Prove or disprove a proposition

Question

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.

Answer 1

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.