Let $u(t) = \Sigma\_{k=1}^n c\_k e^{i \lambda\_k t} (c\_k \in \mathbb C, \lambda\_k \in \mathbb R) $ be an exponential polynomial of order $n$ with purely imaginary exponents. We can assume that the exponents are $\delta$-separated. 


Define $E\_n$ to be the collection of all exponential polynomial of order $n$. i.e.,


$$ E\_n:= \{ u : u(t) = \sum\_{k=1}^n c\_k e^{i \lambda\_k t}, c\_k \in \mathbb C, \lambda\_k \in \mathbb R, and |\lambda\_i - \lambda\_j| > \delta\\ when\\ i \neq j \}. $$ 


Of course $u$ is an almost periodic function i.e., given $\epsilon >0$ there exists $T\_\epsilon$ such that every interval of length $T\_\epsilon$ contains an almost period of $u$. i.e., $\forall x \in \mathbb R,$ $\exists \tau \in (x,x+T\_\epsilon)$ such that


$$ \sup\_{t\in \mathbb R} |u(t)-u(t+\tau)| < \epsilon$$ 


Is it possible to find a bound on the $T\_\epsilon$ for such an exponential polynomial, I added the separation condition hoping that it would lead to an affirmative answer. 


I have this suspicion that the almost period may have to do something with the number theoretical properties of the set of exponents. After all if all the exponents are in a lattice then the function is periodic.


and may I even dare hope to find a bound which will work for the entire class? Some nice subclass maybe ? 


If that is too much to ask then what is a good question to ask ? 


Has some one studied related questions or variations of it ? I would be glad to know. 



How to find the almost period of an exponential polynomial 
You are basically interested in what is called $\epsilon$-dual Characters.


For a set $\Lambda \subset \R^d$ we define 


$$\Lambda^\epsilon := \{ t \in \R^d | \left| e^{2 \pi x \cdot t} -1 \right| < \epsilon \, \forall x \in \Lambda \}\,.$$


In your case $\Lambda := \{ \lambda\_1, .., \lambda\_n \}$ and for all $t \in \Lambda^\epsilon$ you get 


$$\| T\_tu - u\|\_{\infty} < (|c\_1|+...+|c\_n|) \epsilon \,.$$


Here is a short review of what is known about $\Lambda^\epsilon$:


1) If $\Lambda \subset \R^d$ is finite, then $\Lambda^\epsilon$ is relatively dense for all $0< \epsilon <2$. , that is $\Lambda^\epsilon +K\_\epsilon = \R^d$ for some compact $K\_\epsilon$. How to finding this $K$ is exactly waht you need. I will address this question below (\*).


2) If $\Lambda \subset \R^d$ is relatively dense and $\Lambda-\Lambda := \{ x-y | x,y \in \R^d \}$ is uniformly discrete (i.e. $|z\_1-z\_2| > \delta$, for all $z\_1, z\_2 \in \Lambda- \Lambda$) then $\Lambda^\epsilon$ is relatively dense for all $0< \epsilon <2$. Such a set is called a {\bf Meyer set}.




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(\*) How do we actually find $K$ so that $\Lambda^\epsilon +K= \R^d$? This is covered by Meyer in the book mentioned below, when he studies the connection between $\Lambda^\epsilon $ relatively dense and $\Lambda$ being harmnious.


I will explain directly what happens in your case, since in general things are a little more complicated.


First lets observe that if $t= \frac{2pi n}{ \lambda}\, n \in \Z$ then 
$e^{2 \pi \lambda \cdot t} =1 $, which is the reson the problem is easy for latices.


If we only have 2 $\lambda$'s, here is what we need to do:
First if $\lambda\_1/\lambda\_2$ is rational, then the intersection of the latices $ \frac{2pi}{ \lambda\_1}\Z \cap \frac{2pi}{ \lambda\_2}\Z$ is non-empty and any $t$ in here will do.


If $\lambda\_1/\lambda\_2$ is irrational, then the latices $ \frac{2pi}{ \lambda\_1}\Z$ and $\frac{2pi}{ \lambda\_2}\Z$ come arbitrarily close within a fixed gap, and that will do.


More exactly, fix a $\alpha$ so that if $|x-y| < \alpha $ then $ \left| e^{2 \pi x \cdot \lambda\_i} -e^{2 \pi y \cdot \lambda\_i} \right| < \epsilon$, and then the dirichclet theorem tells you that there exists number $N$ so that you can find integers $m,n$, with $m$ in any interval of length $N$ so that:


$$ | m\frac{2pi}{ \lambda\_1} - n\frac{2pi}{ \lambda\_2} | < \alpha \,.$$


Then $m\frac{2pi}{ \lambda\_1}$.




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If you have more than two $\lambda$'s, exactly as above all you have to do is prove that there exists a number $N$ so that within any interval of lenght $N$ there exists a $t$ so that for all $i$ we have: 


$$ | t- n\_i\frac{2pi}{ \lambda\_i} | < \alpha \,.$$


This is what Meyer calls a Harmnious set, and a stronger version of Dirischlet Theorem proves it.




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I recommend to you the following book(s):


-Y. Meyer, Algebraic numbers and Harmonic Analysis.


-R. V. Moody - Meyer Sets and their duals.


