Let $g:\mathbb{R^+}\rightarrow\mathbb{R^+}$ be a function such that $\log g(x)$ is concave, and $\displaystyle{\lim_{x\to\infty}}\frac{g(x+w)}{g(x)} = 1$ for each $w>0$ . Then:
Fact 1: $g(x)$ is increasing;
Fact 2: $\log g(x)$ has derivative $\frac{g^\prime_{-}(x) + g^\prime_{+}(x) }{2g(x)}$ except, possibly, on a countable set, where $g^\prime_{+}(x)$ and $g^\prime_{-}(x)$ are right and left derivatives, respectively;
Fact 3: $\frac{g^\prime_{-}(x) + g^\prime_{+}(x) }{2g(x)}$ is decreasing and non-negative on $\mathbb{R}^+$.
My question is:
Let \begin{align*} F(n) = \sum_{i=1}^n \frac{g^\prime_{-}(i) + g^\prime_{+}(i) }{2g(i)} - \log g(n), \end{align*} prove that the sequence $\{F(n)\}$ converges.
Thanks in advance.
Alternative solution:
Fact 1: Let $f: \mathbb{R} \to \mathbb{R}$ be a convex function. Then, $f(x) - f(y) \ge c(x-y)$ for any real numbers $x, y\in \mathbb{R}$ and $c\in [f'_{-}(y), f'_{+}(y)]$.
See: https://en.wikipedia.org/wiki/Subderivative
https://www.sintef.no/globalassets/project/evitameeting/fho.pdf
A proof of Fact 1 is also given later.
From Fact 1, by letting $f = -\log g$ and $c = \frac{f'_{-}(y) + f'_{+}(y)}{2}$, we have, for any $x, y\in (0, \infty)$, $$-\log g(x) + \log g(y) \ge -\frac{g'_{-}(y) + g'_{+}(y)}{2g(y)}(x-y). \tag{1}$$ By letting $x = i, y = i+1$; $x = i+1, y = i$ respectively, we have, for $i = 1, 2, \cdots$, $$\frac{g'_{-}(i+1) + g'_{+}(i+1)}{2g(i+1)} \le \log g(i+1) - \log g(i) \le \frac{g'_{-}(i) + g'_{+}(i)}{2g(i)}. \tag{2}$$ Let $n < m$ be two positive integers. From (2), we have $$\sum_{i=n}^{m-1} \frac{g'_{-}(i+1) + g'_{+}(i+1)}{2g(i+1)} \le \sum_{i=n}^{m-1} (\log g(i+1) - \log g(i)) \tag{3}$$ and $$\sum_{i=n+1}^{m} (\log g(i+1) - \log g(i)) \le \sum_{i=n+1}^{m} \frac{g'_{-}(i) + g'_{+}(i)}{2g(i)} \tag{4}$$ which results in $$\log \frac{g(m+1)}{g(m)} - \log\frac{g(n+1)}{g(n)} \le \sum_{i=n+1}^{m} \frac{g'_{-}(i) + g'_{+}(i)}{2g(i)} - \log\frac{g(m)}{g(n)} \le 0. \tag{5}$$ Thus, we have $$\log \frac{g(m+1)}{g(m)} - \log\frac{g(n+1)}{g(n)} \le F(m) - F(n) \le 0.\tag{6}$$ Since $\lim_{k\to \infty} \log \frac{g(k+1)}{g(k)} = 0$, we know that $\{F(n)\}$ is a Cauchy sequence. Thus, $\{F(n)\}$ has a limit denoted by $\gamma_g$.
Then, let us prove that $$0 \le \gamma_g + \log g(1) \le \frac{g'_{-}(1) + g'_{+}(1)}{2g(1)}. \tag{7}$$ From (2), we have $$\sum_{i=1}^{n-1} \frac{g'_{-}(i+1) + g'_{+}(i+1)}{2g(i+1)} \le \sum_{i=1}^{n-1} (\log g(i+1) - \log g(i))\tag{8}$$ and $$\sum_{i=1}^{n} (\log g(i+1) - \log g(i)) \le \sum_{i=1}^n \frac{g'_{-}(i) + g'_{+}(i)}{2g(i)}\tag{9}$$ which results in $$\log\frac{g(n+1)}{g(n)} \le \sum_{i=1}^n \frac{g'_{-}(i) + g'_{+}(i)}{2g(i)} - \log g(n) + \log g(1) \le \frac{g'_{-}(1) + g'_{+}(1)}{2g(1)}. \tag{10}$$ By using $\lim_{n\to \infty} \log\frac{g(n+1)}{g(n)} = 0$, the desired result follows.
$\phantom{2}$
Proof of Fact 1: We have, for any $r < t < y < u < v$, $$\frac{f(r) - f(y)}{r- y} \le \frac{f(t)-f(y)}{t-y} \le \frac{f(u)-f(y)}{u-y} \le \frac{f(v)-f(y)}{v-y}.$$ Indeed, for example, $\frac{f(r) - f(y)}{r- y} \le \frac{f(t)-f(y)}{t-y}$ can be written as $f(t) = f(\frac{t-r}{y-r}\cdot y + \frac{y-t}{y-r}\cdot r) \le \frac{t-r}{y-r}f(y) + \frac{y-t}{y-r}f(r)$ which is true by the definition of convex functions.
Thus, $f'_{+}(y) = \lim_{x \to y+} \frac{f(x)-f(y)}{x-y}$ and $f'_{-}(y) = \lim_{x \to y-} \frac{f(x)-f(y)}{x-y}$ both exist, and $f'_{-}(y) \le f'_{+}(y)$; moreover, if $x > y$, we have $f'_{+}(y) \le \frac{f(x)-f(y)}{x-y}$ which results in $f(x) - f(y) \ge f'_{+}(y)(x-y) \ge c(x-y)$ for any $c\in [f'_{-}(y), f'_{+}(y)]$,
and if $x < y$, we have $\frac{f(x)-f(y)}{x-y} \le f'_{-}(y)$ which results in $f(x)-f(y) \ge f'_{-}(y)(x-y) \ge c(x-y)$ for any $c\in [f'_{-}(y), f'_{+}(y)]$. We are done.