$f:\mathbb{R}\to \mathbb{R}$ continuous and $\lim_{h \to 0^{+}} \frac{f(x+2h)-f(x+h)}{h}=0$ $\implies f=$ constant. Let $f:\mathbb{R} \to \mathbb{R}$ be a continuous function with the property that
$$\lim_{h \to 0^{+}} \dfrac{f(x+2h)-f(x+h)}{h}=0$$ for all $x \in \mathbb{R}$. Prove that $f$ is constant.
 A: Hint: for a given $h$ one has
$${f(x + h) - f(x) \over h} = {f(x + h) - f(x + h/2) \over h} + {f(x + h/2) - f(x + h/4) \over h} + ....$$
$$=  {1 \over 2}{f(x + h) - f(x + h/2) \over h/2} + {1 \over 4}{f(x + h/2) - f(x + h/4) \over h/4 } + ....$$
You actually need continuity of $f(x)$ already for the above. Now take limits as $h$ goes to zero in the above carefully and conclude ${\displaystyle \lim_{h \rightarrow 0^+} {f(x + h) - f(x) \over h}} $ is always exists and is equal to zero. Then use this to show $f$ is constant.
A: Fix $\varepsilon>0$.
Assume that there exist $x$ such that $f(x)$ is positive. Consider $h$ so small so that $f(x+2h)$ is positive  Then since $f$ is continuous there exist $h>0$ such that $f(x+h) < f(x)+\varepsilon$
Then $$f(x+2h)-f(x+h)>f(x+2h)-f(x)-\varepsilon$$
So if we take the limit as h tends to zero we get
$$\lim_{h\to0} (f(x+2h)-f(x+h)-(f(x+2h)-f(x)))< \varepsilon$$ but since that is true for all $\varepsilon>0$ it must be that in fact $$\lim_{h\to0} \frac{f(x+2h)-f(x+h)}h=\lim_{h\to0} \frac{f(x+2h)-f(x)}h$$ and by the property we are given by the problem we conclude that the latter is equal to zero therefore by definition of differentiability $f$ has derivative equal to zero at $x$. But $x$ is  just an arbitrary  point with positive $f(x)$. By continuity of $f$ there exist a neighborhood of points around x where we have $f(\text{of those points})>0$ therefore this neighborhood of points has derivative equal to zero at any point and therefore $f$ is constant in that neighborhood so lets say is equal to $c$. Now we consider the closure of this neighborhood, the two points in the closure will have $f(\text{closure points})=c$ by continuity of $f$ and since $c>0$ there exist another neighborhoods with $f(\text{of points})>0$ so they are also constant. Continuing like that we see $f$ is constant.
If no point $x$ such that $f(x)>0$ exist but there exist only $x$ such that $f(x)<0$ with a similar argument proof that $f$ is constant (consider $-f(x)$ and show it is constant)
if none of the two cases exist then $f$ is the zero function which is constant.
