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Here is a question that I have been working on but having trouble with.

Let $f(x)=e^{-|x|^2}$, where $x \in \mathbb{R}^n$ and $|x|$ the usual euclidean norm of $x$.

  1. Prove that for every $\epsilon >0$ there is a positive number $M$ such that $g(x,y):=f(x)g(y)|x-y|^2 < \epsilon$ whenever $|x|^2+|y|^2 >M$. I showed this Using the fact that $e^{-|x|^2}$ goes to zero as norm of $x$ goes to infinity. But I'm having trouble with the 2nd and 3rd part of the question.
  2. Show that $S:=\sup_{x,y\in \mathbb{R}^n}f(x)f(y)|x-y|^2$ is attained at some point in $\mathbb{R}^n \times \mathbb{R}^n$.
  3. Determine the value of S.
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up vote 1 down vote accepted

Let $$g(x,y)=f(x)f(y)|x-y|^2=e^{-(|x|^2+|y|^2)}|x-y|^2.$$

Hint of 2. Take $\varepsilon=S/2$. By problem 1. we get $M$ such that $$|x|^2+|y|^2>M \implies g(x,y)<S/2.$$ So, we just consider the case $|x|^2+|y|^2\le M.$ And

$$\{(x,y):|x|^2+|y|^2\le M\}$$ is compact. So we can apply extreme value theorem.

Hint of 3. Take $x=-y$, and find it.

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Thanks so much for your helpful answer. – Jack Dawkins Dec 31 '12 at 7:01
Can you please explain a little more on the last part. How do you find the S. I mean do you treat $-2e^-|x|^2|x|^2$ as a function from R to R and find the max using single variable calculus? – Jack Dawkins Jan 1 '13 at 0:46
Without loss of generality, we can assume $|x|=|y|$. By triangle inequality, We get $$e^{-|x|^2-|y|^2}|x-y|^2 \le e^{-|x|^2-|y|^2} (|x|+|y|)^2$$ and equality holds iff $x+y=0$. – Hanul Jeon Jan 1 '13 at 7:16

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