Prove or disprove a proposition Prove or disprove: ‎‎Let the function ‎‎$‎g:‎\mathbb{R^+}‎‎‎\rightarrow‎‎\mathbb{R^+}‎$ ‎have ‎the ‎properties ‎that‎‎ for each ‎$‎w>0‎$‎, ‎$‎‎‎‎\displaystyle{\lim_{x\to\infty}}‎\frac{g(x+w)}{g(x)} = 1‎$ and $\log g(x)$ ‎is ‎concave on ‎$‎‎\mathbb{R^+}‎$‎‎‎. ‎Then‎, ‎‎$‎g‎$ ‎is ‎increasing‎. Thanks.
 A: I will prove the problem.$$\\$$
let us define $h(x) \equiv log(g(x))$.
Then first condition becomes clear, by 
$\displaystyle\lim_{x\to\infty}h(x+w)\, - \,h(x) = 0$ for $w>0$. - (1)
Also, $h(x)$ concave. - (2)
$$\\$$
Consider  $\,0<x<y$ .
I will define $ d \equiv y-x >0$.
since h is concave function, 
$$
h(x+d) - h(x) \geq h(x+2d) - h(x+d) \geq ... - (3)$$
Also, for arbitrary $\epsilon >0$, there is $M_{d,\epsilon}\,>0$ such that 
$$
z>M_{d,\epsilon} \Rightarrow |h(z+d) - h(z)| < \epsilon - (4)
$$
$$\\$$Now, assume to the contrary, let us set $(h(y) - h(x))<0$. 
Then there is $\epsilon >0\,$ such that $(h(y) - h(x))<-\epsilon<0$.  -  (5)
Moreover, there is $ N \in \mathbb{N}$ satisfies $x+Nd >M_{d,\epsilon}$.
Then by (4), $\, h(x + Nd) - h(x + (N - 1)d) > -\epsilon$.
This gives following argument, which makes contradiction. (with (5))
$$
h(y) - h(x) = h(x+d) - h(x) \geq ... \geq h(x + Nd) - h(x + (N - 1)d) > -\epsilon
$$
$$\\$$
Thus $(h(y) - h(x))\geq0$.
Since y and x are arbitrary, we actually showed $h(x)$ is monotically increasing.
then $ g = e^h $ also monotonically increasing function naturally.
