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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 $\mathbb{R}$.
Michael Chen
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1 Answers1

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  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 $x_0$.

Jonas Meyer
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  • How do you use the continuity requirement in showing 1? –  Mar 19 '11 at 04:11
  • Continuity isn't needed for 1. – Jonas Meyer Mar 19 '11 at 04:13
  • 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. –  Mar 19 '11 at 04:17
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    @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 04:25
  • @Jonas Meyer: That wasn't clear in your original statement. I'll look this over. –  Mar 19 '11 at 04:27
  • @davidk01: I've reworded the first sentence. I hope that it is clearer. – Jonas Meyer Mar 19 '11 at 04:30
  • @Jonas Meyer: Ok, I understand now. Without the $\forall$ qualifier I was thinking ${x\ :\ \exists y\ |x-y|<\delta\Rightarrow f(y) < f(x)}$. –  Mar 19 '11 at 05:01
  • @Jonas Meyer: Thank you!

    @dadivk01: Thank you for the discussion.

     – Michael Chen Mar 19 '11 at 14:35
  • @JonasMeyer Thank you or your answer. Does your answer generalize to the real valued functions defined on higher dimensional Euclidean spaces $R^n, n> 1?$ – Learning Math Jan 30 '23 at 02:19
  • @JonasMeyer Thank you for this answer! For the first question, if we ask "is the set of strict local maxima of a continuous function $f:R^n \to R$ is countable," then do you think your argument generalize to this case, and we get an affirmative answer? TIA – Learning Math Feb 06 '23 at 13:42
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    @LearningMath: It looks like you already asked as a separate question, which was a good idea! – Jonas Meyer Mar 27 '23 at 06:52