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

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

marked as duplicate by user17762, Clayton, Erick Wong, Namaste, Michael AlbaneseFeb 7 '13 at 1:21

• 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 7 '13 at 0:24
• @lee sorry it misses informations, I'm going to edit my question – user42912 Feb 7 '13 at 0:25
• It appears this question has been asked before on math.SE, so this question will be closed as a duplicate. – Zev Chonoles Feb 7 '13 at 0:28
• @ZevChonoles My question is defined differently than other posts. – user42912 Feb 7 '13 at 1:01

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.