Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am working through a bank of previous exams and couldn't figure a problem out to my satisfaction.

Let $f(x) : \mathbb{R} \to \mathbb{R}\,$ be a continuous function.

  1. Show that $f$ can have at most countably many strict local maxima.
  2. Assume that $f$ is not monotone on any interval. Then show that the local maxima of $f$ are dense in $\mathbb{R}$.
share|cite|improve this question
up vote 7 down vote accepted
  1. For each $\delta>0$, the set of all $x\in\mathbb{R}$ such that $f(y)<f(x)$ for all $y$ with $0<|x-y|<\delta$ is countable. This can be seen by noting that the set contains at most one element of the interval $[k\frac{\delta}{2},(k+1)\frac{\delta}{2}]$ for each integer $k$, and these intervals cover $\mathbb{R}$. The set of strict local maxima is a countable union of such sets, for example taking $\delta=\frac{1}{n}$ as $n$ ranges over the positive integers.

  2. Suppose that $f$ is a continuous function that is not monotone on any interval. We want to show that $f$ has a local maximum in every interval. The argument is the same if the interval is $(0,2)$, so to slightly reduce notation let's work there. Note that not being monotone in any interval means that each interval contains pairs $x_1<x_2$ and $y_1<y_2$ such that $f(x_1)<f(x_2)$ and $f(y_1)>f(y_2)$. So $(0,1)$ contains a pair $x_1<x_2$ such that $f(x_1)<f(x_2)$, and $(1,2)$ contains a pair $y_1<y_2$ such that $f(y_1)>f(y_2)$. Because $f$ is continuous, there is a point $x_0\in [x_1,y_2]$ where $f$ has its maximum value. Because $f(x_2)>f(x_1)$ and $f(y_1)>f(y_2)$, $x_0$ is not an endpoint of $[x_1,y_2]$. Therefore $f$ has a local maximum at $x_0$.

share|cite|improve this answer
How do you use the continuity requirement in showing 1? – davidk01 Mar 19 '11 at 4:11
Continuity isn't needed for 1. – Jonas Meyer Mar 19 '11 at 4:13
Then I don't see how what you've said is true. Why can't I just take $f(y)$ and move it down creating a hole in the function where all the values around it are above the new value where I moved $f(y)$. That would make the entire interval $(y-\delta,y+\delta)$ except for $y$ satisfy your stated condition and $(y-\delta,y+\delta)$ is uncountable. – davidk01 Mar 19 '11 at 4:17
@davidk01: I am saying that for each $\delta>0$, the set $\{x:\text{ for all }$y$, |x-y|<\delta\Rightarrow f(y)<f(x)\}$ is countable. Changing the value at a single point in the domain does not affect this. These are not the points $x$ where $f(x)>f(y)$ for some fixed $y$ close to $x$, but for all $y$ close to $x$. Hence the statement that there can be only one such point in an interval of length less than $\delta$. – Jonas Meyer Mar 19 '11 at 4:25
@Jonas Meyer: That wasn't clear in your original statement. I'll look this over. – davidk01 Mar 19 '11 at 4:27

For #1, prove that you can find disjoint intervals around each local maximum. Then pick a rational number in each interval.

Edit: This answer is wrong but the comments below are instructive, so I'll leave it here.

share|cite|improve this answer
That is not true, by 2. – Jonas Meyer Mar 19 '11 at 3:05
@Jonas Meyer: Why not? Since it is continuous it must be bounded on any small enough interval and in that interval it must obtain a maximum somewhere. Now we can chop out this interval and move on. As long as we keep chopping out non-trivial intervals we should be able to cover the real line with the chopped intervals. – davidk01 Mar 19 '11 at 3:19
@davidk01: In a situation where the strict local maxima are dense, which happens for functions that are monotone on no interval, if you take an interval around a point where $f$ has a local maximum, the interval will contain other points where $f$ has a local maximum, and therefore it will intersect all intervals containing those points. – Jonas Meyer Mar 19 '11 at 3:42
@Jonas Meyer: I still don't see how the density of the local maxima makes a difference? Being bounded and continuous the function must obtain an absolute maximum at some point of whatever bounded, non-trivial interval I'm looking at. So I can always find a single absolute maximum on any non-trivial, bounded interval and since such intervals cover the real line there can only be countably many absolute maximums. I should probably look up a definition of strict local maximum. – davidk01 Mar 19 '11 at 3:57
@davidk01: Continuous functions take on their maxima on closed bounded intervals, but that isn't relevant for the first question. As for how density makes a difference: Being able to cover a set with disjoint intervals each of which contains only one element of the set is a lot stronger than saying that the set is not dense. If a set is dense in $\mathbb{R}$, then every interval contains infinitely many elements of the set. – Jonas Meyer Mar 19 '11 at 4:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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