Bourbaki

Exercise 5, § 3

Functions of a Real Variable, V, 260

Bourbaki marks this one as harder than the rest.

Let ff be a convex real function defined on a neighbourhood of +∞+\infty such that f(x)≫xf(x) \gg x. One says that ff is regularly convex on a neighbourhood of +∞+\infty if, for every convex function gg defined on a neighbourhood of +∞+\infty and such that f∼gf \sim g one also has fd′∼gd′f'_d \sim g'_d.

For every number α>0\alpha > 0 and xx sufficiently large let k(α,x)k(\alpha, x) be the infimum of the numbers (f(y)−(y−x)fd′(y))/f(x)\left( f(y) - (y-x)f'_d(y) \right)/f(x) as yy runs through the set of numbers ≥x\geq x such that fd′(y)≤(1+α)fd′(x)f'_d(y) \leq (1+\alpha)f'_d(x). Similarly let h(α,x)h(\alpha, x) be the infimum of the numbers

(f(z)+(x−z)fd′(z))/f(x) \left( f(z) + (x-z)f'_d(z) \right)/f(x)
as zz runs through the set of numbers ≤x\leq x such that fd′(z)≥(1−α)fd′(x)f'_d(z) \geq (1-\alpha)f'_d(x). Let ψ(α)=lim sup⁡x→+∞k(α,x)\psi(\alpha) = \limsup_{x \to +\infty} k(\alpha, x), φ(α)=lim sup⁡x→+∞h(α,x)\varphi(\alpha) = \limsup_{x \to +\infty} h(\alpha, x). Show that for ff to be regularly convex on a neighbourhood of +∞+\infty it is necessary and sufficient that for all sufficiently small α\alpha one has ψ(α)<1\psi(\alpha) < 1 and φ(α)<1\varphi(\alpha) < 1.

Solution

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Transcribed from the printed volume. Not translated.