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Thread: sup and inf

  1. #1
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    sup and inf

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

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

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

    If $\displaystyle f$ is bounded below so does $\displaystyle g$ then $\displaystyle \inf f\le f\le g,$ where $\displaystyle \inf f\le g\implies \inf f\le \inf g.$ 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 $\displaystyle f,g:A\to\mathbb R$ so that $\displaystyle f\le g,$ then prove that if $\displaystyle g$ is bounded above, so is $\displaystyle f$ and $\displaystyle \inf f\le \sup g.$

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

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

    In such questions, I can suggest you to always mention the domains of the variables, i.e.,
    Assume that $\displaystyle f(t)\leq g(t)$ for all $\displaystyle t\in A$, then
    we have
    $\displaystyle \inf_{t\in A}\{f(t)\}\leq f(t)\leq g(t)$ for all $\displaystyle t\in A$,
    which implies that (dont forget that $\displaystyle \inf_{t\in A}\{f(t)\}$ is constant and is independent of $\displaystyle t$)
    $\displaystyle g$ is bounded below by a constant $\displaystyle \inf_{t\in A}\{f(t)\}$, and thus,
    $\displaystyle \inf_{t\in A}\{f(t)\}\leq\inf_{t\in A}\{g(t)\}\leq g(t)$ for all $\displaystyle t\in A$.
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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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