Continuous large sieve inequality

Question

In Tao's blog we can find this exercise: Let $[M,M+N]$ be an interval for some $M \in {\bf R}$ and $N > 0$, and let $\xi_1,\dots,\xi_J \in {\bf R}$ be $\delta$-separated. For any complex numbers $c_1,\dots,c_J$, show that $$ \int_M^{M+N} |\sum_{j=1}^J c_j e( \xi_j t )|^2\ dt \ll (N + \frac{1}{\delta}) \sum_{j=1}^J |c_j|^2.$$

Since it is well-known that the discrete version of the large sieve inequality holds with $N-1+\frac{1}{\delta}$, I was wondering if the previous inequality can be proved with $N-1+\frac{1}{\delta}$ instead of $N+\frac{1}{\delta}$. I tried to follows the same proof for the case $N+\frac{1}{\delta}$ but I'm not able to obtain the wanted term.

Answer 1

In the continuous version of the problem, it is perfectly possible to have $N < \frac{1}{2}$ and $\delta >2$ so that $N + \frac{1}{\delta}<1$ but the left-hand side is positive, so the $N + \frac{1}{\delta}-1$ version cannot be true.

As to whether the inequality is true with some reasonable assumptions on $\delta$, the inequality doesn't pass dimensional analysis. If we multiply all the $M,N$ by a constant $C$ and divide $\xi_j$ by $C$ for all $j$, we multiply the left hand side by $C$ will multiplying $N$ and $\delta^{-1}$ by $C$ but not multiplying $1$ by $C$. In particular, if $C$ is small, doing this makes the inequality stronger. So if the bound you state holds for $\delta\leq \frac{1}{2}$, say, then taking $C = 1/(2\delta)$ we deduce the stronger inequality $N + \delta^{-1} - \frac{\delta^{-1}}{2}$, but I think this version is not true (or if it is one should just state $N+ \delta^{-1}/2$ rather than the $-1$ version.)

If you are interested for some reason in $\xi_j$ that are $\delta$-separated in the $\mathbf R/\mathbf Z$ sense, then one can easily reduce to the discrete case: For any function $f$ we have

$$\int_{M}^{M+N} f(t)dt = \int_0^1 \sum_{\substack{n \in \mathbb Z\\ n+x \in [M,M+N]}} f(n+x) dx$$ and for $f=|\sum_{j=1}^J c_j e( \xi_j t )|^2$ you can apply the large sieve inequality with the $-1$ to the sum over $n$ and then deduce the large sive inequality with the $-1$ for the integral.