I'm reading the lecture note about minmax theorem in zero-sum game. First are the relevant definitions used in my lecture notes:
A finite zero-sum game, or matrix game, is given by a real-valued matrix $A = (a_{i,j})$ in $\mathbb R^{n \times p}$.
$I = \{1,...,n\}$ and $J = \{1,...,p\}$.
The set of mixed actions of Player 1: $$\Delta(I)=\left\{x= (x_{1}, \ldots, x_{n}) \in \mathbb{R}_{+}^{n} \,\middle\vert\, \sum_{i=1}^{n} x_{i}=1\right\}$$
The set of mixed actions of Player 2: $$\Delta(J)=\left\{y=(y_{1}, \ldots, y_{n}) \in \mathbb{R}_{+}^{n} \,\middle\vert\, \sum_{j=1}^{p} y_{j}=1\right\}$$
A row $i$ in $I$ (column $j$ in $J$) is assimilated to the Dirac measure on $i$ ($j$) and is now called a pure action of Player 1 (Player 2) in the matrix game $A$.
The payoff function is $$\forall x \in \Delta(I), \forall y \in \Delta(J): g(x, y) =\sum_{i=1}^{n} \sum_{j=1}^{p} x_{i} y_{j} g(i, j)$$
The payoff $g(x,y)$ can also be written as the matrix product $xAy$, where $x$ is seen as a row vector and $y$ as a column vector.
The minmax theorem is then stated as follows:
For any matrix $A$ in $\mathbb R^{I \times J}$, $$\min_{y \in \Delta(J)} \max_{x \in \Delta(I)} x A y=\max_{x \in \Delta(I)} \min_{y \in \Delta(J)} x A y$$
My professor goes on to prove the theorem:
My question:
At the end of the proof, my professor said that
Because of the form of $C$, it has to be the case that $y_j \ge 0$ for each $j$
I'm unable to understand this argument. Could you please elaborate more on it? Thank you for your help!
From the definition of $C$, if $z \in C$, then $z' = z - u \in C$ where $u \geq 0$. Therefore, if there exists $y_j < 0$, you can make $u_j$ as large as you want, and therefore the right hand side of your inequality as large as you want, which will violate the inequality. Hence, $y_j$ must be $\geq 0$ for each $j$