There exist a function such that $f\circ f(x)=e^x$?

Question

Based on this question: How to calculate $f(x)$ in $f(f(x)) = e^x$? I would like to know if I can get a function such that $f:\mathbb R \to \mathbb R^+$, defined by $f\circ f(x)=e^x$. My guess is no, but I can't prove it, I need help.

Note that different than the previous question, the function is from $\mathbb R$ to $\mathbb R^+$.

Thanks a lot

I did this before. My result is that you can have a bunch of functions satisfy this formula, hint is that this fucntion must be strictly monotone, and you just need to define it on a little interval to get the whole function. — lee, Feb 07 '13 at 00:24
@lee sorry it misses informations, I'm going to edit my question — user42912, Feb 07 '13 at 00:25
It appears this question has been asked before on math.SE, so this question will be closed as a duplicate. — Zev Chonoles, Feb 07 '13 at 00:28
@ZevChonoles My question is defined differently than other posts. — user42912, Feb 07 '13 at 01:01

score 4 · Answer 1 · edited Apr 13 '17 at 12:58

In fact, you can find such an $f$ that is analytic. From this answer on MathOverflow:

A real-analytic solution in this case was constructed by H. Kneser, "Reelle analytische Lösungen der Gleichung $φ(φ(x))=e^x$ und verwandter Funktional-gleichungen", J. Reine Angew. Math. 187 (1949), 56-67.

I'm afraid I don't know the details of the construction, but hopefully this reference is helpful.

Other relevant threads:

thank you for your answer, did you see my edition? – user42912 Feb 07 '13 at 00:51 — user42912, Feb 07 '13 at 00:51

There exist a function such that $f\circ f(x)=e^x$?

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