Reports of the Academy of Sciences of the USSR

1. Volume 178, No. 3

MATHEMATICS

I. ILYASOV

SUMMATION OF COMPOSITE FUNCTIONS OF THE EULER FUNCTION

(Presented by Academician I. M. Vinogradov on 1 IV 1967)

Let a function \(F(x)\) be continuous on the interval \([0,1]\). It is easily proved (see, for example, \((^{1})\), p. 151) that the limit exists

\[ \lim_{N\to\infty}\frac{1}{N}\sum_{n\le N} F\left(\frac{\varphi(n)}{n}\right). \]

Here \(\varphi(n)\) denotes the Euler function. Denote this limit by \(A(F)\). M. M. Tan proved (see \((^{2})\)) that if \(F(x)\) has an \(r\)-th derivative satisfying on \([0,1]\) a Lipschitz condition of the first order,

\[ |F(x_1)-F(x_2)|<C|x_1-x_2|, \]

then

\[ \frac{1}{N}\sum_{n\le N} F\left(\frac{\varphi(n)}{n}\right) = A(F)+ O\left( \frac{(\ln\ln N)^{r+1}}{(\ln N)^{r+1}} \right). \]

E. V. Novoselov suggested that the factor \((\ln\ln N)^{r+1}\) in this estimate is superfluous, and proved this for the case \(r=0\) (the argument is given in M. M. Tan’s paper \((^{2})\), Lemma 5).

Below I establish that Novoselov’s suggestion is true for \(r=1\), i.e., I prove the theorem:

Theorem. Let the function \(F(x)\) have on the interval \([0,1]\) a derivative satisfying a Lipschitz condition of the first order. Then

\[ \frac{1}{N}\sum_{n\le N} F\left(\frac{\varphi(n)}{n}\right) = A(F)+O\left(\frac{1}{(\ln N)^2}\right). \]

The proof of this assertion is an instance of the idea of a “matryoshka doll”—limit transitions nested one inside another to infinity—an idea that works well in probabilistic number theory. We give the proof.

Lemma 1. Let \(\bar L\ge 2\) be an integer, \(\bar L\mid L\), \(m\) an integer, and \((a,b)\) the greatest common divisor of \(a\) and \(b\). Then

\[ \sum_{\substack{k\equiv m\pmod L\\ k\le N}} \frac{\varphi((k,L))}{(k,L)} = \frac{N}{L} \prod_{p/(L,m)} \left(1-\frac{1}{p}\right) \prod_{\substack{p\mid L\\ p\nmid \bar L}} \left(1-\frac{1}{p^2}\right) + O(\ln L); \]

the constant in the symbol \(O\) is absolute.

The proof is carried out on the basis of the representation

\[ \frac{\varphi(a)}{a} = \sum_{d/a}\frac{\mu(d)}{d} \]

and an interchange of the order of summation.

Lemma 2. Let \(\bar L\ge 2\) be an integer,

\[ \sum_{\substack{k\equiv m\pmod{\bar L}\\ k\le N}} \frac{\varphi(k)}{k} = \frac{6}{\pi^2} \frac{N}{ L\prod_{p\mid L}\left(1-\frac{1}{p^2}\right)} \prod_{p/(m,\bar L)} \left(1-\frac{1}{p}\right) + O(\ln N) \]

with an absolute constant in the symbol \(O\).

The proof begins in the same way as the proof of Lemma 1. The calculations are carried out almost automatically.

Lemma 3. The asymptotic formula is valid

\[ \sum_{\substack{k\equiv m\pmod{\bar L}\\ k\le N}} \frac{\varphi^2(k)}{k^2} = \prod_{p\nmid L} \left( 1-\frac{1}{p} + \frac{1}{p}\left(1-\frac{1}{p}\right)^2 \right) \prod_{p/(\bar L,m)} \left(1-\frac{1}{p}\right)^2 \frac{N}{\bar L} + O(\ln^2 N). \]

This lemma is contained in Tan’s paper \((^{2})\) (Lemma 1, p. 106).

Let the function \(F(x)\) satisfy the conditions of the theorem. By the mean-value theorem and using the Lipschitz condition, we obtain

\[ \left|F(x_1)-F(x_2)-(x_1-x_2)F'(x_2)\right|<C|x_1-x_2|^2. \]

In the assertion of the theorem \(N\to\infty\). We shall construct other parameters, smaller than \(N\), and tending to infinity; the first of these will be \(R_N\), which we now construct. Let \(0<\beta<1\), and let \(T_N\) be the least common multiple of the numbers \(1,2,\ldots,[(1-\beta)\ln N]\). Divide \(N\) by \(T_N\) with remainder

\[ N=R_N+\theta T_N,\qquad 0\leq \theta<1. \]

As in the work of M. M. Tyana, we have

\[ \frac{1}{N}\sum_{m=1}^{N}F\left(\frac{\varphi(m)}{m}\right) = \frac{1}{R_N}\sum_{m=1}^{R_N}F\left(\frac{\varphi(m)}{m}\right) + O\left(\frac{1}{N^{\beta+o(1)}}\right). \]

Denote by \(R_N(m)\) the least positive residue of the division of \(m\) by \(R_N\). By virtue of the periodicity of \(R_N(m)\),

\[ \frac{1}{R_N}\sum_{m=1}^{R_N}F\left(\frac{\varphi(m)}{m}\right) = \lim_{T\to\infty}\frac{1}{T}\sum_{m=1}^{T} F\left(\frac{\varphi(R_N(m))}{R_N(m)}\right), \]

and therefore

\[ A(F)-\frac{1}{R_N}\sum_{m=1}^{R_N}F\left(\frac{\varphi(m)}{m}\right) = \lim_{T\to\infty}\frac{1}{T}\sum_{m=1}^{T} \left( F\left(\frac{\varphi(m)}{m}\right) - F\left(\frac{\varphi(R_N(m))}{R_N(m)}\right) \right). \]

Using Lemma 2, one can calculate that

\[ \lim_{T\to\infty}\frac{1}{T}\sum_{m=1}^{T} \left( \frac{\varphi(m)}{m} - \frac{\varphi(R_N(m))}{R_N(m)} \right) F'\left(\frac{\varphi(R_N(m))}{R_N(m)}\right) = \]

\[ = \frac{1}{R_N}\sum_{k=1}^{R_N} F'\left(\frac{\varphi(k)}{k}\right) \left( \frac{\varphi(k)}{k} - \frac{6}{\pi^2\displaystyle\prod_{p\mid R_N}\left(1-\frac{1}{p^2}\right)} \prod_{p\mid(k,R_N)}\left(1-\frac{1}{p}\right) \right). \]

Hence, one may write the relation

\[ A(F)-\frac{1}{R_N}\sum_{m=1}^{R_N}F\left(\frac{\varphi(m)}{m}\right) = \lim_{T\to\infty}\frac{1}{T}\sum_{m=1}^{T} \left( F\left(\frac{\varphi(m)}{m}\right) - F\left(\frac{\varphi(R_N(m))}{R_N(m)}\right) -\right. \]

\[ \left. - \left( \frac{\varphi(m)}{m} - \frac{\varphi(R_N(m))}{R_N(m)} \right) F'\left(\frac{\varphi(R_N(m))}{R_N(m)}\right) \right) + \]

\[ + \frac{1}{R_N}\sum_{k=1}^{R_N} F'\left(\frac{\varphi(k)}{k}\right) \left( \frac{\varphi(k)}{k} - \frac{6}{\pi^2\displaystyle\prod_{p\mid R_N}\left(1-\frac{1}{p^2}\right)} \prod_{p\mid(k,R_N)}\left(1-\frac{1}{p}\right) \right). \]

From this,

\[ \left| A(F)-\frac{1}{R_N}\sum_{m=1}^{R_N}F\left(\frac{\varphi(m)}{m}\right) \right| \leq \]

\[ \leq \left| \frac{1}{R_N}\sum_{k=1}^{R_N} F'\left(\frac{\varphi(k)}{k}\right) \left( \frac{\varphi(k)}{k} - \frac{6}{\pi^2\displaystyle\prod_{p\mid R_N}\left(1-\frac{1}{p^2}\right)} \prod_{p\mid(k,R_N)}\left(1-\frac{1}{p}\right) \right) \right| + \]

\[ + \lim_{T\to\infty}\frac{C}{T}\sum_{m=1}^{T} \left( \frac{\varphi(m)}{m} - \frac{\varphi(R_N(m))}{R_N(m)} \right)^2. \]

The central point of the work is the estimation of the first expression on the right-hand side of the inequality. We introduce a third parameter \(\overline{R}_N\), tending to infinity; this is a number that we shall choose later and that satisfies the condition

\[ T_N \mid \overline{R}_N \mid R_N . \]

By \(\overline{R}_N(k)\) we shall denote the remainder upon division of \(k\) by \(\overline{R}_N\); furthermore, let

\[ B_{R_N}(k)= \frac{6}{\pi^2 \prod_{p\mid R_N}\left(1-\frac{1}{p^2}\right)} \prod_{p'(k,R_N)} \left(1-\frac{1}{p}\right). \]

We have

\[ \frac{1}{R_N}\sum_{k=1}^{R_N} F'\left(\frac{\varphi(k)}{k}\right) \left(\varphi(k)-B_{R_N}(k)\right)= \]

\[ = \frac{1}{R_N}\sum_{k=1}^{R_N} F'\left(\frac{\varphi(\overline{R}_N(k))}{\overline{R}_N(k)}\right) \left(\frac{\varphi(k)}{k}-B_{R_N}(k)\right)+ \]

\[ +\frac{1}{R_N}\sum_{k=1}^{R_N} \left( F'\left(\frac{\varphi(k)}{k}\right)- F'\left(\frac{\varphi(\overline{R}_N(k))}{\overline{R}_N(k)}\right) \right) \left(\frac{\varphi(k)}{k}-B_{R_N}(k)\right). \]

Hence

\[ \left|A(F)-\frac{1}{R_N}\sum_{m=1}^{R_N} F\left(\frac{\varphi(m)}{m}\right)\right| \leq \left| \frac{1}{R_N}\sum_{k=1}^{R_N} F'\left(\frac{\varphi(\overline{R}_N(k))}{\overline{R}_N(k)}\right) \left(\frac{\varphi(k)}{k}-B_{R_N}(k)\right) \right|+ \]

\[ +\frac{C}{R_N}\sum_{k=1}^{R_N} \left| \frac{\varphi(k)}{k} - \frac{\varphi(\overline{R}_N(k))}{\overline{R}_N(k)} \right| \left| \frac{\varphi(k)}{k}-B_{R_N}(k) \right| + \]

\[ +\varlimsup_{T\to\infty} \frac{C}{T}\sum_{m=1}^{T} \left( \frac{\varphi(m)}{m} - \frac{\varphi(R_N(m))}{R_N(m)} \right)^2 . \]

By Lemmas 2 and 1 we easily obtain

\[ \frac{1}{R_N}\sum_{k=1}^{R_N} F'\left(\frac{\varphi(\overline{R}_N(k))}{\overline{R}_N(k)}\right) \left(\frac{\varphi(k)}{k}-B_{R_N}(k)\right) = O\left(\frac{\overline{R}_N}{R_N}\ln R_N\right) = O\left(\frac{\overline{R}_N}{N}\ln N\right) \]

(for \(R_N\asymp N\)). Now take \(\overline{R}_N=T_N[N^{\beta/2}]\). Then

\[ \frac{1}{R_N}\sum_{k=1}^{R_N} F'\left(\frac{\varphi(\overline{R}_N(k))}{\overline{R}_N(k)}\right) \left(\frac{\varphi(k)}{k}-B_{R_N}(k)\right) = O\left(\frac{\ln N}{N^{\beta/2}}\right). \]

Furthermore,

\[ \frac{1}{R_N}\sum_{k=1}^{R_N} \left| \frac{\varphi(k)}{k} - \frac{\varphi(\overline{R}_N(k))}{\overline{R}_N(k)} \right| \left| \frac{\varphi(k)}{k}-B_{R_N}(k) \right| \leq \]

\[ \leq \left[ \frac{1}{R_N}\sum_{k=1}^{R_N} \left( \frac{\varphi(k)}{k} - \frac{\varphi(\overline{R}_N(k))}{\overline{R}_N(k)} \right)^2 \right]^{1/2} \left[ \frac{1}{R_N}\sum_{k=1}^{R_N} \left( \frac{\varphi(k)}{k}-B_{R_N}(k) \right)^2 \right]^{1/2}. \]

Using Lemma 1, after some calculations we obtain

\[ \frac{1}{R_N}\sum_{k=1}^{R_N} \left( \frac{\varphi(k)}{k}-B_{R_N}(k) \right)^2 = \prod_{p\nmid R_N} \left( 1-\frac{1}{p} + \frac{1}{p}\left(1-\frac{1}{p}\right)^2 \right) - \]

\[ - \prod_{p\nmid R_N} \left(1-\frac{1}{p^2}\right)^2 \prod_{p\mid R_N} \left( 1-\frac{1}{p} + \frac{1}{p}\left(1-\frac{1}{p}\right)^2 \right) + O\left(\frac{\ln^2 R_N}{R_N}\right). \]

Further,

\[ \prod_{p\chi R_N}\left(1-\frac{1}{p}+\frac{1}{p}\left(1-\frac{1}{p}\right)^2\right) -\prod_{p\chi R_N}\left(1-\frac{1}{p^2}\right)^2 \ll C'\int_{(1-\beta)\ln N}^{\infty}\frac{\pi(x)}{x^4}\,dx = \]

\[ =O\left(\frac{1}{\ln^2 N\ln\ln N}\right). \]

Thus,

\[ \frac{1}{R_N}\sum_{k=1}^{R_N}\left(\frac{\varphi(k)}{k}-B_{R_N}(k)\right)^2 = O\left(\frac{1}{\ln^2 N\ln\ln N}\right). \]

Let us now consider

\[ \frac{1}{R_N}\sum_{k=1}^{R_N}\left(\frac{\varphi(k)}{k}-\frac{\varphi(\overline{R}_N(k))}{\overline{R}_N(k)}\right)^2 = \frac{1}{R_N}\sum_{k=1}^{R_N}\left(\sum_{d\mid k}\frac{\mu(d)}{d} -\sum_{d\mid \overline{R}_N(k)}\frac{\mu(d)}{d}\right)^2 . \]

Since \(k\equiv \overline{R}_N(k)\pmod{T_N}\), all divisors not exceeding \((1-\beta)\ln N\) are the same for \(k\) and \(\overline{R}_N(k)\).

\[ \frac{1}{R_N}\sum_{k=1}^{R_N}\left(\frac{\varphi(k)}{k}-\frac{\varphi(\overline{R}_N(k))}{\overline{R}_N(k)}\right)^2 = \frac{1}{R_N}\sum_{k=1}^{R_N}\left( \sum_{\substack{d>(1-\beta)\ln N\\ d\mid k}}\frac{\mu(d)}{d} - \sum_{\substack{d>(1-\beta)\ln N\\ d\mid \overline{R}_N(k)}}\frac{\mu(d)}{d} \right)^2 \ll \]

\[ \ll \frac{1}{R_N}\sum_{k=1}^{R_N}\left( \sum_{\substack{d>(1-\beta)\ln N\\ d\mid k}}\frac{\mu(d)}{d} \right)^2 + \frac{1}{R_N}\sum_{k=1}^{R_N}\left( \sum_{\substack{d>(1-\beta)\ln N\\ d\mid \overline{R}_N(k)}}\frac{\mu(d)}{d} \right)^2 . \]

After elementary, but rather cumbersome, calculations we obtain

\[ \frac{1}{R_N}\sum_{k=1}^{R_N}\left(\frac{\varphi(k)}{k}-\frac{\varphi(\overline{R}_N(k))}{\overline{R}_N(k)}\right)^2 = O\left(\frac{\ln\ln N}{\ln^2 N}\right). \]

This gives

\[ \frac{1}{R_N}\sum_{k=1}^{R_N}\left| \frac{\varphi(k)}{k}-\frac{\varphi(\overline{R}_N(k))}{\overline{R}_N(k)} \right| \left| \frac{\varphi(k)}{k}-B_{R_N}(k) \right| = O\left(\frac{1}{\ln^2 N}\right). \]

The second remainder, i.e. the expression

\[ \lim_{T\to\infty}\frac{C}{T}\sum_{k=1}^{T}\left( \frac{\varphi(k)}{k}-\frac{\varphi(R_N(k))}{R_N(k)} \right)^2 \]

is estimated by the same method as

\[ \frac{1}{R_N}\sum_{k=1}^{R_N}\left( \frac{\varphi(k)}{k}-\frac{\varphi(\overline{R}_N(k))}{\overline{R}_N(k)} \right)^2, \]

and we again obtain an upper bound of order

\[ O\left(\frac{1}{\ln^2 N}\right). \]

The theorem is proved.

Remark. By this method one can verify Novoselov’s hypothesis for any \(r\), but the calculations then become very cumbersome.

Mathematical Institute named after V. A. Steklov
Academy of Sciences of the USSR

Received
29 III 1967

REFERENCES

1. M. Kac, Statistical independences in probability theory, analysis, and number theory, Moscow, 1963.
2. M. M. Tian, Lithuanian Math. Collection, 6, No. 1, 105 (1966).