# Norm and continuous functions

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.

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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
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
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

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?).

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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 $f[\Bbb R^n]$.

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