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It rests on the following lemma. 3. Let C1 ⊇ C2 ⊇ ··· be a descending sequence of nonempty closed convex subsets of a compact metric space (M, d). Then ∞ Cn is nonempty and convex. n=1 Proof. The fact that the intersection is nonempty is immediate from compactness. Suppose x, y ∈ ∞ Cn with x = y. ) By compactness of M the sequence {zn } has a subsequence {znk } which converges to a point z ∈ M and since each of the sets Cn is closed, z ∈ ∞ Cn . Since n=1 the metric d is continuous, 1 d(x, z) = d(y, z) = d(x, y).

Let {xn } be a bounded sequence in C. Then there exists a unique point z ∈ C such that lim sup d 2 (xn , z) + 2cM d 2 (z, x) ≤ lim sup d 2 (xn , x), n→∞ (HEM) n→∞ for any x ∈ C. The next result is the nonlinear version of Theorem 3 of [201]. 30. [113] Let (M, d) be hyperbolic metric space which is 2-uniformly convex. Let C be a nonempty, closed, convex, and bounded subset of M. Let T : C → C be uniformly Lipschitzian with λ (T ) < 1+ 1 + 8cM N(M)2 2 1/2 . Then T has a fixed point in C. 9 More on Convexity Structures Kirk’s fixed point theorem involves some kind of compactness and the normal structure property.

The ρ -distance between f and B is defined as dρ ( f , B) = inf{ρ ( f − g) : g ∈ B}. Let us start with the following easy but important fact. 1. Assume that a sequence of fn ∈ Lρ ρ -converges to an f ∈ M , that is, ρ ( fn − f ) → 0. Then f ∈ Lρ . 1 Proof. Fix ε > 0 and take any 0 < λ < . Observe that 2 1 1 1 1 ε 1 ρ (λ f ) ≤ ρ (2λ ( f − fn )) + ρ (λ fn ) ≤ ρ ( f − fn ) + ρ (λ fn ) ≤ + ρ (λ fn ), 2 2 2 2 2 2 for n ≥ n0 . Since fn0 ∈ Lρ , it follows that there exists λ0 > 0 such that ρ (λ fn0 ) < ε for λ < λ0 .

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A p-adic Property of Fourier Coefficients of Modular Forms of Half Integral Weight by Guerzhoy P.

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