Can exponential sums be small on a whole interval?

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.

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.