Let $F:\mathbb{R}^n_+\rightarrow\mathbb{R}^n, \ (x_1,x_2,\dots,x_n)\mapsto(F_1(x_1,...,x_n),...,F_n(x_1,...,x_n))$ be a vector field defined over the first quadrant of $\mathbb{R}^n$. That is, $\mathbb{R}^n_+=\{(x_1,x_2,...,x_n)\ |\ x_i\geq 0\text{ for all i}\}$. We assume that
- $F_i(\boldsymbol{0})>0$, and
- $\lim_{x_i\to\infty} F_i(\boldsymbol{x})=-\infty$ for all $i=1,2,\dots,n.$
My intuition tells me that such a vector field must vanish at least once inside $\mathbb{R}^n_+$, but I have not been able to prove this. I tried using Brouwer's Theorem and Poincare Miranda Theorem, but always found difficulties at the moment of defining the domains.
I am interested in this problem because I want to maximize a concave function whose gradient has the properties of $F$, so to show the existence of an equilibrium point I need to guarantee the existence of a solution of the equation $F(\boldsymbol{x})=0$. I appreciate any help you can give me.
As I commented, I was required to show the existence of an equilibrium point of a concave function in order to show maximality, but something I missed is that such a function can reach its maximum at the boundary, without the need to have an equilibrium point, even if it is strictly concave.
Consider for example the function $G(\boldsymbol{x})= -||\boldsymbol{x}||^2/2$, defined on the domain $\mathbf{D}=\{\boldsymbol{x}\ | \ ||\boldsymbol{x}||\geq R>0\}$. $G$ is strictly concave because its Hessian matrix is $-\boldsymbol{1}$ ($\boldsymbol{1}$ is the identity matrix), which is negative definite, but it reaches its maximum at any point $\boldsymbol{x}$ with $||\boldsymbol{x}||=R$ (i.e., the maximum is not unique), and these are not critical points of $G$.
In conclusion, a strictly concave function can have multiple maxima if one of them is in the boundary. On the other hand, the maximum is unique if it is inside the domain.
Edit. The example function was changed from a Gaussian to the current one thanks to the clarification of @MathWonk.