# one-side estimates for quasi-trigonometric polynomial

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.

• 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

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).
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.
• 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