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

I'm sorry for the possibly deceiving answer, and I'm ready to adjust it if you wish to change the question, but I think already saw this or a very similar question of yours appearing several times and always wanted to explain in comments why it is trivial, but then each time someone reserved the question very quickly and I couldn't comment any more. (And I couldn't see the answer.)

I want to avoid you paying again and again so I decided to reserve the question and give my answer, yet as I said, I'm ready to adjust if you wish to reformulate the question.

So, very simply, for any such sequence $r$ you can simply DEFINE
$C := \|f\|_\infty / \|r\|_\infty$, and you have the required inequality (with equality),
unless of course $ \|r\|_\infty = 0$, in which case you can take any $C>0$, say $C=1$, to have the inequality with 0 on the right hand side.

Now it should be clear why the answer is trivial ; and it is equally trivial for a reversed inequality ($\le$) and also for strict inequalities (> or <) by just adding a factor 2 or 1/2 in the definition -- except for the case r=0 where you do have equality (0 = 0) in any case, and so it's impossible to have a strict inequality in that case.

Feel free to let me know what you would like to change in order to make the question nontrivial.

EDIT: OK, after I posted this answer, we had some discussions and now it is clear what you actually meant to ask. Before elaborating on that, given the time I'm investing, I just wanted to clarify, to avoid any ambiguity in case of a dispute, that I already did answer the original question as it was formulated when I accepted to answer, namely:

You have given some $\gamma>0$ and a sequence r and a function f(x) satisfying some relation, and asked to prove that there is C > 0 such that $ \|f\|_\infty  \ge C \cdot \|r\|_\infty $.

I did prove this by pointing out that (obviously) $ C := \|f\|_\infty / \|r\|_\infty $ if $r\ne o$, and $C=1$ if $r=o$, is a constant C > 0 which fulfils the desired inequality.

Afterwards, it became clear that the question you mean to ask is the following:

Show that for given $\gamma > 0$, there exists a constant C such that for any sequence $r\in\ell^\infty(\Z)$ such that $\forall x\in\R : f_r(x) := \sum r_n e^{-2\gamma(x-n/2)^2} > 0$, we have $ \|f_r\|_\infty  \ge C \cdot \|r\|_\infty $.

This is of course much more complicated and it is not yet clear whether this can be shown. I'm willing to do my best to make some progress towards that goal but it should be clear that this is a much more difficult question than the one I accepted to (and did) answer.

Let us first note that $f: r\mapsto f_r$ as defined above is (obviously) a linear map.

Remarks: 1) This definition / map depends on the constant gamma, but since that constant is given once for all, we do not show this dependency explicitly, to simplify notation.

2) I do not put $|f(x)|=...$ with an absolute value here. I will rather require the additional condition $f_r \ge 0$, i.e., $\forall x\in\R:f_r(x)\ge 0$, whenever this is relevant. Indeed, you obviously want to have that condition, $f_r\ge0$, when you write $|f(x)|= f_r(x)$, and precisely this makes the interesting but difficult.

We should define the relevant subspace $P := \{ r\in\ell^\infty(\Z) : f_r \ge 0 \}$.

Given the linearity of $f: r\mapsto f_r$, and the homogeneity of norms, the existence of the constant $C>0$ is equivalent to ask whether $\inf\{ ~\|f_r\|_\infty / \|r\|_\infty ~;~ r\in P~\} = \inf\{ \|f_r\|_\infty ~;~ r\in P,~ \|r\|_\infty=1~\} > 0 $.

So we should focus on $r\in P$ and maybe $\|r\|_\infty=1$ (although that may be useful just to fix ideas, but not really essential).

A counter-example to the assertion would be to find a way of constructing a sequence of sequences $r^{(k)} \in P $ for which $\sup f_r/\|r\|_\infty$ is smaller and smaller.
If that ratio tends to zero, we would have a counter example. If not, we might try to understand why, and this could yield an idea to prove the property.

I considered such sequences by taking $r_0 = 1$, then $r_1 = r_{-1} < 0$ as "negative" as possible, and $r_2 = r_{-2}>0$ as small as possible, and so on, as to have still $r\in P$ and a maximum of $f_r$ as small as possible. I found that with only 5 nonzero coefficients $\{ r_{-2}, ..., r_{2}\}$ one can get a ratio $\sup f_r/\|r\|_\infty$ as small as 0.0525 by taking $r_0=2$, $r_1=r_{-1}=-1.493$, and $r_1=r_{-1}=0.5855$.

Taking a few more nonzero coefficients, I also found concrete solutions with lower ratio, approx. 0.025. (Tell me in comments if you're interested in the coefficients or method.)

But while I first had the feeling that it might be possible to find an arbitrarily small ratio, I'm no more sure about that. I think investigation in what prevents to construct this might allow to find a "constructive" proof of impossibility, and/or a formula for (the "optimal" value of) this constant C. But so far I don't have more quantitative results in that sense.