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I am trying to show that for $t$ in $(0, T]$ for some $T > 0$ and $t |y| \leqslant C$ for some $C > 0$ that

$$\tag{1} |x|^M \exp \left (-\frac{D}{1 - \rm e^{-2 t^2}} {|e^{-t^2} x - y|^2}{} \right )$$ is uniformly bounded in $x, y$ (in $\mathbf R^d$) and $t$. The implied constant may depend on $D > 0$ and integers $M$. It is also possible to choose $D > 0$ as long as it does not depend on any of the quantities.

Alright! So let us set $z = t y$ then $|z| \leqslant C$. Right. So we can rewrite $(1)$ as $$\tag{2} |x|^M \exp \left (-\frac{D}{t (1 - \rm e^{-2 t^2})} {|e^{-t^2} t x - z|^2}{} \right ).$$

We could also note that we can just consider real numbers $x$ and $y$ and afterwards take products. That might simplify the analysis a bit. So we "deduce" from $(2)$ that we should find a bound for (abusing $M$ a bit) $$\tag{3} x^{2M} \exp \left (-\frac{D}{t (1 - \rm e^{-2 t^2})} {(e^{-t^2} t x - z)^2}{} \right ).$$

This is where misery starts. I have started expanding the square and then using cases but then stuff starts depending on each other. Ugly.

Does someone have any idea? A suggestion perhaps?

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1 Answer 1

up vote 3 down vote accepted

For $t\in(0,T\;]$, $$ \frac{D}{1-e^{-2t^2}}\ge\frac{D}{1-e^{-2T^2}}\tag{1} $$ If $|x|\ge2(C/t)e^{T^2}$, then $e^{-t^2}|x|>2|y|$, and therefore, $$ \left|e^{-t^2}x-y\right|^2\ge\frac{1}{2}e^{-t^2}|x|\ge\frac{1}{2}e^{-T^2}|x|\tag{2} $$ Thus, $$ \frac{D}{1-e^{-2t^2}}\left|e^{-t^2}x-y\right|^2\ge\left(\frac{D/2\;\;e^{-T^2}}{1-e^{-2T^2}}\right)|x|\tag{3} $$ So that $$ |x|^M\exp\left(-\frac{D}{1-e^{-2t^2}}\left|e^{-t^2}x-y\right|^2\right)\le|x|^M\exp\left(-\left(\frac{D/2\;\;e^{-T^2}}{1-e^{-2T^2}}\right)|x|\right)\tag{4} $$ which is obviously bounded.

However, let $y=e^{-t^2}x$ and $t=C/|x|$. Then, for $|x|>C/T$, we have $t|y|<C$, $t<T$, and $$ |x|^M\exp\left(-\frac{D}{1-e^{-2t^2}}\left|e^{-t^2}x-y\right|^2\right)=|x|^M\tag{5} $$ which can get arbitrarily large.

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