# Constructing a monotonic function

(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 $[0,1]$?

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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

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.