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Monotonicity and continuity

  1. Jan 10, 2009 #1
    Suppose f:A-->R is monotone (ACR: reals)
    and suppose the range of f is an interval, show f is continuous on A.

    By drawing a picture, I can see the conclusion. Since f is monotone, the only type of discontinuity it may have is a jump discontinuity. But since the range of f is an interval, this cannot happen.

    I would like to have a more consistent proof (analytical proof), I mean, I know that since f is monotone, for each point a of A, f(a+) and f(a-) exist. Now f(a+)> or eq. to f(a) and f(a-)< or eq. to f(a). Now how would I show that f(a+)< or eq. to f(a) and f(a-)> or eq. to f(a) using the fact that range(f) is an interval? Hence I would be able to conclude that f(a-)= f(a+)=f(a), therefore, f continuous on A.
     
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  3. Jan 10, 2009 #2

    Dick

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    Just take the direct approach. Pick an x and an epsilon e>0. You want to show that there is a delta d>0 such that |f(x)-f(y)|<e if |x-y|<d, right? Think about x=f^(-1)(f(x)), f^(-1)(f(x)+e) and f^(-1)(f(x)-e). Can you figure out how to define a d using those?
     
  4. Jan 10, 2009 #3
    I would use d such that |f^(-1)(f(x)+e) - f^(-1)(f(x)-e)|> d. Is that correct? How do we use the fact that range of f is an interval?
    I see that you're using f^(-1), in fact the problem has a subproblem that says to use the result to prove the continuity of inverse functions. How do we conclude that?
     
  5. Jan 10, 2009 #4

    Dick

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    If there is a jump discontinuity, then the range isn't an interval. I use that the range is an interval to show that a point like f^(-1)(f(x)+e) even exists. You've probably drawn a picture of a function with a jump discontinuity, so you should be able to see how the proof breaks down. I would say d should be smaller than the maximum of |f^(-1)(f(x)+e)-x| and and |f^(-1)(f(x)-e)-x|. Do you see why? Try doing the subproblem about f^(-1) and see how far you get.
     
  6. Jan 10, 2009 #5
    shouldn't d be smaller than the minimum of |f^(-1)(f(x)+e)-x| and |f^(-1)(f(x)-e)-x| instead?
     
  7. Jan 10, 2009 #6

    Dick

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    Sure. Absolutely. I was just testing you. :)
     
  8. Jan 10, 2009 #7
    oh, good, then, at least I understand what's going on...
     
  9. Jan 10, 2009 #8

    Dick

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    That's what I was checking. Actually 'maximum' was just a typo...
     
  10. Jan 10, 2009 #9
    I am trying for the f^-1 part, but I cannot get anywhere, can you give me another hint?
     
  11. Jan 10, 2009 #10

    Dick

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    What exactly is the f^(-1) part?
     
  12. Jan 10, 2009 #11
    the subproblem that says to use the result to prove the continuity of inverse functions.
     
  13. Jan 10, 2009 #12

    Dick

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    The inverse of a continuous function is not necessarily continuous even if it's monotone and the range is an interval. Can you state the full problem? Do you know something about the domain?
     
  14. Jan 10, 2009 #13
    that's it, that's how it is stated:

    "Suppose f:A-->R is monotone (ACR: reals)
    and suppose the range of f is an interval, show f is continuous on A. Use this result to prove the continuity of inverse functions".
     
  15. Jan 10, 2009 #14

    Dick

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    Let f(x)=x for x in [0,1) and f(x)=x-1 for x in [2,3]. f is continuous. The domain is [0,1)U[2,3] which is a subset of the reals. The range is [0,2]. It's monotone. The inverse has a jump discontinuity at 1. Do you see my problem?
     
  16. Jan 10, 2009 #15
    yeah, I guess, there is a problem or maybe something missing in the question, anyways, thanks for your help.
     
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