I am working on some questions on real analysis and was hoping to get some help regarding my thoughts on these two questions so far:
Let $f:I\to\mathbb{R}$ be a function defined on an open interval $I\subset \mathbb{R}$
"$f$ is monotonic (not strictly), so there exists an open sub-interval on which f is constant."
As $f$ is not strictly monotonic, every $f(x)\geq f(x-1)$ (assuming f is monotonically increasing). As it is explicitly stated that it is not strictly monotonic, there must be $a, b \in I$ such that for every $x\in(a, b)$: $a< f(x)=f(x+1)=…<b$ . Therefore, there exists an open sub-interval $(a, b)$ on which $f$ is constant.
"If $f$ is differentiable, such that $f'\geq0$ or $f'\leq0$ on $I$, and $f'$ disappears only a countable number of times, then f is strictly monotonic"
Aside from the fact that I am only assuming that "disappears" means that $f'$ doesn't need to be continuous, I would assume that if the derivative was greater (less) or equal to zero, and not strictly greater (or less), then there would be an $x$ such that $f(x)=f(x+1)$, which would then imply that the monotonicity is not strict. I am aware that a function is monotonic if its first derivative does not change sign, however I don't see how this implies strict monotonicity.
Best Answer
Without loss of generality we can assume $f$ is monotonically increasing.
So we have:
$\forall x,y \in I: x < y \to f(x) \le f(y)$
Not strictly monotonically increasing:
$\neg \forall x,y \in I: x < y \to f(x) < f(y)$
Which can also be written as:
$\exists x,y \in I: x < y \land f(x) \ge f(y)$
This implies:
$\exists x,y \in I: x < y \land f(x) = f(y)$
Now for any $z \in I$ such that $x < z < y$ we have $f(x) \le f(z) \le f(y)$. And since $f(x) = f(y)$, we have $f(x) = f(y) = f(z)$ and $f$ is constant on interval $(x, y)$.