2. Preliminaries for the proof of Theorem 1·1

Here we collect several key facts, based on the argument of Granville and Soundararajan [ Reference Granville and Soundararajan7 ]. The following is [ Reference Granville and Soundararajan7 , proposition 3·1].

Lemma 2·1. Let $f\in {\mathcal F}$ and define $g(n)=\sum_{d|n}f(d).$ Then

\begin{align*}\sum_{n\le x}\frac{f(n)}{n}=\frac{1}{x}\sum_{n\le x}g(n)+(1-\gamma)\frac{1}{x}\sum_{n\le x}f(n)+O\left(\frac{1}{(\log{x})^{1/5}}\right),\end{align*}

where $\gamma$ denotes Euler’s constant.

We will use a version of Halász’s theorem in the form given by Granville and Soundararajan [ Reference Granville and Soundararajan6 , Theorem 2b].

Lemma 2·2. Let $f\in {\mathcal F}$ and define $M=M(x)$ by

\begin{align*}\max_{|t|\ll \log{x}}\left|F(1+1/\log{x}+it)\right|:=e^{-M}(\log{x}),\end{align*}

where $F(s)=\sum_{n=1}^{\infty}{f(n)}/{n^{s}}$ for $\Re{s}\gt1.$ Then

\begin{align*}\sum_{n\le x}f(n)\ll \frac{(1+M)e^{-M}}{1+|\phi|}x+\frac{1}{(\log{x})^{2-\sqrt{3}+o(1)}}x,\end{align*}

where $\phi=\phi_f(x)$ is a real number in the range $|t|\le \log{x}$ where the function $t\rightarrow |F(1+1/\log{x}+it)|$ attains its maximum.

A result of Hall and Tenenbaum [ Reference Hall and Tenenbaum9 ] allows one to remove the dependence on t in Lemma 2·2 at the cost of a worse upper bound.

Lemma 2·3. For any $f\in {\mathcal F}$ we have

\begin{equation*}\sum_{n\le x}f(n)\ll x\exp\left(-\kappa\sum_{p\le x}\frac{1-f(p)}{p}\right),\end{equation*}

with $\kappa=0.32867$ .

The following estimate is well known, for a proof see [ Reference Granville and Soundararajan5 , Equation 2.1.6].

Lemma 2·4 For any $x\ge 2$ and $t\in \mathbb{R}$ , we have

\begin{align*}\sum_{p\le x}\frac{\Re(p^{it})}{p}=\log\log x+O(1) \quad {if} \quad |t|\le 1/\log{x}\end{align*}

\begin{align*}\sum_{p\le x}\frac{\Re(p^{it})}{p}=\log(1/|t|)+O(1) \quad {if} \quad \frac{1}{\log{x}}\le |t|\le 100\end{align*}

and

\begin{align*}\left|\sum_{p\le x}\frac{\Re(p^{it})}{p} \right|\le \log\log(2+|t|)+O(1) \quad {if} \quad |t|\ge 100.\end{align*}

3. Lower bounds for nonnegative multiplicative functions

We next show how ideas from [ Reference Granville, Koukoulopoulos and Matomäki8 , Reference Matomäki and Shao14 ] may be used to obtain lower bounds for the function g(n) defined in Lemma 2·1. The main result of this section is Proposition 3·3.

The following is due to Matomäki and Shao [ Reference Matomäki and Shao14 , Hypothesis P] which improves on the work of Granville, Koukoulopoulos and Matomäki [ Reference Granville, Koukoulopoulos and Matomäki8 ].

Lemma 3·1 Fix $\lambda\in (0,1)$ . If x is sufficiently large and u,v satisfy

\begin{equation*}1\le u \le v \le \frac{\log{x}}{1000\log\log{x}},\end{equation*}

and ${\mathcal P}$ is a subset of the primes in $(x^{1/v},x^{1/u}]$ with

\begin{equation*}\sum_{\substack{p\in {\mathcal P}}}\frac{1}{p}\ge \frac{1+\lambda}{u},\end{equation*}

then there exists an integer $k\in [u,v]$ such that

\begin{align*}\left|\left\{ (p_1,\dots,p_k)\in {\mathcal P}^{k} \ : \ \frac{x}{2}\le p_1\dots p_k\le x \right\} \right|\ge \pi_v \frac{x}{v^{k}\log{x}},\end{align*}

where $\pi_v$ is a constant with $\pi_v=v^{-o(v)}$ as $v\rightarrow \infty$ . If u is fixed and $v\ge 1000u^2/\lambda^2$ , one can take $k\le e^{-1/u}v$ .

Our next result uses arguments in the spirit of [ Reference Hildebrand12 , p. 218].

Lemma 3·2 Let g be a nonnegative multiplicative function satisfying

(3·1) \begin{align}g(p^{j})\le j+1\end{align}

for each prime power $p^{j}$ . For any real numbers $x,z\ge 1$ satisfying

(3·2) \begin{align}z\le x^{1/C_1}\end{align}

for a sufficiently large absolute constant $C_1$ we have

\begin{align*}\sum_{\substack{n\le x \\ p|n \implies p\le z}}\frac{g(n)}{n}\gg \exp\left(\sum_{p\le z}\frac{g(p)}{p}\right).\end{align*}

Proof. Let $\delta={1}/{\log{z}},$ and consider

\begin{align*}\sum_{\substack{n\le x \\ p|n \implies p\le z}}\frac{g(n)}{n}&\ge \sum_{\substack{n\ge 1 \\ p|n \implies p\le z}}\frac{g(n)}{n}-\sum_{\substack{n\gt x \\ p|n \implies p\le z}}\frac{g(n)}{n} \\&\ge \prod_{p\le z}\left(1+\sum_{j\ge 1}\frac{g(p^{j})}{p^{j}} \right)-\frac{1}{x^{\delta}}\sum_{\substack{n\ge 1 \\ p|n \implies p\le z}}\frac{g(n)}{n^{1-\delta}}.\end{align*}

By (3·1)

\begin{align*}\log\left(\sum_{\substack{n\ge 1 \\ p|n \implies p\le z}}\frac{g(n)}{n^{1-\delta}}\right)&=\sum_{p\le z}\frac{g(p)}{p}+O(1)\end{align*}

which implies that for some absolute constant C

\begin{align*}\sum_{\substack{n\le x \\ p|n \implies p\le z}}\frac{g(n)}{n}&\gg \exp\left(\sum_{p\le z}\frac{g(p)}{p}\right)\left(1-C\exp\left(-\frac{\log{x}}{\log{z}}\right)\right) \gg \exp\left(\sum_{p\le z}\frac{g(p)}{p}\right)\end{align*}

assuming $C_1$ in (3·2) is large enough. This completes the proof.

We are now in a position to establish the main result of this section.

proposition 3·3 Let $\varepsilon\gt0$ be sufficiently small, $f\in {\mathcal F}$ and define $g(n)=\sum_{d|n}f(d).$ For $x\ge 2$ and $0\lt\delta \lt1$ which may depend on x, let ${\mathcal P}_{\delta}$ denote the set

\begin{equation*}{\mathcal P}_{\delta}=\{ p\le x \ \ : \text{prime}, \ \ f(p)\ge -\delta \}.\end{equation*}

Suppose that for some v satisfying

\begin{equation*}\frac{40000}{\varepsilon^2}\le v \le \frac{\log{x}}{1000\log\log{x}},\end{equation*}

we have

(3·3) \begin{align}\sum_{\substack{p\in {\mathcal P}_{\delta} \\ x^{1/v}\le p \le x}}\frac{1}{p}\ge 1+\varepsilon.\end{align}

Then

\begin{align*}\sum_{n\le x}g(n) \gg \varepsilon^4\left(\frac{(1-\delta)}{v}\right)^{v(1+o(1))/e}\exp\left(\sum_{p\le x}\frac{f(p)}{p}\right)x.\end{align*}

Proof. Let $G=\sum_{n\le x}g(n),$ and define

(3·4) \begin{align}{\mathcal A}=\{ \ p \ \ \text{prime} \ : \ p\le x^{1/v} \ \}, \quad {\mathcal B}={\mathcal P}_{\delta}\cap (x^{1/v},x],\end{align}

so that

\begin{align*}G& \ge \sum_{\substack{a\le x^{\varepsilon/5} \\ p|a \implies p\in {\mathcal A}}}g(a)\sum_{\substack{b\le x/a \\ p|b \implies p\in {\mathcal B}}}g(b).\end{align*}

If $p\in {\mathcal P}_{\delta},$ then since f is completely multiplicative, for any integer $j\ge 1$ we have

\begin{equation*}|f(p^{j+1})|\le |f(p^{j})|\end{equation*}

and hence

\begin{equation*}g(p^{\alpha})\ge 1+f(p)+\sum_{2\le j \le \alpha}f(p^{j})\ge 1-\delta\end{equation*}

since

\begin{equation*}\sum_{2\le j \le \alpha}f(p^{j})\ge 0.\end{equation*}

This implies that

(3·5) \begin{align}G\ge \sum_{\substack{a\le x^{\varepsilon/5} \\ p|a \implies p\in {\mathcal A}}}g(a)\sum_{\substack{b\le x/a \\ p|b \implies p\in {\mathcal B}}}(1-\delta)^{\Omega(b)},\end{align}

where $\Omega(n)$ counts the number of prime factors (with multiplicity) of an integer n and we have used the inequality $\omega(n)\le \Omega(n)$ , where $\omega(n)$ counts the number of distinct prime factors of an integer n.

Our next step is to apply Lemma 3·1 to the inner summation over b. This requires verifying for each

(3·6) \begin{align}a\le x^{\varepsilon/5}\end{align}

the following lower bound holds

(3·7) \begin{align}\sum_{\substack{ p\in {\mathcal P}_{\delta} \\ (x/a)^{1/v}\le p \le x/a}}\frac{1}{p}\ge 1+\frac{\varepsilon}{2}.\end{align}

By (3·3) and the fact that

\begin{align*}[x^{1/v},x/a]\subseteq [\left(x/a\right)^{1/v},x/a]\end{align*}

we have

(3·8) \begin{align}\sum_{\substack{ p\in {\mathcal P}_{\delta} \\ (x/a)^{1/v}\le p \le x/a}}\frac{1}{p}\ge \sum_{\substack{ p\in {\mathcal P}_{\delta} \\ x^{1/v}\le p \le x}}\frac{1}{p}-\sum_{\substack{ p\in {\mathcal P}_{\delta} \\ x/a\le p \le x }}\frac{1}{p}\ge 1+\varepsilon-\sum_{\substack{ p\in {\mathcal P}_{\delta} \\ x/a\le p \le x }}\frac{1}{p}\end{align}

and by (3·6)

\begin{align*}\sum_{\substack{ p\in {\mathcal P}_{\delta} \\ x/a\le p \le x }}\frac{1}{p}&\le \log\log{x}-\log\log{(x/a)}+o(1) \\ & \le \log\log{x}-\log\log{(x^{1-\varepsilon/5})}\le \frac{\varepsilon}{4}+o(1)\cdot\end{align*}

Combining the above with (3·8) implies (3·7). By Lemma 3·1, for each a satisfying (3·6), if v satisfies

(3·9) \begin{align}v\ge \frac{40000}{\varepsilon^2},\end{align}

then there exists an integer $k\le {v}/{e}$ such that

\begin{equation*}\left|\left\{ (p_1,\dots,p_k)\in {\mathcal P}_{\delta}^{k} \ : \ \frac{x}{2a}\le p_1\dots p_k \le \frac{x}{a} \right\} \right| \ge v^{o(v)}\frac{x}{v^{k}a\log{x}}.\end{equation*}

Hence

\begin{align*}\sum_{\substack{b\le x/a \\ p|b \implies p\in {\mathcal B}}}(1-\delta)^{\Omega(b)}&\ge v^{o(v)}\left(\frac{(1-\delta)}{v}\right)^{k}\frac{x}{a \log{x}} \\ & \ge \left(\frac{(1-\delta)}{v}\right)^{v(1+o(1))/e}\frac{x}{a \log{x}},\end{align*}

which combined with (3·5) implies

(3·10) \begin{align}G\ge \left(\frac{(1-\delta)}{v}\right)^{v(1+o(1))/e}\frac{x}{\log{x}}\sum_{\substack{a\le x^{\varepsilon/5} \\ p|a \implies p\in {\mathcal A}}}\frac{g(a)}{a}.\end{align}

Recalling (3·4)

\begin{align*}\sum_{\substack{a\le x^{\varepsilon/5} \\ p|a \implies p\in {\mathcal A}}}\frac{g(a)}{a}=\sum_{\substack{a\le x^{\varepsilon/5} \\ p|a \implies p\le x^{1/v}}}\frac{g(a)}{a}.\end{align*}

If $\varepsilon$ is sufficiently small then by (3·9) the condition that $C_1$ in (3·2) is sufficiently large is satisfied. Hence we may apply Lemma 3·2 to deduce that

\begin{align*}\sum_{\substack{a\le x^{\varepsilon/5} \\ p|a \implies p\in {\mathcal A}}}\frac{g(a)}{a}\gg \exp\left(\sum_{p\le x^{\varepsilon^2/40000}}\frac{g(p)}{p}\right).\end{align*}

Using (3·10) gives

\begin{align*}G\gg \left(\frac{(1-\delta)}{v}\right)^{v(1+o(1))/e}\frac{x}{\log{x}}\exp\left(\sum_{p\le x^{\varepsilon^2/40000}}\frac{g(p)}{p}\right),\end{align*}

and the result follows after noting

\begin{align*}\sum_{p\le x^{\varepsilon^2/40000}}\frac{g(p)}{p}&\ge \sum_{p\le x}\frac{g(p)}{p}+2\log{\varepsilon}+O(1) \\&\ge \log\log{x}+\sum_{p\le x}\frac{f(p)}{p}+4\log{\varepsilon}+O(1).\end{align*}

4. Proof of Theorem 1·1

First note we may assume that

(4·1) \begin{align}\sum_{n\le x}\frac{f(n)}{n}\le -\frac{1}{(\log{x})^{1/6}},\end{align}

since otherwise, for any constant C there exists $x_0$ such that if $x\ge x_0$ then

\begin{align*}\sum_{n\le x}\frac{f(n)}{n}\gt -\frac{1}{(\log{x})^{1/6}}\gt -\frac{C(\log\log\log{x})^2}{(\log\log{x})}.\end{align*}

By (4·1) and Lemma 2·1, for sufficiently large x we have

\begin{align*}\sum_{n\le x}\frac{f(n)}{n}&\ge \frac{1}{x}\sum_{n\le x}g(n)+(1-\gamma)\frac{1}{x}\sum_{n\le x}f(n)-\frac{1}{2(\log{x})^{1/6}} \\&\ge \frac{1}{x}\sum_{n\le x}g(n)+(1-\gamma)\frac{1}{x}\sum_{n\le x}f(n)+\frac{1}{2}\sum_{n\le x}\frac{f(n)}{n},\end{align*}

which implies that

(4·2) \begin{align}\sum_{n\le x}\frac{f(n)}{n}\ge 2\left(\frac{1}{x}\sum_{n\le x}g(n)-\frac{(1-\gamma)}{x}\left|\sum_{n\le x}f(n)\right|\right).\end{align}

Hence from Lemma 2·2, there exists an absolute constant $C_1$ such that

(4·3) \begin{align}\sum_{n\le x}\frac{f(n)}{n}\ge 2\left(\frac{1}{x}\sum_{n\le x}g(n)-\frac{C_1(1+M)e^{-M}}{1+|t|}\right),\end{align}

where M is defined by

(4·4) \begin{align}\left|F(1+1/\log{x}+it)\right|=e^{-M}(\log{x}),\end{align}

for some $|t|\ll \log{x}.$ Let $\varepsilon\gt0$ be small, define

(4·5) \begin{align}\varepsilon_1=\frac{C}{\log\log\log{x}}\end{align}

for a suitable absolute constant C and put

(4·6) \begin{align}\delta=1-\varepsilon_1, \quad v=\frac{\log\log{x}}{\log\log\log{x}}.\end{align}

Consider the set

\begin{align*}{\mathcal P}_{\delta}=\{p\le x \ : \ \ p \ \text{prime}, \ f(p)\ge -\delta\}.\end{align*}

We distinguish between two cases. Either

(4·7) \begin{align}\sum_{\substack{x^{1/v}\le p \le x \\ p\in {\mathcal P}_{\delta}}}\frac{1}{p}\le 1+\varepsilon\end{align}

or

(4·8) \begin{align}\sum_{\substack{x^{1/v}\le p \le x \\ p\in {\mathcal P}_{\delta}}}\frac{1}{p}\gt 1+\varepsilon.\end{align}

Suppose first that (4·7) holds and consider

\begin{align*}\Re\left(\sum_{p\le x}\frac{1-f(p)p^{-it}}{p}\right)&\ge \Re\left(\sum_{\substack{x^{1/v}\le p \le x \\ p\not \in {\mathcal P}_{\delta}}}\frac{1-f(p)p^{-it}}{p}\right)+O(1) \\& \ge \Re\left(\sum_{\substack{x^{1/v}\le p \le x \\ p\not \in {\mathcal P}_{\delta}}}\frac{1+p^{-it}}{p}\right)-\sum_{\substack{x^{1/v}\le p \le x \\ p\not \in {\mathcal P}_{\delta}}}\frac{|1+f(p)|}{p}+O(1).\end{align*}

If $p\not \in {\mathcal P}_{\delta},$ then recalling (4·6)

\begin{equation*}|1+f(p)|\le 1-\delta = \varepsilon_1,\end{equation*}

which implies for such values of p

\begin{align*}\Re\left(1-f(p)p^{-it}\right)&=\Re\left(1+p^{-it}\right)-\Re\left((f(p)+1)p^{-it}\right) \\&\ge \Re\left(1+p^{-it}\right)-\varepsilon_1\end{align*}

and consequently by (4·7), we have

(4·9) \begin{align}\Re\left(\sum_{p\le x}\frac{1-f(p)p^{-it}}{p}\right)&\ge \Re\left(\sum_{\substack{x^{1/v}\le p \le x }}\frac{1+p^{-it}}{p}\right)-\varepsilon_1 \log{v}+O(1).\end{align}

We next consider two subcases depending on the size of t relative to x. Suppose first that

\begin{align*}|t|\ge 100.\end{align*}

By Lemma 2·4

\begin{align*}\left|\Re\left(\sum_{\substack{x^{1/v}\le p \le x }}\frac{p^{-it}}{p}\right)\right|\le 2\log\log{|t|}+O(1)\end{align*}

and hence from (4·9)

\begin{align*}\Re\left(\sum_{p\le x}\frac{1-f(p)p^{-it}}{p}\right)&\ge (1-\varepsilon_1) \log{v}+O(\log\log{|t|}).\end{align*}

After recalling (4·5) and (4·6), we get

\begin{align*}e^{-M}=\frac{1}{\log{x}}\left|F(1+1/\log{x}+it)\right|\ll \frac{(\log{|t|})^{O(1)}}{v^{1-\varepsilon_1}}\ll \frac{(\log\log\log{x})(\log{|t|})^{O(1)}}{\log\log{x}}\end{align*}

which when combined with (4·3), (4·4) and the bound

\begin{equation*}1+M\ll \log\log\log{x},\end{equation*}

gives

(4·10) \begin{align}\sum_{n\le x}\frac{f(n)}{n}\ge -\frac{C(\log\log\log{x})^2}{(\log\log{x})}\frac{(\log{|t|})^{O(1)}}{|t|}\ge -\frac{C'(\log\log\log{x})^2}{(\log\log{x})}\end{align}

for some absolute constant C’. If $|t|\le 100$ then by Lemma 2·4

\begin{equation*}\Re\left(\sum_{\substack{x^{1/v}\le p \le x }}\frac{p^{-it}}{p}\right)\ge -C'\end{equation*}

for some absolute constant C’. Hence from (4·9)

\begin{align*}\Re\left(\sum_{p\le x}\frac{1-f(p)p^{-it}}{p}\right)&\ge (1-\varepsilon_1)\log{v}+O(1).\end{align*}

After recalling (4·5) and (4·6), we get

\begin{align*}e^{-M}=\frac{1}{\log{x}}\left|F(1+1/\log{x}+it)\right|\ll \frac{1}{v^{1-\varepsilon_1}}\ll \frac{(\log\log\log{x})}{\log\log{x}}\end{align*}

which when combined with (4·3), (4·4) and the fact that $g(n)\ge 0$ , gives

(4·11) \begin{align}\sum_{n\le x}\frac{f(n)}{n}\ge -\frac{C'(\log\log\log{x})^2}{(\log\log{x})}\end{align}

for some absolute constant C ^′. This completes our analysis of the case when (4·7) holds. Suppose next that (4·8) holds. By (4·3) and Proposition 3·3

\begin{align*}\frac{1}{x}\sum_{n\le x}g(n)\gg \varepsilon^4\left(\frac{(1-\delta)}4{v}\right)^{v(1+o(1))/e}(\log{x})\exp\left(-\sum_{p\le x}\frac{1-f(p)}{p}\right) .\end{align*}

Recalling (4·6), we have

\begin{align*}\left(\frac{(1-\delta)}{v}\right)^{v(1+o(1))/e}\log{x}&\gg \exp\left(-\frac{v(1+o(1))\log{(v/\varepsilon_1)}}{e}\right)(\log{x})^{1-1/e} \\&\gg (\log{x})^{1-1/e+o(1)}.\end{align*}

Upon applying Lemma 2·3 and (4·2) we deduce

\begin{align*}\sum_{n\le x}\frac{f(n)}{n}\ge C_1(\log{x})^{1-o(1)}\exp\left(-\sum_{p\le x}\frac{1-f(p)}{p}\right)-C_2\exp\left(-\kappa\sum_{p\le x}\frac{1-f(p)}{p}\right),\end{align*}

for some absolute constants $C_1,C_2$ . Optimising the function

\begin{equation*}H(u)=C_1(\log{x})^{1-o(1)}\exp\left(-u\right)-C_2\exp\left(-\kappa u\right)\end{equation*}

subject to the restriction $u\ge 0$ implies

\begin{align*}\sum_{n\le x}\frac{f(n)}{n}\ge -\frac{c}{(\log{x})^{1/100}}\ge -\frac{C(\log\log\log{x})}{\log\log{x}},\end{align*}

and combined with (4·10) we complete the proof.

5. Preliminaries for the proof of Theorem 1·2

We next present some preliminary results required for the proof of Theorem 1·2. The following is known as Bernstein’s inequality.

Lemma 5·1 Let $X_1,\dots,X_n$ be a sequence of independent random variables satisfying

\begin{equation*}P(|X_i|\le M)=1, \quad \mathbb{E}(X_i)=0, \quad 1\le i \le n.\end{equation*}

For all $t\gt0$ we have

\begin{align*}P\left(\sum_{i=1}^{n}X_i\ge t\right)\le \exp\left(-\frac{t^2/2}{\sum_{i=1}^{n}\mathbb{E}(X_i^2)+Mt/3}\right).\end{align*}

For $1\le z \le w$ define

\begin{equation*}\Psi(w,z)=\sum_{\substack{n\le w \\ p|n \implies p\le z}}1.\end{equation*}

The following is due to Canfield, Erdös and Pomerance [ Reference Canfield, Erdös and Pomerance4 , Corollary p. 15].

Lemma 5·2 Let $1\le z\le w$ be real numbers and define

\begin{equation*}u=\frac{\log{w}}{\log{z}}.\end{equation*}

If

\begin{equation*}u\le z^{1-o(1)} \quad \text{and} \quad u\rightarrow \infty\end{equation*}

then

(5·1) \begin{align}\Psi(w,z)= \frac{w}{u^{u(1+o(1))}}.\end{align}

Recall that ${\mathcal F}_1$ denotes the space of completely multiplicative functions f satisfying $f(p)=\pm 1$ for $p\le x.$ Let $\mu$ denote the counting measure on ${\mathcal F}_1$ , normalised so that

\begin{equation*}\mu({\mathcal F}_1)=1\end{equation*}

and let $\mathbb{E}$ denote expectation with respect to $\mu$ . We will make considerable use of the majorant principle.

Lemma 5·3 Let q be an even integer and $a_n$ and $b_n$ be two sequences of real numbers satisfying

\begin{equation*}|a_n|\le b_n.\end{equation*}

Then

\begin{align*}\mathbb{E}\left[\left(\sum_{n\le x}a_nf(n)\right)^{q}\right]\le \mathbb{E}\left[\left(\sum_{n\le x}b_nf(n)\right)^{q}\right].\end{align*}

Proof. Recall that $f(p)=\pm 1$ with probability $1/2$ . If q is an even integer, then for any real numbers $a_n$ we have

\begin{align*}\mathbb{E}\left[\left(\sum_{n\le x}a_nf(n)\right)^{q}\right]=\sum_{\substack{1\le n_1,\cdots,n_{q} \le x \\ \exists s\in \mathbb{N} \ : \ n_1n_2\cdots n_{q}=s^2}}a_{n_1}\ {\cdots}\ {a_{n_{q}}}.\end{align*}

Hence if $|a_n|\le b_n$ then

\begin{align*}\mathbb{E}\left[\left(\sum_{n\le x}a_nf(n)\right)^{q}\right]\le \sum_{\substack{1\le n_1,\cdots,n_{q} \le x \\ \exists s\in \mathbb{N} \ : \ n_1n_2\ {\cdots}\ n_{q}=s^2}}b_{n_1}\cdots b_{n_q}=\mathbb{E}\left[\left(\sum_{n\le x}b_nf(n)\right)^{q}\right].\end{align*}

A particular consequence of Lemma 5·3 is as follows.

Corollary 5·4 Let q be an even integer and $c_n$ a sequence of nonnegative real numbers. For any ${\mathcal A},{\mathcal B}\subseteq\mathbb{N}$ satisfying ${\mathcal A}\subseteq {\mathcal B}$ , we have

\begin{align*}\mathbb{E}\left[\left(\sum_{\substack{n\le x \\ n\in {\mathcal A}}}c_nf(n)\right)^{q}\right]\le \mathbb{E}\left[\left(\sum_{\substack{n\le x \\ n\in {\mathcal B}}}c_nf(n)\right)^{q}\right].\end{align*}

Proof. This follows from Lemma 5·3 taking

\begin{equation*}a_n=\begin{cases}c_n \quad \text{if $n\in {\mathcal A}$} \\ 0 \quad \text{otherwise} \end{cases}\end{equation*}

and

\begin{equation*}b_n=\begin{cases}c_n \quad \text{if $n\in {\mathcal B}$} \\ 0 \quad \text{otherwise} \end{cases}\end{equation*}

after noting the condition $a_n\le b_n$ follows from the assumption that ${\mathcal A}\subseteq {\mathcal B}$ .

Our main input for Theorem 1·2 is the following.

Corollary 5·5. For any $M\gt0$ , there exists a constant $C\gt0$ such that for sufficiently large x

\begin{align*}\mu\left(\left\{ f\in {\mathcal F}_1 \ : \ \left|\sum_{n\le x}\left\{\frac{x}{n}\right\}f(n)\right|\ge \frac{M}{\log{x}}x\right\} \right)\le \exp \left(-\exp\left(\frac{(\log \log \log{x})\log{x}}{C\log\log{x}}\right)\right).\end{align*}

Corollary 5·5 is a consequence of the following moment inequality.

proposition 5·6. For any $C \gt 0$ , positive real number x sufficiently large in terms of C and any even integer q, we have

\begin{align*}\left(\mathbb{E}\left[\left(\sum_{n\le x}f(n)\right)^{q} \right]\right)^{1/q}& \ll x^{7/8}\exp\left(\left(q\exp\left(-\frac{(\log\log\log{x})\log{x}}{C\log\log{x}}\right)+2\log\log{x}\right)\right) \\&\quad+\left( \frac{x}{(\log{x})^{C/4-1}}\right).\end{align*}

We first show that Proposition 5·6 implies Corollary 5·5 which in turn implies Theorem 1·2. We then prove Proposition 5·6 in Section 5·3.

5·1. Deduction of Theorem 1·2 from Corollary 5·5:

Let $g(n)=\sum_{d\vert n}f(d)$ and apply (1·8) to get

\begin{equation*}\sum_{n\le x}\frac{f(n)}{n}=\frac{1}{x}\sum_{n\le x}g(n)+\frac{1}{x}\sum_{n\le x}f(n)\left\{\frac{x}{n}\right\}.\end{equation*}

By Lemma 5·1, the probability that

\begin{equation*}\sum_{p\le x}f(p)\le -\frac{x}{2\log x}\end{equation*}

is bounded by

\begin{equation*}O\left(\exp \left(-\frac{x}{100\log x}\right)\right).\end{equation*}

Since $g(n)\ge 0$ for all $n\ge 1,$ this implies that for sufficiently large x the probability that

\begin{equation*}\frac{1}{x}\sum_{n\le x}g(n)\ge \frac{1}{x}\sum_{p\le x} (1+f(p))\ge \frac{1}{2\log x}\end{equation*}

is equal to

\begin{equation*}1-O\left(\exp\left(-\frac{x}{100\log x}\right)\right).\end{equation*}

After applying Corollary 5·5 with $M=1/2$ the desired result follows.

5·2. Deduction of Corollary 5·5 from Proposition 5·6

By Lemma 5·3, for any even integer q

(5·2) \begin{align}\mathbb{E}\left[\left(\sum_{n\le x}\left\{\frac{x}{n}\right\}f(n)\right)^{q} \right]\le \mathbb{E}\left[\left(\sum_{n\le x}f(n)\right)^{q} \right].\end{align}

We next use Markov’s inequality, which states that for a nonnegative random variable X and real number $y\gt0$ we have

\begin{equation*}P(X\ge y)\le \frac{E(X)}{y}.\end{equation*}

Applying this with

\begin{equation*}X=\left|\sum_{n\le x}\left\{\frac{x}{n}\right\}f(n)\right|^{q}\end{equation*}

and

\begin{equation*}y=\left(\frac{M}{\log{x}}x\right)^{q}\end{equation*}

we get

\begin{align*}\mu\left(\left\{ f\in {\mathcal F}_1 \ : \ \left|\sum_{n\le x}\left\{\frac{x}{n}\right\}f(n)\right|\ge \frac{M}{\log{x}}x\right\} \right)\le \left(\frac{M}{\log{x}}x\right)^{-q}\mathbb{E}\left[\left(\sum_{n\le x}\left\{\frac{x}{n}\right\}f(n)\right)^{q} \right]\end{align*}

which by (5·2) implies that

\begin{align*}\mu\left(\left\{ f\in {\mathcal F}_1 \ : \ \left|\sum_{n\le x}\left\{\frac{x}{n}\right\}f(n)\right|\ge \frac{M}{\log{x}}x\right\} \right)\le \left(\frac{M}{\log{x}}x\right)^{-q}\mathbb{E}\left[\left(\sum_{n\le x}f(n)\right)^{q} \right].\end{align*}

Applying Proposition 5·6 with the choice of parameter

\begin{equation*}q=\exp\left(\frac{(\log\log\log{x})(\log{x})}{C\log\log{x}}\right)\end{equation*}

completes the proof of Corollary 5·5.

5·3. Proof of Proposition 5·6

Let

(5·3) \begin{align} \varepsilon=\frac{\log\log\log{x}}{4C\log\log{x}},\end{align}

where C is a sufficiently large absolute constant. Since

\begin{align*}\sum_{\substack{m\le x}}f(m)=\sum_{\substack{n\ell\le x \\ p|n \implies p\ge x^{\varepsilon} \\ p|\ell \implies p \lt x^{\varepsilon}}}f(\ell)f(n)=\sum_{j\le \log{x}+1}\sum_{\substack{n\ell\le x \\ p|n \implies p\ge x^{\varepsilon} \\ p|\ell \implies p \lt x^{\varepsilon} \\ e^{j}\le \ell \lt e^{j+1}}}f(\ell)f(n),\end{align*}

by Minkowski’s inequality and the majorant principle to ignore the condition $n\ell \le x$ we get

\begin{align*} \mathbb{E}\left[\left(\sum_{n\le x}f(n)\right)^{q} \right]^{1/q}\le \sum_{j\le \log{x}+1}\mathbb{E}\left[\left(\sum_{\substack{n\le x/e^{j} \\ p|n \implies p\ge x^{\varepsilon}}}f(n)\sum_{\substack{\ell \le e^{j+1} \\ p|\ell \implies p\le x^{\varepsilon}}}f(\ell)\right)^{q} \right]^{1/q}\end{align*}

and hence

(5·4) \begin{align} \mathbb{E}\left[\left(\sum_{n\le x}f(n)\right)^{q} \right]^{1/q}\le S_1+S_2,\end{align}

where

(5·5) \begin{align}S_1=\sum_{j\le \log{x}/2}\mathbb{E}\left[\left(\sum_{\substack{n\le x/e^{j} \\ p|n \implies p\ge x^{\varepsilon}}}f(n)\sum_{\substack{\ell \le e^{j+1} \\ p|\ell \implies p\lt x^{\varepsilon}}}f(\ell)\right)^{q} \right]^{1/q}\end{align}

and

\begin{align*}S_2=\sum_{\log{x}/2\lt j\le \log{x}+1}\mathbb{E}\left[\left(\sum_{\substack{n\le x/e^{j} \\ p|n \implies p\ge x^{\varepsilon}}}f(n)\sum_{\substack{\ell \le e^{j+1} \\ p|\ell \implies p\lt x^{\varepsilon}}}f(\ell)\right)^{q} \right]^{1/q}.\end{align*}

Consider first $S_2$ . Taking a maximum over the outer summation, we see that there exists $j\ge \log{x}/2$ such that

\begin{align*}S_2\le (\log{x})\mathbb{E}\left[\left(\sum_{\substack{n\le x/e^{j} \\ p|n \implies p\ge x^{\varepsilon}}}f(n)\sum_{\substack{\ell \le e^{j+1} \\ p|\ell \implies p\lt x^{\varepsilon}}}f(\ell)\right)^{q} \right]^{1/q}.\end{align*}

Define $\beta$ by $x^{\beta}=e^{j+1}$ , and note that $\beta\ge 1/2.$ Since f is completley multiplicative, $|f(n)|\le 1$ for each $n\in \mathbb{N}$ and hence

\begin{align*}\sum_{\substack{n\le x/e^{j} \\ p|n \implies p\ge x^{\varepsilon}}}f(n)\sum_{\substack{\ell \le e^{j+1} \\ p|\ell \implies p\lt x^{\varepsilon}}}f(\ell)\ll x^{1-\beta}\Psi(x^{\beta},x^{\varepsilon}).\end{align*}

This implies

(5·6) \begin{align}S_2\ll (\log{x})x^{1-\beta}\Psi(x^{\beta},x^{\varepsilon}).\end{align}

We apply Lemma 5·2 with

\begin{equation*}w=x^{\beta}, \quad z=x^{\varepsilon}.\end{equation*}

Define

\begin{equation*}u=\frac{\log{w}}{\log{z}}=\frac{\beta}{\varepsilon},\end{equation*}

and note that by (5·3) $u\rightarrow \infty$ and

\begin{align*}u\le \frac{10C\log\log{x}}{\log\log\log{x}}\le x^{10C\log\log\log{x}/\log\log{x}}\le x^{\varepsilon/2}.\end{align*}

Combining (5·1) and (5·6) gives

\begin{align*}S_2&\ll (\log{x})\left(\frac{\varepsilon}{\beta}\right)^{(1+o(1))\beta/\varepsilon}x.\end{align*}

By (5·3), if x if sufficiently large in terms of C then

\begin{align*}\left(\frac{\varepsilon}{\beta}\right)^{(1+o(1))\beta/\varepsilon}\ll \frac{1}{(\log{x})^{C}}\end{align*}

and hence

\begin{equation*}S_2\ll \frac{x}{(\log{x})^{C}}.\end{equation*}

Consequently, using (5·4)

(5·7) \begin{align} \mathbb{E}\left[\left(\sum_{n\le x}f(n)\right)^{q} \right]^{1/q}\ll S_1+\frac{x}{(\log{x})^{C-1}}.\end{align}

We now turn our attention to estimating $S_1.$ We recall (5·5) and write

\begin{align*}S_1=\sum_{j\le \log{x}/2}\mathbb{E}\left[\left(\sum_{\substack{n\le x/e^{j} \\ p|n \implies p\ge x^{\varepsilon} \\ \ell \le e^{j+1} \\ p|\ell \implies p\lt x^{\varepsilon}}}(\ell n^{3/4})\frac{f(n\ell)}{n^{3/4}\ell}\right)^{q} \right]^{1/q}.\end{align*}

If $n,\ell$ satisfy conditions in the above summation, we have

\begin{align*}\ell n^{3/4}\ll e^{j}\left(\frac{x}{e^{j}}\right)^{3/4}\ll x^{3/4}e^{j/4}\end{align*}

and hence Lemma 5·3 implies

\begin{align*}S_1\ll x^{3/4}\sum_{j\le \log{x}/2}e^{j/4}\mathbb{E}\left[\left(\sum_{\substack{n\le x/e^{j} \\ p|n \implies p\ge x^{\varepsilon} \\ \ell \le e^{j+1} \\ p|\ell \implies p\lt x^{\varepsilon}}}\frac{f(n\ell)}{n^{3/4}\ell}\right)^{q} \right]^{1/q}.\end{align*}

By Corollary 5·4

\begin{align*}\mathbb{E}\left[\left(\sum_{\substack{n\le x/e^{j} \\ p|n \implies p\ge x^{\varepsilon} \\ \ell \le e^{j+1} \\ p|\ell \implies p\lt x^{\varepsilon}}}\frac{f(n\ell)}{n^{3/4}\ell}\right)^{q} \right]^{1/q}\ll \mathbb{E}\left[\left(\sum_{\substack{n\le x \\ p|n \implies x^{\varepsilon}\le p\le x }}\frac{f(n)}{n^{3/4}}\sum_{\substack{\ell \ge 1 \\ p|\ell \implies p\lt x^{\varepsilon}}}\frac{f(\ell)}{\ell}\right)^{q} \right]^{1/q}\end{align*}

and hence by another application of Corollary 5·4

\begin{align*}S_1&\ll x^{7/8}\mathbb{E}\left[\left(\sum_{\substack{n\le x \\ p|n \implies x^{\varepsilon}\le p\le x }}\frac{f(n)}{n^{3/4}}\sum_{\substack{\ell \ge 1 \\ p|\ell \implies p\lt x^{\varepsilon}}}\frac{f(\ell)}{\ell}\right)^{q} \right]^{1/q} \\& \ll x^{7/8}\mathbb{E}\left[\left(\prod_{x^{\varepsilon}\le p\le x}\left(1-\frac{f(p)}{p^{3/4}}\right)^{-1}\right)^{q} \right]^{1/q}\mathbb{E}\left[\left(\prod_{p\le x^{\varepsilon}}\left(1-\frac{f(p)}{p}\right)^{-1}\right)^{q} \right]^{1/q}.\end{align*}

We next expand the expectation as a product to get

\begin{align*}\mathbb{E}\left[\left(\prod_{x^{\varepsilon}\le p\le x}\left(1-\frac{f(p)}{p^{3/4}}\right)^{-1}\right)^{q} \right]&\le \exp(O(q))\mathbb{E}\left[\left(\prod_{x^{\varepsilon}\le p\le x}\left(1+\frac{f(p)}{p^{3/4}}\right)\right)^{q} \right] .\end{align*}

Observe that

\begin{align*}\mathbb{E}\left[\left(\prod_{p\le x}\left(1+\frac{f(p)}{p^{3/4}}\right)\right)^{q} \right]=\prod_{p\le x}\frac{1}{2}\left( \left(1+\frac{1}{p^{3/4}}\right)^{q}+\left(1-\frac{1}{p^{3/4}}\right)^{q} \right)\end{align*}

and since for any real number a

\begin{equation*}\frac{\exp(a)+\exp(-a)}{2}\le \exp(a^2),\end{equation*}

the above implies

\begin{align*}\prod_{x^{\varepsilon}\le p\le x}\frac{1}{2}\left( \left(1+\frac{1}{p^{3/4}}\right)^{q}+\left(1-\frac{1}{p^{3/4}}\right)^{q} \right)&\le \prod_{x^{\varepsilon}\le p \le x}\frac{\exp(q/p^{3/4})+\exp(-q/p^{3/4})}{2} \\&\le \prod_{x^{\varepsilon\le p \le x}}\exp\left(\frac{q^2}{p^{3/2}}\right) \\&\le \exp\left(\frac{q^2}{x^{\varepsilon/2}}\right).\end{align*}

The above inequalities imply that

\begin{align*}\mathbb{E}\left[\left(\prod_{x^{\varepsilon}\le p\le x}\left(1-\frac{f(p)}{p^{3/4}}\right)^{-1}\right)^{q} \right]&\le \exp\left(\frac{q^2}{x^{\varepsilon/2}}+O(q)\right).\end{align*}

After recalling (5·3), the latter yields

(5·8) \begin{align}\mathbb{E}\left[\left(\prod_{x^{\varepsilon}\le p\le x}\left(1-\frac{f(p)}{p^{3/4}}\right)^{-1}\right)^{q} \right]&\le \exp\left(q^2\exp\left(-\frac{(\log\log\log{x})(\log{x})}{4C\log\log{x}}\right)+O(q)\right).\end{align}

We finally deduce the bound

\begin{align*}\mathbb{E}\left[\left(\prod_{p\le x^{\varepsilon}}\left(1-\frac{f(p)}{p}\right)^{-1}\right)^{q} \right]&\ll \exp\left(O(q)+q\sum_{p\le x^{\varepsilon}}\frac{1}{p}\right) \\ &\le \exp\left(O(q)+q\log \log{x}\right).\end{align*}

Combining the above with (5·5) and (5·8), we get

\begin{align*}S_1\ll x^{7/8}\exp\left(\left(q\exp\left(-\frac{(\log\log\log{x})(\log{x})}{4C\log\log{x}}\right)+\log\log{x}\right)\right),\end{align*}

and the result follows from (5·7) after renaming the constant C.