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First let $z, w \in \mathbf R^d$ with $|z - w| < \min\{1, |z|^{-1}\}$. Further let $0 < t < r < \infty$. I wish to obtain an inequality of the form

$$\exp \left (- \frac{|e^{-t^2} z - w|^2}{1 - e^{-2t^2}} \right )\exp \left (-\alpha\frac{|e^{-t^2} w - z|^2}{1 - e^{-2t^2}} \right ) \lesssim \frac1{|z - w|^d}$$ where our friend $\alpha > 0$ is very very large. $a \lesssim b$ means that there is a constant $C > 0$ independent on the usual things (in this case only $z$, $w$ and $t$) such that $a \leq C b$.

So my question is if such an inequality works and if so how do we derive it? I've looked at this for a while, but I can't seem to figure this out. Any suggestions would be great!

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"independent on the usual things" actually is not entirely clear in this context. It should clearly be independent of $z,w$. Can it be dependent on $t,r$? How'bout $\alpha$? –  Willie Wong Aug 22 '11 at 14:51
    
@Willie: That is a good remark. I've added it (only independent on $z$, $w$ and $t$). –  Jonas Teuwen Aug 22 '11 at 14:53
    
The LHS is at most 1 (as the exponential of a negative real number) and the RHS is at least 1 (because |z-w|<1). Hence C=1. –  Did Aug 22 '11 at 15:18
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@Didier: Whoah. That is horrible. Thanks. –  Jonas Teuwen Aug 22 '11 at 15:20
    
Oh, $\alpha$ can be assumed to be positive? I had assumed that you meant it is large in modulus. –  Willie Wong Aug 22 '11 at 16:09

1 Answer 1

up vote 4 down vote accepted

The LHS is at most 1 (as the exponential of a negative real number) and the RHS is at least 1 (because |z-w|<1). Hence the inequality holds with C=1.

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