Is $f$ non decreasing?

Question

Suppose $f:\mathbb{R} \to \mathbb{R}$ is a function satisfying the conditions $2f(x)\le f(x+h)+f(x+2h)\quad \forall x\in \mathbb{R}$ and $h\ge0 $. Then is it true that $f$ is non decreasing?

I noticed that if $f$ is differentiable then this will be true since we can take all terms on one side and divide by $h$ and then let $h\to 0$ we'll see $3f'\ge0$, but what happens otherwise?

Answer 1

Suppose $f$ is not non-decreasing, so that there exist $a,b$ with $a < b$ and $f(a) > f(b)$. Then $f$ is unbounded on every finite interval $(s,t)$ with $b \le s < t$.

To keep things simple, suppose $a=0, b=1, f(0)=1, f(1)=0$; what follows can easily be adapted for the general case.

Putting $x=0,h=\frac12$ in the inequality gives $f(\frac12) \ge 2$.
Putting $x=\frac12,h=\frac14$ in the inequality gives $f(\frac34) \ge 4$.
Putting $x=\frac34,h=\frac18$ in the inequality gives $f(\frac78) \ge 8$.

And so on. So $f$ is unbounded on every non-empty interval $(1-\epsilon,1)$.

Now let $(s,t)$ be any finite interval with $1 \le s < t$. Then $f$ is unbounded on $(s,t)$.

To see this, let $u = s+t-1$ and $\epsilon = t - s$. Then for any $x \in (1-\epsilon,1)$ we can put $h = \frac12(u-x)$, so that $x+h \in (s,t)$ and $x+2h = u$. The inequality tells us that $f(x+h) \ge 2f(x)-f(u)$; but $f(u)$ is fixed, and $f(x)$ is unbounded. So $f$ is unbounded on $(s,t)$.

Answer 2

Let us write the expression as $$A= f(x+h)+f(x+2h)-2f(x)$$ with the conditions $A \geq 0$ and $h \geq 0$.

Now expand, as you almost did, the function as Taylor series around $h=0$. We then have $$A=3 h f'(x)+\frac{5}{2} h^2 f''(x)+O\left(h^3\right)$$ So, if the conditions are satisfied, at least at the first order they imply that $f'(x) \geq 0$ and $f(x)$ is not decreasing.

As Dario commented, the above assumes that $f(x)$ is a continuous function and as Tony K. commented that $f(x)$ is differentiable.