Prove an inequality by using the integral test or another way

Question

‎Let the function ‎‎$‎g:‎\mathbb{R^+}‎‎‎\rightarrow‎‎\mathbb{R^+}‎$ ‎have ‎the ‎properties ‎that‎‎ for each ‎$‎w>0‎$‎, ‎ $$ \lim_{x\to\infty}‎\frac{g(x+w)}{g(x)} = 1‎ $$ and $\log g(x)‎$ ‎is ‎concave‎. ‎Now, since ‎$‎\log g‎$ ‎is ‎concave, ‎so ‎it ‎has ‎derivative‎ $$ \big(\log g(x)\big)^\prime=\frac{g^\prime_-(x)+g^\prime_+(x)}{2g(x)} $$ except, possibly, on a countable set, where ‎$‎g^\prime_{+}(x‏)‎$ ‎and ‎‎$‎g^\prime_{-}(x)‎$ ‎are ‎right ‎and ‎left ‎derivatives, ‎respectively. Also, since $$ ‎‎‎‎\displaystyle{\lim_{x\to\infty}}‎\frac{g(x+w)}{g(x)} = 1‎ $$ for each ‎$‎w>0‎$ and $\log g(x)‎$ ‎is ‎concave, ‎then ‎‎$‎g‎$ ‎is ‎increasing. ‎Now, ‎‎‎‎‎‎‎‎‎my ‎questions ‎is:‎‎ ‎‏‎

How should I prove the following inequality?‎ $$ 0\leq\gamma_g+\log g(1)\le \frac{g^\prime_-(1)+g^\prime_+(1)}{2g(1)}, ‎$$ where $$ \sum_{i=1}^n \frac{g^\prime_-(i)+g^\prime_+(i)}{2g(i)}\to \gamma_g\text{ as }n\to\infty. $$

Hint. Note that we can using the proof of the integral test for the convergence of infinite series, for the proof inequality (but I can't prove it).

Answer 1

The hypothesis imply that $\log g$ is absolutely continuous on compact sets, i.e., there is a function $h:\mathbb{R}^+\to\mathbb{R}$ such that $$ \log g(x) = \log g(a) + \int_a^x h(t)\; dt, $$ for all $a,x\in \mathbb{R}^+$. After redefining $h$ on a set of measure zero if necessary, we can assume that $$ h(x) = \frac{g'_-(x)+g'_+(x)}{2g(x)} $$ for all $x\in\mathbb{R}^+$. Again, the hypothesis imply that $h$ is decreasing and non-negative. Now, $$ h(i+1)\leq \int_i^{i+1} h(t)\; dt \leq h(i) $$ and then $$ \sum_{i=2}^n h(i) \leq \int_1^n h(t)\; dt \leq \sum_{i=1}^{n-1} h(i), $$ so we obtain $$ \sum_{i=2}^n h(i) - \log g(n) + \log g(1) \leq 0 \qquad \text{ and } \qquad 0\leq \sum_{i=1}^{n-1} h(i)- \log g(n) + \log g(1) $$ that we can rewrite as $$ \sum_{i=1}^n h(i) - \log g(n) + \log g(1) \leq \frac{g'_-(1)+g'_+(1)}{2g(1)} \qquad \text{ and } \qquad 0\leq \sum_{i=1}^{n} h(i)- \log g(n) + \log g(1) - h(n). $$ Now, $h$ is decreasing and non-negative, so the limit $L=\lim_{n\to\infty} h(n)$ exists and $L\geq 0$, so $$ \gamma_g+\log g(1)\leq \frac{g'_-(1)+g'_+(1)}{2g(1)} \qquad \text{ and } \qquad 0\leq \gamma_g+\log g(1)-L, $$ that is $$ 0\leq L\leq \gamma_g+\log g(1) \leq \frac{g'_-(1)+g'_+(1)}{2g(1)} $$ as desired.

Answer 2

Too long for comment

First, let us clarify the problem. We rephrase the problem as follow (cf. the OP). I put an image snipped from the article later.

Let $g: (0, \infty) \to (0, \infty)$ be a function such that $\log g(x)$ is concave, and $\lim_{x\to \infty}\frac{g(x+w)}{g(x)} = 1$ for each fixed $w > 0$. Then:
Fact 1: $g(x)$ is increasing;
Fact 2: $\frac{g'_{-}(x) + g'_{+}(x)}{2g(x)}$ is decreasing and non-negative on $(0, \infty)$, where $g'_{+}(x)$ and $g'_{-}(x)$ are the right and left derivatives;
Fact 3: $\log g(x)$ has derivative $\frac{g'_{-}(x) + g'_{+}(x)}{2g(x)}$ except, possibly, on a countable set.
Let $$F(n) = \sum_{i=1}^n \frac{g'_{-}(i) + g'_{+}(i)}{2g(i)} - \log g(n) .$$ Prove that the sequence $\{F(n)\}$ converges. Also, by denoting $\lim_{n\to\infty} F(n) = \gamma_g$, prove that $$0 \le \gamma_g + \log g(1) \le \frac{g'_{-}(1) + g'_{+}(1)}{2g(1)}.\tag{1}$$

Image snipped from the article