polynomial function is a proper map

Assume that $\mathbb{K}$ is either the field of real numbers or the field of complex numbers and let $W:\mathbb{K}\to\mathbb{K}$ be a polynomial function in one variable over $\mathbb{K}$ with positive degree.

. Polynomial function $W:\mathbb{K}\to\mathbb{K}$ is a proper map, i.e. for any compact subset $K\subseteq\mathbb{K}$ the preimage $W^{-1}(K)$ is compact.

Proof. Assume that

 $W(x)=\sum_{k=0}^{m}a_{k}\cdot x^{k},$

where $m=\mathrm{deg}(W)\geq 1$ is the degree of $W$.

Recall that $K\subseteq\mathbb{K}$ is compact if and only if $K$ is closed and bounded. Since polynomial functions are continous, it is sufficient to show that preimage of a bounded set is bounded. So assume that $K$ is bounded and $W^{-1}(K)$ is not bounded. Take a sequence $\{x_{n}\}_{n=1}^{\infty}\subseteq K$ such that

 $\lim_{n\to\infty}\|x_{n}\|=+\infty,$

where $\|x\|$ denotes the Euclidean norm of $x\in\mathbb{K}$.
Recall that for any $x,y\in\mathbb{K}$ we have $\|x+y\|\geq\|x\|-\|y\|$. Thus we have:

 $\|W(x)\|=\|\sum_{k=0}^{m}a_{k}\cdot x^{k}\|\geq\|a_{m}\cdot x^{m}\|-\sum_{k=0}% ^{m-1}\|a_{k}\cdot x^{k}\|=\|a_{m}\|\cdot\|x\|^{m}-\sum_{k=0}^{m-1}\|a_{k}\|% \cdot\|x\|^{k}.$

Let

 $V(x)=\|a_{m}\|\cdot x^{m}-\sum_{k=0}^{m-1}\|a_{k}\|\cdot x^{k}.$

Then $V$ is a real polynomial of degree $m$ and the leading coefficient of $V$ is positive, which implies that

 $\lim_{x\to+\infty}V(x)=+\infty.$

Now for each $n\in\mathbb{N}$ we have

 $\|W(x_{n})\|\geq V(\|x_{n}\|),$

but $V(\|x_{n}\|)$ tends to infinity, therefore $\|W(x_{n})\|$ tends to infinty. Contradiction, since for each $n\in\mathbb{N}$ we have that $W(x_{n})\in K$ and $K$ is bounded. $\square$

Corollary 1. Polynomial functions on $\mathbb{K}$ are closed maps.

Proof. Note that $\mathbb{K}$ is compactly generated Hausdorff space and therefore every proper and continous map $f:\mathbb{K}\to\mathbb{K}$ is closed. Thus (due to proposition) polynomial functions are closed. $\square$

Corollary 2. Assume that $W:\mathbb{K}\to\mathbb{K}$ is a polynomial function such that $W(x)\neq 0$ for any $x\in\mathbb{K}$. Let $f:\mathbb{K}\to\mathbb{K}$ be a map defined by the formula

 $f(x)=\frac{1}{W(x)}.$

Then $f$ is bounded.

Proof. We wish to show that there exists $M>0$ such that for all $x\in\mathbb{K}$ the inequality $\|f(x)\|\leq M$ holds. Since polynomial functions are closed maps, then the image $\mathrm{Im}(W)$ of $W$ is a closed subset of $\mathbb{K}$. Therefore $\mathbb{K}\setminus\mathrm{Im}(W)$ is open and it contains $0$, thus there exists $\epsilon>0$ such that the ball around $0$ with radius $\epsilon$ has empty intersection with $\mathrm{Im}(W)$. This means that for all $x\in\mathbb{K}$ we have that $\|W(x)\|\geq\epsilon>0$. Now for $M=\epsilon^{-1}$ and for any $x\in\mathbb{K}$ we have:

 $\|f(x)\|=\bigg{\|}\frac{1}{W(x)}\bigg{\|}=\frac{1}{\|W(x)\|}\leq\frac{1}{% \epsilon}=M$

which completes the proof. $\square$

Title polynomial function is a proper map PolynomialFunctionIsAProperMap 2013-03-22 18:30:49 2013-03-22 18:30:49 joking (16130) joking (16130) 8 joking (16130) Theorem msc 12D99 ProperMap PolynomialFunction