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Let $\phi$ the Euler totient, and $f : [0,1] \to \mathbb{C}$ piecewise continuous. Prove that $$ \lim_{n\to\infty} \frac{1}{\phi(n)} \sum_{\substack{1\leq k \leq n\\ \gcd(k,n)=1}} f\Bigl(\frac{k}{n}\Bigr) = \int_{0}^{1} f(x)\,dx $$

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    I can see the sum of reimann but I don't know how I can get to the answer – 2017-02-09
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    what is pgcd or rather it is gcd?? – 2017-02-09
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    @Srijit: translated from french :) Of course it is gcd, and "continuous in pices" should be replaced by "piecewise continuous". – 2017-02-09
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    In essence, what you need to prove is that, if $j$ and $k$ are both relatively prime to $n$, and if there are no such numbers between $j$ and $k$, i.e., $j$ and $k$ are “neighbours”, then $|j-k|/n<\delta_n$ where $\delta_n\to0$ when $n\to\infty$. – 2017-02-09
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    what is $x$ you mean $k$ ?I don't how this could help us with the problem – 2017-02-09
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    I already edited the $x$. It was a typo. It helps because it means the mesh size of the partition given by the numbers $k/n$ with $\gcd(k,n)=1$ goes to zero. – 2017-02-09
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    @Seginus: Could you give some contextual information about this problem ? Did you find this question by yourself or as an exercise in some textbook ? – 2017-02-09

2 Answers 2

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The Jacobstahl function is defined by $j(n)$ = the maximal distance between consecutive integers coprime to $n$.

It can be shown that (see here) that $j(n)=O(\ln^2(n))$, which proves that the mesh size of the partition has limit $0$ as $n\to\infty$, hence the conclusion.

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    That's nice (+1), but I guess a simpler proof might give a coarser estimate that's still good enough? Even an outrageously loose estimate like $j(n)=O(n/\ln n)$ would suffice for the present purpose. – 2017-02-09
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Possible sketch of an elementary (though not simple) proof

Let $f$ be a piecewise continuous, complex valued function defined on $[0,1]$.

Consider, for all $n\ge1$ :

$$F(n)=\sum_{k=1}^nf\left(\frac kn\right)\quad\mathrm{and}\quad F^\star(n)=\sum^\star f\left(\frac kn\right)$$

where $\displaystyle{\sum^\star}$ denotes $\displaystyle{\sum_{\substack{1\le k\le n\cr gcd(k,n)=1}}}$

It can be shown that $F$ and $F^\star$ are Möbius transforms of each other. I can post the details if required.

Using Möbius inversion formula, we see that :

$$F^\star(n)=\sum_{d\mid n}\mu(d)F\left(\frac nd\right)$$

where $\mu$ is the Möbius function. Recall that $\mu(1)=1$, $\mu(n)=(-1)^r$ if $n$ is the product of $r$ pairwise distinct primes and $\mu(n)=0$ otherwise.

I would like to prove that, whenever $f$ is a step function, we have :

$$\lim_{n\to\infty}\frac{1}{\varphi(n)}F^\star(n)=\int_0^1f\tag{1}$$

where $\varphi$ is Euler's totient function.

If this is true, then (1) will also hold in the general case (that is : when $f$ is piecewise continuous), because every piecewise continuous function on $[0,1]$ can be uniformly approximated on this interval by step functions.