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Sup of a function

  1. Mar 4, 2004 #1
    I don't have background in analysis, but was looking for
    a simple explanation(proof?) of this statement,

    Over a compact set, a differentiable function V Rn-> R, with V<0 in that set, then sup(V)<0 in that set


    Actually I'm not certain if I interpreted the statement right, so maybe the statement as it is might be wrong/incomplete.
     
  2. jcsd
  3. Mar 5, 2004 #2

    NateTG

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    You mean
    [tex]f:\Re^n\rightarrow \Re[/tex]
    and [tex]f(v) < 0[/tex]
    ?

    Consider the possibility that the image of the compact set [tex]S[/tex] might be [tex](-1,0)[/tex] which certainly has sup [tex]0[/tex].
     
  4. Mar 5, 2004 #3

    matt grime

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    The image of a compact set under a continuous (diffble) map is compact.
    So considering (0,1) as the image won't get you very far.


    Moreover it is bounded and attains its bounds, hence there is some element in the set with V(x)=sup{V(y)}

    this is negative by hypothesis
     
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