Trying to understand $\max_{x} \max_{y} f(x,y) = \max_{x, y} f(x,y)$

Question

I am reading this post where @Yanko gives an answer that proves

$$\max_{x} \max_{y} f(x,y) = \max_{x, y} f(x,y)$$ for a function $f(x,y)$. Please let me quote the answer

Let $(x_0,y_0)$ be any two elements. Then clearly we have that $f(x_0 ,y_0) \leq\max_y f(x_0 ,y)$

Moreover we notice that $\max_x \max_y f(x,y)$ is the maximum over elements of the form $\max_y f(x',y)$ for some $x'$. In particular this implies that $f(x_0 ,y_0)\leq \max_y f(x_0 ,y)\leq \max _x \max_y f(x,y)$. As $x_0,y_0$ were chosen arbitrarily we may conclude that $\max_{x,y} f(x,y)\leq \max_x\max_y f(x,y)$ and the inequality in the other direction is obvious.

I am trying to understand the last sentence, i.e.,

As $x_0,y_0$ were chosen arbitrarily we may conclude that $\max_{x,y} f(x,y)\leq \max_x\max_y f(x,y)$ and the inequality in the other direction is obvious.

Could you please someone provide some more details on how this? Also, does this proof hold for non-separable functions $f(x,y)$? Could you please give an example?

Answer 1

By definition of max, it should be immediate that

$$\max_{x}\max_{y}f(x,y)\leq \max_{x,y}f(x,y).\quad (1)$$

Now note we have the following for any $(x_0,y_0)$:

$$ f(x_0,y_0)\leq \max_y f(x_0,y)\leq \max_x\max_yf(x,y).\quad (2)$$

Choosing $(x_0,y_0)\in \arg\max_{x,y}f(x,y)$ in $(2)$ gives

$$ \max_{x,y}f(x,y)\leq \max_x\max_yf(x,y).\quad (3)$$

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$(1)$ and $(3)$ give the desired result.