# 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. • answered • 259 views •$10.00