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Given $f_1(x)$, $f_2(x)$, $x\in \mathbb{R}^d$, two convex functions, we define the following problem:

$\underset{x\in C}{{\rm minimize}}\,{\rm max}\left(f_{1}\left(x\right),f_{2}\left(x\right)\right)$,

where $C$ is a convex set. This problem should be easy for d=1 because even in the case that the solution is in the set where $f_1(x)=f_2(x)$, this set is convex for sure. However this set could not be convex for $d>1$.

Could anyone point me to any clue about how to solve the problem for $d>1$ (e.g. alternating-like algorithm)?

Thanks.

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The maximum of two convex functions is convex for any $d$. This can be shown using the definition $f(tx_1+(1-t)x_2)\ge tf(x_1)+(1-t)f(x_2)$. –  Rahul Feb 10 '13 at 19:05
    
Sure, that is clear. What I want to know is how to get the minimum of this convex function. I think the resultant problem is easy for $d=1$, but it is not clear to me how to solve it for bigger d. –  bernard Feb 10 '13 at 19:10
    
Ah, I misread; I thought you were not sure the function itself was convex. The set $f_1(x)=f_2(x)$ may not be convex even when $d=1$, for example when $f_1(x)=1$ and $f_2(x)=x^2$. Anyway, sorry for not helping. –  Rahul Feb 10 '13 at 20:00

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