Mathematics 442C
Suggested solutions to exercise sheet 1
1. Suppose that an → a and bn → b. Let M = 1 + kak + supn≥1 kbn k. Since (bn )
converges, we have 0 < M < ∞, and kak < M . Let ε > 0. There is n0 ∈ N
ε
ε
such that kan − ak < 2M
and kbn − bk < 2M
for all n ≥ n0 . Hence if n ≥ n0 ,
we have
kan bn − abk = k(an − a)bn + a(bn − b)k
≤ k(an − a)bn k + ka(bn − b)k by the triangle inequality
≤ kan − ak kbn k + kak kbn − bk since A is a Banach algebra
ε
ε
· M + 2M
· M = ε.
< 2M
So an bn → ab as n → ∞.
2. (a) Recall that
A(D) = {f ∈ C(D) : f is analytic on D}.
We claim that A(D) is a unital Banach subalgebra of C(D), hence a
unital Banach algebra. Clearly, A(D) ⊆ C(D), and the unit of C(D)
(the constant function 1) is in A(D). Moreover, if f, g ∈ A(D), then
• for λ ∈ C we have λf ∈ C(D) and (λf )|D = λ · f |D is analytic, so
λf ∈ A(D);
• f + g ∈ C(D) and (f + g)|D = f |D + g|D is the sum of two analytic
functions, so is analytic, hence f + g ∈ A(D);
• f g ∈ C(D) and (f g)|D = (f |D )(g|D ) is the product of two analytic
functions, so is analytic, hence f g ∈ A(D);
• if hn ∈ A(D) and hn → h ∈ C(D) then hn |T is a sequence of analytic functions which converges uniformly to h|T , which is therefore
analytic, so h ∈ A(D).
Hence A(D) is a closed unital subalgebra of C(D), i.e. a unital Banach
subalgebra of C(D).
(b) Recall that
C0 (R) = {f : R → C : f is continuous and lim f (x) = 0}
x→±∞
with the uniform norm. Let BC(R) denote the Banach algebra of
bounded continuous functions R → C. We claim that C0 (R) is a Banach
subalgebra of BC (R). If f ∈ C0 (R) then f is bounded since f (x) → 0
as x → ±∞. So C0 (R) ⊆ BC(R). If fn ∈ C0 (R) and fn → f then
f ∈ BC (R). If ε > 0, then there is n ∈ N such that kf − fn k < 2ε . Since
fn (x) → 0 as x → ±∞, there is R > 0 such that |fn (x)| < 2ε whenever
|x| > R. Hence, for |x| > R we have
|f (x)| = |f (x) − fn (x) + fn (x)| ≤ |f (x) − fn (x)| + |fn (x)| <
This shows that f (x) → 0 as x → ±∞, so f ∈ C0 (R).
ε
2
+
ε
2
= ε.
Using basic properties of limits, it is now easy to check that C0 (R) is a
subalgebra of BC(R), and we have just shown that it is closed. Hence
it is a Banach subalgebra of BC(R).
Suppose that C0 (R) has an identity element e such that ef = f e = f
for all f ∈ C0 (R). For every x ∈ R, there is some function fx ∈ C0 (R)
with fx (x) = 1 (for example, fx (t) = (1 + |x − t|)−1 will do). Thus
e(x) = e(x)fx (x) = (efx )(x) = fx (x) = 1.
So e(x) = 1 for every x ∈ R. But then e(x) →
6
0 as x → ∞, so
e 6∈ C0 (R), which is a contradiction. Thus C0 (R) is non-unital.
(c) Since dim X > 2, there are two linearly independent vectors x, y ∈ X.
The subspace M = Cy is closed and x 6∈ M , so there is α ∈ X ∗ with
α(x) = 1 and α(y) = 0, by [FA 3.12]. Similarly, there is β ∈ X ∗ with
β(x) = 0 and β(y) = 1.
Let S, T : X → X be defined by
S(z) = α(z)y
and T (z) = β(z)x for z ∈ X.
Observe that S(x) = y and S(y) = 0; and T (y) = x and T (x) = 0.
Since α and β are linear, it is easy to check that S and T are linear. If
z ∈ X and kzk ≤ 1 then kS(z)k = kα(z)yk = |α(z)| kyk ≤ kαk kyk, so
S is bounded; so S ∈ B(X). Similarly, T ∈ B(X). Now
ST x = S(0) = 0 and T Sx = T y = x.
So ST x 6= T Sx, so ST 6= T S. Hence B(X) is not abelian.
Remark: if we restrict S and T to span{x, y} and represent these linear
maps as matrices
with respect to the basis (x, y), then we obtain 01 00
and 00 10 . Our strategy above was to model these non-commuting
matrices with bounded linear maps X → X.
3. (a) If a and b are identity elements for an algebra A then a = ab = b.
(b) We have La (b + c) = a(b + c) = ab + ac = La b + La c and La (λb) =
a(λb) = λLa b. So La is linear. Moreover, kLa bk = kabk ≤ kak kbk for
all b ∈ A, so La is bounded (with kLa k ≤ kak).
(c) Let a ∈ A. Clearly, |a| ≥ 0, and if |a| = 0 then La = 0 so ab = 0 for all
b ∈ A. Hence a = a1 = 0. Moreover, if λ ∈ C then
|λa| = kLλa k = sup kλabk = |λ| sup kabk = |λ| |a|,
kbk≤1
kbk≤1
and if b ∈ A then
|a + b| = kLa+b k = kLa + Lb k ≤ kLa k + kLb k = |a| + |b|.
Hence | · | is a norm.
As remarked in (b), |a| = kLa k ≤ kak for all a ∈ A; and if m = k1k−1
then km1k = 1 and so
|a| = kLa k ≥ kLa (m1)k = mkak.
Hence | · | is equivalent to k · k. Finally, L1 is the identity operator on A,
so |1| = kL1 k = 1.
2
4. (a) If 1 is the unit of A then 1 ∈ Inv A and 1a = a1 = a for all a ∈ Inv A.
So A has an identity element.
Multiplication is associative on A, so it is associative on Inv A.
If a ∈ Inv A then there is b ∈ A with ab = ba = 1. Thus b ∈ Inv A, and
b is the inverse element for A.
If a, b ∈ Inv A then there are c, d ∈ Inv A with ac = ca = 1 = bd = db.
Then dc ∈ A and abdc = 1 = dcab. So ab ∈ Inv A.
Hence Inv A is a group.
(b) Suppose that ba = 1 and ac = 1. By associativity, we have b = b(ac) =
(ba)c = c, so b = c. Hence ab = ba = 1, so a is invertible with inverse b.
(c) Suppose that a = bc = cb.
If b, c are invertible, then so is a (because Inv A is a group). Conversely,
suppose that a is invertible. Then ad = 1 and da = 1 for some d ∈ A,
so bcd = cbd = 1 and dbc = 1. So c is left invertible and right invertible;
so c is invertible by (b). Hence b = ac−1 is the product of two invertible
elements, so b is invertible.
(d) Apply (c) n − 1 times.
(e) First, let’s work out what S ∗ is. We have
hS ∗ en , em i = hen , Sem i = hen , em+1 i = δn,m+1 .
Observe that δ1,m+1 = 0 for all m ≥ 1, and if n > 1 then δn,m+1 6=
0 ⇐⇒ m = n − 1, so S ∗ en = en−1 if n > 1.
Hence S ∗ Sen = S ∗ en+1 = en , so S ∗ S = I, so S ∗ S is invertible.
Since ker S ∗ 6= {0} (it contains e1 6= 0), S ∗ is not injective, so is not
invertible. Moreover, for all x ∈ H, we have hSx, e1 i = hx, S ∗ e1 i = 0,
so Sx 6= e1 . So S is not surjective, so is not invertible.
5. Let f ∈ BC (X). If 0 ∈ f (X) then there are xn ∈ X with f (xn ) → 0. If
f ∈ Inv BC (X), there is g ∈ BC (X) with f g = 1. Hence f (xn )g(xn ) = 1, so
g(xn ) = f (xn ) → ∞ as n → ∞, so g is not bounded. This is a contradiction,
so f 6∈ Inv BC (X).
Conversely, if 0 6∈ f (X) then, in particular, f (x) 6= 0 for all x ∈ X.
Let C∗ = C \ {0}. Then we may view f as a continuous map X → C∗ , and
i : z 7→ z −1 , C∗ → C is continuous. Let g = i ◦ f , which is a composition
of continuous functions, so is continuous. If g is unbounded then there are
xn ∈ X with |g(xn )| → ∞, and then f (xn ) = g(xn )−1 → 0, so 0 ∈ f (X)
which is a contradiction. So g ∈ BC (X), and g(x) = f (x)−1 , so gf = f g = 1.
Thus f ∈ Inv BC (X) with inverse g.
6. If A is abelian, then clearly Inv A is commutative and so is an abelian group.
Conversely, suppose that Inv A is abelian. If a, b ∈ A with kak < 1 and
kbk < 1 then 1 − a and 1 − b are invertible by Theorem 1.2.7. Hence
1 − a − b − ab = (1 − a)(1 − b) = (1 − b)(1 − a) = 1 − a − b − ba,
1
1
so ab = ba. If c, d ∈ A are non-zero, then 2kck
c and 2kdk
d have norm 21 , so
they commute; hence c and d commute. Finally, 0c = 0 = c0 for any c ∈ A.
So A is abelian.
3