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.

Having the definition: A function $f:\mathbb{R^n}\rightarrow \mathbb{R^n}$ is proper if $\|f(x)\|$ tends to $\infty$ when $\|x\|$ tends to $\infty$. I have to show :

a)If $f:\mathbb{R^n}\rightarrow \mathbb{R^n}$ is proper and continuous , the inverse image of a compact set is compact;

b) If $f:\mathbb{R^n}\rightarrow \mathbb{R^n}$ is proper and continuous , show that $f$ attains its minimum.

I am really stuck (I can't get started...).Some explanation is welcome.Thanks.

share|improve this question
1  
In b) you probably want that $\lVert f(x) \rVert$ reaches its minimum. –  t.b. Sep 12 '12 at 22:31
    
@t.b. the exercises says only about $f$... –  HipsterMathematician Sep 12 '12 at 22:31
1  
The identity map $f\colon \mathbb{R} \to \mathbb{R}$ is a counterexample to that assertion. (and if $n \gt 1$ I don't know what it means for a function $f \colon \mathbb{R}^n \to \mathbb{R}^n$ to reach its minimum). –  t.b. Sep 12 '12 at 22:32
1  
It's not a linguistic issue: what is the minimum of a function $f\colon \mathbb{R}^2 \to \mathbb{R}^2$? –  t.b. Sep 12 '12 at 22:47

2 Answers 2

up vote 2 down vote accepted

For a), suppose $K$ is compact. Then $K$ is closed, so $f^{-1}[K]$ is closed by continuity. Furthermore, $K$ is bounded. Can $f^{-1}[K]$ be unbounded? If $f^{-1}[K]$ is closed and bounded, then...

For b), let $M$ be such that $\|x\|>M\implies \|f(x)\|>f(0)$. Note that the closed ball around $0$ of radius $M$ is compact, so $f$ reaches its minimum on this set. The minimum on this set is the global minimum of $f$ (why?).

share|improve this answer
    
is there anything to do with the fact that in a compact a functions attains its max and min? –  HipsterMathematician Sep 12 '12 at 22:30
    
@MeAndMath The statement "Note that the closed ball around 0 of radius M is compact, so f reaches its minimum on this set." is true because functions on compact sets attain their minimums, if that is what you are asking. –  Alex Becker Sep 12 '12 at 22:35

HINT: You know that a set in $\Bbb R^n$ is compact if and only if it is closed and bounded, and you know that if $K$ is closed, $f^{-1}[K]$ is closed. For (a), therefore, you need only show that if $K$ is bounded, then $f^{-1}[K]$ is bounded. Try supposing that $f^{-1}[K]$ is not bounded, and then use the properness of $f$.

In (b), do you mean that $\|f(x)\|$ attains its minimum? If so, see if you can see how to apply (a) by looking at $f$ on the inverse image of an appropriate compact subset of $f[\Bbb R^n]$.

share|improve this answer

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.