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I was thinking, inspired by mathlinks, precisely from this post, if there exists a continuous real function $f:\mathbb R\to\mathbb R$ such that $$f(f(x))=e^x.$$ However I have not still been able to come up with an answer. I would like to share this problem with you.

I'm not aware of its level, though i wouldn't classify it as homework, so I'm not giving it the homework tag. If anybody feels that this problem is in reality easy or looks like an homework, please feel free to add that tag.

EDIT for those interested in the complex case, I've found this on MathOverflow. It's the first answer. Wow! link

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Wikipedia's Functional square root article links to a journal article (in German) about that very problem. –  Henning Makholm Sep 19 '11 at 19:34
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I could be wrong, but I think the "homework" tag is reserved for... uh... homework! :) –  The Chaz 2.0 Sep 19 '11 at 19:34
    
One important thing to notice about equations is that all functions in equations are input to the problem. Solving the equation gives you single number as output, but you needed to provide all functions as input. Now you're trying to figure out which function works. The problem is that every function does work, it just changes which value of x is the root –  tp1 Sep 19 '11 at 19:36
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@tp1, for a functional equation the convention is that the equation has to hold for all (relevant) $x$. Also, it is not true that there is always an $x$ that works. For example, if $f$ is the constant zero function, then $f(f(x))=e^x$ has no solution for $x$. –  Henning Makholm Sep 19 '11 at 19:42
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see the MO answer by bo198214 in the MO link above for information on restricting to the real line, including a sunmmary of the article by Kneser mentioned by Henning Makholm. –  Will Jagy Sep 19 '11 at 20:48
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4 Answers

up vote 10 down vote accepted

If $f$ is a solution, it is injective, hence increasing or decreasing. If $f$ was surjective, $\mathrm{exp}$ would be too, so $f$ is not. If $f$ was decreasing, it would have a fixed point, and $\mathrm{exp}$ too, so $f$ is increasing. The set $f(\mathbb{R})$ is equal to $]a,b[$ for some $a<b$, possibly infinite. If $b \neq + \infty$, $\exp (x) < b$ for all $x$, contradiction. So $b = + \infty$ and since $f$ is not surjective, $a \neq - \infty$. If $a \geq 0$, $\exp (x) \geq f(a)>0$ for all $x$, and that's impossible ($x=\log f(a)-1$ for example). So $a<0$. Since $\lim_{- \infty} f = a$, $\lim_{- \infty} \exp = f(a)$, and so $f(a)=0$.

Now the whole function $f$ can be reconstructed from $f|_{]- \infty, a]}$. Assume we only have a continuous, increasing and surjective function $f: ]-\infty,a] \rightarrow ]a,0]$. We can extend it to a continuous function on $\mathbb{R}$ such that $f \circ f = \exp$. For example, if $a < x \leq 0$, $x=f(y)$ for some unique $y \in ]-\infty,a]$, and so $f(x)=\exp y = \exp f^{-1}(x)$ ($f^{-1} : ]a,0] \rightarrow ]-\infty,a]$). It is easily checked that this defines a continuous, increasing and surjective function $f : ]-\infty,0] \rightarrow ]a,\exp a]$ (only the continuity at $a$ is not completely obvious, but easy). We can go on to extend $f$ on $]0,\exp (a)]$, $]\exp (a),1]$, etc.

So there are infinitely many continuous solutions, and a recipe that gives them all: pick $a<0$, and "pick" a continuous, increasing and surjective function $f : ]-\infty,a] \rightarrow ]a,0]$. There is a unique continuous extension of $f$ such that $f \circ f = \exp$.

If you want $f$ to be $C^k$, $C^{\infty}$, analytic; you need to take $f$ that is so on $]-\infty,a]$, and check that it is so around $a$ when you do the first extension (the behavior of $f$ on the right of $a$ is controlled by its behavior at $-\infty$, because $f(a+\epsilon)=\exp f^{-1} (a+\epsilon)$). Have fun!

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+1 Very nice solution! –  Sasha Sep 19 '11 at 21:39
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This is a large and difficult topic. I put a number of references at LINK where one should begin with the obituary of Baker. Over the real line, the difficulty is cause by fixed points. So, while there are infinitely many complex numbers such that $e^z = z,$ for all real numbers $x$ we have $e^x > x$ and so $e^x \neq x.$ So my memory, and this should be in Baker's papers, and I believe is originally due to Kneser, is that there is a solution along the real line to $f(f(x)) = e^x,$ such that $f$ is real analytic, meaning it is analytic in a strip around the real line in the complex plane, but the "strip" may have varying width. As there are no real fixed points, one may use the easier Schroder equation method.

Note that the same question goes badly wrong if the target function decreases near a fixed point. So, there is a solution to $g(g(x)) = \sin x$ which is $C^\infty$ and actually analytic except possibly at integer multiples of $\pi/2.$ In comparison, there is no such thing for $h(h(x)) = \cos x,$ as $\cos x$ is decreasing at the fixed point in $\mathbf R.$

The most flexible method for constructing these functions in the difficult case is apparently due to J. Ecalle. I do not have his papers. But the method is given in the book I call KCG,

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thanks for all your references +1 –  uforoboa Sep 19 '11 at 23:42
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Just one addendum to the present answers. Sometimes it seems to me, that the problem of non-uniqueness of the selection of some version of the half-iterate is taken too abstract. But in my view there should some natural argument, for instance some smoothness.
To show the effect of different initializations, which are so nicely worked out by @Plop , I made some pictures using excel. To have more information in the graphs I used however the function $ \small f(x)=h(h(x)) = b^x $ with $\small b=\exp(0.5)$ instead of $\small b=e $. These pictures are in (see: Excel_presentation) ; it is just a quick presentation; use only the first seven sheets. Here are three of that pictures so that you have an idea what I'm talking about: enter image description here

It shows the functions $ \small f(x)=x, f(x)=b^x, f(x)=b^{b^x}, f(x)=b^{b^{b^x}} $ as strong curved lines, and the estimated 0,5,1.5,2.5 - iterated versions as far as computable from the initial guess (that a-value in Plop's post). For this the dottest lines were used to construct the subsequent values of the orbit. (If I recall right then this iterative construction was already used by G.H.Hardy, but I don't know whether he was the first one)
The goal of a good choice for a is then to make the magenta line for the half-iterate as smooth as possible.

But how to define/operationalize "smooth" with such a discrete orbit? One idea is to take the distances to the according coordinates of the mirrored function $\small f°^{-0.5}(x) $, see the next picture.

enter image description here

The distances show a sequence of erratically changing lengthes. This is shown in the next image.

enter image description here

We see, that a better choice of the initial value of a could provide a "smoother" version; instead of initializing with $\small (0,f° ^{0.5}(x))=(0,0.4) $ we use $\small (0,f° ^{0.5}(x))=(0,0.58) $ and get a far better curve of distances:

enter image description here

A set of 3 initial guesses is in the provided link of the Excel-html-output as given in the first link . The first three pictures are of the first type, then there is one picture explaining the distance-idea, and then the next three pictures are of the latter type. The following pictures are not of interest, they are just working material. You can use the tabstrip at the bottom of the image to navigate.

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There are infinitely many such continuous functions. You can even add infinitely differentiable.

However there is no entire function over $\mathbb C$ verifying $\forall z\in \mathbb C, f(f(z))=e^z$.

A more interesting question is whether or not there exist an analytic function over $\mathbb R$ such that $\mathbb R$ verifying $\forall x\in \mathbb R , f(f(x))=e^x$. The answer is positive but difficult I think.

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could you give me an example of a continuous function? –  uforoboa Sep 19 '11 at 19:42
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Is there a proof or a reference for these facts? I was looking for these facts some time back... –  Srivatsan Sep 19 '11 at 19:43
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Please substantiate claims in your answer. –  Sasha Sep 19 '11 at 19:43
    
see my answer above, with links to references –  Will Jagy Sep 19 '11 at 20:22
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