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Statement

Setting (p. 584). $\mathscr F$ is the class of completely multiplicative $f$ with $-1\le f(m)\le1$ for all $m$, and $c$ is the infimum of the Theorem. The paper defines

$$c_0=\liminf\Bigl\{\frac1x\sum_{m\le x}f(m):\ f\in\mathscr F\Bigr\}, \quad(x\to\infty),\qquad(11)$$

so that $c\le c_0$, and says the inequality may be strict. For a set $E$ of primes, possibly depending on $x$, let $f(p)=-1$ for $p\in E$ and $f(p)=+1$ otherwise, so $f(m)=(-1)^{\Omega(m,E)}$ (12-13). With $E=\{p:\ x^{1/t}<p\le x\}$ for a fixed $t>1$, the limit of $\frac1x\sum_{m\le x}f(m)$ as $x\to\infty$ exists (14), is an upper bound for $c_0$, and is written $R(t)$.

Inequality (15) (p. 584, quoted). "for all $t>1$ we have"

$$R(t)\geqq R(1+\sqrt e)=-.656999\ldots \qquad(15)$$

The paper's stated aim (p. 584) is the consequence $c_0\le-0.656999\ldots$, and it concludes (pp. 587-588) that $R(1+\sqrt e)$ is the global minimum of $R(t)$, the value $-0.656999\ldots$ coming from numerical integration. It leaves the value of $c$, and whether $c<c_0$, open.

Source. R. R. Hall, Proof of a conjecture of Heath-Brown concerning quadratic residues, Proc. Edinburgh Math. Soc. (2) 39 (1996), 581-588, doi:10.1017/S0013091500023324: Section 2 (The value of c), pp. 584-588. The edition read is identified on the source card.

Read depth. Claims checked: the definitions, (11) to (15) and the conclusion on pp. 587-588 were read clause by clause on the printed pages, and the argument of pp. 585-588 was followed; the inner sum in (25), whose treatment the paper omits as standard, and the numerical integration were not checked. Nothing here is independently reviewed.

Proof pointer

Pp. 585-588. Lemma 3 (p. 585) compares $\sum_{m\le x}f(m)$ with $S(x,E)=\sum_{m\le x}(-1)^{\omega(m,E)}$, with an error the print writes as $O\bigl(1/(p_0\log p_0)\bigr)$ in (17), where $p_0$ is the least element of $E$; the proof's last line (21) carries a factor $x$. Since $p_0\to\infty$, the limit may be computed from $S(x,E)$. Expanding $(-1)^{\omega(m,E)}$ over squarefree divisors with prime factors in $E$ gives $R(t)=\sum_{k\ge0}(-2)^kF_k(t)$ (28), with $F_k$ the integrals (27). These satisfy $tR'(t)=-2R(t-1)$ for $t>1$, with $R(t)=1$ on $(0,1]$ (30). An adjoint equation and its inner product (31-33) show that $\lvert R(t)\rvert<R^*(t)=\max\{\lvert R(u)\rvert:\ t-1\le u\le t\}$ for $t>3$, so the extreme values occur on $[1,3]$. There $R(t)=1-2\log t$ on $[1,2]$ and $R(t)=1-2\log t+4\int_2^t\frac{\log(u-1)}u\,du$ on $(2,3]$ (34), with the minimum at $t=1+\sqrt e$.

Dependencies

The Theorem supplies the definition of $c$ and of $\mathscr F$. External input named by the paper: Iwaniec's inner product for differential-difference equations (Recent progress in analytic number theory, 1981).

Later work

Granville and Soundararajan proved the matching lower bound: their Corollary 1 gives $\sum_{n\le x}f(n)\ge(\delta_1+o(1))x$ with $\delta_1=-0.656999\ldots$ for real completely multiplicative $f$ with values in $[-1,1]$.

Bears on

• Problem 121: background only. The paper says nothing about products of integers that are squares, and the bound does not concern the problem's $F_k(N)$. The problem page cites the paper's bounds on the least mean value of a completely multiplicative $f$ with values $\pm1$ as bounding the related $F(N)$ (no odd number of elements multiplying to a square), a different question from the problem's.