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Is there such a continuous function?

  1. Jan 23, 2008 #1
    Is there a continuous function f(x) defined on [tex](-\infty,+\infty)[/tex] such that f(f(x))=[tex]e^{-x}[/tex]?
    My opinion is "no", and here is how i think:
    first of all if such a function exists, it should be a "one-to-one" function, that is for every y>0, there should be exaclty one x such that f(x)=y.
    Thus by the "one-to-one" property of f(x), for every [tex]x_{1}>x_{2}[/tex], either [tex]f(x_{1})>f(x_{2}) or f(x_{1})<f(x_{2})[/tex], but not [tex]f(x_{1})=f(x_{2})[/tex].
    However we notice that in both case, [tex]f(f(x_{1}))>f(f(x_{2}))[/tex], and this is contradicting to the fact that [tex]e^{-x_{1}}<e^{-x_{2}}[/tex].
    So I conclude that no such function f(x) exists, let along any "continuous" function.
    Am I right so far?

    And why emphasize "continuous" since I haven't find anything to do with continuity in my prove?
  2. jcsd
  3. Jan 23, 2008 #2


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    How do you justify that f must be 1-to-1?
  4. Jan 23, 2008 #3
    I think you nearly got it, but the continuity seems to be needed. We know that [itex]e^{-x}[/itex] is strictly decreasing everywhere. Now assume that [itex]f[/itex] is (not necessarily strictly) increasing on some open interval. It follows that [itex]f\circ f[/itex] is also increasing on this interval. Consequently, this cannot be the right [itex]f[/itex]. Now assume that the function is (not necessarily strictly) decreasing on an open interval. Again then, [itex]f\circ f[/itex] is increasing on this interval and [itex]f[/itex] cannot be the function.

    As a result [itex]f[/itex] has to be a function which on any open interval is neither decreasing nor increasing.

    Now I am missing a hard argument, but I can hardly imagine such a function being continuous. Hmm, actually why not, but it is certainly not differentiable. (Anyone able to fill this gap?)

    But I don't think the "neither decreasing nor increasing" rules out the existance of a function [itex]f[/itex] in general, i.e. one which does not have to be continuous.

  5. Jan 23, 2008 #4


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    It's not hard to show that continuity => f is strictly increasing, or decreasing, by the fact that f must be 1-1 (as f(f(x)) is 1-1) and by applying the intermediate value theorem.
    For non-continuous functions it's harder, you'd have to look at the periodic points of exp(-x). I still think it is not possible though (how many periods of size 2 does exp(-x) have?)

    Edit: I don't think exp(-x) has any periodic points, except for the unique fixed point, and there will exist non-continuous solutions for f.
    Last edited: Jan 23, 2008
  6. Jan 23, 2008 #5
    to answer EnumaElish's question: because exp(-x) is 1-1, and if f(x) is not, i.e. there are distinct x1 and x2 such that f(x1)=f(x2)=y, then f(y)=exp(-x1)=exp(-x2), obviously this is impossible since f(x) has to be a function.

    I think continuity is needed here to show that f is strictly increasing or decreasing on the whole real line. (gel, how do you justify this by using intermediate value theorem?)

    And without continuity I guess we can't get the "neither decreasing nor increasing" rule of f(x) on any open inverval. Am I right here? So I imagine such a non-continuous f(x) does exist, can anybody help me justify that?

    Thank you guys so much!
  7. Jan 23, 2008 #6
    one thing i couldn't understand that

    it comes out that


    f(inverse)(x1)>f(inverse)(x2) as you said



    f(y1)>f(y2) ..... x1>x2



    which contradict what you assumed
  8. Jan 26, 2008 #7
    Well, although I demonstrated f can be neither increasing nor decreasing on any open interval, I also expressed my doubt that this rules out a continuous function already. The best I came up with until now is a rough idea how it may be possible to prove that such an f cannot be differentiable in any point. But this does not rule out continuity.

  9. Jan 26, 2008 #8


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    We know that f is 1-1, and assume that it is continuous. Suppose that x<y<z and f(y) > f(z) > f(x). Then, by the intermediate value theorem there would have to be a number a between x and y such that f(a)=f(z), which contradicts the 1-1 property of f. Similarly, IVT rules out f(y) > f(x) > f(z), f(y) < f(z) < f(x) and f(y) < f(x) < f(z). The only remaining possibilities are that f(x) < f(y) < f(z) or f(x) > f(y) > f(z), so f has to be either increasing or decreasing on any set of three points x,y,z. I'll leave you finish off and show that this means it is either increasing or decreasing everywhere.
  10. Jan 26, 2008 #9


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    A (non-continuous) f does exist, but I don't know if there's a simple way to express it. I'll try to explain how it can be done. First let a be the unique number such that exp(-a)=a. We will set f(a)=a. The hard part is defining f for x != a.
    Let g(x)=exp(-x), so its inverse [itex]g^{-1}(x)=-\log(x)[/itex] exists for all positive x.

    You can show that if you pick any x != a then repeatedly applying g^{-1} will eventually give a number y <= 0. This means that for every x !=a there is a unique integer n>=0 and real y <= 0 satisfying [itex]x=g^n(y)[/itex].

    Now split the non-positive real numbers into two disjoint sets A,B such that there is an invertible map u:A->B. for example, you can have [itex]A=(-\infty,-1][/itex] and [itex]B=(-1,0][/itex].

    Finally, if [itex]x=g^n(y)[/itex] for y in A, set [itex]f(x)=g^n(u(y))[/itex]. Alternatively if y is in B, set [itex]f(x)=g^{n+1}(u^{-1}(y))[/itex].

    You can fill in the gaps in this construction, although you might find it rather tricky. I don't know if f can be constructed in any simpler way.
    Last edited: Jan 26, 2008
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