When I was reading the paper of Winfried Kohnen, YukKam 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 nonzero 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
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.

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