polynomial function is a proper map

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

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

Proof. Assume that

$$W(x)=\sum _{k=0}^m{a}_k\cdot x^k,$$

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

Recall that $K\subseteq 𝕂$ 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 ${\{x_n\}}_{n=1}^{\mathrm{\infty }}\subseteq K$ such that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n\parallel =+\mathrm{\infty },$$

where $\parallel x\parallel$ denotes the Euclidean norm of $x\in 𝕂$.
Recall that for any $x,y\in 𝕂$ we have $\parallel x+y\parallel \ge \parallel x\parallel -\parallel y\parallel$. Thus we have:

$$\parallel W(x)\parallel =\parallel \sum _{k=0}^m{a}_k\cdot x^k\parallel \ge \parallel a_m\cdot x^m\parallel -\sum _{k=0}^{m-1}\parallel a_k\cdot x^k\parallel =\parallel a_m\parallel \cdot {\parallel x\parallel }^m-\sum _{k=0}^{m-1}\parallel a_k\parallel \cdot {\parallel x\parallel }^k.$$

Let

$$V(x)=\parallel a_m\parallel \cdot x^m-\sum _{k=0}^{m-1}\parallel a_k\parallel \cdot x^k.$$

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

$$\underset{x\to +\mathrm{\infty }}{lim}V(x)=+\mathrm{\infty }.$$

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

$$\parallel W(x_n)\parallel \ge V(\parallel x_n\parallel ),$$

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

Corollary 1. Polynomial functions on $𝕂$ are closed maps.

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

Corollary 2. Assume that $W:𝕂\to 𝕂$ is a polynomial function such that $W(x)\ne 0$ for any $x\in 𝕂$. Let $f:𝕂\to 𝕂$ 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 $x\in 𝕂$ the inequality $\parallel f(x)\parallel \le M$ holds. Since polynomial functions are closed maps, then the image $\mathrm{Im}(W)$ of $W$ is a closed subset of $𝕂$. Therefore $𝕂\setminus \mathrm{Im}(W)$ is open and it contains $0$, thus there exists $ϵ>0$ such that the ball around $0$ with radius $ϵ$ has empty intersection with $\mathrm{Im}(W)$. This means that for all $x\in 𝕂$ we have that $\parallel W(x)\parallel \ge ϵ>0$. Now for $M={ϵ}^{-1}$ and for any $x\in 𝕂$ we have:

$$\parallel f(x)\parallel =\parallel \frac{1}{W(x)}\parallel =\frac{1}{\parallel W(x)\parallel }\le \frac{1}{ϵ}=M$$

which completes the proof. $\mathrm{\square }$

Title	polynomial function is a proper map
Canonical name	PolynomialFunctionIsAProperMap
Date of creation	2013-03-22 18:30:49
Last modified on	2013-03-22 18:30:49
Owner	joking (16130)
Last modified by	joking (16130)
Numerical id	8
Author	joking (16130)
Entry type	Theorem
Classification	msc 12D99
Related topic	ProperMap
Related topic	PolynomialFunction