By D. Cox, E. Snell

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**Additional resources for Applied Sstatistics - Principles and Examples**

**Sample text**

Indeed, suppose X is uniformly convex with the modulus of convexity δX . Take any functional x∗ ∈ X ∗ . Without loss of generality, we can assume that x∗ ∈ SX ∗ . For any n ∈ N, consider the slice S x∗ , δX 1 n x ∈ BX : x∗ (x) ≥ 1 − δX = Then, for any two points x, y ∈ S x∗ , δX 1 − δX 1 n ≤ x∗ x+y 2 1 n x+y x+y = . 2 2 Since X is uniformly convex, we conclude that x − y ≤ diam S x∗ , δX n1 ≤ n1 . Then, the intersection n∈N . , we have ≤ x∗ S x∗ , δX 1 n 1 n, and, consequently, 1 n consists of exactly one point z, as it is a descending family of nonempty, closed sets with diameters converging to zero, and it must be the case that x∗ (z) = 1.

1 x−y . α1 In particular, k T 2 ≤ min 1 1 1 , , α12 α2 α1 = min 1 1 , α1 α2 . 2) Now we find an interesting part of the story. It occurs that the above estimate is not sharp! To see it, first observe that, for all x, y ∈ C, we have α1 T 2 x − T 2 y + α2 T 3 x − T 3 y ≤ T x − T y . Multiplying both sides of the above inequality by α1 and adding α2 T 2 x − T 2 y to both sides, we obtain α12 + α2 T 2 x − T 2 y + α1 α2 T 3 x − T 3 y ≤ α1 T x − T y + α2 T 2 x − T 2 y ≤ x−y . Consequently, T 2x − T 2y ≤ 1 x−y .

500000000] ∞ It is not easy to guess the shape of the sequence {bn }n=0 in general case of k > 0. Indeed, as we shall see later, for any T ∈ L((α1 , α2 ), k), we have k(T n ) ≤ bn , where bn = k √ 2n+1 ∆ n α1 + √ ∆ n+1 − α1 − √ ∆ n+1 , On the Lipschitz constants for iterates of mean lipschitzian mappings 47 with ∆ = α12 + 4α2 k. Despite being complicated and sophisticated, the above-mentioned boundary is sharp! Nevertheless, this is still only the case of multi-index α of length n = 2. It may suggest that, in general, we should find a recurrence formula ∞ for the sequence {bn }n=0 , provided it exists!