Consider a function $f:\mathcal{X\times \mathcal{Y}}\rightarrow [0,1]$ where
- $\mathcal{X}\equiv \{x_1,x_2,x_3,x_4\}$.
- $\mathcal{Y}\equiv \{y_1,y_2\}$.
- $g(x)\equiv \sum_{y\in \mathcal{Y}}f(x,y)\in (0,1)$ $\forall x\in \mathcal{X}$
- $h(y)\equiv \sum_{x\in \mathcal{X}}f(x,y)\in (0,1)$ $\forall y\in \mathcal{Y}$
- $\sum_{x\in \mathcal{X}}g(x)=1$
- $\sum_{y\in \mathcal{Y}}h(y)=1$
- $g(x_1)>g(x_2)>g(x_3)>g(x_4)$
By the relations above $$ g(x_1)=\max_{x\in \mathcal{X}} g(x) $$ Is it correct that $$ g(x_1)=\max_{x\in \mathcal{X}}f(x,y_1)+\max_{x\in \mathcal{X}}f(x,y_2) $$ ?
My thoughts: Note that, by definition, $$ g(x_1)=\max_{x\in \mathcal{X}} (f(x,y_1)+f(x,y_2)) $$ I know that in general the maximum of sums is, at most, sum of maximums (e.g., Elegant proof that maximum of sums is, at most, sum of maximums). However, I'm wondering whether the above relations are sufficient for such relation to hold with equality. I cannot find a counter-example where the equality is not satisfied. Could you help?
Let $f(x_1, y_1) = f(x_1, y_2) = \frac{1}{6}$, $f(x_2, y_1) = \frac{1}{4}$,$f(x_3, y_2) = \frac{1}{4} - \epsilon$, $f(x_2, y_2) = f(x_3, y_1) = 0$, $f(x_4, y_1) = f(x_4, y_2) = \frac{1}{12} + \frac{\epsilon}{2}$, and let $\epsilon = \frac{1}{100}$ Note that this satisfies all your conditions, but $g(x_1) = \frac{1}{3} < \frac{1}{4} + \frac{1}{4} - \frac{1}{100} = \max_x f(x, y_1) + \max_x f(x, y_2)$