TITLE: Can exponential sums be small on a whole interval?
QUESTION [8 upvotes]: This is almost certainly routine to an analyst, so forgive me in advance.
Let $\alpha_i\in \mathbb{R}$. Consider the functional $$\varphi: L^1[0.9A,A]\to \mathbb{C}$$ via $$f\mapsto \sum_i \hat{f}(\alpha_i),$$ where by '$\hat{f}$' I mean the Fourier transform of $f$ regarded as a function on $\mathbb{R}$ (via extension by zero).
What can one say about the norm of $\varphi$?
I am wondering about the case where all $|\alpha_i| = O(1)$, $A$ is large, and the number of $\alpha_i$ is also large (comparable to $A$). As the title suggests, I'm hoping for a lower bound on $||\varphi||$. The thing that seemed most natural to me was to write this as integrating against a sum of exponentials (say $g(t)$, so that $||g||_\infty = ||\varphi||$) and then calculate the $L^2$ norm of $g$, but I can't bound the off-diagonal terms well enough.
The idea is that if, for some $t$ in the interval, $g(t)$ is very small, then some small-order derivative of $g$ should be bounded below so that $g((1\pm \delta)t)\gg_\delta A^{-O(1)}$ maybe. One could get a lower bound on some derivative via the nonvanishing of a Vandermonde determinant, but I think that would require too many derivatives.
Perhaps I am mistaken and there is an example with $||\varphi||\ll \exp(-c A)$?
Anyway, thanks much!

REPLY [6 votes]: I think I can prove a polynomial bound using complex analysis. This really seems like it shouldn't work but it seems to.
The bound is that for all $\alpha_1, \dots, \alpha_n \in \mathbb R$ 
$$ \sup_{t \in [A,B]}\left| \sum_{i=1}^n e( \alpha_i t) \right| \geq n^{1-\frac{\pi}{4 \arctan {\sqrt{\frac{B}{A}}}-\pi}}$$
So in the special case $[.9A,A]$ this is $n^{-28.8313 \dots}$.
We may freely translate the $\alpha_i$, therefore, we assume $\min \alpha_i=0$. Form the holomorphic function
$$f(z) = \sum_{i=1}^n e(\alpha_i z)$$
We have the following estimates:
In the whole upper half plane, $|f(z)| \leq n$, as all the $\alpha_i$ are nonnegative so $e(\alpha_i z) \leq 1$.
On the imaginary line, $|f(z)| \geq 1$, as some $\alpha_i$ is zero giving $e(\alpha_i z) =1$ and the rest of the terms are positive reals.
Assume that for $z$ on the real line in the interval $[A,B]$, $|f(z)|<C$. (Then the same holds for $[-B,-A]$, which helps to get a better constant but isn't essential).
$f$ is holomorphic so $\log | f(z)|$ is subharmonic. Hence by the mean value theorem for subharmonic functions and the various bounds we stated on $f$:
$$ 0 \leq \log |f(it)| \leq\frac{t}{2\pi} \int_{-\infty}^{\infty} \frac{\log |f(s)|}{t^2+s^2}ds \leq \frac{t}{2\pi} \int_{-\infty}^{\infty}  \frac{\log n}{t^2+s^2}ds+ 2\frac{t}{2\pi} \int_{A}^{B}  \frac{\log C - \log n}{t^2+s^2}ds$$
$$ = \log n + \frac{\theta}{\pi}  (\log C - \log n)$$
where $\theta$ is the hyperbolic angle between $A$ and $B$ from $it$. So we have:
$$0 \leq \log n + \frac{\theta}{\pi} (\log C - \log n)$$
$$ \frac{\theta}{\pi} (\log n - \log C) \leq \log n$$
$$ \log C \geq (1- \frac{\pi}{\theta}) \log  n$$
$$C \geq n^{1-\pi/\theta}$$
as desired.
The optimal value of $\theta$ is a hyperbolic trigonometry problem which I may or may not have gotten the right answer to. Anyways it's some explicit formula.
Tao's example starting with, $\alpha_1=0$, $\alpha_2= \frac{1}{A+B}$ has an upper  bound of the form $n^{\frac{\log \left|1- e\left(\frac{A}{A+B}\right)\right|}{\log 2}}$. My exponent is far from this as $A$ goes to $B$ so sharpness is an interesting question.