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Math Help - sup and inf

  1. #1
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    sup and inf

    Let f,g:A\to\mathbb R so that f\le g, then prove that if g is bounded above, so is f and \inf f\le \sup g.

    First part is easy, now since \inf f\le f\le g, we have that \inf f\le g\implies \inf f\le\sup g. Is this correct?

    Now suppose f is bounded below, we have \inf f\le \inf g.

    If f is bounded below so does g then \inf f\le f\le g, where \inf f\le g\implies \inf f\le \inf g. Is it correct?

    Thanks!
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  2. #2
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by Connected View Post
    Let f,g:A\to\mathbb R so that f\le g, then prove that if g is bounded above, so is f and \inf f\le \sup g.

    First part is easy, now since \inf f\le f\le g, we have that \inf f\le g\implies \inf f\le\sup g. Is this correct?
    Correct
    Quote Originally Posted by Connected View Post
    Now suppose f is bounded below, we have \inf f\le \inf g.

    If f is bounded below so does g then \inf f\le f\le g, where \inf f\le g\implies \inf f\le \inf g. Is it correct?
    Correct

    In such questions, I can suggest you to always mention the domains of the variables, i.e.,
    Assume that f(t)\leq g(t) for all t\in A, then
    we have
    \inf_{t\in A}\{f(t)\}\leq f(t)\leq g(t) for all t\in A,
    which implies that (dont forget that \inf_{t\in A}\{f(t)\} is constant and is independent of t)
    g is bounded below by a constant \inf_{t\in A}\{f(t)\}, and thus,
    \inf_{t\in A}\{f(t)\}\leq\inf_{t\in A}\{g(t)\}\leq g(t) for all t\in A.
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  3. #3
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    Yes of course, it's just to save writing, that's all!
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