Let $f(x)=e^{-a|x|}$ and $a>0$ a constant. Suppose we take a linear combination

$s(x) = \sum_{i=1}^n \alpha_i f(x-x_i)$

where $\alpha_i\in\mathbb{R}$ and $-r< x_1< \ldots < x_n< r$. Is there a constant $C>0$, independent of $\alpha_i$ for all $i$ and n such that

$ \int_{-\infty}^\infty s(x)^2dx \leq C \int_{-r}^r s(x)^2 dx $

Thank you in advance.

Worked example: I don't think this helps in the general case but I thought it was worth writing down anyway.

Take $n=2$, $r=1$, $x_1=-.5$ and $x_2=.5$. So

$s(x) = Ae^{-|x-.5|} + Be^{-|x+.5|}$

Using Mathematica and optimising for the constants $A$ and $B$ we find that

$ \int_{-\infty}^\infty s(x)^2dx \leq 1.3854 \int_{-1}^1 s(x)^2 dx $

with equality when $B=-A=0.211483$

Taking $ s(x) = e^{-|x-1|} -Ae^{-|x-1+\epsilon|} $ and using mathematica there is good numerical evidence for $ \int_{-\infty}^\infty s(x)^2dx \leq \mathcal{O}\left(\epsilon^{-1}\right) \int_{-1}^1 s(x)^2 dx $