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Complex variables: ML inequality

  1. Oct 13, 2007 #1
    I'm studying the proof of the [tex]ML[/tex] inequlity from complex analysis. I don't know what they did in one step of the proof and I was wondering if anyone can explain the step to me.

    First of all the theorem says:

    Let f(z) be continuous on a contour C. Then

    [tex]|\int_{c}f(z)dz| \leq ML[/tex]

    Where [tex]L[/tex] is the length of C and [tex]M[/tex] is an upper bound for |f| on C.

    After a few steps we have:

    [tex]|\int_{c}f(z)dz| \leq \int^{b}_{a} |f(z(t))||z'(t)|dt[/tex]

    And the text says " |f| is bounded on C, [tex]|f(z)| \leq M[/tex] on C, where [tex]M[/tex] is constant, then:

    [tex]|\int_{c}f(z)dz| \leq M\int^{b}_{a} |z'(t)|dt[/tex]

    How did they get the [tex]M[/tex] out of the integral?
    Last edited: Oct 13, 2007
  2. jcsd
  3. Oct 13, 2007 #2


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    [tex]\int^{b}_{a} |f(z(t))||z'(t)|dt \leq \int_a^b M|z'(t)|dt = M \int_a^b |z'(t)|dt [/tex]
  4. Oct 14, 2007 #3
    So it's okay that M is an upper bound for |f(z)|, that's the same thing as an upper bound for |f(z(t))|? Okay thanks, I think I get it now.
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