# Proving that a function doesn't have global minimum in a specific domain

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:

$$0 $$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 $$d=0.9$$:

• Computing the gradient of $f$ is always a good idea. Maybe it has a nice shape Commented Apr 28 at 13:24
• @HyperbolicPDEfriend $\frac{\partial f}{\partial p_1} = \frac{4}{1+d} ((1+2 p_1^{1+d} - 2 p_2^{1+d}) + (1+d) p_1^d (1+2 p_1 - 2 p_2))$. What is the shape of this? Commented Apr 28 at 13:34
• The math is messy, but you can show $(\partial^2_{p_1p_1}f)(\partial^2_{p_2p_2}f )-(\partial^2_{p_1p_2}f)^2 < 0$ on the whole domain. Therefore all critical points are saddle points. Commented Apr 28 at 13:48
• The last step is to use $(a +b)^2 = (a-b)^2 + 4ab$ to make this the sum of two terms that are always negative. Commented Apr 28 at 14:57
• I'll post my answer for this question's case. the $k>2$ case should probably be a separate question. Commented Apr 28 at 15:28

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.
• Thanks, this doesn't change the outcome but $p_2>p_1$ Commented Apr 28 at 15:58
• @AmirhosseinRezaei Extended the result to $p_2 > p_1$ and $d \ge 1$. Is the case $0 < d < 1$ of interest to your problem? Commented Apr 29 at 13:20
• @AmirhosseinRezaei OK, found proof for all $d$. Commented Apr 29 at 14:49