Lower bound for sum of Hecke eigenvalues











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Let $lambda$ be weakly multiplicative, $lambda(n)geq0$, $p$ prime and $S(x)=sum_{nleq x}lambda(n)log(frac{x}{n})$ for real $x$.



How can I show $S(x)gg left(sum_{pleq sqrt{x/3}}lambda(p)right)^2-left(sum_{pleq sqrt{x/3}}1right)$?



Here is the background:
The question is coming from IKS, section 3. $lambda(n)$ are the eigenvalues of a newform $f$ of level $N$. All above sums are chosen such that $(n,N)=1$ or $pnotmid N$, thus giving multiplicativity.



In Xu, section 3.1 something similar happens. Here the hints $lambda(p)geq1$ and $|lambda(n)|leqsigma_0(n)$ (divisor function) are given.



My approach so far starts estimating $log(x/n)geq1$ but I could not figure out the strange bound of $sqrt{x/3}$...










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    up vote
    1
    down vote

    favorite












    Let $lambda$ be weakly multiplicative, $lambda(n)geq0$, $p$ prime and $S(x)=sum_{nleq x}lambda(n)log(frac{x}{n})$ for real $x$.



    How can I show $S(x)gg left(sum_{pleq sqrt{x/3}}lambda(p)right)^2-left(sum_{pleq sqrt{x/3}}1right)$?



    Here is the background:
    The question is coming from IKS, section 3. $lambda(n)$ are the eigenvalues of a newform $f$ of level $N$. All above sums are chosen such that $(n,N)=1$ or $pnotmid N$, thus giving multiplicativity.



    In Xu, section 3.1 something similar happens. Here the hints $lambda(p)geq1$ and $|lambda(n)|leqsigma_0(n)$ (divisor function) are given.



    My approach so far starts estimating $log(x/n)geq1$ but I could not figure out the strange bound of $sqrt{x/3}$...










    share|cite|improve this question









    New contributor




    Nodt Greenish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Let $lambda$ be weakly multiplicative, $lambda(n)geq0$, $p$ prime and $S(x)=sum_{nleq x}lambda(n)log(frac{x}{n})$ for real $x$.



      How can I show $S(x)gg left(sum_{pleq sqrt{x/3}}lambda(p)right)^2-left(sum_{pleq sqrt{x/3}}1right)$?



      Here is the background:
      The question is coming from IKS, section 3. $lambda(n)$ are the eigenvalues of a newform $f$ of level $N$. All above sums are chosen such that $(n,N)=1$ or $pnotmid N$, thus giving multiplicativity.



      In Xu, section 3.1 something similar happens. Here the hints $lambda(p)geq1$ and $|lambda(n)|leqsigma_0(n)$ (divisor function) are given.



      My approach so far starts estimating $log(x/n)geq1$ but I could not figure out the strange bound of $sqrt{x/3}$...










      share|cite|improve this question









      New contributor




      Nodt Greenish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      Let $lambda$ be weakly multiplicative, $lambda(n)geq0$, $p$ prime and $S(x)=sum_{nleq x}lambda(n)log(frac{x}{n})$ for real $x$.



      How can I show $S(x)gg left(sum_{pleq sqrt{x/3}}lambda(p)right)^2-left(sum_{pleq sqrt{x/3}}1right)$?



      Here is the background:
      The question is coming from IKS, section 3. $lambda(n)$ are the eigenvalues of a newform $f$ of level $N$. All above sums are chosen such that $(n,N)=1$ or $pnotmid N$, thus giving multiplicativity.



      In Xu, section 3.1 something similar happens. Here the hints $lambda(p)geq1$ and $|lambda(n)|leqsigma_0(n)$ (divisor function) are given.



      My approach so far starts estimating $log(x/n)geq1$ but I could not figure out the strange bound of $sqrt{x/3}$...







      number-theory eigenvalues-eigenvectors estimation modular-forms upper-lower-bounds






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      Nodt Greenish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




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      Check out our Code of Conduct.






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