# 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.

Proposition^{}. 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 $m=\mathrm{deg}(W)\ge 1$ 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 ${\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}}\subseteq K$ such that

$$\underset{n\to \mathrm{\infty}}{lim}\parallel {x}_{n}\parallel =+\mathrm{\infty},$$ |

where $\parallel x\parallel $ denotes the Euclidean norm of $x\in \mathbb{K}$.

Recall that for any $x,y\in \mathbb{K}$ 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 $\mathbb{K}$ 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. $\mathrm{\square}$

Corollary 2. Assume that $W:\mathbb{K}\to \mathbb{K}$ is a polynomial function such that $W(x)\ne 0$ 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 $\parallel f(x)\parallel \le M$ holds. Since polynomial functions are closed maps, then the image $\mathrm{Im}(W)$ of $W$ is a closed subset of $\mathbb{K}$. Therefore $\mathbb{K}\setminus \mathrm{Im}(W)$ is open and it contains $0$, thus there exists $\u03f5>0$ such that the ball around $0$ with radius $\u03f5$ has empty intersection^{} with $\mathrm{Im}(W)$. This means that for all $x\in \mathbb{K}$ we have that $\parallel W(x)\parallel \ge \u03f5>0$. Now for $M={\u03f5}^{-1}$ and for any $x\in \mathbb{K}$ we have:

$$\parallel f(x)\parallel =\parallel \frac{1}{W(x)}\parallel =\frac{1}{\parallel W(x)\parallel}\le \frac{1}{\u03f5}=M$$ |

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 |