5

Let $I$ be an interval and let $f$ : $I \rightarrow\mathbb{R}$ be a function. If $f$ is monotonically increasing as well as monotonically decreasing on $I$, then show that $f$ is constant on $I$.

My approach: Consider $x,y\in I$ such that $x\neq y$. Assume that $x>y$. Since $f$ is monotonically increasing on $I$, so $f(x)\geq f(y)$. Also $f$ is monotonically decreasing on $I$. Thus we have $f(x)\leq f(y)$. Hence $f(x)=f(y)$. But from here how I can show that $f$ is constant on $I$?

urt43as
  • 345
  • 1
  • 8
  • 4
    Now fix one $c\in I$ and show that $f(x)=f(c)$ for all $x\in I$. –  Dec 31 '22 at 06:57
  • You should assume that the order on the domain of $f$ is total, otherwise the statement is false. But your notation $I$ suggest a real interval, in which case there is no problem: the usual order on $\Bbb R$ (and on any subset) is total. – Anne Bauval Dec 31 '22 at 07:08
  • 1
    You proved that for any $x,y\in I,$ $f(x)=f(y).$ This is sufficient to claim that $f$ is constant. – Anne Bauval Dec 31 '22 at 07:14
  • 2
    Ok. Suppose we fix a point, say $\alpha\in I$. Now let $x\in I$ be any arbitrary such that $x\neq c$. Assume that $x>c$. Now $f$ is monotonically increasing on $I$. Thus $f(x)\geq f(c)$. Again, $f$ is monotonically decreasing on $I$. Thus $f(x)\leq f(c)$. So we have $f(x)=f(c)$ where $x\in I$ is arbitrary and $c\in I$ is fixed. This implies that $f$ is constant. Will it work? – urt43as Dec 31 '22 at 07:22
  • @JSN Yes, that's a complete proof. – Bertrand Wittgenstein's Ghost Dec 31 '22 at 07:45

1 Answers1

0

Here's a different approach:

A monotonically increasing function has nonnegative derivative at all points. A monotonically decreasing function has nonpositive derivative at all points. Therefore, the function in question must have derivative $0$ at all points. Since only constant functions have a derivative of $0$ at all points, the function must be constant.

mathlander
  • 3,997