Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook