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Jensen's inequality

  1. Dec 12, 2008 #1
    Is the Jensen's inequality in complex analysis related to the one in measure theory, or did Jensen just go around finding inequalities?
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  3. Dec 12, 2008 #2

    mathman

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    It looks like the complex analysis inequality is a special case, where ln|f| is the convex function of the measure theory theorem. By dividing by 2pi, the measure on the unit circle is normalized to 1.
     
  4. Dec 13, 2008 #3
    Only ln |f| isn't convex on the real axis - exp is convex - and f is complex, not real.
    It looks tantalizingly close, so I wonder if it can be twisted somehow.
    Laura
     
  5. Dec 13, 2008 #4

    mathman

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    As I read the theorem, ln(x) has to be convex (which it is), not ln|f|. Ln corresponds to phi in the general theorem. The only requirement on |f| is that it be L1 with respect to the measure.
     
    Last edited: Dec 13, 2008
  6. Dec 17, 2008 #5
    Convex means that if you draw a line between 2 points on the graph of [tex]\phi[/tex]
    then the graph between those 2 points is below or on the line. Ln isn't convex but its inverse exp is.
    Laura
     
  7. Dec 17, 2008 #6

    mathman

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    Convex can be convex down or convex up. The main idea is that a straight line connecting any two points on the curve does not cross the curve. For example, circles are convex.
     
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