Exercise 2, § 1

Let g be a continuous increasing function on ]0, +∞[.

a) Show that if $\liminf_{x \to +\infty} g(x)/x = 0$ there exists at most one convex solution of the equation $u(x + 1) - u(x) = g(x)$ taking a given value at a given point (remark that for all $h > 0$ the function $v(x) = u(x + h) - u(x)$ is an increasing function satisfying the equation $v(x + 1) - v(x) = g(x + h) - g(x)$, and apply exerc. 1 a)).

b) Show that if $g$ is concave on ]0, $+\infty[$ there exists a convex solution of the equation $u(x + 1) - u(x) = g(x)$; there exists a unique convex solution of this equation taking a given value at a given point if and only if $\lim_{x \to +\infty} g(x)/x = 0$ (cf. exerc. 1 b)).

c) Suppose from here on that $g$ is increasing and concave on ]0, $+\infty[$, and that $\lim_{x \to +\infty} g(x)/x = 0$. For every pair of numbers $h > 0, k > 0$ and every finite real function $f$ defined on ]0, $+\infty[$ one puts

for $x > k$. Show that if $u$ is a convex solution of the equation $u(x + 1) - u(x) = g(x)$, one has $\lim_{x \to +\infty} \Delta(u(x); h, k) = 0$ for all $h > 0$ and $k > 0$ (use the expression for $u'_r$ from exerc. 1 b) of VII, p. 325).

d) With the notation of c) show that there exists a constant $\alpha$ such that one has $u(x) = v(x) + \alpha$, with

where we have put further, one has (Remark that $v(x + 1) - v(x) = g(x)$ and that $\lim_{x \to +\infty} \Delta(v(x); h, k) = 0$ irrespective of $h > 0$ and $k > 0$.)