Bourbaki marks this one as harder than the rest.
Let f be an (H) function, convex on a neighbourhood of +∞, such that f(x)≫x. For every α>0 let xα be the point such that f′(xα)=(1+α)f′(x).
a) If f is of order +∞ relative to x show that
xα−x∼log(1+α)f′(x)f(x)
(use prop. 2 (V, p. 251) and 4 (V, p. 255), applying the Taylor formula to logf′(x)). Show that (f(xα)−(xα−x)f′(xα))/f(x) tends to (1+α)(1−log(1+α)) as x tends to +∞.
b) If f is of order r>1 relative to x show that
xα∼(1+α)1/(r−1)x (remark that if
f1 is of order 0 relative to
x one has
f1(kx)∼f1(x) for every constant
k, by prop. 4 of V, p. 255). Deduce that
(f(xα)−(xα−x)f′(xα))/f(x) tends to
r(1+α)−(r−1)(1+α)r/(r−1) as
x tends to
+∞.
c) Suppose finally that f is of order 1 relative to x. Putting f(x)=xf1(x) show that
xα−x∼log(1+α)f1′(x)f1(x) (consider the inverse function of
f1, which is of order
+∞ relative to
x). Deduce that
(f(xα)−(xα−x)f′(xα))/f(x) tends to
−∞ as
x tends to
+∞.
Similarly let xα′ be the point such that f′(xα′)=(1−α)f′(x) (for 0<α<1). State the formulae analogous to the previous ones that express the principal part of x−xα′ and the limit of
(f(xα′)+(x−xα′)f′(xα′))/f(x) as
x tends to
+∞.
Conclude from these results that f is a regularly convex function on a neighbourhood of +∞ (V, p. 260, exerc. 5).