Timeline for A lower bound on the $L^2$ norm of a Dirichlet polynomial
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Oct 30, 2014 at 16:25 | comment | added | H.Flip | Forgive my limited capaicty if the mistakes are making. I think $dominate$ maybe depends on the choices of $a_m$. Note the integral with respect to $t$ is $\le |\log^{-1}(n/m)|$. For off-diagonal terms, $m,n$ can be restricted to $n>m,|m-n|=\Delta$ with $\Delta\le m\le M$. The off-diagonal terms can be bounded by $\ll \sum_{m\le M}m |a_m| \sum_{\Delta\le m}|a_{m+\Delta}|/\Delta$ which is big if $a_m>0$ for any $m$, and one of $a_m$ for $M<m\le 2M$ is sufficiently large, $T/\epsilon^2$, say. | |
| Oct 29, 2014 at 13:37 | comment | added | Matt Young | Thanks for the reference @H.Flip, it confirms my remark that it is necessary to control the corellations $\sum_n a_n a_{n+h}$ to get a lower bound. Why do you say you don't think the diagonal terms dominate when $M \leq \varepsilon T$? With a smoothing factor in $T$, both $m$ and $n$ must be close to each other. | |
| Oct 26, 2014 at 4:11 | comment | added | H.Flip | I search on Google that (the paper matwbn.icm.edu.pl/ksiazki/aa/aa84/aa8426.pdf by Goldston and Gonek) in the truncated interval $[\alpha T,\beta T]$ the asymptotic (Theorem 1) was obtained under the restrictions (6),(7),(8). This shows that the low bound is closely related to the propeties of the real sequences $a_n.$ I also feel that when $M\le\epsilon T $, "the diagonal terms dominate" may be inaccuate. | |
| Oct 25, 2014 at 18:39 | vote | accept | Matt Young | ||
| Oct 25, 2014 at 18:39 | comment | added | Matt Young | I think your example indicates that to get a reasonable lower bound one needs to require cancellation in $\sum_n a_n a_{n+h} $ for nonzero $h$. This clarifies the situation. | |
| Oct 25, 2014 at 15:27 | comment | added | Lucia | It's a bit unclear to me what you're looking for. For example, consider a smoothed version (smooth the sum over $n$) of the example (and use Poisson summation). Seems to me that can be very small for all $T\le t\le 2T$. | |
| Oct 25, 2014 at 14:21 | comment | added | Matt Young | Thanks this is helpful. I still wonder what kind of lower bound is true, even if it is smaller than what one might expect looking only at the diagonal terms. | |
| Oct 25, 2014 at 6:20 | history | answered | Lucia | CC BY-SA 3.0 |