(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 $[0,1]$?



Constructing a monotonic function
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.


