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Inequality in the unit disc

  1. Aug 22, 2014 #1
    hi there,

    I am trying to prove the following inequality:
    let [tex]z\in \mathbb{D}[/tex] then

    [tex]\left| \frac{z}{\lambda} +1-\frac{1}{\lambda}\right|<1[/tex] if and only if [tex]\lambda\geq1.[/tex]

    The direction if [tex]\lambda>1[/tex] is pretty easy, but I am wondering about the other direction.

    Thanks in advance
     
    Last edited: Aug 22, 2014
  2. jcsd
  3. Aug 22, 2014 #2

    mathman

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    You need to clarify. For z = 1, the expression = 1, independent of λ.
     
  4. Aug 27, 2014 #3
    In fact [tex]z[/tex] is inside the disc that means [tex]|z|<1[/tex].
     
  5. Aug 27, 2014 #4

    mathman

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    z = a+bi. Restate question. |a+bi+λ-1|2 < λ2
    L.H.S. = (a+λ-1)2+b2=|z|2+2a(λ-1)+(λ-1)2<1+2(λ-1)+(λ-1)22
     
  6. Aug 27, 2014 #5
    thank, but how could that imply that lambda is bigger than one? On the other hand how did you get the last inequality, why you consider a is positive?
     
  7. Aug 30, 2014 #6

    FactChecker

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    If z=0 and lamda=0.9999 then the LHS ~ 0.0001
     
  8. Aug 31, 2014 #7

    mathman

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    2a(λ-1) < 2(λ-1) requires λ > 1 for a > 0. Also |a| < 1
     
  9. Aug 31, 2014 #8

    FactChecker

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    The statement the OP is trying to prove is not true. See #6. z=0 and lamda = 0.9999 is a counterexample.
     
  10. Sep 1, 2014 #9

    mathman

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    Your counterexample is wrong. λ ≥ 1 is a condition.
     
  11. Sep 1, 2014 #10

    FactChecker

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    The OP said "If and only if". lambda is less than 1 but the equation is much less than 1 (nearly 0)
     
  12. Sep 2, 2014 #11

    mathman

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    You misread the op. λ≥1 is the condition. For λ < 1 the expression does not hold for all z, with |z| < 1.
     
  13. Sep 2, 2014 #12
    Maybe it was not clear in the question that the inequality is for all z in the disc, that was my fault, I am sorry. Thank you very much mathman for the solution.
     
  14. Sep 2, 2014 #13

    FactChecker

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    Oh. Well, that's different. Never mind. https://www.youtube.com/watch?v=V3FnpaWQJO0
     
  15. Sep 2, 2014 #14
  16. Sep 3, 2014 #15
    I thought I got it, but it seems not yet :confused:.

    We will start like that,
    let
    [tex]|z+\lambda-1|^2 <|\lambda|^2[/tex], fro all z in the disc. Let [tex]z=a+ib[/tex], hence

    [tex]a^2+b^2+2a(\lambda-1)+(\lambda-1)^2<\lambda^2.[/tex]
    How could that mean 2a(λ-1) < 2(λ-1)?
    Thx.
     
  17. Sep 4, 2014 #16

    mathman

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    Since [itex]a^2+b^2 < 1, |a| < 1[/itex]
     
  18. Sep 4, 2014 #17
    How could that help :redface:.
     
  19. Sep 5, 2014 #18

    mathman

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    Grade school arithmetic!!! If [itex]|a| \ge 1[/itex], then [itex] a^2 \ge |a| \ge 1 [/itex] contradicting [itex] a^2 + b^2 < 1 [/itex]
     
  20. Sep 9, 2014 #19
    of course that is a grade school arithmetic, but it was not my question.

    My question is:

    how
    [tex] a^2+b^2 +2a( \lambda -1)+(
    \lambda-1)^2\leq\lambda^2[/tex]

    implies

    2a(λ-1) < 2(λ-1)?
     
  21. Sep 9, 2014 #20

    mathman

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    By itself it doesn't. |z| < 1 is the needed condition.
     
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