**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.