
#1
Jan2308, 08:30 AM

P: 17

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 "onetoone" function, that is for every y>0, there should be exaclty one x such that f(x)=y. Thus by the "onetoone" 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? 



#3
Jan2308, 04:09 PM

P: 148

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. Harald. 



#4
Jan2308, 06:38 PM

P: 532

is there such a continuous function?
It's not hard to show that continuity => f is strictly increasing, or decreasing, by the fact that f must be 11 (as f(f(x)) is 11) and by applying the intermediate value theorem.
For noncontinuous 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 noncontinuous solutions for f. 



#5
Jan2308, 08:20 PM

P: 17

to answer EnumaElish's question: because exp(x) is 11, 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 noncontinuous f(x) does exist, can anybody help me justify that? Thank you guys so much! 



#6
Jan2308, 09:03 PM

P: 158

one thing i couldn't understand that
if x1>x2 then assume f(x1)>f(x2) it comes out that fof(x1)>fof(x2) but f(inverse)(x1)>f(inverse)(x2) as you said f(in)(x1)=y1 f(in)(x2)=y2 y1>y2 f(y1)>f(y2) ..... x1>x2 then fof(y1)<fof(x2)............f(x1)<f(x2) which contradict what you assumed 



#7
Jan2608, 05:45 AM

P: 148

Harald. 



#8
Jan2608, 07:01 AM

P: 532





#9
Jan2608, 07:40 AM

P: 532

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 nonpositive 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. 


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