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Math Help - inequality of supremums

  1. #1
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    inequality of supremums

    Hi all,

    If  f,g and h are 3 continous functions on a set A,
    that verify:

    \sup({f(t),t\in A})<\sup({h(t),t\in A})+\sup({g(t),t\in A})

    (beware the < and not \leq)

    can I say that there exists x\in A such that:
    f(x)<g(x)+h(x)

    thank you

    Deubelte
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  2. #2
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    Let A=\mathbb{R} now define f(x)=4,~g(x)=1,~\&~h(x)=2.
    Is that statement still true.
    You must have left something out of the statement of the question.
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  3. #3
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    Quote Originally Posted by deubelte View Post
    If  f,g and h are 3 continuous functions on a set A,
    that verify:

    \sup({f(t),t\in A})<\sup({h(t),t\in A})+\sup({g(t),t\in A})

    (beware the < and not \leq)

    can I say that there exists x\in A such that:
    f(x)<g(x)+h(x)
    No. Take A = \mathbb{R} and define g(x) = \frac1{(x-10)^2+1}, h(x) = \frac1{(x+10)^2+1} and f(x) = 1.5 (constant).

    Then \sup\{g(x)\} = 1 (attained when x=10) and \sup\{h(x)\} = 1 (attained when x=10), so 1.5 = \sup\{f(x)\} < \sup\{g(x)\}+\sup\{h(x\} = 2.

    But x\geqslant0\;\Rightarrow\; h(x)\leqslant0.01, and x\leqslant0\;\Rightarrow\; g(x)\leqslant0.01. So g(x)+h(x) is always less than 1.01, which is less than f(x).
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  4. #4
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    I have an other question. supposing we have two continuous functions
    f and g
    such that:
    \sup(g(t),t\in A)<\sup(h(t),t\in A)

    can I say that:

    there exists x in A such that:
    1) g(x)\leq h(x)

    2) forall\epsilon there exists x such that:
    g(x)-\epsilon<h(x)


    thx
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  5. #5
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    Quote Originally Posted by deubelte View Post
    I have an other question. supposing we have two continuous functions
    f and g
    such that:
    \sup(g(t),t\in A)<\sup(h(t),t\in A)

    can I say that:

    there exists x in A such that:
    1) g(x)\leq h(x)

    2) \forall\epsilon there exists x such that:
    g(x)-\epsilon<h(x)
    Choose k with \sup\{g(t),t\in A\}<k<\sup\{h(t),t\in A\}. Then there exists x in A such that h(x)>k. But g(x)<k, and therefore g(x)<h(x).
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  6. #6
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    Ok, so I guess that my second statement is also true ??


    Merci l'Anglais.
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