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Math Help - sup of the sum of functions

  1. #1
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    sup of the sum of functions

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

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

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

    Well since \inf g\le g\implies f+\inf g\le f+g so \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 f,g:A\to\mathbb R be bounded above, then f+g is bounded above and \sup(f+g)\le \sup f+\sup g.

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

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

    Well since \inf g\le g\implies f+\inf g\le f+g so \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
    \sup_{t\in A}\{f(t)+g(t)\}\leq\sup_{t\in A,\ s\in A}\{f(t)+g(s)\} (releasing the variable of g might greater values)
    .................................. =\sup_{t\in A}\{f(t)\}+\sup_{s\in A}\{g(s)\} (as the variables of f and g are un related, we can split it)
    .................................. =\sup_{t\in A}\{f(t)\}+\sup_{t\in A}\{g(t)\}. ( \sup does not depend on the letter of the variable)
    Probably, this will help you understand it easier.
    You can do the same for \inf.
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