TITLE: Iwaniec & Kowalski partial sums of multiplicative functions
QUESTION [7 upvotes]: In their famous book Analytic Number Theory, Iwaniec and Kowalski include a section (1.6) on sums of multiplicative functions. The text seems to me unclear (or even wrong?) at some points. I would like some help to clarify the following points below.
For a multiplicative function $f$ and a prime $p$ they define
$$\sigma_p(f)=\sum_{\nu=0}^\infty f(p^\nu)p^{-\nu}\text{.}$$
Moreover $\mathcal{M}_f(x)$ denotes the partial sum $\sum_{n\leq x}f(n)$.

After devising an upper bound (1.79) for $\mathcal{M}_f(x)$ in terms of the $\sigma_p$'s, they attempt to apply it to the function $f(m)=\tau_k(m)^\ell$, where $\tau_k(m)$ is the number of ways to write $m$ as the product of $k$ positive integers (so that $\tau_2$ is the usual divisor counting function.) In order to do this, they state the bound
$$\sigma_f(p)\leq\left(1+\frac{1}{p}\right)^{k^\ell}\left(1+\frac{1}{p^2}\right)^c$$
where $c=c(k,\ell)$ is a positive constant. Where does this come from?
A bit later in the text, they devise a strategy to obtain an upper bound for $\mathcal{M}_f(x)$ when $f$ is non-negative and "completely sub-multiplicative", meaning that $f(mn)\leq f(m)f(n)$ for any $m,n\geq1$. They then apply the obtained bound to the case where $f$ is the characteristic function of the sums of two squares. However, this function is not completely sub-multiplicative! Indeed, $f(3)=0$, but $f(9)=1>f(3)^2$. Is this a clear mistake in the text or am I missing something?

Thanks in advance for any help.

REPLY [3 votes]: Joshua gave nice argument, let me provide my perspective on 1, with several arguments. The bottom line is that the constant $k^{\ell}$ comes from $f(1)=k^{\ell}$, and that very little information on $f$ is needed in order to derive this bound.
Claim 1: Let $F(x) = \sum_{\nu} a_{\nu} x^{\nu} = 1+a_{1}x+\ldots$ be a power series with radius of convergence $r>1/2$. Then
$$F(1/p) \le \left(1+\frac{1}{p}\right)^{a_1} \left(1+\frac{1}{p^2}\right)^{C}$$
for large enough $C$ depending on $F$.
Proof: Consider $$G(x) = \frac{F(x)(1+x)^{-a_1}-1}{x^2}.$$ It is analytic in $|x|<\min\{r,1\}$. Letting $C= \max\{1 , \max_{[0,1/2]}|G(x)|\}$ we find that
$$F(x) =  (1+x)^{a_1}(x^2 G(x)+1) \le (1+x)^{a_1}(1+Cx^2) \le (1+x)^{a_1}(1+x^2)^C$$
for $x \in [0,1/2]$. We now specialize to $x=1/p$. $\blacksquare$
Just apply the claim with $a_{\nu} = f(p^{\nu})$. In your case, however, there is much more structure. For instance, $\sigma_p(f)$ is a rational function of $1/p$, due to the following.
Claim 2: For a non-zero polynomial $P$ we have that $(\sum_n P(n) x^n )(1-x)^{\deg P +1}$ is a polynomial in $x$.
Proof: By linearity it suffices to consider $P_d(t)=\binom{t+d}{t}$ in which case $(\sum_n P_d(n) x^n) (1-x)^{d+1} \equiv 1$ by the binomial series.
This does not directly help us find an explicit value of $C$. Here is a strategy to get hold on such $C$, in the spirit of Joshua's argument.
Claim 3: $\binom{n+k-1}{n}^{\ell} \le \binom{n+k^{\ell}-1}{n}$.
Proof I: Since $\binom{a+b}{a} = \prod_{i=1}^{a} \left(1+\frac{b}{i}\right)$, the inequality can be written as $\prod_{i=1}^{n} \left(1+\frac{k-1}{i}\right)^{\ell} \le \prod_{i=1}^{n} \left(1+\frac{k^{\ell}-1}{i}\right)$. It suffices to show $\left(1+\frac{k-1}{i}\right)^{\ell} \le 1+\frac{k^{\ell}-1}{i}$ for each $i$. This follows from Hölder: if $X$ is a random variable assuming the values $1$ with probability $1-1/i$ and $k$ with probability $1/i$, the inequality just says $(\mathbb{E}X)^{\ell} \le \mathbb{E}(X^{\ell})$. Proof II: We can deduce the claim from a repeated application of the inequality $\binom{n+k_1-1}{n}\binom{n+k_2-1}{n}\le \binom{n+k_1 k_2-1}{n}$. This has a combinatorial interpretation (see Max's comment to this answer).
Claim 4: For all $0\le x \le 1/2$ we have $(1-x)^{-A} \le (1+x)^{A} (1+x^2)^C$ for $C \ge \max\{-A,5A/3\}$. Proof: The inequality simplifies as $(1-x^2)^{A}(1+x^2)^C \ge 1$, or $f(t)=A\log(1-t)+C\log(1+t) \ge 0$ where $t \in I:=[0,1/4]$. Note $f(0)=0$ and that $f$ increases in $I$ if $C$ is large enough, namely $f'=-\frac{A}{1-t}+\frac{C}{1+t} = \frac{C-A-t(A+C)}{1-t^2} \ge 0$ for $t \le (C-A)/(C+A)$, which is $\ge 1/4$ for $C \ge \max\{-A,5A/3\}$.
By Claim 3 and the binomial series, $\sigma_p(f) = \sum_{n \ge 0}\binom{n+k-1}{n}^{\ell} p^{-n} \le \sum_{n \ge 0}\binom{n+k^{\ell}-1}{n} p^{-n} = \sum_{n \ge 0}\binom{n+k^{\ell}-1}{k^{\ell}-1} p^{-n} = (1-1/p)^{-k^{\ell}}$. This is small enough by Claim 4 with $x=1/p$, and we can take $c=\frac{5}{3}k^{\ell}$. $\blacksquare$

Regarding question 2: I do not know what the authors had in mind. I do suggest reading Selberg's elementary, and not sufficiently well known, for $\sum_{n \le x} \alpha(n) \sim Cx/\sqrt{\log x}$ where $\alpha$ is the indicator function of integers divisible only by primes of the form $4k+1$. This is essentially the claim you need. It is proved in 'Lectures on Sieves' in his collected papers, second volume, pages 183-185.