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Let $p(x)$ be a real polynomial that is bounded below. Prove that there is a real number $x_0$ such that $p(x) ≥ p(x_0)$ for all $x$.

This is listed on some practice problems for a contest math group I'm in. The question seems really easy but I'm not sure how to solve it.

One thought I had would be to decompose $\Bbb R$ into a countably infinite union of closed intervals, and then apply the extreme value theorem to each of those, and finally to order the infimum using the axiom of choice (inside the set because it is closed) and choose the smallest one (which may be repeated). But I don't know if this is correct or elegant.

For the countable union of closed intervals, I could take the set {... ,[0,1] [1,2], [2,3], [3,4], ...} which would cover $\Bbb R$, and I believe is countable. Hints appreciated.

It's also worth mentioning that with calculus you could describe all the types of intervals that are possible in a polynomial of degree $n$ and proceed from there, but I haven't fleshed that out either.

IntegrateThis
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2 Answers2

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Hint: $p$ has finitely many critical points. Consider a compact interval containing them.

lhf
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  • Ok. so f cannot be decreasing after its lefmost or rightmost critical point, so take a compact interval containing all the critical points, and since polynomials are continuous, they take compact (i.e. closed and bounded) intervals to closed and bounded intervals? – IntegrateThis Oct 01 '18 at 00:02
  • Also is my solution completely off the rails? – IntegrateThis Oct 01 '18 at 00:02
  • @IntegrateThis Your generic approach is never going to be enough by itself: the function $e^x$ is bounded below and continuous but does not attain a minimum anywhere. Any successful argument must appeal to something more than just continuity. – Erick Wong Oct 04 '18 at 20:14
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Let $b$ be the glb for for $f(\mathbb R)$. Since $f$ is a polynomial, it follows easily from the hypotheses, that $\lim_{x\to \pm \infty}=\infty$, so there is an interval $[-n,n]$ such that $f(x)>b$ whenever $x\in [-n,n]^c.$ Since $[-n,n]$ is compact, $f$ attains a minimum on $[-n,n]$, which in fact, is easily seen to be $b$

Matematleta
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