1
$\begingroup$

When I was reading the paper of Winfried Kohnen, Yuk-Kam Lau and Igore E. Shparlinski (ON THE NUMBER OF SIGN CHANGES OF HECKE EIGENVALUES OF NEWFORMS), I found this result (which is Theorem 2 of the paper): There are absolute constants $\eta <1$ and $A>0$ such that, $$S_{f}^{\pm}(x+x^{\eta})- S_{f}^{\pm}(x)>0,$$ whenever $x\geq (kN)^{A}$ and where $$S_{f}^{\pm}(x):=\sum_{\substack{n \leq x\\\gcd(n,N)=1; \ \lambda_f(n)\gtrless 0}} 1$$ $\lambda_f(n)$ is the $n$th normalized Hecke eigenvalues and $f$ is a non-zero cusp form of even integral weight $k \geq 2$ and level $N \geq 1.$ After that, the authors introduced the integer $T_f(x)$ as the number of sign changes in the sequence $(\lambda_f(n))$ taken for consecutive positive integers $n \leq x$ with $\gcd(n,N)=1.$ From Theorem 2, they deduced the following Corollary: There are absolute constants $\kappa >0$ and $A>0$ such that $$T_f(x)>x^{\kappa},$$ whenever $x\geq (kN)^{A}.$ Their deduction is based on splitting the interval $[1,x]$ into $x^{1-\eta}$ intervals of length $x^{\eta}.$ I tried to understand this deduction but I couldn't. Could you explain to me this deduction? Many thanks, Khadija

$\endgroup$
3
$\begingroup$

The authors' formulation is a bit sloppy, but the argument is correct and straightforward.

Below, a "sign change" will mean a sign change of the sequence $(\lambda_f(n))$ restricted to $\gcd(n,N)=1$.

The first display guarantees that in any interval $[x,x+x^\eta]$ with $x\geq (kN)^A$, there are $n^-$ and $n^+$ coprime with $N$ such that $\lambda_f(n^-)<0<\lambda_f(n^+)$. Hence, in any such interval, at least one sign change occurs.

Now let $x\geq(kN)^A$, more precisely let $x\geq 2(kN)^A$ for safety. Define recursively the increasing sequence $(x_k)$ by $$x_1:=(kN)^A,\qquad x_{k+1}:=x_k+x_k^\eta\quad\text{for}\quad k\geq 1.$$ The intervals $[x_k,x_{k+1}]$ are pairwise disjoint apart from their endpoints, and by the above, in each of them a sign change occurs. As $(x_k)$ tends to infinity, there is a smallest $K\geq 1$ such that $x_{K+2}>x$. That is, $$ (kN)^A=x_1<x_2<\dots<x_K<x_{K+1}\leq x<x_{K+2}. $$ By the above, in $[1,x]$ there are at least $K$ sign changes, so it suffices to show that $K\gg x^{1-\eta}$. However, this is clear, because $$ x/2\leq x-(kN)^A<x_{K+2}-x_1=\sum_{k=1}^{K+1}(x_{k+1}-x_k)=\sum_{k=1}^{K+1}x_k^\eta\leq (K+1)x^\eta\leq 2K x^\eta.$$ Comparing the two sides, we get $K>x^{1-\eta}/4$, and we are done.

$\endgroup$
1
  • 1
    $\begingroup$ Thanks @GH from Mo! Your explanation is quite clear. I am so grateful to you as you frequently answered my questions. $\endgroup$ – Khadija Mbarki Sep 1 '15 at 19:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.