Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.
share|improve this question

1 Answer 1

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.

share|improve this answer
    
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 $x+y=0$. –  tetori Jan 1 '13 at 7:16

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.