By A.G. Buckley, J-.L. Goffin

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**Additional info for Algorithms for Constrained Minimization of Smooth Nonlinear Functions**

**Example text**

In general, for n = 2, 3, . . , we have T nx = 1 − (−α2 )1 1 − (−α2 )2 1 − (−α2 )3 x , x , xn+3 , . . n+1 n+2 1 − (−α2 )n+1 1 − (−α2 )n+2 1 − (−α2 )n+3 Let us observe that, for any x = (x1 , x2 , . . ) ∈ 1, we get α1 T x + α2 T 2 x ∞ = α1 i=1 ∞ 1 − (−α2 )i 1 − (−α2 )i |x | + α |xi+2 | i+1 2 1 − (−α2 )i+1 1 − (−α2 )i+2 i=1 ∞ = |x2 | + α1 i=1 ∞ = |x2 | + α1 i=1 ∞ = |x2 | + 1 − (−α2 )i+1 1 − (−α2 )i + α2 i+2 1 − (−α2 ) 1 − (−α2 )i+2 |xi+2 | i=1 ≤ x . ∞ 1 − (−α2 )i+1 1 − (−α2 )i |x | + α |xi+2 | i+2 2 1 − (−α2 )i+2 1 − (−α2 )i+2 i=1 |xi+2 | .

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. Consequently, for each 1 < p < ∞, the spaces Lp [0, 1] and p are reflexive.

Converging to zero with a standard sup norm, x = (x1 , x2 , . . ) = sup |xi | . i=1,2,... Let S : B → B be a mapping defined by Sx = S(x1 , x2 , . . ) = (1, x1 , x2 , . . ). It is easy to verify that S is an isometry, and so it is nonexpansive. However, the mapping S is fixed point free. Indeed, Sx = x implies that xi = 1 for i = 1, 2, . . But x = (1, 1, . . , 1, . . ) ∈ / c0 . Thus, Fix(S) = ∅. p. for short) if each nonexpansive mapping S : C → C has at least one fixed point. p. for short.