Can exponential sums be small on a whole interval? 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!
 A: One can indeed make $\|\phi\|$ exponentially small in $A$ (though the example I have requires an exponentially large number of frequencies $\alpha_i$).
We use the dual formulation, that is we find a sum of exponentials $g$ whose $L^\infty$ norm is small on $[0.9A, A]$.  Firstly, if we consider the sum of just two exponentials $g_0(t) = 1 + e^{\pi i/A}$, then there is significant cancellation on $[0.9A,A]$, in particular $\|g_0\|_{L^\infty} \leq c < 1$ for some absolute constant $c<1$.  If we then raise $g_0$ to the power $A$ (let's say $A$ is an integer) then we get a sum of exponentials $g_0^A$ whose frequencies are still bounded by 1, and whose sup norm decays exponentially fast in A.
I would expect exponential decay to be best possible, in particular the Vandermonde determinant approach probably gives such a bound.  EDIT: it appears that one has the more general bound $\|\mu\|_1 \ll \exp(CA) \| \hat \mu \|_{L^\infty([0.9A,A])}$ for any nonnegative measure supported on $[-1,1]$, by using a variant of the real-variable proof of Hardy's uncertainty principle (see e.g. https://terrytao.wordpress.com/2009/02/18/hardys-uncertainty-principle/ ) to use $L^\infty$ control of $\hat \mu$ on $[0.9A,A]$ and a "doubling bound" to control $\hat \mu(0) = \|\mu\|_1$ with an exponential loss in constants, after first normalising $\|\mu\|_1=1$ to control error terms.
A: 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.
