2
$\begingroup$

If $f:\mathbb{R}\to\mathbb{R}$ and every point takes a local maximum value, it's a fact that the local maximum values of a real function can only have countable, so if we assume $f$ is continuous we have $f$ must be constant. My question is, if $f$ isn't continuous, can we prove there must be some interval that $f$ is constant on it?

$\endgroup$
  • $\begingroup$ Can I assume that $f$ is differentiable? $\endgroup$ – Randall Aug 15 '17 at 4:00
  • $\begingroup$ Well, yea. If every point of the domain is the local maximum, we have $f'(x) = 0$ for $\forall x \in \mathbb{R}$. Which makes $f(x) = constant$ for the whole domain. $\endgroup$ – Aniruddha Deshmukh Aug 15 '17 at 4:00
  • $\begingroup$ @AniruddhaDeshmukh what if f isn't even continuous? $\endgroup$ – Idele Aug 15 '17 at 4:02
  • $\begingroup$ @hctb Assuming it is continuous and differentiable in $\mathbb{R}$. $\endgroup$ – Aniruddha Deshmukh Aug 15 '17 at 4:05
  • $\begingroup$ @AniruddhaDeshmukh Maybe i didn't make my question clear, i edited it. $\endgroup$ – Idele Aug 15 '17 at 4:14
2
$\begingroup$

I think your conclusion is right. I've written a proof, please help me check if it's right.

Since "local maximum values can only be countable", we assume they are $\{a_n\}_n$. And let $F_n=\{f=a_n\}$. Then $\mathbb{R}=\bigcup_{n\geq1}F_n$.

Due to Baire's theorem, there is a $n_0$ such that $F_{n_0}$ is dense in an open interval (expressed as $U$).

Because $\{f=a_{n_0}\}$ is dense in $U$, it's easy to prove that $f(x)\geq a_{n_0}$ in $U$.

Assume that $x_0\in\{f=a_{n_0}\}$ is not an interior point of $\{f=a_{n_0}\}$ in $U$. In other words, $ \exists\{x_n\}_n\bigcap\{f=a_{n_0}\}=\emptyset$ such that $x_n\to x_0$. However, it can't be correct because $x_0$ is a local maximum.

Then we know $\{f=a_{n_0}\}$ has an interior point $x_0$ and we arrive at your conclusion. What's more, since $x_0$ is arbitrary, we know that $F_{n_0}\cap U$ is open too.

$\endgroup$
  • $\begingroup$ Which baire theorem are you refering? $\endgroup$ – Idele Aug 15 '17 at 7:57
  • $\begingroup$ @hctb: Baire's category theorem $\endgroup$ – XIAODA QU Aug 15 '17 at 8:28
  • $\begingroup$ but we can't say F_n is closed :( $\endgroup$ – Idele Aug 15 '17 at 8:30
  • 2
    $\begingroup$ @hctb: Actually we don't need that. Note that my conclusion is that $F_n$ is density in an open interval. You can pick the closure of $F_n$ if you're really worried. $\endgroup$ – XIAODA QU Aug 15 '17 at 8:40
  • $\begingroup$ i see… thanks ! $\endgroup$ – Idele Aug 15 '17 at 8:44
0
$\begingroup$

Suppose that $ f\colon {\mathbb R}\to{\mathbb R}$ is a function such that it has a local maximum at each point of $[a,b]$ but it is not constant on any interval.

By induction we can construct a sequence of points $x_i$ and closed intervals with the properties:

  • $x_i$ is a local maximum on $I_i$;
  • $f(x_{i+1})<f(x_i)$;
  • $I_{i+1}\subseteq I_i$;
  • $\operatorname{diam} (I_i)\searrow 0$.

Inductive step: Since $f$ is not constant on $I_i$, there is a point $x_{i+1}\in I_i$ such that $f(x_{i+1})<f(x_i)$. Since $f$ has a local maximum at $x_{i+1}$, there is a neighborhood of $x_{i+1}$ on which the values are at most $f(x_{i+1})$. We can take a smaller closed interval $I_{i+1}$ in this neighborhood such that, at the same time, $I_{i+1}\subseteq I_i$ and $\operatorname{diam} (I_i) \le 2^{-i}$.

By Cantor Intersection Theorem there is a unique point $x\in\bigcap\limits_{i\in\mathbb N} I_i$. Clearly, $\lim\limits_{i\to\infty} x_i=x$. We also have $f(x)<f(x_i)$ for each $i\in\mathbb N$ (since $x\le f(x_{i+1})<f(x_i)$). So $x$ is not a local maximum, which is a contradiction.


This is the approach I have taken when trying to solve Exercise 10.S from the book van Rooij, Schikhof: A Second Course on Real Functions. See also here.

$\endgroup$

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.