Must $f: \mathbb{Z} \to \mathbb{R}_{\geq 0}$ (monotone non-decreasing) be constant if $\frac{f(n-1) + f(n+1)}{2} \leq f(n)$ for all $n$? Suppose $f$ is a non-negative function defined on the integers. Suppose $f$ satisfies $f(n-1) \leq f(n)$ for any $n$, as well as $\frac{f(n-1) + f(n+1)}{2} \leq f(n)$.
It seems clear to me that $f$ should be a constant, but playing around with the two inequalities hasn't gotten me close to showing this (or showing this is false). Would appreciate any guidance.
 A: You have for all $n$ an equivalent formulation of your condition:
$$ 0 \le f(n+1)-f(n) \le f(n) - f(n-1)$$
hence by induction you can prove that for all $n \le m \in \mathbb{Z}$ you have
$$f(m)-f(m-1) \le f(n) - f(n-1).$$
Since your sequence is increasing and bounded below by $0$, you have that
$$L= \inf_{n \in \mathbb{Z}} f(n) = \lim_{n \to -\infty} f(n)$$
exists and is a non-negative real number. In particular,
$$\lim_{n \to -\infty} (f(n)-f(n-1)) = 0$$
Now, if you suppose by contradiction that there exists some index $a$ such that
$$f(a) - f(a-1) = 2 \varepsilon > 0$$
you have the contradiction
$$0 < 2 \varepsilon = f(a) - f(a-1 ) \le f(n) - f(n-1) \le \varepsilon$$
for $-n$ big enough ($n$ is meant to be negative and $n \to - \infty$).
A: Note that if $f(n)=f(n+1)$ for some $n$ then $f$ is constant. If $f(n+1)-f(n)=x>0$ then $f(n)-f(n-1)\geq x$. So inductivly we will arrive at some integer $k$ for which $f(k)$ is negative. But this is a contradiction.
A: Since $f(n)\ge 0$  for every $n\in \Bbb Z$ and $f$ is increasing we have that exists the limit $$\lim_{n \to -\infty} f(n)=l\ge0.$$ This implies that $|f(n+1)-(n)|=f(n)-f(n+1)$ must go to $0$ as $n$ goes to $-\infty$. But from the given inequality we have 
$$f(n)-f(n+1)\ge f(n-1)-f(n)$$
so the difference between two consecutive terms is increasing as $n$ goes to $-\infty$ (actually for every $n$). So it must be constantly $0$ and $f(n)=f(n+1)$ for every $n$.
