Let $A$ be a subset of $\mathbb{R}$. Let $f$ be a function $A\rightarrow\mathbb{R}$. By "increasing" I mean nonstrictly increasing. But the case of strictly increasing is also interesting.
Are the following statements equivalent to each other:
For every $a\in A$ there is a $b\in\mathbb{R}$ such that $b>a$ and $f|_{A\cap [a;b[}$ is increasing.
For every $a\in A$ there is a $b\in\mathbb{R}$ such that $b>a$ and $f(x)\geq f(a)$ for every $x\in A\cap [a;b[$.
If not equivalent, do they become equivalent if $f$ is continuous? (We can for simplicity consider the variant if $A$ is an open interval.)
First, obviously $A$ must be a connected set (even if you assume $f$ to be as regular as you want, not just continuous); so it must be an interval - open or closed or half-open, half-closed, bounded or half-bounded or all of $\mathbb R$.
NO:
Let $f(x) = 0 \text{ for } x\le0, f(x) = 1 \text{ for } x\ge 1$, and $f(x) = x - 1/2^n$ for $x \in [1/2^n, 1/2^{n-1})$ for $n \in \mathbb N, n \ge 1$. 2. is true for $a=0$, but 1. isn't.
YES: for $f$ continuous
Take any $a$ and let $c = \sup \{b > a \mid f(x) \ge f(a) \forall x \in [a,b)\}$. Then $c$ is the right endpoint (finite or infinite) of $A$. Indeed, by continuity $f(a) \le f(c)$, and if $c$ is not already the right endpoint of $A$, apply condition 2. to $c$. Any new "$b$" greater than $c$ that satisfies condition 2. for $c$ will also satisfy it for $a$ also, but $c$ was supposed to be the $\sup$ - contradiction. This proves that $f$ is (globally) non-decreasing on all of $A$.