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Thread: inequality of supremums

  1. #1
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    inequality of supremums

    Hi all,

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

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

    (beware the $\displaystyle <$ and not $\displaystyle \leq$)

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

    thank you

    Deubelte
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  2. #2
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    Let $\displaystyle A=\mathbb{R}$ now define $\displaystyle f(x)=4,~g(x)=1,~\&~h(x)=2$.
    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 $\displaystyle f,g $ and $\displaystyle h$ are 3 continuous functions on a set A,
    that verify:

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

    (beware the $\displaystyle <$ and not $\displaystyle \leq$)

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

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

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

    can I say that:

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

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


    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
    $\displaystyle f$ and $\displaystyle g$
    such that:
    $\displaystyle \sup(g(t),t\in A)<\sup(h(t),t\in A)$

    can I say that:

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

    2) $\displaystyle \forall\epsilon$ there exists x such that:
    $\displaystyle g(x)-\epsilon<h(x)$
    Choose k with $\displaystyle \sup\{g(t),t\in A\}<k<\sup\{h(t),t\in A\}$. Then there exists x in A such that $\displaystyle h(x)>k$. But $\displaystyle g(x)<k$, and therefore $\displaystyle g(x)<h(x)$.
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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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