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Let $f(x,y): \mathbb R^2 \to \mathbb R$ be a polynomial such that for all $x,y, f(x,y) > 0$. By the Axiom of Completeness, $f$ has a lower bound. Does $f$ reach this lower bound?

This question is posed by Vladimir Arnold. It seemed to me "obvious" that $f$ attains its minimum, though I couldn't prove it.

I couldn't prove it for a good reason - it's not true! There are some simple examples of polynomials $f$ that do not reach their lower bound.

I've already seen an example of such, so my question is not "Can you provide a counterexample?" but rather "How would you go about exploring this question?". As such, I'm specifically not posting any counterexamples, but rather asking for how you would search for one.

Thus, this question asks:

Find a polynomial $f$ that has a lower bound but does not attain it, and explain how you found it.

SRobertJames
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    I guess one approach would be to fix a $y$ and find the minimum of the polynomial in $x$ (which is achieved), and then ensure that this minimum behaves as a decaying function of $y$, say e.g., $1/y$. – Pranay Dec 06 '24 at 16:33
  • The image set of a polynomial on $\mathbb R^2$ can be an open ray. – MPW Dec 06 '24 at 16:36

1 Answers1

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The easiest way for a polynomial to be non-negative (meaning that $f(x, y) \ge 0$) is if it's a sum of squares of polynomials. (In fact by Hilbert's 17th problem it must be a sum of squares of rational functions; we don't need this result but it's useful motivation.) So, let's consider a polynomial of the form

$$f(x, y) = g(x, y)^2 + h(x, y)^2.$$

Suppose that the lower bound of $f$ is $0$. If $f$ reaches this lower bound at a point $(x_0, y_0)$, then $g = h = 0$ at this point. So, if this doesn't happen then the zero sets $\{ g = 0 \}$ and $\{ h = 0 \}$ are disjoint. However, if $f$ gets arbitrarily close to zero then there must be zero sets of the form $\{ g = \varepsilon_1 \}, \{ h = \varepsilon_2 \}$ where $\varepsilon_1, \varepsilon_2$ are arbitrarily small which intersect; since zero sets "vary continuously" (I don't want to be too precise about what this means because it isn't necessary for this argument), this is more or less equivalent to saying that the zero sets $\{ g = 0 \}$ and $\{ h = 0 \}$ get arbitrarily close.

So we need to find two polynomials $g, h$ such that their zero sets don't intersect but get arbitrarily close. We can take $\{ g = 0 \}$ to be any hyperbola and $\{ h = 0 \}$ to be either of the asymptotes of the hyperbola. Taking $g = 1 - xy, h = x$ produces the famous

$$f(x, y) = (1 - xy)^2 + x^2$$

which has been discussed previously here and on MO.

Qiaochu Yuan
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