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Math Help - Maximality with respect to inclusion

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    Maximality with respect to inclusion

    Let g(x)=max_{i=1,...,m} f_{i}(x) where f_{i} are continuous concave functions and
    let X = \{x: a_{j}^T x \geq b_{j} \} ; M(x) = \{i: f_{i}(x) = g (x) \} and J(x) = \{j: a_{j}^T x = b_{j} \}.


    We define a "special" point to be a point \hat{x} for which the set M(\hat{x}) \cup J(\hat{x}) is maximal, i.e., there is no point y \in X for which M(\hat{x}) is properly contained in M(y) and J(\hat{x}) is properly contained in J(y).



    Assume x^* is the minimizer of g (i.e.  g(x^*) \leq g(x) for any x \in X ).

    We show there exists a "special point" \hat{x} such that

     M(x^*)   \subseteq M(\hat{x}) and  J(x^*)   \subseteq J(\hat{x}).


    I am trying to prove by contradiction. So assume there is no "special" point \hat{x} for which  M(x^*)   \subseteq M(\hat{x}). Let  \hat{x} be a "special point". Hence, there exist an  i \in M(x^*) such that  i \notin M(\hat{x}).
    So we have  f_{i}(x^*) = g(x^*) \leq g(\hat{x}) \neq f_{i}(\hat{x})

    I think the contradiction comes from the maximality of M(\hat{x}). But I am not sure how to prove this.
    I also think that the part with  J(x^*)   \subseteq J(\hat{x}) can be proven in a similar way, but I am not sure how.
    Last edited by math8; November 26th 2012 at 10:58 AM.
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