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 



Need an explanation of a deduction
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.


