By Constantin Niculescu

Convex capabilities play a major function in just about all branches of arithmetic in addition to different parts of technology and engineering. This e-book is an intensive advent to modern convex functionality idea addressed to everybody whose learn or instructing pursuits intersect with the sector of convexity. It covers a wide number of matters, from the only genuine variable case (with all its mathematical gemstones) to a couple of the main complicated issues resembling Choquet's idea, the Prékopa-Leindler variety inequalities and their ramifications, in addition to the variational procedure of partial differential equations and convex programming. Many effects are new and the total publication displays the authors’ personal event, either in educating and learn. The publication can function a reference and resource of suggestion to researchers in different branches of arithmetic and engineering and it could even be used for graduate courses.

**Read Online or Download Convex Functions and their Applications: A Contemporary Approach (CMS Books in Mathematics) PDF**

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N ≥ 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. Prove that b eif (t) dt ≤ 4 2/λ for all a, b ∈ R. a [Hint: Use integration by parts on intervals around the point where f vanishes. 4 An Upper Estimate of Jensen’s Inequality An important topic related to inequalities is their precision.

I) The inequality is immediate. 5, Exercise 1 (ii)). Since the function h(x) = f (x)−xy is convex and −f ∗ (y) = − sup{xy − f (x) | x ∈ I} = inf{f (x) − xy | x ∈ I}, the equality −f ∗ (y) = f (x) − xy holds true if and only if 0 ∈ ∂h(x), that is, if y ∈ ∂f (x). (ii) Letting x ∈ I, we have f ∗ (y) − xy ≥ −f (x) for all y ∈ I ∗ . Thus ∗ f (y) − xy = −f (x) occurs at a point of minimum, which by the above discussion means y ∈ ∂f (x). In other words, y ∈ ∂f (x) implies that the function h(z) = f ∗ (z) − xz has a (global) minimum at z = y.

Compute ∂f , ∂f ∗ and f ∗ for f (x) = |x|, x ∈ R. 3. Prove that: (i) the conjugate of f (x) = |x|p /p, x ∈ R, is f ∗ (y) = |y|q /q, (ii) y ∈ R (p > 1, 1 1 + = 1); p q the conjugate of f (x) = (1 + x2 )1/2 , x ∈ R, is the function f ∗ (y) = −(1 − y 2 )1/2 , y ∈ [−1, 1]; (iii) the conjugate of f (x) = ex , x ∈ R, is the function f ∗ (y) = y log y − y for y > 0 and f ∗ (0) = 0; (iv) the conjugate of f (x) = − log x, x > 0, is the function f ∗ (y) = −1 − log(−y), y < 0. 4. (A minimization problem) Let f : R → R be a convex function such that lim|x|→∞ f (x) = ∞.