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Prove f is not piecewise continuous on [-1,1]

  1. Jul 14, 2010 #1
    1. The problem statement, all variables and given/known data

    Let f(x) = x sgn(sin(1/x)) if x != 0
    f(x) = 0 if x = 0
    on the interval I=[-1,1]

    Now I 'm asked to show that f(x) is not piecewise continuous on I and later, I must show that f is integrable on I.


    3. The attempt at a solution

    I am completely lost here and don't know how to proceed...
    I think we should first divide the partition in some regular subpartitions so that x=0 is excluded (the only possible discontinuity?):
    P1=[-1, -e/100] and P2=[e/100, 1], and then try to show that continuity on the interval P3=[-e/100, e/100] is impossible but that we still have uniform continuity on P1 and P2?
    Should I be looking in that direction? Or should I look at it in a complete different way?

    Thank you for the help!
     
  2. jcsd
  3. Jul 14, 2010 #2

    CompuChip

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    Intuitively, it is not piecewise continuous because sin(1/x) changes sign arbitrarily often as you let x tend to zero (1/x grows without bound, so there is always an x arbitrarily close for which the sign of the sine has flipped).

    So maybe you can use this for a rigorous proof. For example, suppose that sin(1/x) is piecewise continuous. Then you can list all points of discontinuity and there are finitely many, for example x1 < x2 < ... < xn.
    If you can show that there must be some 0 < x0 < x1, such that sin(1/x0) has a sign opposite from sin(1/x1), you are done (you have listed them all, and found another one, which is a contradiction).
     
  4. Jul 14, 2010 #3

    lanedance

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    I assume you mean the sign function
    http://en.wikipedia.org/wiki/Sign_function

    your function will be discontinous everytime sin(1/x) = 0, or when 1/x = n.pi,

    so when x = 1/(n.pi)

    if you definition for piecewise continuous contains discontinuous at only finitely many discontinuities, it sould be easy to show this is contradicted
     
  5. Jul 14, 2010 #4

    lanedance

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    right, so just like Compuchip said ;)
     
  6. Jul 14, 2010 #5
    Thank you very much for your help guys!
     
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