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) ∞.
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