My question is how to prove the following assertion:
If $f:\mathbb{R}\to\mathbb{R}$ is a concave function such that $\lim\limits_{x\to\infty}(f(x)-f(x-1))=0$ then $f$ increasing.
I know that if $f$ is differentiable then the mentioned hypothesis ($f'$ is decreasing) and using M.V.T imply that firstly $\lim\limits_{x\to\infty}f'(x)=0$ also $f'\geq0$ and $f$ is increasing. but here we do not have the differentiability.
Anyone can help me. Thanks.
Hint. Since $f$ is concave it follows that for any $x_0\in \mathbb{R}$ the function $$x\to R(x,x_0):=\frac{f(x)-f(x_0)}{x-x_0}$$ is decreasing in $\mathbb{R}\setminus\{x_0\}$. Here it is a reference for the convex function $-f$.
Now assume that $f$ is not increasing. Then there are $x_0,x_1$ such that $x_0<x_1$ and $f(x_1)<f(x_0)$. It follows that $$0>R(x_1,x_0)\geq R(x_0+k+1,x_0)\geq R(x_0+k+1,x_0+k)$$ where $k$ is any positive integer such that $x_0+k>x_1$.
Can you take it from here?