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Thread: sup of the sum of functions

  1. #1
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    sup of the sum of functions

    Let $\displaystyle f,g:A\to\mathbb R$ be bounded above, then $\displaystyle f+g$ is bounded above and $\displaystyle \sup(f+g)\le \sup f+\sup g.$

    Okay, I think there's no much to do here, we have $\displaystyle f\le \sup f$ and $\displaystyle g\le \sup g$ so $\displaystyle f+g$ is bounded above and besides $\displaystyle f+g\le \sup f+\sup g\implies \sup(f+g)\le \sup f+\sup g.$ Is it correct?

    Now if $\displaystyle g$ is bounded below, then $\displaystyle \sup f+\inf g\le \sup(f+g).$

    Well since $\displaystyle \inf g\le g\implies f+\inf g\le f+g$ so $\displaystyle \sup f+\inf g\le\sup(f+g),$ and we're done. Is it correct?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Connected View Post
    Let $\displaystyle f,g:A\to\mathbb R$ be bounded above, then $\displaystyle f+g$ is bounded above and $\displaystyle \sup(f+g)\le \sup f+\sup g.$

    Okay, I think there's no much to do here, we have $\displaystyle f\le \sup f$ and $\displaystyle g\le \sup g$ so $\displaystyle f+g$ is bounded above and besides $\displaystyle f+g\le \sup f+\sup g\implies \sup(f+g)\le \sup f+\sup g.$ Is it correct?

    Now if $\displaystyle g$ is bounded below, then $\displaystyle \sup f+\inf g\le \sup(f+g).$

    Well since $\displaystyle \inf g\le g\implies f+\inf g\le f+g$ so $\displaystyle \sup f+\inf g\le\sup(f+g),$ and we're done. Is it correct?
    This looks good to me.
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  3. #3
    Senior Member bkarpuz's Avatar
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    Think about the following
    $\displaystyle \sup_{t\in A}\{f(t)+g(t)\}\leq\sup_{t\in A,\ s\in A}\{f(t)+g(s)\}$ (releasing the variable of $\displaystyle g$ might greater values)
    ..................................$\displaystyle =\sup_{t\in A}\{f(t)\}+\sup_{s\in A}\{g(s)\}$ (as the variables of $\displaystyle f$ and $\displaystyle g$ are un related, we can split it)
    ..................................$\displaystyle =\sup_{t\in A}\{f(t)\}+\sup_{t\in A}\{g(t)\}$. ($\displaystyle \sup$ does not depend on the letter of the variable)
    Probably, this will help you understand it easier.
    You can do the same for $\displaystyle \inf$.
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