3 Basic Fourier estimates

Lemma 3.1 \(L^2\) integral estimate

Let \(\xi _1,\dots ,\xi _R\) be real numbers that are \(1/N\)-separated. Then for any interval \(I\) of length \(T\), and any sequence \(a_1,\dots ,a_R\) of complex numbers one has

\[ \int _I |\sum _{r=1}^R a_r e(\xi _r t)|^2\ dt = (T + O(N)) \sum _{r=1}^R |a_r|^2. \]

Proof ▶

We adapt the proof of [ 149 , Theorem 9.1 ] . Without loss of generality we may normalize \(\sum _{r=1}^R |a_r|^2=1\). From the Plancherel identity we have

\begin{equation} \label{art} \int _\mathbf{R}|\sum _{r=1}^R a_r e(\xi _r t)|^2 |\hat\psi ((t-t_0)/N)|^2\ dt = N \end{equation}

whenever \(t_0 \in \mathbf{R}\) and \(\psi \) is a smooth function supported on \([-1/4, 1/4]\) of \(L^2\) norm \(1\). By suitable choice of \(\psi \), this implies that

\begin{equation} \label{trap} \int _J |\sum _{r=1}^R a_r e(\xi _r t)|^2\ dt \ll N \end{equation}

whenever \(J\) is an interval of length \(N\). If one integrates 1 for all \(t_0 \in I\), we see that

\[ \int _I |\sum _{r=1}^R a_r e(\xi _r t)|^2\ dt = T - \int _\mathbf{R}|\sum _{r=1}^R a_r e(\xi _r t)|^2 \left(\int _I |\hat\psi ((t-t_0)/N)|^2\ dt_0 - 1_I(t)\right)\ dt. \]

Since \(\hat\psi \) is rapidly decreasing and has \(L^2\) norm \(1\), one can compute

\[ \int _I |\hat\psi ((t-t_0)/N)|^2\ dt_0 - 1_I(t) \ll (1 + \mathrm{dist}(t, \partial I) / N)^{-10} \]

and hence by 2 and the triangle inequality

\[ \int _\mathbf{R}|\sum _{r=1}^R a_r e(\xi _r t)|^2 \left(\int _I |\hat\psi ((t-t_0)/N)|^2\ dt_0 - 1_I(t)\right)\ dt \ll N \]

giving the claim.