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

  • Thread starter maupassant
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  • #1
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Homework Statement



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.


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!
 

Answers and Replies

  • #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).
 
  • #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
 
  • #4
lanedance
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right, so just like Compuchip said ;)
 
  • #5
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Thank you very much for your help guys!
 

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