# A lower bound on infinite sum of exponential functions (corrected version)

Let $\gamma >0$. Consider a sequence $r=(r_n)_{n \in \mathbb{Z} } \in \ell^\infty(\mathbb{Z})$ and  a function $f$ such that $|f(x)|=\sum\limits_{n \in \mathbb{Z}} r_n \exp \left (-2 \gamma \left (x- \frac{n}{2} \right)^2\right)$ for $x \in \mathbb{R}$.

Prove that there exists a constant $C>0$ such that $||f||_\infty \geq C \cdot ||r||_\infty$.

• Are you assuming that r_n \geq 0 for all n?

• Are you sure the absolute value of f is given by that equation? Because all the r_n could be negatives, so it would be an absurd.

• The sequence is not arbitraty, i want to consider a sequence r such that the sum is always non-negative

• Oh, it looks the other way round : the function f is defined as a function of r. It looks complicate to define r as a function of f -- one would have to restrict ourselves on the range (image) of the function F : $\ell^\infty \to \R^\R ~;~~ r \mapsto f = f(r)$ as defined ... how actually ? because you write |f(x)| ... maybe f(x) would be enough/better? (do you maybe mean: f(x) := ... \ge 0$? Please advise. • And yet, in any case, whether r is defined in terms of f (if that's possible), or conversely, you always have the given solution, C := |f| / |r|. So I still don't really get how to make this question nontrivial. • I can define f as a function of r as well, that makes more sense. But it‘s important that the sum is always non-negative. And i need a constant C>0 such that the inequality holds for all f, i.e. C needs to be independent of f and r. • OK, now I andserstand and the question makes sense, so C must depend only on$\gamma$, but then be valid for any sequence r [such that f(x) >= 0 for all x ?] • I think one can let$C := sup { \| f[r] \|_\infty ~;~~ r \in B_1(\ell^\infty ) } $; ~ I'm busy right now but let me some time to elaborate. • Er, actually we need exactly that but with "inf", not sup, because$ r \mapsto f[r] $is indeed a linear map, easy to see. So the question is, do we have an inf strictly positive, and I fear that the answer might well be "no". We need to construct a counter example, i.e., r with sup |r_n| =1 such that sup|f[r] is as small as we wish. I think that shouldn't be very difficult. Again I have to work this afternoon but I think I can work it out by tomorrow . (Maybe you can do it yourself!) • It might be sufficient to consider r with r_0 = 1 and r_k = -c with 0 • Ouch, all after the "less than" sign has disappeared ! I went on : c sufficiently small to have your constraint f ≥ 0, and conveniently chosen k • If we let$C:= inf{ ||f[r]||_\infty ; ||r|| \leq 1\}$, then C=0. so that wouldn't help, but I am not sure if C=0 when we only allow ||r|| =1 • No, not |r| ≤ 1, but = 1 (or ≥ 1 if you wish but given linearity it doesn't matter, it's the same as ≠ 0. • (oh sorry for the typo, I meant = 1 indeed) yes, that's exactly equivalent to your question : do we have that this inf > 0 or not ? • I started trying to construct a sum of exponentials with alternating signs of r_n as to get f > 0 but max f as small as possible for given sup r_n. • In case you didn't see this yet (but maybe it's the preprint you're trynig to understand...), you might be interested in https://arxiv.org/abs/1605.00165 • I'm sorry for not commenting for a week, I was a bit preoccupied. but I really appreciate your time and patience, you helped me a lot, so thanks :) • kein Problem, war interessant, aber ich hatte auch keine Zeit mehr, da weiterzumachen ... vielleicht bei nächster Gelegenheit. The answer is accepted. Join Matchmaticians Affiliate Marketing Program to earn up to a 50% commission on every question that your affiliated users ask or answer. • answered • 645 views •$10.00