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I am interested in the scaling of

$$F(x_1,x_4)=\int_{\mathbb R^2} e^{-\vert x_1 -x_2 \vert -\varepsilon \vert x_2 -x_3 \vert- \vert x_3 -x_4 \vert } \ dx_2 dx_3 $$

In particular, I suspect that

$$F(x_1,x_4) \le C \varepsilon^{-n} e^{-{\varepsilon}\vert x_1 -x_4\vert}$$ for some universal $C>0$ and $n \ge 0$.

But this is really only based on pure heuristic and I do not know which $n$ could be optimal here.

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    $\begingroup$ This integral can be performed exactly. I'll write it down in a little while if nobody else does - first, dinner ... $\endgroup$ May 10, 2021 at 0:39
  • $\begingroup$ @MichaelEngelhardt I did mean that, thank you. $\endgroup$
    – Kung Yao
    May 10, 2021 at 0:42

1 Answer 1

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$\newcommand\ep\varepsilon$$F(x_1,x_4)$ is $8/\ep$ times a value of the convolution of two copies of a pdf with maximum value $1/2$ and a pdf with maximum value $\ep/2$. So, $$F(x_1,x_4)\le(8/\ep)\min(1/2,1/2,\ep/2)=4\min(1/\ep,1)$$ for all real $x_1,x_4$.


The straightforward integration gives $$F(x_1,x_4)=2\frac{\ep e^{-\left| x\right| } \left(\ep^2 (\left| x\right| +1)-\left| x\right| -3\right)+2 e^{-\ep \left| x\right| }}{\left(1-\ep^2\right)^2}$$ for $\ep\in(0,1)\cup(1,\infty)$, with $$F(x_1,x_4)=\frac{1}{2} e^{-|x|} \left(x^2+3|x|+3\right)$$ for $\ep=1$, where $x:=x_4-x_1$, for all real $x_1,x_4$.

In particular, for each $\ep_*\in(0,1)$ and all $\ep\in(0,\ep_*]$, $$F(x_1,x_4)\le C(\ep_*)e^{-\ep\left|x_4-x_1\right| }$$ for some real $C(\ep_*)>0$ depending only on $\ep_*$ and all real $x_1,x_4$.

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  • $\begingroup$ are you sure there is no decay as $\vert x_1 -x_4 \vert$ become large? $\endgroup$
    – Kung Yao
    May 10, 2021 at 1:30
  • $\begingroup$ @KungYao : I did not say that there would be no decay -- but the original version of your question allowed even for growth. $\endgroup$ May 10, 2021 at 1:53
  • $\begingroup$ I reproduce this result. $\epsilon $ has to satisfy the constraint $1+\epsilon >0$ for $F(x_1,x_4)$ to converge. $\endgroup$ May 10, 2021 at 2:41
  • $\begingroup$ @MichaelEngelhardt : Thank you for your comment. $\endgroup$ May 10, 2021 at 13:40
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    $\begingroup$ Using $|x|+\epsilon |x-w| \geq (1-\epsilon)|x| +\epsilon |w|$, for $\epsilon <1$ twice (one for the integral in $dx_2$, the other for $dx_3$), one gets $F(x_1,x_4) \leq \frac{C}{(1-\epsilon)^2 }e^{-\epsilon |x_1-x_4|}$. $\endgroup$ May 10, 2021 at 13:53

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