# Convex functions and their applications - download pdf or read online

By Niculescu C.P., Persson L.-E.

ISBN-10: 0387243003

ISBN-13: 9780387243009

Thorough advent to an immense quarter of arithmetic comprises fresh effects comprises many routines

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**Sample text**

N ∈ [0, 1], with n k=1 λk = 1. 5 is indicated in Exercise 6. 6 (The complete form of Chebyshev’s inequality) Let (X, Σ, µ) be a ﬁnite measure space, let g : X → R be a µ-integrable function and let ϕ be a nondecreasing function given on an interval that includes the image of g and such that ϕ ◦ g and g · (ϕ ◦ g) are integrable functions. Then for every primitive Φ of ϕ for which Φ ◦ g is integrable, the following inequalities hold true: 0 ≤ M1 (Φ ◦ g) − Φ(M1 (g)) ≤ M1 (g · (ϕ ◦ g)) − M1 (g)M1 (ϕ ◦ g).

2. Infer from the integral form of the arithmetic-geometric-harmonic mean inequality that L(a, b) < I(a, b) < A(a, b) for all a, b > 0, a = b. Here L(a, b), I(a, b), A(a, b) are the logarithmic, the identric and respectively the arithmetic mean of a and b. 3. 7 the discrete form of Hardy’s inequality: ∞ n=1 1 n n p 1/p ak k=1 < p p−1 ∞ apk 1/p , k=1 for every sequence (an )n of nonnegative numbers (not all zero) and every p ∈ (1, ∞). 4. (The P´ olya–Knopp inequality; see [99], [130]) Prove the following limiting case of Hardy’s inequality: for every f ∈ L1 (0, ∞), f ≥ 0 and f not identically zero, ∞ exp 0 1 x ∞ x log f (t) dt dx < e 0 f (x) dx.

Proof. The ﬁrst inequality is that of Jensen. The second can be obtained from f (M1 (g)) ≥ f (g(x)) + (M1 (g) − g(x)) · ϕ(g(x)) for all x ∈ X by integrating both sides over X. 5 Let f be a convex function deﬁned on an open interval I and let ϕ : I → R be a function such that ϕ(x) ∈ ∂f (x) for all x ∈ I. Then n 0≤ n λk f (xk ) − f k=1 n ≤ λk xk k=1 n λk xk ϕ(xk ) − k=1 n λk xk k=1 λk ϕ(xk ) k=1 for all x1 , . . , xn ∈ I and all λ1 , . . , λn ∈ [0, 1], with n k=1 λk = 1. 5 is indicated in Exercise 6.

### Convex functions and their applications by Niculescu C.P., Persson L.-E.

by William

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