Let $f: \mathbb{N} \to \mathbb{C}$ be a multiplicative arithmetic function (meaning that $f(mn) = f(m)f(n)$ whenever $m$ and $n$ are relatively prime). Suppose that for every $\epsilon >0,$ that $$f(n) \ll_\epsilon n^\epsilon,$$ here I am using Vinogradov’s notation, this means that $|f(n)| \leq C_\epsilon n^\epsilon$ for some constant $C_\epsilon$ which depends on $\epsilon.$.

Suppose further that for each prime $p, f(p)=g $ for some $g \geq 1.$ This $g$ is thus independent of $p.$.

Consider now, for a prime p, the identity of formal power series $$1+gX+f(p^2)X^2 + \cdots = (1-X)^{-g} h_p(X).$$ It is then easy to see that $h_p(0)=1,h_p’(0)=0.$.

Question. Let now $b_p(k)$ be the coefficient of $X^k$ in $h_p(X).$ I have seen it claimed that the coefficients $b_p(k)$ are bounded by a polynomial in $k$ that is independent of $p.$
Could someone explain why this holds?

  • $\begingroup$ I don't think that $h'_p(0)=1$. Are you sure that your definition is correct? $\endgroup$ – FXV Jun 29 at 15:37
  • $\begingroup$ Let $f(p) = g,f(p^k) = \log p$ then $f(n) = \prod_{p^k \| n}f(p^k) = O(n^\epsilon)$ even if $ f(p^k)$ isn't bounded by a polynomial of $k$. If you mean a polynomial of $p^k$ then it is immediate from $f(n) = O(n)$. Also (for a multiplicative function) $f(n) = O(n^\epsilon)$ iff $f(p^k) = O((p^k)^\epsilon)$ $\endgroup$ – reuns Jun 29 at 16:00
  • $\begingroup$ @reuns I’d be careful with epsilons in the last statement. If $f(p^k) = 2p^k$ then it isn’t quite true that $f(n) = O(n)$, but it’s close. $\endgroup$ – Erick Wong Jun 29 at 17:14
  • $\begingroup$ @FXV Typo, thanks! Now fixed. $\endgroup$ – inequalitynoob2 Jun 29 at 17:14
  • $\begingroup$ @reuns I really want a polynomial in $k,$ not in $p^k.$ Are you giving a counterexample? I am not sure I understood your comment. $\endgroup$ – inequalitynoob2 Jun 29 at 17:15

Here's my naive approach.

Let $a(x) =1+gX+f(p^2)X^2 + \cdots =\sum_{n-0}^{\infty} a_n x^n $ so that $(1-x)^{-g} h_p(x) =a(x) $.


$\begin{array}\\ h_p(x) &=a(x)(1-x)^g\\ &=\sum_{n=0}^{\infty} a_n x^n \sum_{j=0}^g \binom{g}{j}(-1)^{g-j}x^{g-j}\\ &=\sum_{n=0}^{\infty} a_n x^n \sum_{j=0}^g \binom{g}{j}(-1)^{j}x^{j}\\ &=\sum_{n=0}^{\infty} x^n\sum_{i=0}^{\min(n, g)} a_{n-i} \binom{g}{i}(-1)^{i}\\ \end{array} $


$\begin{array}\\ |\sum_{i=0}^{\min(n, g)} a_{n-i} \binom{g}{i}(-1)^{i}| &\le \sum_{i=0}^{\min(n, g)} |a_{n-i} \binom{g}{i}|\\ &\le \sum_{i=0}^{\min(n, g)} g^if(p^{n-i})\\ &\le g^g\sum_{i=0}^{\min(n, g)} f(p^{n-i})\\ &\le g^g\sum_{i=0}^{\min(n, g)} C_{\epsilon}(n-i)^{\epsilon}\\ &\le g^{g+1}C_{\epsilon}n^{\epsilon}\\ \end{array} $

  • 2
    $\begingroup$ Why is $f(p^{n-i}) \leq C_\epsilon (n-i)^\epsilon?$ We only have that $f(n) \leq C_\epsilon n^\epsilon, $ so rather $f(p^{n-i}) \leq C_\epsilon (p^{n-i})^\epsilon.$ Am I missing something? $\endgroup$ – inequalitynoob2 Jun 29 at 18:46
  • $\begingroup$ Naw. Just my usual carelessness. I'll leave it up in case someone can make it right. $\endgroup$ – marty cohen Jun 29 at 21:27

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