0
$\begingroup$

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 $x\leq y\implies f(x)\leq f(y)$.

$\endgroup$
1
  • $\begingroup$ I've edited the question for clarity $\endgroup$ – porton Jul 27 '16 at 15:09
2
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ I had in mind "nonstrict" monotonicity rather than strict monotonicity as in your proof. Can the proof be modified for nonstrict monotonicity? $\endgroup$ – porton Jul 27 '16 at 16:00
  • $\begingroup$ The proof is for non-strict monotonicity. It can be adapted for strict monotonicity by replacing $f(x) < f(x')$ by $f(x) \leq f(x')$ and similarly for $f(x'') < f(x')$. (Perhaps it becomes clearer if you read $f(x) < f(x')$ as $f(x) \not \geq f(x')$ instead, and similarly for $f(x'') < f(x')$.) $\endgroup$ – Mees de Vries Jul 27 '16 at 16:09
  • $\begingroup$ It seems that your proof does not work for nonstrictly increasing $f$: You advised (in a now deleted comment) that we can take $\varepsilon = f (x') - f (x) > 0$; by continuity there exists a point $x''$ sufficiently close to $x$ such that $| f (x) - f (z) | < \varepsilon$. But we can't take $\varepsilon=0$. $\endgroup$ – porton Jul 29 '16 at 19:15
  • $\begingroup$ Gotcha: I take all $\varepsilon > f (x') - f (x) \geq 0$ and then it is easy to reconstruct the proof $\endgroup$ – porton Jul 30 '16 at 13:39
  • $\begingroup$ Actually, the proof works (as written) for non-strict monotonicity, although you are right it needs slightly more modification for the strict case than I previously stated. $\endgroup$ – Mees de Vries Aug 2 '16 at 22:22
0
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ I've edited the question. Saying "monotone" I meant "monotone increasing": $x\leq y \implies f(x)\leq f(y)$ $\endgroup$ – porton Jul 27 '16 at 15:10
0
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ The hypothesis says "there exists some $\varepsilon$ such that. . . . " and not "for every $\varepsilon$ we have that. . . ." $\endgroup$ – Daron Jul 27 '16 at 15:04
  • $\begingroup$ I formulated my question wrongly. We should assume that $[a;a+\epsilon[$ is increasing for some $\epsilon>0$ rather than for every $\epsilon>0$ $\endgroup$ – porton Jul 27 '16 at 15:04
  • $\begingroup$ I am editing my answer. $\endgroup$ – P Vanchinathan Jul 27 '16 at 15:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.