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!



Can exponential sums be small on a whole interval?
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.


