# A lower bound on the $L^2$ norm of a Dirichlet polynomial

The Question. Suppose $0 < \alpha < \beta$ are fixed, and $a_n$ is an arbitrary sequence of real numbers. Is it known how to bound from below \begin{equation*} \int_0^{T} \Big| \sum_{\alpha T < n < \beta T} a_n n^{-it} \Big|^2 dt? \end{equation*}

What I know. The mean value theorem for Dirichlet polynomials will tell us \begin{equation*} \int_0^{T} \Big| \sum_{n \leq N} a_n n^{-it} \Big|^2 dt = (T + O(N)) \sum_{n \leq N} |a_n|^2, \end{equation*} and the implied constant is absolute. This takes care of $\beta$ small enough.

If $\beta$ is large, and $\alpha$ is small, then one can use the method of Rudnick and Soundararajan (http://arxiv.org/abs/math/0601498), as follows. By Cauchy-Schwarz, \begin{equation*} \Big|\int_0^{T} \Big(\sum_{n \leq N} a_n n^{-it} \Big) \Big(\sum_{m \leq M} a_m m^{it} \Big) dt \Big|^2 \leq \int_0^{T} \Big| \sum_{n \leq N} a_n n^{-it} \Big|^2 dt \int_0^{T} \Big| \sum_{n \leq M} a_n n^{-it} \Big|^2 dt. \end{equation*} Choosing $M$ small enough, say $M \leq \varepsilon T$, allows the computation of the left hand side above (the diagonal terms dominate). The mean value theorem for Dirichlet polynomials treats the sum over $m \leq M$ on the right hand side. Rearranging, one gets a lower bound \begin{equation*} \int_0^{T} \Big| \sum_{n \leq N} a_n n^{-it} \Big|^2 dt \gg T \sum_{n \leq \varepsilon T} |a_n|^2. \end{equation*} However, if $a_n = 0$ for $n \leq \varepsilon T$, then this method breaks down.

There may be no such bound if $\alpha$ and $\beta$ are suitably large. For example, using the first derivative test (see e.g. Titchmarsh Chapter 4) one can see that if $T\le t\le 2T$ then $$\sum_{10T\le n \le 20T} n^{it} = O(1).$$ Therefore $$\int_T^{2T} \Big| \sum_{10T \le n\le 20 T} n^{it} \Big|^2 dt \ll T$$ whereas one might have initially expected a diagonal contribution of size $T^2$.
(If you're really interested in the interval $[0,T]$ rather than the dyadic $[T,2T]$ then twist the above example by $\chi_{-4}$ say. )
• 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 $T\le t\le 2T$. – Lucia Oct 25 '14 at 15:27
• 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. – Matt Young Oct 25 '14 at 18:39
• 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. – H.Flip Oct 26 '14 at 4:11
• 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. – Matt Young Oct 29 '14 at 13:37