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Math Help - f bounded above implies -f bounded below

  1. #1
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    f bounded above implies -f bounded below

    Let f:A\to\mathbb R. Show that f is bounded above, then \underset{x\in A}{\mathop{\inf }}\,(-f(x))=-\underset{x\in A}{\mathop{\sup }}\,(f(x)).

    Okay, so I first showed that -f is bounded below, that's easy, but as for second part, I want to know whether my work is correct or not:

    Since f is bounded above, then it has supremum, so f\le \sup f (from this and the rest, sup and inf are taken for x\in A), and on the other hand \inf(-f)\le -f, since f and -f are bounded above and below, respectively, then it exists a k\in\mathbb R so that f\le k and -k\le -f, namely we have k=\sup f and \inf(-f)=-k, hence \underset{x\in A}{\mathop{\inf }}\,(-f(x))=-\underset{x\in A}{\mathop{\sup }}\,(f(x)).

    If it's not correct, what's the way to do this?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Connected View Post
    Let f:A\to\mathbb R. Show that f is bounded above, then \underset{x\in A}{\mathop{\inf }}\,(-f(x))=-\underset{x\in A}{\mathop{\sup }}\,(f(x)).

    Okay, so I first showed that -f is bounded below, that's easy, but as for second part, I want to know whether my work is correct or not:

    Since f is bounded above, then it has supremum, so f\le \sup f (from this and the rest, sup and inf are taken for x\in A), and on the other hand \inf(-f)\le -f, since f and -f are bounded above and below, respectively, then it exists a k\in\mathbb R so that f\le k and -k\le -f, namely we have k=\sup f and \inf(-f)=-k, hence \underset{x\in A}{\mathop{\inf }}\,(-f(x))=-\underset{x\in A}{\mathop{\sup }}\,(f(x)).

    If it's not correct, what's the way to do this?
    Looks fine to me.
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