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Let $f(x,y)$ be a function such that $f:\mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$.
Now we have to maximize $f$ over $x$ and minimize it over $y$ $i.e.\ $
$$\underset{x}{\text{max}}\: \underset{y}{\text{min}}\:f(x,y)$$
Now Let $x^*$ be the value of x which maximizes this function for some $y$.
Then My professor says $$f(x^*,y) \leq \underset{x}{\text{max}}\:f(x,y) $$
I didn't understand this step. I mean doesn't $\underset{x}{\text{max}}\:f(x,y)$ mean find value of $x$ which maximizes $f(x,y)$, which is same as $f(x^*,y)$ as this is the value we get when we found that $x$ which maximized this function.
I can't really get this.

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Please, define $x^*$ with a mathematical formula. Your sentence is rather unclear: what function? Why "for some $y$"? However, remember that $a \leq a$ for every real number $a$. –  Siminore Aug 25 '12 at 8:22
    
What was the step that came after this? If a professor puts up a seemingly obvious restatement of the current equation, in my experience it usually means there's a trick coming. –  Trevor Alexander Mar 18 '13 at 9:47

1 Answer 1

$\max_{x} f(x,y)$ means the maximum value of $f$ as $x$ is varied. $\arg\max$ is the notation used for denoting the value of $x$ which gave the maximum value of $f$. The equation $f(x^*,y) \leq \max_x f(x,y)$ (where $x^* = \arg \max_x f(x,y)$) is like saying $a \leq a$, which is trivially true. I think the following might be more useful: $f(x,y) \leq f(x^*,y) \forall x$.

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