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Math Help - showing a functional is a norm

  1. #1
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    Sep 2009
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    showing a functional is a norm

    having trouble showing whether
    J(y)=suplyl is a norm on the space B[0,infinity) where B is the space of bounded functions on interval [0,infinity)

    I have no trouble showing llfll=0 iff f=0 but the other three conditions have me stuck any help would be really appreciated
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  2. #2
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    Joined
    Aug 2010
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    Hi,

    let f and g be 2 functions in B[0,infinity].
    Then, \forall x \in [0,\infty [, |f(x)+g(x)| \leq |f(x)|+|g(x)| \leq J(f)+J(g)
    Finally, J(f+g) \leq J(f)+J(g)

    moreover,

    let f be a function in B, let k be a real number.
    Then  \forall x \in [0,\infty [, |kf(x)| \leq J(k.f) and \forall x \in [0,\infty [, |kf(x)| \leq |k|.J(f), so J(k.f)=|k|.J(f)
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