Given any function $\varphi : X \times Y \rightarrow \mathbb{R}$ it holds that
$$\max_{y}\min_x \varphi(x,y) \le \min_y\max_y \varphi(x,y).$$
The proof I'm trying to read states the following thing: 1)the above inequality can be proven by considering that $$For \,\, all \,\, x,y \qquad \min_x \varphi(x,y) \le \max_y \varphi(x,y)$$
(Why is this true?)
- Then I can take $\max_y$ on left-hand side since the above ineq is true for all $y$ values, getting:
$$\max_y\min_x \varphi(x,y) \le \max_y \varphi(x,y)$$
3)Finally I can take the $\min_x$ on the 'inner' part of the equality, getting $$\max_y\min_x \varphi(x,y) \le \max_y (\min_x)\varphi(x,y)$$
Are these steps correct? Also why is the first inequality (1) true? Thanks
I think your statement isn't well stated. Maybe you want prove the following?: $\max_\limits {y}\min_\limits x \varphi(x,y) \le \min_\limits x\max_\limits y \varphi(x,y)$ (note that on the RHS you have taken the minimum on $y$ after taking the maximum on $y$ too, which doesn't make much sense).
For the first part, note that for all $x,y$ we have $\min_\limits x \varphi(x,y) \le\varphi(x,y)$ and $\varphi(x,y) \le \max_\limits y \varphi(x,y)$, so for all $x,y$ we have $\min_\limits x \varphi(x,y) \le \max_\limits y \varphi(x,y)$.
The second step seems ok, but it should be pointed out that the inequality holds for all $x$.
The third step is just taking minimum on $x$, not on the inner part or anything, it's just like you did with the maximum in the second step.