Hence θ is surjective.
Hence θ is an isometric isomorphism of Banach algebras.
2
8. (a) We have S ∗ S = I, so σ(S ∗ S) = {1}. Also, SS ∗ e1 = 0, so SS ∗ is
not invertible. Hence 0 ∈ σ(SS ∗ ). By Exercise 1, σ(SS ∗ ) \ {0} =
σ(S ∗ S) \ {0} = {1}, so σ(SS ∗ ) = {0, 1}.
(b) We have T (λI − S ∗ ) = I, so T = λ−1 (I + T S ∗ ). Thus T e1 = λ−1 e1 ,
and for n ≥ 2 we have T en = λ−1 (en + T en−1 ). Solving this recurrence
relation (by induction, if you like) gives
T en = λ−1 en + λ−2 en−1 + · · · + λ−n e1 .
(c) If x =
P
n≥1
xn en ∈ H then
X
X
kS ∗ xk2 = k
xn en−1 k2 =
|xn |2 ≤ kxk2 .
n≥2
n≥2
So kS ∗ xk ≤ kxk; hence kS ∗ k ≤ 1, and so σ(S ∗ ) ⊆ D.
If λ ∈ D and λI − S ∗ is invertible, then T (λI − S ∗ ) = I for some
T ∈ B(H). By (b), we have
kT en k2 = |λ|−2 + |λ|−4 + · · · + |λ|−2n ≥ n
so kT en k → ∞ as n → ∞. So T is not bounded, which is a contradiction. So λI − S ∗ is not invertible, so λ ∈ σ(S ∗ ).
Hence σ(S ∗ ) = D.
(d) If T ∈ Inv B(H) then T R = RT = I for some R ∈ B(H). Hence
I = I ∗ = (T R)∗ = (RT )∗ , so I = R∗ T ∗ = T ∗ R∗ . So T ∗ ∈ Inv B(H).
Now if T ∗ ∈ Inv B(H) then T = (T ∗ )∗ ∈ Inv B(H).
Hence λ ∈ σ(S) ⇐⇒ λI − S 6∈ Inv B(H) ⇐⇒ (λI − S)∗ 6∈
Inv B(H) ⇐⇒ λI − S ∗ 6∈ Inv B(H) ⇐⇒ λ ∈ D ⇐⇒ λ ∈ D.
So σ(S) = D.
P
xk
9. (a) We have ex = ∞
n=0 k! for every real number x, and pn (x) is the nth
partial sum of this series.
(b) We have
X (x + y)k
X 1 k X xℓ y m
pn (x + y) =
=
.
xr y k−r =
k!
k!
r
ℓ!
m!
0≤k≤n
0≤r≤k≤n
0≤ℓ,m≤n,
ℓ+m≤n
So
pn (x)pn (y) − pn (x + y) =
X xj y k
X xℓ y m
X xj y k
−
=
.
j!
k!
ℓ!
m!
j!k!
0≤j,k≤n
0≤ℓ,m≤n
1≤j,k≤n,
ℓ+m≤n
Hence the coefficient of xj y k is either 0 or
1
j! k!
j+k>n
> 0.
(c) We have pn (x) → ex , pn (y) → ey and pn (x + y) → ex+y . Hence
qn (x, y) → ex ey − ex+y = 0 as n → ∞.
P
ak
1
1
ak
k
k
(d) The series ∞
k=0 k! converges absolutely, since k k! k = k! ka k ≤ k! kak ,
P∞ kakk
P∞ ak
= ekak < ∞. Hence pn (a) converges to
so
k k! k ≤
k=0 k!
P∞ k=0
ak
k=0 k! as n → ∞.
3
P
(e) Since ab = ba and qn (x, y) = 0≤j,k≤n tjk xj y k for some tjk ≥ 0 by (b),
we have
X
tjk aj bk kpn (a)pn (b) − pn (a, b)k = 0≤j,k≤n
≤
X
tjk kaj bk k by the triangle inequality
0≤j,k≤n
≤
X
kakj kbkk
0≤j,k≤n
= qn (kak, kbk).
(f) We have
kea eb − ea+b k = lim kpn (a)pn (b) − pn (a + b)k ≤ lim qn (kak, kbk) = 0
n→∞
n→∞
by (e) and (c). Hence ea eb = ea+b .
(g) If a ∈ A then a commutes with −a. Hence e−a ea = ea e−a = ea−a =
e0 = 1, so e−a = (ea )−1 and ea is invertible. Since a product of invertible
elements is invertible, the result follows.
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