$f(x+1)=f(x)+\ln(x+1)$ , $f $ convex

Question

let $f$ a convex function on the iterval $I=]-1,+\infty[$ :

$f(x+1)=f(x)+\ln(x+1)$

I have proved that for all natural numbers:

$f(x)=f(x+n)-\ln(\prod_{k=1}^{n} (x+k))$

and I have proved that:

$\lim_{x\to+\infty} f'_+(x) -f'_-(x)=0$

How to prove that $f$ is differentiable on the interval $I$

i think that we should prove that:

$\lim_{x\to a} f'_+(x) -f'_-(x)=0$ for all $a \in I$

Please help me with this question

Answer 1

Let $f(x)=ln(g(x))$ $$ g(x+1)=(x+1)g(x) $$

But since f is logarithmically complex convex and follows the above equation it must be a variant of $\Gamma(x)$ function by Bohr-Mollerup theorem

Edit: just saw @gonçalo's comment after posting