Max–min inequality

Define ${\displaystyle g(z)\triangleq \inf _{w\in W}f(z,w)\ .}$ For all ${\displaystyle z\in Z}$, we get ${\textstyle g(z)\leq f(z,w)}$ for all ${\displaystyle w\in W}$ by definition of the infimum being a lower bound. Next, for all ${\textstyle w\in W}$, ${\displaystyle f(z,w)\leq \sup _{z\in Z}f(z,w)}$ for all ${\textstyle z\in Z}$ by definition of the supremum being an upper bound. Thus, for all ${\displaystyle z\in Z}$ and ${\displaystyle w\in W}$, ${\displaystyle g(z)\leq f(z,w)\leq \sup _{z\in Z}f(z,w)}$ making ${\displaystyle h(w)\triangleq \sup _{z\in Z}f(z,w)}$ an upper bound on ${\displaystyle g(z)}$ for any choice of ${\displaystyle w\in W}$. Because the supremum is the least upper bound, ${\displaystyle \sup _{z\in Z}g(z)\leq h(w)}$ holds for all ${\displaystyle w\in W}$. From this inequality, we also see that ${\displaystyle \sup _{z\in Z}g(z)}$ is a lower bound on ${\displaystyle h(w)}$. By the greatest lower bound property of infimum, ${\displaystyle \sup _{z\in Z}g(z)\leq \inf _{w\in W}h(w)}$. Putting all the pieces together, we get

${\displaystyle \sup _{z\in Z}\inf _{w\in W}f(z,w)=\sup _{z\in Z}g(z)\leq \inf _{w\in W}h(w)=\inf _{w\in W}\sup _{z\in Z}f(z,w)}$

which proves the desired inequality. ${\displaystyle \blacksquare }$

• Boyd, Stephen; Vandenberghe, Lieven (2004). Convex Optimization (PDF). Cambridge University Press.

• Minimax theorem