By Constantin Niculescu

ISBN-10: 0387243003

ISBN-13: 9780387243009

Thorough advent to an immense quarter of arithmetic includes fresh effects contains many workouts

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**Extra info for Convex Functions and their Applications: A Contemporary Approach**

**Example text**

10. 4 (The Hardy–Littlewood–P´ olya inequality) Suppose that f is a convex function on an interval I and consider two families x1 , . . , xn and y1 , . . , yn of points in I such that m m xk ≤ for m ∈ {1, . . , n} yk k=1 k=1 and n n xk = k=1 yk . k=1 If x1 ≥ · · · ≥ xn , then n n f (xk ) ≤ k=1 f (yk ), k=1 while if y1 ≤ · · · ≤ yn this inequality works in the reverse direction. Proof. We shall concentrate here on the ﬁrst conclusion (concerning the decreasing families), which will be settled by mathematical induction.

N} and let p > 1. Prove that the maximum of n n ak xpk n p bk ykp k=1 k=1 ck xk yk k=1 for xk ∈ [mk , Mk ] and yk ∈ [mk , Mk ] (k ∈ {1, . . , n}) is attained at a 2n-tuple whose components are endpoints. 3. Assume that f : I → R is strictly convex and continuous and g : I → R is continuous. For a1 , . . , an > 0 and mk , Mk ∈ I, with mk < Mk for k ∈ {1, . . , n}, consider the function n h(x1 , . . , xn ) = n ak f (xk ) + g k=1 n ak xk k=1 ak k=1 n deﬁned on k=1 [mk , Mk ]. Prove that a necessary condition for a point (y1 , .

9 Suppose that I is an open interval. A real-valued function f 2 is convex on I if and only if f is continuous and D f ≥ 0. Accordingly, if a function f : I → R is convex in the neighborhood of each point of I, then it is convex on the whole interval I. 2 Proof. If f is convex, then clearly D f ≥ D2 f ≥ 0. 3. 2 Now, suppose that D f > 0 on I. If f is not convex, then we can ﬁnd a 2 point x0 such that D f (x0 ) ≤ 0, which will be a contradiction. In fact, in this case there exists a subinterval I0 = [a0 , b0 ] such that f ((a0 +b0 )/2) > (f (a0 )+ f (b0 ))/2.

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