By Constantin Niculescu
Thorough advent to a huge region of mathematics
Contains contemporary results
Includes many workouts
Read Online or Download Convex Functions and Their Applications. A Contemporary Approach PDF
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Additional resources for Convex Functions and Their Applications. A Contemporary Approach
6 (The Jensen–Steﬀensen inequality) Let xn ≤ xn−1 ≤ · · · ≤ x1 be points in [a, b] and let p1 , . . , pn be real numbers such that the k partial sums Sk = i=1 pi verify the relations 0 ≤ Sk ≤ S n and Sn > 0. Then every convex function f deﬁned on [a, b] veriﬁes the inequality f Proof. Put x ¯=( n k=1 1 Sn n pk xk ≤ k=1 1 Sn n pk f (xk ). k=1 pk xk )/Sn and let S¯k = Sn − Sk−1 = n Sn (x1 − x ¯) = n (xj−1 − xj )S¯j ≥ 0 pi (x1 − xi ) = i=1 and j=2 n−1 Sn (¯ x − xn ) = n−1 pi (xi − xn ) = i=1 n i=k (xj − xj+1 )Sj ≥ 0, j=1 pi .
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 , . . , yn ) to be a point of maximum is that at most one component yk is inside the corresponding interval [mk , Mk ]. 5 The Subdiﬀerential In the case of nonsmooth convex functions, the lack of tangent lines can be supplied by support lines. See Fig. 3. Given a function f : I → R, we say that f admits a support line at x ∈ I if there exists a λ ∈ R such that 30 1 Convex Functions on Intervals f (y) ≥ f (x) + λ(y − x), for all y ∈ I.
Let f : [0, 2π] → R be a convex function. Prove that an = 2π 1 π f (t) cos nt dt ≥ 0 for every n ≥ 1. 0 8. (J. L. W. V. Jensen ) Prove that a function f : [0, M ] → R is nondecreasing if and only if n n λk f (xk ) ≤ k=1 n λk f k=1 xk k=1 for all ﬁnite families λ1 , . . , λn ≥ 0 and x1 , . . , xn ∈ [0, M ], with n k=1 xk ≤ M and n ≥ 2. This applies to any continuous convex function g : [0, M ] → R, noticing that [g(x) − g(0)]/x is nondecreasing. 9. (van der Corput’s lemma) Let λ > 0 and let f : R → R be a function of class C 2 such that f ≥ λ.