# Reversing minimax function

Let $$g(x) = \inf_{a \in A} \sup_{b \in B} f(a,b,x).$$ When it is true that $$g^{-1}(x) = \inf_{a \in A} \sup_{b \in B} f^{-1}(a,b,x)\ ?$$ where $f^{-1}(a,b,x)$ means that $f^{-1}(a,b,f(a,b,x)) =x$ i.e. we reverse $x$.

• What are your assumptions on the function $f$? You seem to be abusing notation writing $f(x)$ as well as $f(a,b,x)$. – Rasmus Jan 21 '15 at 8:54
• I can assume that $f$ is contiuoues, but I prefer not to – nir Jan 21 '15 at 8:58
• I think I need to assume monoticity of $f(a,b,x)$ as a function of $x$ – nir Jan 21 '15 at 9:19

It seems that we should assume monoticity, I'll demonstrate with only one optimizasion, i.e. on $a \in A$ and will abuse the notation bt changing $\inf$ and $\sup$ when ever i'll need.

Let $$y=g(x)=f(a_x,x).$$

Then $$x = g^{-1}(y) = f^{-1}(a_x,y)$$ and we need to prove that $$f^{-1}(a_x,y) = \inf_{a \in A} f^{-1}(a,y)$$ i.e. $$f^{-1}(a_x,y) \leq f^{-1}(a,y), \forall a\in A$$

Applying $f(a,\cdot)$ to both side and using reverse! monoticity

$$f(a,f^{-1}(a_x,y)) \geq y, \forall a\in A$$

Or

$$f(a,x) \geq y, \forall a\in A$$ which is what we needed.