By Constantin Niculescu

Thorough advent to an immense zone of arithmetic includes fresh effects comprises many workouts

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N − 2 (in the unknowns ak ) to prove that the general form of a convex sequence a = (a0 , a1 , . . , an ) with a0 = an = 0 is given by the formula n−1 cj wj , a= j=1 where cj = 2aj − aj−1 − aj+1 and wj has the components wkj = (ii) k(n − j)/n, j(n − k)/n, for k = 0, . . , j for k = j, . . , n. Prove that the general form of a convex sequence a = (a0 , a1 , . . , an ) n is a = j=0 cj wj , where cj and wj are as in the case (i) for j = 1, . . , n − 1. The other coeﬃcients and components are: c0 = a0 , cn = an , wk0 = (n − k)/n and wkn = k/n for k = 0, .

3, f+ (a) ≤ f+ (x) ≤ f (y) − f (x) ≤ f− (y) ≤ f− (b) y−x for all x, y ∈ [a, b] with x < y, hence f |[a,b] veriﬁes the Lipschitz condition with L = max{|f+ (a)|, |f− (b)|}. 8 If fn : I → R (n ∈ N) is a pointwise converging sequence of convex functions, then its limit f is also convex. Moreover, the convergence is uniform on any compact subinterval included in int I, and (fn )n converges to f except possibly at countably many points of I. 24 1 Convex Functions on Intervals Since the ﬁrst derivative of a convex function may not exist at a dense subset, a characterization of convexity in terms of second order derivatives is not possible unless we relax the concept of twice diﬀerentiability.

5 The Subdiﬀerential 29 Exercises 1. (Kantorovich’s inequality) Let m, M, a1 , . . , an be positive numbers, with m < M . Prove that the maximum of n f (x1 , . . , xn ) = n ak xk ak /xk k=1 k=1 for x1 , . . ,n} 2 ak − min k∈X ak . k∈ X Remark. The following particular case 1 n n xk k=1 1 n n k=1 1 xk (M + m)2 (1 + (−1)n+1 )(M − m)2 − 4M m 8M mn2 ≤ represents an improvement on Schweitzer’s inequality for odd n. 2. Let ak , bk , ck , mk , Mk , mk , Mk be positive numbers with mk < Mk and mk < Mk for k ∈ {1, .