2
$\begingroup$

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 $x$.

$\endgroup$
  • $\begingroup$ What are your assumptions on the function $f$? You seem to be abusing notation writing $f(x)$ as well as $f(a,b,x)$. $\endgroup$ – Rasmus Jan 21 '15 at 8:54
  • $\begingroup$ I can assume that $f$ is contiuoues, but I prefer not to $\endgroup$ – nir Jan 21 '15 at 8:58
  • $\begingroup$ I think I need to assume monoticity of $f(a,b,x)$ as a function of $x$ $\endgroup$ – nir Jan 21 '15 at 9:19
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.