Wirsing translation

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E. Wirsing, "Das asymptotische verhalten von summen über multiplikative funktionen. II." Acta Mathematica Academiae Scientiarum Hungaricae Tomus 18 (3-4), 1967, pp. 411-467.

English Translation by: Google Translator



In I we have the asymptotic behavior of the sum [math] \sum_{n \leq x} \lambda (n)[/math] for nonnegative multiplicative functions [math] \lambda[/math] essentially under the condition

[math] (1.1) \sum_{p\leq x}\lambda(p)\log(p)\sim\tau x \mbox{ (p prime)}[/math]

Determine:

[math] (1.2) \sum_{n\leq x}\lambda(n)\sim\frac{e^{-ct}}{\Gamma(\tau)}\frac{x}{\log x}\prod_{p\leq x}\left(1+\frac{\lambda(p)}{p}+\frac{\lambda(p^{2})}{p^{2}}+\cdots\right)[/math]

([math] c[/math] is the Euler-) constant. Special rates are the same type Delange [3]. The same result (1.2) is here under the much weaker assumption

[math] (1.3) \sum_{p\leq x}\lambda(p)\frac{\log p}{p}\sim\tau\log x[/math]

However, with the additional. Call [math] \lambda(p)= O(1)[/math] and only for [math]\tau\gt0[/math] are shown (Theorem 1.1). The terms of [math] \lambda(p^{v}) (v\geq2)[/math] are thieves than I, but we want them in the introduction . neglect The same result for complex-function [math] \lambda[/math], we get only if [math] \lambda[/math] by [math] |\lambda|[/math] not significantly different, namely, if [math]\sum\frac{1}{p}(|\lambda(p)-Re\lambda(p)|)[/math] converges (Theorem 1.1.1). The special case [math]\tau=1,|\lambda|\leq1,\sum\frac{1}{p}(1-\lambda(p))[/math] is convergent was proved by Delange [4] easier to by Rényi [8].

An interesting counterpart to give Erdos and Rényi [7]: convergence [math]\sum\frac{1}{p}(1-\lambda(p)),\sum\frac{1}{p^{2}}\lambda(p)^{2}[/math] and [math]\sum_{p}\sum_{v\geq2}\frac{1}{p^{v}}\lambda(p^{v})[/math] and for each [math]\epsilon\gt0[/math]

[math]\liminf_{x\to\infty}\sum_{x\ltp\leq(1+\epsilon)x}\lambda(p)\frac{\log p}{p}\gt0,[/math]

then (1.2) (with [math]\tau=1[/math]). Here [math]\lambda[/math] will be restricted to the bottom.

If [math]\lambda^{*}[/math] is another multiplicative function [math]|\lambda^{*}|\leq\lambda[/math], so we could in I with the conditions (1.1) and

[math](1.4) \sum_{p\leq x}\lambda^{*}(p)\log p\sim\tau^{*}x[/math]

n is the sum [math]\sum_{n\leq x}\lambda^{*}(n)[/math] up to [math]o(\sum_{n\leq x}\lambda(n))[/math] identify and, in particular

[math]\sum_{n\leq x}\lambda^{*}(n)=o(\sum_{n\leq x}\lambda(n))[/math]

show if [math]\sum_{p}\frac{1}{p}(\lambda(p)-Re\lambda^{*}(p))[/math] diverges. In the event [math]\lambda=1[/math] see Delange [3]. In the present paper we obtain such results without (1.4), some of which (1.1), some of them already with (1.3), if the range of values of [math]\lambda^{*}[/math] is suitably restricted. In particular, we prove (Theorem 1.2.2): Do (1.3), [math]\lambda(p)=O (1)[/math], [math]|\lambda^{*}|\leq\lambda[/math] is [math]\lambda^{*}[/math] real-valued, the average of [math]\lambda^{*}[/math] exists regarding [math]\lambda[/math]:

[math](1.5) \lim_{x\to\infty}(\sum_{n\leq x}\lambda^{*}(n))(\sum_{n\leq x}\lambda(n))^{-1}[/math]

and has the value

[math](1.6) \lim_{x\to\infty}\prod_{p\leq x}\left(1+\frac{\lambda^{*}(p)}{p}+\frac{\lambda^{*}(p^2)}{p^2}+\cdots\right)\left(1+\frac{\lambda(p)}{p}+\frac{\lambda(p^2)}{p^2}+\cdots\right)^{-1}.[/math]

If one specifically for [math]\lambda[/math] is the constant 1 and allowed [math]\lambda^{*}[/math] only values [math]\pm1[/math], then this is the solution of a known problem of Wintner [9] (there) with a supposed proof, see Erdos [6].

For complex-valued functions [math]\lambda^{*}[/math], the situation is more complicated. The example [math]\lambda(n)=1,\lambda^{*}(n)=n^{i}[/math] shows that the existence of (1.6), the foltg of (1.5) does not if only (1.1), [math]\lambda(p)=O(1)[/math] and [math]|\lambda^{*}|\leq\lambda[/math] requires. It is then that is [math]\sum_{n\leq x}\lambda^{*}(n)\sim x^{1+i}(1+i)^{-1}[/math] while (1.6) has the value 0. But if [math]|\lambda^{*}|\leq\lambda[/math] vershärft to the following claim:

[math]\lambda^{*}(n)=\epsilon(n)\lambda(n),|\epsilon(n)|\leq1.[/math]

There are a number [math]e^{i\phi}[/math] of magnitude 1, which is "not an accumulation point of the sequence [math](\epsilon(p))[/math], then (Theorem 1.2.1 follows):

[math](1.7) (\sum_{n\leq x}\lambda^{*}(n))(\sum_{n\leq x}\lambda(n))^{-1}\to0\mbox{, resp.}\sim\prod_{p\leq x}\left(1+\frac{\lambda^{*}(p)}{p}+\cdots\right)\left(1+\frac{\lambda(p)}{p}+\cdots\right)^{-1},[/math]

depending on the product does not tend to 0 or.

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