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Give an example of $f : \mathbb R → [0, \infty) $ so that every $r \in \mathbb Q$ is a strict local minimum for $f$.

Strict local minimum means there is a vicinity $V$ of $r$ such that $f(y) > f(r) ,\ \forall y \in V-\{r\}$


My attempt

So far, none. My feeling is there isn't such a function, mainly because of the density of $\mathbb Q$ in $\mathbb R$. Suppose I define $f$ like this: $f(x) = 0$ for $x \in \mathbb Q$ and $f(x) = 1$ for $x \not \in \mathbb Q$. Every rational $r$ does not map to a strict local minimum for $f$ only because of the other rationals present in every vicinity of $r$. So $f$ cannot be constant on $\mathbb Q$, but how to define it is beyond my imagination.

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    $\begingroup$ My first Google hit for "local minimum at all rational numbers" was en.wikipedia.org/wiki/Thomae%27s_function which – I think – can be modified to fit your needs by considering $1 - f(x)$. $\endgroup$
    – Martin R
    May 12 '16 at 18:48
  • $\begingroup$ @MartinR more of a meta topic, but is Googling advised? $\endgroup$
    – ahorn
    May 12 '16 at 18:56
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    $\begingroup$ @MartinR Yes, you are right, 1 - (Thomae's function) is such an example $\endgroup$
    – user261263
    May 12 '16 at 19:00
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Here's a pretty direct hint. If there were only rationals, the function could be defined as the denominator in lowest terms, because everything nearby enough would have a larger denominator. If we want to also define it on the irrationals this wouldn't work. However, we can compress the positive integers into $[0,1)$ in an order preserving way, then we can make the function $1$ on the irrationals.

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  • $\begingroup$ I like this answer better because it explains some of the reasoning. $\endgroup$ May 13 '16 at 0:24
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I think that the function that sends any irrational number on $1$ and sends a rational $\frac{p}q$ (where the fraction is irreducible) on $1-\frac1q$ does the job. Given any rational number $\frac{p}q$, you can always find a neighborhood so that any rational number has a denominator bigger than $q$

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    $\begingroup$ The function needs to be nonnegative. $\endgroup$ May 12 '16 at 18:43
  • $\begingroup$ @MattSamuel: Then add $1$ ... $\endgroup$
    – Martin R
    May 12 '16 at 18:49
  • $\begingroup$ @H. Potter perhaps $1$ and $1-\frac 1 q$ $\endgroup$
    – ahorn
    May 12 '16 at 18:50
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    $\begingroup$ @MattSamuel There are many ways to change the function so that it becomes nonnegative. Find your favorite one :) $\endgroup$
    – H. Potter
    May 12 '16 at 19:01
  • $\begingroup$ Can you make it nonnegative (with 1 - whatever), to accept your answer, please? $\endgroup$
    – user261263
    May 12 '16 at 19:08

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