4 MART ABEL
means that both ρ(a),ρ(b) ∞. The rest of the proof coincides with the proof of
Theorem 2.2 (i) in [2, pp. 157–158]. No topology is needed in this proof.
The next proposition uses the ideas of the proof of Theorem 2.5 part (ii) from
[2, pp. 160–161]. We would like to point out that in case of an element a of a
Banach algebra A, it is known that ρ(a) ∞ for every a ∈ A. Hence, if A is a
Banach algebra and a, b ∈ A, it is always true that σ(a) ∪ σ(b) = C.
Proposition 3.2. Let A be a unital algebra and a, b ∈ A such that
σ(a) ∪ σ(b) = C and σ(a + x) = σ(b + x) for every x ∈ A with ρ(x) ∞. Take any
λ ∈ C \ (σ(a) ∪ σ(b)). Then σ((a −
λeA)−1x)
= σ((b −
λeA)−1x)
for every x ∈ A
with ρ(x) ∞.
Proof. Fix any λ ∈ σ(a) ∪ σ(b) and α ∈ C\{0}. Take any x ∈ A with
ρ(x) ∞ and define y := α−1x. Then ρ(y) = |α|−1ρ(x) ∞. Notice, that
λeA − (a + y) ∈ Inv(A)
is equivalent to the fact that λ ∈ σ(a + y) = σ(b + y). The last is equivalent to the
fact that
λeA − (b + y) ∈ Inv(A).
Hence, we get that
λeA − (a + y) ∈ Inv(A) if and only if λeA − (b + y) ∈ Inv(A).
Since λ ∈ σ(a) ∪ σ(b), then λeA − a, λeA − b ∈ Inv(A) and there exist
(λeA −
a)−1,
(λeA −
b)−1
∈ A. Therefore, the last ”if and only if condition” is
equivalent to the fact that
(λeA − a)(eA − (λeA −
a)−1y)
∈ Inv(A)
if and only if
(λeA − b)(eA − (λeA −
b)−1y)
∈ Inv(A).
From this we see that
α ∈ σ((λeA −
a)−1x)
if and only if α ∈ σ((λeA −
b)−1x).
It is also clear that 0 ∈ σ((λeA − a)−1x) if and only if 0 ∈ σ((λeA − b)−1x). Hence,
we get that σ((λeA −a)−1x) = σ((λeA −b)−1x) for every x ∈ A with ρ(x) ∞.
Now, using the last corollary, we are able to prove Theorem 2.2 part (ii) in a
different way than the one provided in [2, pp. 157–158].
Corollary 3.3. Let A be a unital semisimple algebra and a, b ∈ A such that
ρ(a) ∞ or ρ(b) ∞. Then a = b if and only if σ(a + x) = σ(b + x) for every
x ∈ A with ρ(x) ∞.
Proof. It is clear that from a = b follows that σ(a + x) = σ(b + x) for every
x ∈ A with ρ(x) ∞.
To show the other implication, suppose that ρ(a) ∞. Notice, that
σ(θA) = {0}. Hence, ρ(θA) = 0 ∞. Taking x = θA in the condition of Corollary
3.3, we obtain that σ(a) = σ(b). Hence, also ρ(b) ∞. In the same way we get
from ρ(b) ∞ that ρ(a) ∞. Therefore, σ(a) ∪ σ(b) = C.
Fix any λ ∈ σ(a) ∪ σ(b) Then, by Proposition 3.2, we know that the equality
σ((λeA −
a)−1x)
= σ((λeA −
b)−1x)
holds for all x ∈ A with ρ(x) ∞.
