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This problem looks very difficult )= Construct a continuous function, such that it set of strictly local maximum points, is the set of rationals.

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Let $f(x)=-\sqrt{x(1-x)}$ for $x\in[0,1]$, and let $g(x) = f(\text{the fractional part of }x)$.

Then $g(x)$ is a continuous function with very sharp local maxima at every integer.

Now, $h(x)=\sum_{k=1}^\infty a_k g(k!x)$, for some appropriate coefficients $a_k$ that make everything converge, should have the specified property.

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    Would you care to elaborate a bit why $h$ would satisfy that property?2011-10-17
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    After step $n$ of the summation, all rationals with denominator $n$ are local maxima. Furthermore, all remaining terms of the series are 0 at these points, and no term is ever positive, so one of these local maxima cannot stop being one as we to to the limit.2011-10-17
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    (Of course one still needs to prove that the sum is continuous, and the _all_ local maxima are rational, but I don't want to give everything away, because the original question sounded a bit like homework).2011-10-17
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    How can i prove the continuity of the sum?2011-10-28
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    @August: Choose the $a_k$s such that it converges uniformly.2011-10-28