1

Please help to prove that there is no polynomial $p\colon \mathbb{R}^n \to (0,+\infty)$.

1 Answers1

3

Although it may be counterintuitive, yes, there is; for example, $$f(x,y) = (1-xy)^2+x^2,$$ which I found here: https://math.stackexchange.com/a/3822, which in turn comes from a comment in math overflow.

You can see that $f(x,y)>0$ everywhere, and we cannot have $f(x,y)=0$. On the other hand, for any $\epsilon > 0$, you can find $x,y$ such that $f(x,y) = \epsilon$. (Specifically, $f(\sqrt{\epsilon}, \frac1{\sqrt{\epsilon}}) = \epsilon$.)

BCLC
  • 13,459
Théophile
  • 24,627
  • why would it be counterintuitive? – BCLC Mar 17 '21 at 04:35
  • 1
    @JohnSmithKyon It is not true for a polynomial in one variable, for example, so it's a good demonstration that adding a dimension can have interesting results. – Théophile Mar 17 '21 at 04:37
  • 1
    oh wow that was mindblowing. thanks for the info. is this some commonly known elementary result in analysis? or algebra? or even just calculus or something? – BCLC Mar 17 '21 at 04:39
  • 1
    @JohnSmithKyon I'm surprised too! You'll find it on this list of common false beliefs in mathematics. I think it also comes down to the way we learn calculus, where polynomials are the "well-behaved" functions. There are any number of exercises like "find the global extrema of this polynomial", and it easily becomes an instinct that they must exist if it's bounded. – Théophile Mar 17 '21 at 04:44
  • 1
    @JohnSmithKyon A polynomial is continuous,the set of values it takes is connected. A non-constant polynomial $f$ of one variable must have $\lim_{x\to\pm\infty}|f(x)|=\infty$. If $\lim_{x\to-\infty}f(x)=-\infty$ and $\lim_{x\to\infty}f(x)=\infty$ then $f$ takes all real values (and the set $(-\infty,\infty)$ of values is closed). Now assume $\lim_{x\to-\infty}f(x)=\infty=\lim_{x\to\infty}f(x)$. There are $a,b$ with $f(x)>f(0)$ if $x<a$ or $x>b$. Then $f$ achieves its minimum $m$ on the closed interval $[a,b]$, $m$ also global (absolute) minimum. The set of values of $f$ is $[m,\infty)$,closed. – Mirko Mar 17 '21 at 05:08
  • 1
    @Mirko right thanks for the info! – BCLC Mar 17 '21 at 05:21