2

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?

Michael Hardy
  • 1
Idele
  • 1,968
  • Can I assume that $f$ is differentiable? – Randall Aug 15 '17 at 04:00
  • 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. – Aniruddha Deshmukh Aug 15 '17 at 04:00
  • @AniruddhaDeshmukh what if f isn't even continuous? – Idele Aug 15 '17 at 04:02
  • @hctb Assuming it is continuous and differentiable in $\mathbb{R}$. – Aniruddha Deshmukh Aug 15 '17 at 04:05
  • @AniruddhaDeshmukh Maybe i didn't make my question clear, i edited it. – Idele Aug 15 '17 at 04:14
  • @hctb But here, you will need an interval where all the points will take the local maximum value. Otherwise how can you say that every point on the domain will take local maximum? Because if your function is discontinuous at say $x = a$, then we will have a value at $a$ and some different value in its neighbourhood. So, take any interval containing that point, the function will never have local maximum at all its points. – Aniruddha Deshmukh Aug 15 '17 at 04:18
  • @AniruddhaDeshmukh: See this function, $f(x)$ is equal to 0 at any positive x and is equal to 1 at any other x. This function takes the local maximum values at every point. – Selene Aug 15 '17 at 04:45
  • @XIAODAQU Everypoint where? If you take whole $\mathbb{R}$, then not all values take local maximum. – Aniruddha Deshmukh Aug 15 '17 at 04:54
  • @AniruddhaDeshmukh you should take a moment check the function again patiently:) – Idele Aug 15 '17 at 05:00
  • @AniruddhaDeshmukh: Note that ${x>0}$ is open.$f(x)=0$ when $x>0$, and $f(x)=1$ when $x\leq0$. – Selene Aug 15 '17 at 05:01
  • @XIAODAQU What I am saying is suppose you take an interval $\left[-a, a \right]$, then the local maximum would be 1 which will be taken by points only less than or equal to zero. So, in this interval, not "all" points take the local maximum. Now, since 'a' is arbitrary, it can take any value however large or small. So, if it is really really large, we can take the interval $\left(-\infty, \infty \right)$ which is $\mathbb{R}$. Again, here we may conclude that not all points on the domain take the local maximum. – Aniruddha Deshmukh Aug 15 '17 at 05:05
  • @AniruddhaDeshmukh i think you misunderstand local maximum , see https://en.wikipedia.org/wiki/Maxima_and_minima?wprov=sfsi1 – Idele Aug 15 '17 at 05:08
  • @hctb: May I ask how would you prove the first sentence that "local maximum values can only have countable"? It seems not that obvious to me. – Selene Aug 15 '17 at 05:10
  • @hctb I agree with both of you. I was really confused. I mistook "there exists some $\epsilon$" for "given any $\epsilon$". You are write, the discontinuous function does take local maximum at every point of its domain. However, we cannot generalize it for all discontinuous functions. If you want a counter example, try the Dirichlet's function $f(x) = 1$ if x is rational otherwise $f(x) = 0$. This will not take the local maximum at all points. In fact, according to me, it will be really really difficult to even define the local extrema for this function! – Aniruddha Deshmukh Aug 15 '17 at 05:16
  • @XIAODAQU Sure! We can have a injection from the set of local maximum values to the countable set $\mathbb{Q}\times\mathbb{Q}$. If $b$ is a loxal maximum, assume $f(a)=b$ then we have a small enough positive real number$\varepsilon$ that $f(a)$ is the maximum of f on$(a-\varepsilon,a+\varepsilon)$ now take two rational number p,q in this interval on each side of a , we can prove from $b$ to (p,q) is the injection we need.(And of course in the proof we use the axiom of choice) – Idele Aug 15 '17 at 05:17
  • @hctb: What a smart proof! – Selene Aug 15 '17 at 05:27

2 Answers2

3

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.

Selene
  • 1,123
  • 5
  • 15
1

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.

Martin Sleziak
  • 56,060