Page 96 - Textos de Matemática Vol. 44

86 M. KARAMANLIS AND P. J. PSARRAKOS
As a consequence,
∥χ−μψ+λψ∥ ≥ 1−ε21∥ψ∥|λ| > 1−ε2∥ψ∥|λ|,
Thus, for every complex number λ ̸= 0,
2 2 2
∥χ−μψ+λψ∥− 1−ε2∥ψ∥|λ| ≥ 1−ε1 − 1−ε2
∀λ∈C.
∥ψ∥|λ| > 0.
Since χ is not a scalar multiple of ψ, it follows that ∥χ − μψ + λψ∥ > 0, and
hence, the continuous function f (λ) = ∥χ − μψ + λψ∥ − 1 − ε2 ∥ψ∥ |λ| takes
only positive values in the disk D(0, 1). Thus, inf f (λ) = min f (λ) > 0. λ∈D(0,1) λ∈D(0,1)
This means that we can consider a real δ > 0 such that
2 2
δ≤min min f(λ), 1−ε1− 1−ε2 ∥ψ∥ . λ∈D(0,1)
For every λ ∈ C with |λ| > 1, we have
2 2 2
∥χ−μψ+λψ∥− 1−ε2∥ψ∥|λ| ≥ 1−ε1 − 1−ε2 ∥ψ∥ ≥ δ, and thus,
2 δ ≤ inf ∥χ−μψ+λψ∥− 1−ε2∥ψ∥|λ| .
λ∈C
As a consequence, for every ξ ∈ D 0, δ ,
∥ψ∥
∥χ−(μ+ξ)ψ+λψ∥ ≥ ∥χ−μψ+λψ∥−∥ξψ∥ ≥ 1−ε2∥ψ∥|λ|, ∀λ∈C.
Hence, D 0, δ ⊆ Fε2 (χ;ψ), and the proof is complete. ∥ψ∥ ∥·∥
Corollary 3.2. Let χ, ψ ∈ X with ψ ̸= 0, and suppose that χ is not a scalar multiple of ψ. Then, for any ε ∈ (0, 1), F ε (χ; ψ) has a non-empty interior.
Proposition3.3. Letχ,ψ∈X withψ̸=0.Then,χ=aψforsomea∈Cif and only if Fε (χ;ψ) = {a} for every ε ∈ [0,1).
Proof. If χ = aψ for some a ∈ C, then (2.2) yields Fε (χ;ψ) = Fε (aψ;ψ)
∥·∥
∥·∥
μ∈C: ∥(a−λ)ψ∥≥ 1−ε2∥ψ∥|μ−λ|, ∀λ∈C μ∈C: |a−λ|≥ 1−ε2|μ−λ|, ∀λ∈C .
∥·∥
= =
∥·∥
It is apparent that a ∈ Fε (aχ;ψ). Furthermore, for any μ ̸= a and λ = a, we √ ∥·∥
have0=|a−λ|< 1−ε2|μ−λ|,i.e.,μ∈/Fε (aψ;ψ). ∥·∥