Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

(I'm not sure how to tag this; I'm tagging it math.CO because that's where it arose, but math.FA or something might be appropriate as well. Feel free to edit or comment on what this should be.)

I have a polynomial in a bunch of variables $P(x_1, x_2, \ldots, x_n, y_1, y_2, \ldots, y_m)$. I know that if we evaluate the polynomial at

$x_1 = x_2 = \ldots = x_n = 1, y_1 = y_2 = \ldots = y_m = 0$

then the result is greater than the evaluation at

$x_1 = x_2 = \ldots = x_n = 0, y_1 = y_2 = \ldots = y_m = 1$.

Then are there necessarily functions

$f_1, \ldots, f_n$, with $f_i \in C^1[0,1], f_i(0) = 0, f_i(1) = 1$

and

$g_1, \ldots, g_m$, with $g_j \in C^1[0,1], g_j(0) = 1, g_j(1) = 0$

such that

$Q(t) = P(f_1(t), f_2(t), \ldots, f_n(t), g_1(t), \ldots, g_m(t))$

is monotonically decreasing on $[0,1]$?

share|improve this question
    
I'm embarrassed at how trivial it is to see the necessary and sufficient condition -- I just got bogged down in the complexity of the application, I think. –  Harrison Brown May 9 '11 at 14:08

1 Answer 1

up vote 3 down vote accepted

The answer is no. Is suffices to construct a polynomial such that these points are local minima. An example when $n=1$ is $$x^2(y-1)^2(\epsilon+y^2(x-1)^2)$$ with $0<\epsilon$ small.

share|improve this answer
    
I have a sneaking suspicion that I asked the question wrong and this situation doesn't apply; nevertheless, I think I have a better intuition for the problem now. –  Harrison Brown May 9 '11 at 13:59

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.