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

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

    User Avatar
    Science Advisor

    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

    User Avatar
    Science Advisor

    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

    User Avatar
    Science Advisor
    Gold Member
    2017 Award

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

    mathman

    User Avatar
    Science Advisor

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

    FactChecker

    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    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

    User Avatar
    Science Advisor

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

    FactChecker

    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    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

    User Avatar
    Science Advisor

    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

    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    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

    User Avatar
    Science Advisor

    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

    User Avatar
    Science Advisor

    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

    User Avatar
    Science Advisor

    By itself it doesn't. |z| < 1 is the needed condition.
     
  22. Sep 9, 2014 #21
    Yes, the question is how is that?
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook