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

Revision as of 16:22, 2 February 2010

E. Wirsing, "Das asymptotische verhalten von summen über multiplikative funktionen. II." Acta Mathematica Academiae Scientiarum Hungaricae Tomus 18 (3-4), 1978, pp. 411-467.

English Translation by: Google Translator



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

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

Determine

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

However, with the additional. Call [math]\displaystyle{ \lambda(p)= O(1) }[/math] and only for [math]\displaystyle{ \tau\gt 0 }[/math] are shown (Theorem 1.1). The terms of [math]\displaystyle{ \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]\displaystyle{ \lambda }[/math], we get only if [math]\displaystyle{ \lambda }[/math] by [math]\displaystyle{ |\lambda| }[/math] not significantly different, namely, if [math]\displaystyle{ \sum\frac{1}{p}(|\lambda(p)-Re\lambda(p)|) }[/math] converges (Theorem 1.1.1). The special case [math]\displaystyle{ \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]\displaystyle{ \sum\frac{1}{p}(1-\lambda(p)),\sum\frac{1}{p^{2}}\lambda(p)^{2} }[/math] and [math]\displaystyle{ \sum_{p}\sum_{v\geq2}\frac{1}{p^{v}}\lambda(p^{v}) }[/math] and for each [math]\displaystyle{ \epsilon\gt 0 }[/math]

[math]\displaystyle{ \liminf_{x\to\infty}\sum_{x\lt p\leq(1+\epsilon)x}\lambda(p)\frac{\log p}{p}\gt 0, }[/math]

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

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

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

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

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

show if [math]\displaystyle{ \sum_{p}\frac{1}{p}(\lambda(p)-Re\lambda^{*}(p)) }[/math] diverges. In the event [math]\displaystyle{ \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]\displaystyle{ \lambda^{*} }[/math] is suitably restricted. In particular, we prove (Theorem 1.2.2): Do (1.3), [math]\displaystyle{ \lambda(p)=O (1) }[/math], [math]\displaystyle{ |\lambda^{*}\leq\lambda|\lt / math\gt is \lt math\gt \lambda^{*} }[/math] real-valued, the average of [math]\displaystyle{ \lambda^{*} }[/math] exists regarding [math]\displaystyle{ \lambda }[/math]:

[math]\displaystyle{ (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]\displaystyle{ (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]\displaystyle{ \lambda }[/math] is the constant 1 and allowed [math]\displaystyle{ \lambda^{*} }[/math] only values [math]\displaystyle{ \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]\displaystyle{ \lambda^{*} }[/math], the situation is more complicated. The example [math]\displaystyle{ \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]\displaystyle{ \lambda(p)=O(1) }[/math] and [math]\displaystyle{ |\lambda^{*}|\leq\lambda }[/math] requires. It is then that is [math]\displaystyle{ \sum_{n\leq x}\lambda^{*}(n)\sim x^{1+i}(1+i)^{-1} }[/math] while (1.6) has the value 0. But if [math]\displaystyle{ |\lambda^{*}|\leq\lambda }[/math] vershärft to the following claim:

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

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

[math]\displaystyle{ (1.7) (\sum_{n\leq x}\lambda^{*}(n))(\sum_{n\leq x}\lambda(n))^{-1}\to0\mbox{ bzw.}\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.

...