**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.



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


