# max min is less than min max proof

I saw the following proof that max min of a function is $$\leq$$ than min max of a function on Max Min of function less than Min max of function, pasted below for your reference

Let $$f(x_{0}, y_{0}) = \max_x\min_y f(x, y)$$ and $$f(x_{1}, > y_{1}) = \min_y\max_x f(x, y)$$.

By this definition the problem is to prove that $$f(x_{0}, y_{0}) \leq > f(x_{1}, y_{1})$$ provided that they exist.

By definition of min and max function we have:

$$\min_yf(x, y) = f(x, y_{0}) \leq f(x, y) \forall y$$. Here $$\min_yf(x, > y)$$ would be a function only of x.

$$\max_xf(x, y_{0})=f(x_{0}, y_{0}) \geq f(x, y_{0})\forall x$$. Here $$\max_xf(x, y_{0})$$ is a scalar.

$$\max_xf(x, y)=f(x_{1}, y) \geq f(x,y) \forall x$$. Here $$\max_xf(x_{1}, y)$$ is a function of y.

$$\min_yf(x_{1}, y)=f(x_{1},y_{1}) \leq f(x_{1}, y) \forall y$$. Here $$\min_yf(x_{1}, y)$$ is a scalar.

From all this equation you get the following inequalities:

$$f(x, y_{0}) \leq f(x_{0},y_{0}) \leq f(x,y) \leq f(x_{1}, y_{1}) \leq > f(x_{1}, y)$$

My confusion is on the last line, where did the $$f(x_{0},y_{0}) \leq f(x,y) \leq f(x_{1}, y_{1})$$ come from? I don't see how this conclusion was made from the prior steps.

$$f(x_0, y_0) \leq f(x, y)$$ because $$f(x, y_0) \leq f(x, y)$$ for all $$x$$ and $$y$$, and $$f(x_0, y_0) \geq f(x, y_0)$$ since $$x_0$$ is a maximizer. Similar logic applies to $$f(x_1, y_1)$$.