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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 $

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Did you try to see what happens in the following kind of situation ? $a=r=1$ and $s(x)=\exp(-|x-1|)-a(\epsilon)\exp(-|x-1+\epsilon)$ and optimize in $a(\epsilon)$ for the second integral ? (An obvious note (I hope I am right) if you assume some uniform room between $-r$ and $x_1$ and between $x_N$ and $r$ then this is obvious). – camomille Apr 3 '11 at 20:14
And did you try when $x_1$ and $x_2$ are close to $r$ ? As I said earlier, if you have $-r+\delta < x_1 < ... < x_N < r-\delta$ then you can find a bound (that depends also on $\delta$). – camomille Apr 4 '11 at 9:16
So if we assume in addition that $-r+\delta< x_1< \ldots < x_N < r-\delta$ then how do we go about proving the inequality in this case? – alext87 Apr 4 '11 at 9:21
On each $[r+(k-1)\delta,r+k\delta[$, $k\ge 1$, you have $s(x)^2=s(x-k\delta)e^{-2ak\delta}$. So the integral on this interval is less than $e^{-2ak\delta}$ times the integral on $[-r,r]$. You sum etc. – camomille Apr 4 '11 at 9:44
Great. So working through with this I get the constant to be something like $\coth(a\delta)$ which is like $1/(a\delta)$ for $a\delta<<1$ which fits the numerical evidence. – alext87 Apr 4 '11 at 10:21

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