1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Homework Helper

    I assume you mean the 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


    User Avatar
    Homework Helper

    right, so just like Compuchip said ;)
  6. Jul 14, 2010 #5
    Thank you very much for your help guys!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook