This originated from an textbook exercise (recently posted to math.stackexchange https://math.stackexchange.com/questions/62883/quadratic-characters-and-liouvilles-function with no success) but I think it has research level implications so please bear with me.

I'm working through the problems in Montgomery and Vaughan's Multiplicative Number Theory. In Section 11.2 'Exceptional Zeros', Exercise 9a says that for a quadratic character $\chi$, show that for all $k\ge 0, x\ge1 $ $$ \sum_{n<x}\frac{\chi(n)}{n}(1-n/x)^k \ge \sum_{n<x}\frac{\lambda(n)}{n}(1-n/x)^k, $$ where $\lambda$ is Liouville's function. This is elementary. In part b, under the hypothesis that there exists a $k$ such that $$ \sum_{n<x}\frac{\lambda(n)}{n}(1-n/x)^k\ge0 $$ for all $x\ge 1$, (no such $k$ is known to exist), one is to show that for all quadratic $\chi$ and all $\sigma>0$ $$ L(\sigma,\chi)>0. $$ I expect one is meant to use a Mellin transform with Cesaro weighting, S 5.1 in Montgomery and Vaughan. The difficulty is that $\chi(n)/n$ are the Dirichlet series coefficients of $L(s+1,\chi)$, not $L(s,\chi)$. Thus (5.18) gives $$ L(\sigma+1,\chi)>0 $$ for all $\sigma>0$.

Question: Show, as the exercise states, that if there exists $k$ such that for all $x\ge1$ $$ \sum_{n<x}\frac{\lambda(n)}{n}(1-n/x)^k\ge0 $$ then for all quadratic $\chi$ and for $\sigma>0$, $$ L(\sigma,\chi)>0. $$

By a theorem of Landau, the positivity hypothesis for some $k$ and all $x$ would imply the Riemann Hypotheses as well; I don't see how to use this (and I think Landau's theorem is beyond the scope of the book.)

Alternately, the method of part a will show that $$ \sum_{n<x}\chi(n)(1-n/x)^k \ge \sum_{n<x}\lambda(n)(1-n/x)^k $$ so under the hypothesis that there exists a $k$ such that for all $x\ge1$ $$ \sum_{n<x}\lambda(n)(1-n/x)^k\ge0 $$ one gets $L(\sigma,\chi)>0$ for all quadratic $\chi$.

The reason one cares which version of part a is used, is that the numerics for small $k$ and moderate $x$ indicate that positivity is at least plausible for the original part a. It is not plausible for the revised version.

Moreover, several conjectures in the literature were known to imply the Riemann Hypothesis, but have since been disproved. Polya conjectured that $$ \sum_{n<x}\lambda(n)\le0 $$ for $x\ge2$, and Turan conjectured that $$ \sum_{n<x}\frac{\lambda(n)}{n}\ge0 $$ for $x\ge 1$. Both where disproved by Haselgrove, and subsequently Odlyzko disproved the Mertens conjecture by similar methods (more or less the Explicit Formula plus numerical values for low lying zeta zeros.) The same techniques should show, as long as $k$ is not too large, that there exists an $x$ such that $$ \sum_{n<x}\frac{\lambda(n)}{n}(1-n/x)^k<0. $$

Updates: (1) Theorem 1.7 (Landau) in Montgomery and Vaughan, has a stronger version: Satz 454 in Landau's Vorlesungen uber Zahlentheorie. From this one deduces that if there exists a $k$ such that for all $x$ $$ \sum_{n<x}\frac{\lambda(n)}{n}(1-n/x)^k \ge 0, $$ then the Riemann Hypothesis follows. But you can't get this directly from Theorem 1.7, and in any case I don't see that it applies to the problem at hand.

(2) I obtained the original paper of Bateman and Chowla that Montgomery and Vaughan reference. There it is proven that for all $x\ge 1$ $$ \sum_{n<x}\frac{\lambda(n)}{n} \ge 0 $$ if and only if for all quadratic $\chi$ and all $x\ge 1$ $$ \sum_{n<x}\frac{\chi(n)}{n} \ge 0. $$ (The proof generalizes from $k=0$ Cesaro weighting to any $k$.) Bateman and Chowla mention that former inequality and Landau's theorem implies the Riemann Hypothesis, but do not make any claim about exceptional zeros of quadratic character $L$-functions.

NB: Let me reiterate that I have no hopes that such a $k$ exists, as Peter Humphries points out this would contradict the Linear Independence Hypothesis.

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    $\begingroup$ A few observations: have you managed to track down the reference that Montgomery and Vaughan give for this question? I had a quick Google search but it doesn't seem to be available online. Also, Landau's theorem is in the book (see Theorem 1.7 or Lemma 15.1), unless you're talking about some other theorem. The other thing to note is that $\sigma_c = -1$ for $L(s+1,\chi)$, so even though (5.18) won't work (as we require $\sigma > \max\{0,\sigma_c\}$, some variant of this might be able to overcome this issue. $\endgroup$ Sep 12, 2011 at 3:37
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    $\begingroup$ Also, the Linear Independence Hypothesis implies that $\liminf_{x \to \infty} \sqrt{x} \sum_{n \leq x} \frac{\lambda(n)}{n} \left(1 - \frac{n}{x}\right)^k = - \infty$ (and also that $\limsup_{x \to \infty} \sqrt{x} \sum_{n \leq x} \frac{\lambda(n)}{n} \left(1 - \frac{n}{x}\right)^k = \infty$) for every $k \geq 0$; see Ingham's paper "On Two conjectures in the Theory of Numbers". This is where the method comes from in both Haselgrove's disproof of Polya's and Turan's conjectures and Odlyzko's disproof of the Mertens conjecture. $\endgroup$ Sep 12, 2011 at 3:47
  • $\begingroup$ @Peter: I've requested the Bateman-Chowla paper via Interlibrary Loan, but J Indian Math Soc is difficult to obtain. I've got Ingham's paper. Regarding your first remark, replacing $\sigma>\max\{0,\sigma_c\}$ with $\sigma>\sigma_c$ in all the relevant theorems seems very difficult. $\endgroup$
    – Stopple
    Sep 12, 2011 at 20:15
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    $\begingroup$ Just a side comment: Landau's theorem is in the M-V book, chapter 15. $\endgroup$ Sep 15, 2011 at 1:28

1 Answer 1


I am now very dubious that one can deduce anything about the sign of $L(\sigma,\chi)$ for $0<\sigma<1$ from the positivity of $$ \sum_{n<x}\frac{\chi(n)}{n}(1-n/x)^k. $$ Here's why. For simplicity consider the case $k=0$, and $\chi$ odd, i.e. complex quadratic. Bateman and Chowla (in the article M&V cite) point out that $L(1,\chi)>d^{-1/2}$, while Summation by Parts and the Polya-Vinograov inequality bounds the tail of the infinite series by $$ \sum_{x<n}\frac{\chi(n)}{n}<\frac{5}{3}\cdot \frac{d^{1/2}\log(d)}{x}. $$ Thus $$ \sum_{n<x}\frac{\chi(n)}{n}>d^{-1/2}-\frac{5}{3}\cdot \frac{d^{1/2}\log(d)}{x}. $$ The above is positive for $$ x>\frac53 d\log(d),\quad\text{or}\quad d<\frac{3x}{5W(3x/5)}, $$ where $W(x)$ is the Lambert function, that is, the inverse function of $x=w\exp(w)$. Thus assuming one could show that $$ \sum_{n<x}\frac{\lambda(n)}{n}>0 $$ for some very large $x$ (Haselgrove's disproof of Turan's conjecture suggests that $x=\exp(853)$ might be possible), one would be able (if the premise of the problem were correct) to rule out the possibility of an exceptional zero for a very large collection of $d$. For example, with $x=\exp(853)$ one would get all $d<2\cdot 10^{367}.$ A similar argument with fixed $k>0$ would give still more $d$.

  • $\begingroup$ Interesting. Are you still yet to hear anything from the authors? It is, of course, entirely possible that they're mistaken. $\endgroup$ Oct 7, 2011 at 1:12
  • $\begingroup$ Also Borwein et al.'s paper "Sign changes in the sums of the Liouville function" (cecm.sfu.ca/personal/pborwein/PAPERS/P208.pdf) finds the first counterexample to Turan's conjecture. $\endgroup$ Oct 7, 2011 at 1:15
  • $\begingroup$ @Peter: Thanks, I have the Borwein paper but forgot about it. Their result would then give 'no exceptional zero for $d<1.5\times 10^{12}$'. But still one could expect better results for $k>0$. Still no word from the authors, but then, I often ignore email about the exercises in my own book. $\endgroup$
    – Stopple
    Oct 7, 2011 at 3:28

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