Consider the following function:
$$f(p_1,p_2) = \frac{2}{(1 + d)}(1 + 2 p_1 - 2 p_2) (1 + 2 p_1^{1 + d} - 2 p_2^{1 + d})$$
with the following domain:
$$0<p_1<p_2<1$$ $$d>0$$
My first question is, how can I show analytically that the function $f$ doesn't have a global minimum in its domain? As an example, here's the plot of the function for $d=10$:
The case for $d=0.9$:
Since the domain is open, any global extremum must be a critical point. So if we can show that all critical points in the domain must be saddle points, then there is no global extremum. In two dimensions, a critical point is a saddle point if and only if Hessian determinant $|H| = (\partial_{11}f)(\partial_{22}f) - (\partial f_{12})^2$ is negative. So if we can determine a subset of the domain in which all critical points must lie and show the Hessian determinant is negative on that subset, we have shown all critical points must be saddles.
We can get a condition on the critical points from $\partial_1 f + \partial_2 f = 4(1 +2 p_1 - 2 p_2)(p_1^d-p_2^d)$. For this to be zero with $p_2 > p_1$, we must have $p_2 = 1/2 + p_1$. Calculating the second derivatives is a little annoying, but they come out to nice expressions along $p_2 = 1/2 + p_1$. We have \begin{eqnarray} \partial_{11} f &=& 4p_1^{d-1}\left[4p_1 + d(1-2p_2 + 2p_1) \right] = 16p_1^d\\ \partial_{22} f &=& 4p_2^{d-1}\left[4p_2 - d(1-2p_2 + 2p_1)\right] = 16p_2^d\\ \partial_{12}f &=& -8(p_1^d + p_2^d). \end{eqnarray} From which we get that at any critical point, we have $$ |H| = 256p_1^dp_2^d - 64(p_1^d + p_2^d)^2 = -64(p_2^d - p_1^d). $$ This is clearly always negative, so all critical points are saddles and there are no extrema in the domain.
