In their paper Sign changes of Hecke eigenvalues, Matomaki and Radziwill showed that (Theorem 1.2 of the paper) for a large enough $x$ , the number of sign changes of sign changes of in the non-vanishing sequence of Hecke eigenvalues $(\lambda_f(n))_{n\leq x}$ is $\asymp \frac{x}{k(x)},$ where $$k(x)=\prod_{\substack{p \leq x\\ \lambda_f(p)=0}} \left(1+\frac{1}{p}\right).$$ In the proof of this result, they used the fact that there is a positive proportion of $x$ in $[X,2X]$ such that $$\left|\sum_{x\leq n \leq x+h k(x)} \operatorname{sgn}(\lambda_f(n))w_{n}\right|< \sum_{x\leq n \leq x+h k(x)} w'_{n}$$ once $h$ is large enough but satisfying some condition. They say that since $w'_n \leq w_n$ for every $n$ then for every $x$ satisfying the upper bound then a sign change of $\lambda_f(n)$ must occur in the interval $[x, x+h k(x)]$. From which they say that there are $\gg \frac{X}{k(X)}$ sign changes.

I have two questions:

1) I could not understand how they get their deduction about the lower bound of sign changes.

2) Why they used the same proof to deduce an upper bound although they did not mention it!

Did I misunderstand something?

  • $\begingroup$ Thanks for accepting my response officially. I edited it slightly, because we cannot take $h=1$. Instead, $h$ has to be sufficiently large but fixed. $\endgroup$
    – GH from MO
    Mar 4, 2017 at 11:15
  • $\begingroup$ @GH from MO You are welcome! Thanks to you Sir! I thought to tell you that $h=1$ can not be true but you change it. Thanks again! $\endgroup$ Mar 4, 2017 at 11:18

1 Answer 1


The proof of Theorem 1.2 relies on Propositions 3.4 and 3.5. In particular, if $h$ is sufficiently large but fixed, the bound you quote is true for a positive proportion of $x\in\mathbb{N}\cap[X,2X]$. Call such an $x$ nice.

Consider a maximal family $F$ of nice points $x\in[X,2X]$ such that any two of them differ by more than $hk(2X)$. Then any nice $x\in[X,2X]$ is within distance $hk(2X)$ from some $x\in F$, hence the cardinality of $F$ satisfies $hk(2X)|F|\gg X$, i.e., $|F|\gg X/k(2X)$. Shrink $F$ slighly to $G:=F\cap[X,2X-hk(2X)]$. Using that $k(2X)\ll\log X$, it is clear that $$|G|\gg X/k(2X)\gg X/k(X).$$ Now $[x,x+hk(x)]$ with $x\in G$ are pairwise disjoint subintervals of $[X,2X]$, because $k(x)\leq k(2X)$, and in each such subinterval $\lambda_f(n)$ changes sign. So the number of sign changes of $\lambda_f(n)$ in $[X,2X]$ is at least $|G|$ which is $\gg X/k(X)$.

Regarding your second question, note that the number of sign changes of $\lambda_f(n)$ in $[X,2X]$ cannot exceed the number of $n\in[X,2X]$ with $\lambda_f(n)\neq 0$. However, the number of such $n$'s is $\ll X/k(2X)\leq X/k(X)$ by part (i) of Lemma 3.1 in the paper, so we are done.


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