Functions with property $1<2f(x)-f(x-1)-f(x+1)<2$ Let $x>0$ and let $f(x)$ be an increasing function on the domain $x>0$. Are there any $f(x)$ with the property  $$1<2f(x)-f(x-1)-f(x+1)<2?$$
Since $f(x-1)+f(x+1)<2f(x)$, $f(x)$ is a concave function. But I don't know how to construct $f(x)$ satisfying above inequality. Do you know how to find such $f(x)$ or are there some references on this kind of problems? Thanks!!
 A: Write the condition in the form
$$
1 < \left[ {f(x) - f(x - 1)} \right] - \left[ {f(x + 1) - f(x)} \right] < 2.
$$
Summing this over the integers from $2$ to $n$ gives
$$
n-1 < f(2) - f(1) + f(n) - f(n + 1) < 2n-2.
$$
This implies that for sufficiently large $n$ (more precisely, $n > f(2) - f(1)+1$),
$$
0 < f(n) - f(n + 1),
$$
i.e., $f$ cannot be an increasing function.
A: Intuitively, if we set $2f(x)-f(x-1)-f(x+1)=c$ for some constant $c$, our function will look something like $a+bx-c\frac{x^2}2$, and won't be able to be increasing (this is a linear recurrence relation, for which there are many references). Your question is slightly different, but can be solved by essentially the same techniques.
Consider
$$\sum_{i=1}^n (2f(x+i)-f(x+i-1)-f(x+i+1)).$$
On one hand, this at least $n$ and on the other hand it's
$$2\sum_{i=1}^n f(x+i)-\sum_{i=1}^n f(x+i-1)-\sum_{i=1}^n f(x+i+1)=f(x+1)+f(x+n)-f(x)-f(x+n+1)\leq f(x+1)-f(x),$$
since $f$ should be increasing. However, $f(x+1)-f(x)$ can't be arbitrarily large, a contradiction.
(We never used the upper bound of $2$ for the given constraint.)
A: Here is a positive result complementing the negative result shown in the other answers : while it is true that there is no solution on $[0,\infty)$, on the other hand there is a solution on $[0,T]$ for every $T\gt 0$.
Indeed, let $f(x)=\frac{3}{4}\big(-x^2+2Tx\big)$. Then $f'(x)=\frac{3}{2}(T-x)$ so $f$ is increasing on $[0,T]$. On the other hand, $2f(x)-f(x-1)-f(x+1)=\frac{3}{2}$ for every $x$.
