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After I've read a problem from [1] (it is a Spanish journal) I've consider the next question that seems curious to me, since now I don't know how get such simple example.

Question. We denote for integers $n\geq 1$ the Möbius function as $\mu(n)$, see the definition of this arithmetic function for example from this MathWorld. Can you provide me an (simple) example of a (real) continuous function $f$ over $[0,1]$ satisfying $$\int_0^x f(u)du\geq \sum_{n=1}^\infty\frac{\mu(n)}{n^2}x^{n-1}$$ for each $x\in[0,1]$? Many thanks.

I tried simple functions as constants, monomials and exponentials, I'm not even asking for a function such that the difference $RHS-LHS$ is small, I would like to know just an example. And it is just a curiosity. Then please, how to find it?

References:

[1] PROBLEMA 272, by Marcel Chiriţă, La Gaceta de la RSME, Vol. 19 (2016), Núm. 1, page 111.

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  • $\begingroup$ Everyone, many thanks for your patience. $\endgroup$ – user243301 Jan 31 '18 at 22:49
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No solution, as $\int_0^0 f(x){\mathrm d}x=0$, but the RHS evaluates to $1$ for $x=0$.

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  • $\begingroup$ I am going to wait more answers, but many thanks. And this is the code for Wolfram Alpha online calculator for (a toy model of) our function in RHS plot sum mu(n)/n^2x^(n-1), from n=1 to 1000, for 0<x<1 $\endgroup$ – user243301 Jan 31 '18 at 20:48
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It's impossible. If f is continuous over the compact set $[0,1]$ so is bounded and since it is continuous it is Reimann integrable therefore$$\lim_{x\to 0^+}\int_{0}^{x}f(u)du=0$$also using $\epsilon-\delta$ approach we conclude that if $x<\delta=\dfrac{\epsilon}{\epsilon+1}$ therefore$$|\sum_{n=1}^\infty\frac{\mu(n)}{n^2}x^{n-1}-1|=|\sum_{n=2}^\infty\frac{\mu(n)}{n^2}x^{n-1}|\le\sum_{n=2}^\infty\frac{|\mu(n)|}{n^2}x^{n-1}\le\sum_{n=2}^\infty\frac{1}{n^2}x^{n-1}\le\sum_{n=2}^\infty x^{n-1}\le\sum_{n=2}^\infty (\dfrac{\epsilon}{\epsilon+1})^{n-1}=\dfrac{\dfrac{\epsilon}{\epsilon+1}}{1-\dfrac{\epsilon}{\epsilon+1}}=\epsilon$$which yields to $$\lim_{x\to0^+}\sum_{n=1}^\infty\frac{\mu(n)}{n^2}x^{n-1}=1$$which is a contradiction to $\lim_{x\to 0^+}\int_{0}^{x}f(u)du=0$ therefore no such $f(x)$ exists.

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  • $\begingroup$ Many thanks for all these details, I've a problem is that I misunderstood the answer below, now I undestand that the answer below is an answer (I read no solution understanding that it was a comment, instead of a full answer). Thus I am going to accept the answer below as an answer, and I say to you I'm sorry and to the other user. Many thanks for the patience. $\endgroup$ – user243301 Jan 31 '18 at 22:43
  • $\begingroup$ That's very kind of you $\endgroup$ – Mostafa Ayaz Jan 31 '18 at 22:48
  • $\begingroup$ I was stuck, and sometimes there are details that I don't see. Many thanks one more time for your attention and help. $\endgroup$ – user243301 Jan 31 '18 at 22:50
  • $\begingroup$ No mind at all... $\endgroup$ – Mostafa Ayaz Jan 31 '18 at 22:51

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