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Bijective & continuous -> differentiable?

  1. Aug 3, 2009 #1
    Is a bijective continuous function:[a,b]->[f(a),f(b)] differentiable?
    I think it has to be.
    continuity between two distinct values of f(a) and f(b): it gotta take all the values between f(a) and f(b) at x in [a,b], by the intermediate value theorem.
    if f is bijective, at [a,b], f(x) can't go up and then down at [a,b]. it has to be monotonically increasing for it to be bijective.
    so a function can only be non-differentiable at a set of points of measure zero. (like the vertical tangent of f(x)=x^1/3 at 0)
  2. jcsd
  3. Aug 3, 2009 #2


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    The function f defined by:
    f(x) = x when 0 <= x <= 1
    f(x) = 2x - 1 when 1 <= x <= 2
    on [0, 2] is a counter-example to the original statement "bijective & continuous => differentiable".

    I suspect you can use this even to make a continuous bijection that is differentiable only on a set of measure zero, by taking a monotonously increasing function like above, of which the slope increases at every non-rational number.
  4. Aug 3, 2009 #3
  5. Aug 4, 2009 #4


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    Consider y(x)= x if [itex]0\le x< 1[/itex], 2x if [itex]1\le x\le 2[/itex]. That is a bijective from [0, 2] to [0, 4] and is continuous. It is NOT differentiable at x=1.
  6. Aug 4, 2009 #5


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    It's not continuous, it has a jump at x = 1. If you shift it down by 1, like I did in my example, it works out.
  7. Aug 5, 2009 #6
    i mean... differentiable at all but a "small" (smaller than a real interval) set of points
    like a finite set - like the one compuchip mentioned
  8. Aug 6, 2009 #7


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    First, I can think of a continuous bijection that fails to be differentiable on an uncountable set of measure zero, viz. the Cantor set. It is

    [tex] f(x) = x + c(x) [/tex]​

    where [itex] c(x) [/itex] is the Cantor function.

    Second, that Wikipedia article links to Minkowski's question mark function, which is claimed to have a zero derivative on the rationals but is not differentiable on the irrationals.
  9. Aug 6, 2009 #8
    the question mark function -
    ouch, i was wrong!
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