Lemma: For any function of two variables, say f ( x, y ) , we have max min f ( x, y ) ≤ min max f ( x, y ) , x
y
x
(1) y
with the equality, if and only if there exist x0 and y0 such that min f ( x, y0 ) = f ( x0 , y0 ) = max f ( x0 , y ) . x
y
Proof: Note that for any x0 and y0 we have min f ( x, y0 ) ≤ f ( x0 , y0 ) ≤ max f ( x0 , y ) . x
(2) y
Minimizing over x0 , we get min f ( x, y0 ) ≤ min max f ( x0 , y ) , and then maximizing over y0 , we have x
x0
y
max min f ( x, y0 ) ≤ min max f ( x0 , y ) , which is equivalent to (1). If we have two equalities in (2), then we y0
x
x0
y
will also obtain an equality in (1) using the above arguments. Let’s now assume the equality in (1), and let x0 , be such that min max f ( x, y ) = max f ( x0 , y ) , and y0 , be such that max min f ( x, y ) = min f ( x, y0 ) . x
Clearly, y
y
x
y
x
max min f ( x, y ) = min f ( x, y0 ) ≤ f ( x0 , y0 ) ≤ max f ( x0 , y ) = min max f ( x, y ) y
x
x
y
x
y
and from the assumed equality in (1), all terms in (3) are equal, which proves our necessary condition.
(3)