Please help to prove that there is no polynomial $p\colon \mathbb{R}^n \to (0,+\infty)$.
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Please clarify your question. What do you mean by a polynomial on $R^n$? – RobertTheTutor Mar 17 '21 at 04:26
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What about $p(x) = 1 + \sum_j x_j^2$? – asahay Mar 17 '21 at 04:26
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1@asahay That does not produce any values in $(0,1)$. – Théophile Mar 17 '21 at 04:27
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@Théophile you're assuming OP means that $image(p)=(0,\infty)$ rather than $image(p) \subseteq (0,\infty)$ ? – BCLC Mar 17 '21 at 04:31
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1@JohnSmithKyon Yes, I assume that "all positive and only positive values" means that the image is $(0, \infty)$. – Théophile Mar 17 '21 at 04:34
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Related: https://math.stackexchange.com/q/279497/42969, https://math.stackexchange.com/q/2481978/42969, https://math.stackexchange.com/q/1493380/42969 – Martin R Mar 17 '21 at 05:42
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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, $f(\sqrt{\epsilon}, \frac1{\sqrt{\epsilon}}) = \epsilon$.)
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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
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1oh 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
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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
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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
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