# A function $f(x)$ such that $f(f(x))=\ln(x)$

How to find a continuous function $f(x)$ for $x > 12$, such that $f(f(x))=\ln(x)$? Preferably analytic too.

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You can rewrite this as $e^{f(x)}=f(e^x)$ - maybe this helps. (let $y=e^x$, then $f(f(f(y)))=f(\ln y)=\ln(f(y)$ whence my equality). –  Yulia Kuznetsova Jan 9 '12 at 16:38
But this will increase number of possible $f$'s since this equation has solution $f(x)=x$ while the initial one doesn't. –  Yulia Kuznetsova Jan 9 '12 at 16:55
such function would be half-iterate of $\ln x$; its inverse would be half-iterate of $\exp x$. For half-iterate of $\exp$, see mathoverflow.net/questions/12081 –  sdcvvc Jan 9 '12 at 17:32
I suppose there is no such $f$ because the equation $f(x)=\exp(f(\ln x))$ has only the solution $f(x)=x$. I cannot prove this but I can show some estimate for $f$: Suppose that such $f$ exists and is continuous. Let $e_n$ be the sequence $e_0=1$, $e_{n+1}=\exp(e_n)$. Let $f_n$ be the maximum of $f$ on $I_n=[e_n,e_{n+1}]$. Then for $x\in I_n$ we have $\ln x\in I_{n-1}$ so $f(x)=\exp(f(\ln x))\le \exp(f_{n-1})$, and $f_n\le \exp(f_{n-1}$. You start with $x>12$, so let my $n$ start with 3. Let $f_3=a$, then $f_n\le \exp(\dots \exp(a)\dots)$ with $\exp$ repeating $n-3$ times. –  Yulia Kuznetsova Jan 9 '12 at 17:33
Since $x>\exp(\dots\exp(1)\dots)$ with $n-1$ exponents, we get $f_n\le (\ln(\ln x))^a$. Sorry, no more time now. –  Yulia Kuznetsova Jan 9 '12 at 17:33