Recently I've came across such problem: give a polynomial $P(x,y)$, with $\inf_{\mathbb{R}^2} P=0$, but there is no point on the plane where $P=0$. I couldn't solve it after a day, and seriously doubt whether such a function exists, however its source claims that there is. Is that really possible?

  • $\begingroup$ Is this in the real plane $\mathbb{R}^2$, or could it perhaps be over the rationals? $\endgroup$
    – hardmath
    Jan 15, 2013 at 19:51
  • $\begingroup$ The real plane. $\endgroup$
    – aplavin
    Jan 15, 2013 at 19:51
  • $\begingroup$ I edited "$in f(P)$ to $\inf f(P)$, as that seems to be the more common notation. Nice question, by the way. $\endgroup$ Jan 15, 2013 at 19:51

1 Answer 1


$P(x,y)=(1-xy)^2+x^2$ has this property. Clearly $P>0$ and also the sequence $(x_n,y_n)=(1/n,n)$ shows that $\inf P=0$.

  • 2
    $\begingroup$ Very nice example. $\endgroup$ Jan 15, 2013 at 19:54
  • $\begingroup$ Thanks! Did you hear about the answer before or created it now? $\endgroup$
    – aplavin
    Jan 15, 2013 at 20:02
  • 3
    $\begingroup$ I read about it somewhere before --- I'm not sure where exactly. If I recall correctly it is a counter-example to the claim that $\text{im } P(x,y)$ is closed. $\endgroup$ Jan 15, 2013 at 20:05
  • $\begingroup$ @JpMcCarthy: sorry, but what is $im$? $\endgroup$
    – aplavin
    Jan 15, 2013 at 20:12
  • $\begingroup$ Sorry. The image of the real plane under a polynomial in two variables. I suppose one would think it would be closed. $\endgroup$ Jan 15, 2013 at 20:16

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