127 Let us now prove the converse. Suppose (i) and (ii) are satisfed. Fix any ε > 0 and let δ > 0 satisfy (i). Then g(δ )= sup f (x) ≤ ℓ + ε. x∈B0(x0;δ )∩D This implies limsup f (x)= inf g(δ ) ≤ ℓ + ε. δ >0 x→x0 Since ε is arbitrary, we get limsup f (x) ≤ ℓ. x→x0 Again, take any ε > 0. Given δ > 0, let xˆ be as in (ii). Then ℓ − ε < f (xˆ) ≤ sup f (x)= g(δ ). x∈B0(x0;δ )∩D This implies ℓ − ε ≤ inf g(δ )= limsup f (x). δ >0 x→x0 It follows that ℓ ≤ limsupx →x0 f (x). Therefore, ℓ = limsupx →x0 f (x). □ Corollary 5.3.2 Suppose ℓ = limsupx →x0 f (x), where ℓ is a real number. Then there exists a sequence {xk} in D such that {xk} converges to x0, xk ̸ = x0 for every k, and lim f (xk)= ℓ. k→∞ Moreover, if {yk} is a sequence in D that converges to x0, yk ̸ = x0 for every k, and limk →∞ f (yk)= ℓ ′ , then ℓ ′ ≤ ℓ. 1 Proof: For each k ∈ N, let ε k = k . By Condition (i) of Theorem 5.3.1, there exists δk > 0 such that f (x) <ℓ + εk for all x ∈ B0(x0;δk) ∩ D. (5.3) ′ 1 ′ ′ Let δ k = min{δk, k }. Then δk ≤ δk and limk →∞ δk = 0. From Condition (ii) of Theorem 5.3.1, there ′ exists xk ∈ B0(x0;δk) ∩ D such that ℓ − εk < f (xk). Moreover, f (xk) <ℓ + εk by (5.3). Therefore, {xk} is a sequence that satisfes the conclusion of the corollary. Now let {yk} be a sequence in D that converges to x0, yk ̸ = x0 for every k, and limk →∞ f (yk)= ℓ ′ . For any ε > 0, let δ > 0 be as in Condition (i) of Theorem 5.3.1. Since yk ∈ B0(x0;δ ) ∩ D when k is suffciently large, we have f (yk) <ℓ + ε for such k. This implies ℓ ′ ≤ ℓ + ε. It follows that ℓ ′ ≤ ℓ. □
RkJQdWJsaXNoZXIy NTc4NTAz