# Zeros of Gradient of Positive Polynomials.

It was asked in the Putnam exam of 1969, to list all sets which can be the range of polynomials in two variables with real coefficients. Surprisingly, the set $(0,\infty )$ can be the range of such polynomials. These don't attain their global infimum although they are bounded below. But is it also possible that such polynomials with range $(0,\infty )$ also have a non zero gradient everywhere?

-
What's the question? –  Daniel Moskovich Sep 8 '10 at 1:43
@Will $x^2+(xy-1)^2$. @Daniel The question is, can there be such a polynomial with nonvanishing gradient? I don't see why not, but I haven't thought about it very hard. –  David Speyer Sep 8 '10 at 2:36
Here is a sketch of a proof that it is impossible for the gradient to be everywhere nonzero. If all level surfaces are bounded, the gradient has to have a zero somewhere: either it vanishes somewhere on a nonempty level surface or one has a smooth bounded level surface and then the gradient vanishes somewhere in the interior. If there is an unbounded level surface, then, assuming the gradient is everywhere nonzero there are lines through the origin along which the function infinitely increases and lines along which it infinitely decreases. –  algori Sep 8 '10 at 2:58
Algori, I don't follow your last sentence at all. If you really think you have an argument, could you write it up in detail as an answer? This question was also posted on Art of Problem Solving, but no one there made any progress: artofproblemsolving.com/Forum/viewtopic.php?f=73&t=357702 –  JBL Sep 8 '10 at 3:17
If algori's approach can be fixed up it is subtle. I'm looking up Poincare-Bendixson and related matters. With no critical point we do have a global nonzero vector field, the gradient. With no periodic level curve it is tempting to predict that the integral curves of the gradient field foliate the plane and therefore so do the level curves, and all meet orthogonally. But it needs work. –  Will Jagy Sep 8 '10 at 3:54

$(1+x+x^2y)^2+x^2$

-
fedja, beautiful! –  Will Jagy Sep 14 '10 at 2:49
I knew an answer would have appeared here, containing only a polynomial! –  Pietro Majer Sep 14 '10 at 18:36
Hope it's not spoiling if I post a link to fedja's explanation in a different thread mathoverflow.net/questions/38639/thinking-and-explaining/… –  j.c. Sep 14 '10 at 21:54

Too long for a comment. I've got to wonder just how difficult this is.

Anyway, one thing did work out, at least locally: We have a polynomial function $F(x,y)$ that is assumed to have a nonvanishing gradient. Then the vector field $$\left( \frac{\partial F}{\partial x}, \; \frac{\partial F}{\partial y} \right)$$ has integral curves that foliate the plane. On the other hand, the integral curves of $$\left( \frac{ - \partial F}{\partial y}, \; \frac{\partial F}{\partial x} \right)$$ are level curves as well, and foliate the plane. We know that if one of these level curves is a simple closed curve, using the Jordan Curve theorem it has an interior. If the function is constant on this it has zero gradient within, otherwise it achieves its maximum or minimum within and again has a critical point.  After this I'm stuck.

In particular, I simply don't see what polynomial does for us. Homogeneous polynomial would be different. There is a conjecture of Thom about local behavior that he apparently settled for homogeneous polynomials only. I would like to say that the picture that is being built up resembles that for $F(x,y) = e^x$ and that is absurd for a polynomial. Well, perhaps.

I've also got to wonder how much the OP knows.

-
I didn't do much.. Looking into the fact that, for polynomials of degree 2 except a small class, the levels curves are smymmetric with respect to some point (h,k),and this point turns out to be a point where gradient is zero. I wonder if anything can be said about polynomials of higher degrees,with their gradient vanishing only at a single point.It works for the example atleast.. –  Anonymous Sep 9 '10 at 8:57