On the Liouville Function in Short Intervals

Let $\lambda$ denote the Liouville function; that is, the completely multiplicative function defined by $\lambda(p):=-1$, for all primes $p$. It is well-known that the Prime Number Theorem (PNT) is equivalent to the fact that $\lambda$ exhibits some cancellation in its partial sums; more precisely, that

as $x\rightarrow\infty.$

The Möbius Randomness Law (see page 338 of [IK04]) tells us that $\{\lambda(n)\}_{n}$ should behave like a sequence of independent random variables, taking on the values $\pm 1$ with equal probability. As a result, we expect “square-root cancellation” in the partial sums for the Liouville function; that is, for any $\epsilon>0$,

as $x\rightarrow\infty.$ In fact, the above estimate is equivalent to the Riemann Hypothesis (RH); see Theorem 14.25(C) of [THB86]. Furthermore, it is possible to quantity $\epsilon$ in terms of $x$; see [Sou09].

The Liouville function also exhibits cancellation in short intervals, provided that the length of the interval is sufficiently large. Motohashi [Mot76] and Ramachandra [Ram76], independently, proved that

provided $h>x^{\frac{7}{12}+\epsilon}$. Assuming RH, this can be improved to $h>x^{\frac{1}{2}}(\log x)^{A}$, for some suitable constant $A>0$; see [MM09]. Recently, Matomäki and Teräväinen [MT19] improved Motohashi and Ramachandra’s results to include the larger range $h>x^{\frac{11}{20}+\epsilon}.$

By relaxing the condition that our estimates hold for all short intervals, we can get results that hold for smaller values of $h$. In unpublished work of Gao¹¹1Strictly speaking, Gao’s result, and those stated above his, were initially proven for the Möbius function, $\mu$, but the proofs extend to the Liouville function with little effort., it is shown that

assuming RH, for some large constant $A>0$.

The preceding results should be compared with what is known for primes in short intervals. More precisely, an equivalent form of the PNT is given by

where $\Lambda$ denotes the von Mangoldt function (defined to be $\log p$ if $n$ is a power of the prime $p$ and $0$ otherwise). Furthermore, the Riemann Hypothesis is equivalent to the following estimate:

in particular, RH implies that

provided $h>x^{\frac{1}{2}}(\log x)^{2+\epsilon}.$

Again restricting ourselves to results that hold almost everywhere, Selberg [Sel43] proved that if RH holds, then

so that almost all short intervals contain the correct number of primes.

It is important to note that both Gao and Selberg obtain square-root cancellation in their estimates. In each case, they relate the short sum to a contour integral of some Dirichlet series and then shift this contour to the edge of the critical strip, picking up any poles along the way. For Selberg, the corresponding Dirichlet series is $\zeta^{\prime}(s)/\zeta(s)$ and, assuming RH, the only pole will be at $s=1$. Near the half-line, we still need to consider the contributions from the non-trivial zeros, $\rho=1/2+i\gamma$, of $\zeta(s)$, as $\zeta^{\prime}(s)/\zeta(s)$ is large for $s\approx\rho$. Fortunately, the behaviour of $\zeta^{\prime}(s)/\zeta(s)$ for $s\approx\rho$ is well understood (it is even conjectured that the zeros of $\zeta(s)$ are simple). For Gao, the corresponding Dirichlet series is $\zeta(2s)/\zeta(s)$, which also has poles at $s=\rho$, but the residues in this case are equal to $\zeta(2\rho)/\zeta^{\prime}(\rho)$. Since very little is known about $1/\zeta^{\prime}(\rho)$, we need to proceed in an indirect manner. The key idea is to use a sum over primes to approximate $\zeta(s)$ and then avoid “clusters” of zeros of $\zeta(s)$ on the half-line: near these regions, $1/\zeta(s)$ is large and the contour is chosen so that $1/\zeta(s)$ is not too large, at least in some sense; see [MM09], to get an idea of Gao’s proof.

We now turn our attention to the breakthrough work of Matomäki and Radziwiłł, where they relate the average value of $1$-bounded multiplicative functions in short intervals to the corresponding average value in large intervals:

In the case of the Liouville function, Theorem 1.1 implies that

for almost all $x\in[X,2X]$; in particular,

for any $h=h(X)\rightarrow\infty$ as $X\rightarrow\infty.$ Although Matomäki and Radziwiłł do not get square-root cancellation, their results are unconditional, hold for all $h$ going to infinity, and avoid the complex analytic approach used by both Gao and Selberg; instead, they employ a clever decomposition of the corresponding Dirichlet polynomials, which is done by restricting to integers which have prime factors from certain convenient ranges. For a detailed account of the work in [MR16] restricted to the Liouville function, see [Sou16].

In this paper, we apply a simple variation of the methods developed in [MR16] in order to prove the following:

Note that Theorem 1.2 shows square-root cancellation for the Liouville function in almost all short intervals, provided $(\log X)^{6+\epsilon}\leq h\leq\exp\left(\sqrt{\left(\frac{1}{2}-o(1)\right)\log X\log\log X}\right)$. Of course, Theorem 1.2 gives an upper bound on the exceptional set of $x\in[X,2X]$ for which square-root cancellation does not hold (via Chebyshev’s Inequality). By splitting the short sum into shorter intervals, Theorem 1.2 can also be used to obtain some cancellation for larger $h$; namely,

We present here a collection of standard results, which we use freely throughout our paper. The first result is a quantitative version of Perron’s Formula, which serves as an approximation to the indicator function on $(1,\infty)$. The second result relates the mean-square of a Dirichlet polynomial to its sum of squares and this will be our main tool in proving Theorem 1.2.

We need two more preliminary results. The first is a pointwise bound on $\sum_{X\leq n\leq 2X}\lambda(n)/n^{\frac{1}{2}+it},$ which allows us to remove the contribution from the small values of $t$ in our average value (this can be thought of as the analogous result to Lemma 1 in [MR15]). The second is an analogue of Lemma 2 in [MR16]: we need a pointwise bound on sums of the form $\sum_{P\leq p\leq 2P}1/p^{\frac{1}{2}+\frac{1}{\log X}+it},$ for large values of $t$.

In this section, we reduce our problem to that of bounding the mean square of a Dirichlet polynomial. We begin with the following standard lemma, which essentially follows from Perron’s Formula (together with a few other tricks):

Using Lemma 2.3, we can remove the contribution from the small values of $t$ in the RHS of Lemma 3.1: for any $\epsilon>0$,

which follows by bounding the integrand pointwise. Thus, Theorem 1.2 follows from Lemma 3.1 once we show that

for $T\geq X/h$, where we have assumed, w.l.o.g., that $h\leq X^{\frac{1}{2}}$ (this is done to avoid the degenerate case when $T=X/h\leq X^{1/2}$).

For $T>X$, the Mean Value Theorem (Lemma 2.2) immediately gives the desired bound:

It remains to prove that (3.1) holds for $X/h\leq T\leq X$. Our main tool is still the Mean Value Theorem, but this is only winning if we can shorten the length our sum: we should think of $T\approx X/h$ and recall that the MVT is best possible if the length of the corresponding Dirichlet polynomial is of size $N\approx T$; see Remark 2.1. Since our Dirichlet polynomial has length $X$, our goal is to split the sum over $X\leq n\leq 4X$ into two sums, one of which has length $\approx X/h$ and another which can be handled separately. This is done by considering the integers $X\leq n\leq 4X$ which have a prime factor $p>h$: by writing such integers as $n=pm$, and then factoring out the sum over $p$, the sum over $m\approx X/p$ that remains has length $\ll X/h$; using the MVT on the sum over $m$ is now winning. In Section 5, we deal with the remaining integers, all of whose primes factors are $\leq h$. Fortunately for us, there are few of these so-called $h$-smooth integers, at least for small $h$, so that the MVT can be applied directly to give us what we want. Getting square-root cancellation for larger $h$ will require some new ideas; this is ongoing work.

For the integers $X\leq n\leq 4X$ which have at least one prime factor $p>h$, we have the following: