Proposition 1

There exists one and only one convex function $g$ defined on ]0, +∞[ that satisfies the equation (3) and takes the value 0 for $x = 1$.

First we show that if there is a function $g$ satisfying the conditions stated then it is well-determined on the interval ]0, 1], and consequently on the interval ]0, +∞[. Indeed, for every integer $n > 1$ the gradient of the line joining the point $(n, g(n))$ to the point $(x, g(x))$ is an increasing function of $x$, since $g$ is convex (I, p. 27, prop. 5); one thus must have, for $0 < x \leq 1$,

that is, by (3),

Now, by (3),

Moreover, one can write $\log n = \sum_{k=2}^n \log \frac{k}{k-1}$ so the inequality (4) can be written

Let us put, for every $n \geq 2$,

(5)

and

For $0 < x \leq 1$ one then has

(6)

Since $\log \frac{n}{n-1}$ tends to 0 as $n$ tends to $+\infty$ one deduces from (6) that if a solution $g$ exists then it is necessarily equal, on $]0, 1]$, to the limit of the $g_n(x)$.

Now one deduces immediately from (5) that for $x$ fixed and $> 0$ one has

as $n$ tends to $+\infty$, which proves that the series with general term $u_n(x)$ converges for every $x > 0$. Each of the functions $u_n(x)$ being convex on $]0, +\infty[$, as is $-\log x$, the function $g(x) = -\log x + \sum_{n=2}^{\infty} u_n(x)$ is convex on this interval (I, p. 27, prop. 2 and prop. 4); finally, one has $u_n(1) = 0$, whence $g(1) = 0$, and whence in other words, $g$ satisfies equation (3) of VII, p. 306.