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My teacher gave me this problem:

Decide whether there is a polynomial of two variables bounded below with its set of values being an open subset of $\mathbb{R}$.

That means I can think of this function:

$$f(x,y)=Ax^2+Bxy+Cy^2+D$$

The fact that it is bounded below means that both $A$ and $B$ must be positive, right? But at this point I am stuck and I have no idea how to proceed. My guess is that it is unfeasible, because I can't think of a function that would satisfy all the requirements.

SlowerPhoton
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1 Answers1

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Consider the polynomial $f : \mathbb{R}^2 \to \mathbb{R}$ given as $f(x,y) = (1-xy)^2 + x^2$.

$\DeclareMathOperator{\Ima}{Im}$The image of $f$ is $\langle 0, +\infty \rangle$, an open set which is bounded from below.

Indeed, we have $f(x,y) \ge 0$. Assume $0 = f(x,y) = (1-xy)^2 + x^2$. This gives $x = 0$ and $xy = 1$, a contradiction. Thus, $\Ima f \subseteq \langle 0, +\infty \rangle$.

Furthermore, for every $\varepsilon > 0$, we have:

$$f\left(\sqrt{\varepsilon}, \frac1{\sqrt{\varepsilon}}\right) = \left(1 - \sqrt{\varepsilon}\cdot \frac1{\sqrt{\varepsilon}}\right)^2 + \varepsilon = \varepsilon$$

So $\Ima f =\langle 0, +\infty \rangle$.

The example is from this answer.

mechanodroid
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  • A Putnam Contest Q was to classify the types of images of polynomials $p:\Bbb R^2\to \Bbb R.$ The open half-line was the one that most contestants missed, or "proved" that it couldn't occur. – DanielWainfleet Oct 21 '17 at 04:28
  • @DanielWainfleet What are the other ones? The closed half-line and $\mathbb{R}$ are obvious. Could it be a bounded interval? – mechanodroid Oct 21 '17 at 08:27
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    A one-point set (for a constant polynomial). – DanielWainfleet Oct 21 '17 at 16:56
  • It is not a closed interval because it doesn't contain '0'. It is more than obvious, you can apply the definition of an open set directly to it. Am I missing something? – SlowerPhoton Oct 22 '17 at 07:37
  • @SlowerPhoton In this example, the image is an open (unbounded) interval, yes. However, if you take $p(x,y) = x^2 + y^2$, then the image is $[0,+\infty\rangle$. – mechanodroid Oct 22 '17 at 09:29