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 nonnegative
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yes you are right, I din't make it clear that the sequence r depends on the function f. i.e. f is an element of a space of all functions for which there is a bounded sequence r such that the function is equal to the sum above. (In this case, there exists a bounded sequence r such that the absolute value of f is equal to that sum). and now I want to find a constant C sucht that the inequality holds for all such f. (so r depends on f and C does not depend on either f or r).

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 nonnegative. 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 supf[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.
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