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Let $f(x)=\Re(\sum_{k=1}^n a_k e^{i\lambda_k x})$ for $0 < \lambda_1 < \lambda_2 < \dots < \lambda_n$ and some complex $a_1$, $a_2$, $\dots$, $a_n$. What is the best (in some sense) estimate for $\inf_{[-M,M]} f(x)$ for large $M$ (in particular, for $M=+\infty$). For example, is it true that $\inf f(x)\leq -C|a_1|$ for some absolute $C$?

The best estimate I was able to get contains $\sum |a_i|$ in denominator, but I would like to have the one which does not tend to zero when we add many new terms.

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To the first question: Of course no. Take $n = 2$ and $|a_2| \gg |a_1|$. One might wonder if one can estimate it in terms of $\max_{j} |a_j|$, but I would doubt it. Idea of proof: Use some Parseval-type argument to get some control on $\|f\|_{L^2}$, conclude by it being large, that $f$ must be large ... – Helge Aug 2 '10 at 17:25
Also one should note that it makes no difference, if one wants to estimate $|\inf f(x)|$ or $\|f\|_{\infty}$, since one can multiply all $a_j$ by some scalar. – Helge Aug 2 '10 at 17:27
Take $n=2$ and $a_2\gg a_1$ Sorry, but in this case $\inf f$ behaves like $-|a_2| < -|a_1|$. Am I missing smth? I also do not see how to derive estimates for infimum of a real part via lower estimates of $\|f\|_{\infty}$, even by multiplying all $a_j$ to some scalar. Say, if $\sum a_j e^{i\lambda_j x}$ always takes values in some large disk, it does not imply that it always takes values with large positive (or large negative, as I ask) real part. – Fedor Petrov Aug 3 '10 at 6:50
up vote 4 down vote accepted

Let's do the case of full line, which is nice and clean. Without loss of generality, $\lambda_1=1$. Then you can ignore all non-integer $\lambda$s because if $f=g+h$ where $g$ includes all integer frequences and $h$ all non-integer ones, then $\inf \Re g$ is the same as the infimum of $\lim_{T\to+\infty}\Re \frac 1{2T}\int_{-T}^T gP$ where $P$ runs over positive $2\pi$-periodic trigonometric polynomials with integral $1$ over the period. Note that the corresponding limit for $h$ is $0$, and that this integral infimum (with $f$ instead of $g$, of course) estimates the infimum of $\Re f$ from above for any $f$ so $\inf\Re f\le\inf\Re g$. Now, once we are in the periodic setting, we can go to the circle and see that our problem is equivalent to the following: given an analytic function $F$ in the unit disk with $F(0)=0$ and $\Re F(z)\ge -1$, estimate $|F'(0)|$ from above. The answer now is given by the conformal mapping (Schwarz lemma).

You can do similar things for finite intervals, but you need some separation assumptions, etc. to get something meaningful. I'll not go into that now just because I'm not sure what exactly you want there and what you can afford to assume.

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Just a hint for a result somehow in the direction you are looking at, assuming generic conditions. For a generic choice of $\lambda:=(\lambda_1\,\dots,\lambda_n),$ the vector $\big(\frac{2\pi}{\lambda_1}\,\dots,\frac{2\pi}{\lambda_n}\big)$ generates a dense additive subgroup in $\mathbb{R}^n/\mathbb{Z}^n$. In such a situation we have an equality: $$\sup_\mathbb{R} f=-\inf_\mathbb{R} f=\|f\|_{\infty,\mathbb{R}}= |a_1|+\dots+|a_n|,$$

and in particular the answer to the second question is affirmative with $C=1$.

Rmk: here the assumption on $\lambda$ is "generic" both in topological and measurable sense: precisely, it holds for a $G_\delta$ set of full Lebesgue measure: the complement is a countable union of codimension 1 submanifolds.

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Thanks, of course generic case is OK, the question is rather about $\lambda_j=j$ or like this. In generic case, the estimates for $M$, which (almost) realises infimum, do depend on arithmetical properties of $\lambda$'s, so it is hard to deduce something by limiting procedure. – Fedor Petrov Aug 3 '10 at 6:52
Yep. Then possibly one has first to treat separately the case where $\lambda_k$ are rational multiples of each other (essentially the one you wrote), which reduces to bounds for a polymonial on the unit circle. In general, I guess one should partition the index set in classes according whether $\lambda_h/\lambda_h$ is rational or not. Then one should consider the closure of the additive group generated by the vector of periods (it has to be a linear vector space + a discrete lattice), and consider the worse case for getting the bounds. – Pietro Majer Aug 3 '10 at 10:37

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