Maximin optimization So I have the following optimization problem.
$$\max_{x\in C} \min(f^{\top}x,g^{\top}x)$$
Here $C$ is some bounded constraint set defined by linear inequalities only.
Suppose that $z^{*} = \max_{x\in C} \min(f^{\top}x,g^{\top}x)$. Let $x^{*}\in C$ denote an optimal solution. I was wondering under which conditions the following is true.
$$z^{*} = f^{\top}x^{*} = g^{\top}x^{*} \qquad (\triangle)$$


Non-example $1$: $C = \{x \in \mathbb{R}^2: 0 \leq x \leq 1\}$, $f = \begin{pmatrix}1 \\ 1\end{pmatrix}$ and $g = \begin{pmatrix}1 \\ 2\end{pmatrix}$. 
Non-example $2$: $C = \{x \in \mathbb{R}^2: 0 \leq x \leq \begin{pmatrix}1 \\ 2\end{pmatrix}\}$, $f = \begin{pmatrix}1 \\ 0\end{pmatrix}$ and $g = \begin{pmatrix}0 \\ 1\end{pmatrix}$
Example $1$: $C = \{x \in \mathbb{R}^2: 0 \leq x \leq 1\}$, $f = \begin{pmatrix}1 \\ 1\end{pmatrix}$ and $g = \begin{pmatrix}1 \\ -1\end{pmatrix}$


Based on these toy examples, I get the feeling that at least one component of $f$ and $g$ should be opposite signed for $(\triangle)$ to be true.
 A: This is an answer to a special case of the problem in question due to the discussion in the comments:
We consider
\begin{align}
\tag{1}
\max \quad \min(f^T x, g^T x) \quad\text{s.t.}\quad x \ge 0, e^T x = 1.
\end{align}
Using the KKT conditions or Lagrangian multiplier techniques, you will get a characterization:
We can rewrite the problem equivalently to
$$ 
\tag{2}
\max \{ z \mid  z \le f^T x, z \le g^T x, x \ge 0, e^T x = 1 \}, 
$$
which is a convex problem. That is, the KKT conditions are sufficient and necessary. They are
$$ \begin{bmatrix} -1 \\ 0 \end{bmatrix} + \begin{bmatrix} 1 \\ -f \end{bmatrix} \lambda + \begin{bmatrix} 1 \\ -g \end{bmatrix} \mu + \begin{bmatrix}0 \\ -I \end{bmatrix} \nu + \begin{bmatrix}0\\ e\end{bmatrix}\rho = 0, $$
$$ z - f^T x \le 0, z - g^T x \le 0, -x \le 0, e^T x - 1 = 0$$
$$ \lambda, \mu, \nu \ge 0, $$
$$ \lambda(z - f^T x) = 0,  \mu(z - g^T x) = 0, \nu^T x = 0. $$
Simplify and adding the fact that $f^T x = g^T x$, we obtain
$$ \nu = \rho e - \lambda f - (1-\lambda) g, $$
$$ (f-g)^T x = 0, x \ge 0, e^T x = 1, $$
$$ \lambda \in[0,1], \nu \ge 0, $$
$$ \nu^T x = 0. $$
Using $\nu^T x = 0$ and $e^T x = 1$, we obtain
$$ \rho = f^T x. $$
That is, we obtain following sufficient condition for a maximizer of (1):
$$ f^T x e - \lambda f - (1-\lambda) g \ge 0, $$
$$ (f-g)^T x = 0, e^T x = 1, x \ge 0, \lambda\in [0, 1]. $$
A non example: 
Consider
$$ C = [0, 1]^2 $$ and
$$ f = \begin{bmatrix}-1 \\ 0 \end{bmatrix}, \qquad g = \begin{bmatrix}0 \\ 1 \end{bmatrix}. $$
Then, for $x\in C$ we have
$$ \min(f^T x, g^T x) = \min(-x_1, x_2) = -x_1, $$
hence
$$ z^* = \max_{x\in C} \min(f^T x, g^T x) = \max_{x_1 \in [0,1]} -x_1 = 0. $$
Now, the set of maximizers is
$$ X^* = \{ (0, x_2) \mid x_2 \in [0,1] \}. $$
So, for any $x^*\in X^*\setminus \{ 0 \}$ we have
$$ z^* = f^T x^* = 0 < g^T x^*. $$
