If a continuous real function $f$ on an interval is such that for each $a$ on its domain there is $\epsilon>0$ such that $[a;a+\epsilon[$ is monotone increasing, does it follow that $f$ is monotone increasing on the interval?
By "monotone increasing" I mean the implication $x\leq y\implies f(x)\leq f(y)$.
On
Without loss of generality let us assume you want to show the function is increasing.
Take two points $x,y$ with $x<y$ on your (unspecified) interval. We have o show $f(x)< f(y). $ These two points lie in the interval $[a, a+\epsilon[$ for some $\epsilon >0$. There your hypothesis gives it is increasing. QED.
EDIT: As I understood the hypothesis wrongly my answer above is useless. Let it be there.
On
The answer is no. Consider something like the absolute-value function on the interval $[-1,1]$. For each $x < 0$ the function is monotone-decreasing on $[x,x/2)$ and for each $x>0$ the function is monotone-increasing on $[x,1)$.
I'm not sure about the following sharpening of the question:
If a continuous real function $f$ on an interval is such that for each a on its domain there is ϵ>0 such that [a;a+ϵ[ is monotone increasing (decreasing), does it follow that f is monotone increasing (decreasing) on the interval?
For convenience, let $f$ be defined on $[a,b]$. For some $x' \leq b$, $f$ is monotone on $[a,x']$. Let $x$ be the supremum of these, so that for at least every $a < x' < x$, we have $f$ monotone on $[a,x']$. By continuity then, $f$ is monotone on all of $[a,x]$; for if $f(x) < f(x')$ for some $a < x'< x$, then there is an $x''$ near $x$ with $f(x'') < f(x')$.
Now if $x \neq b$, we can find $\epsilon$ so that $f$ is monotone on $[x,x+\epsilon)$ contradicting that $x$ was the supremum. Thus $x = b$ and $f$ is monotone.