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Are there monotonically increasing functions $f:\mathbb{R}\to\mathbb{R}$ such that $f(f(x)) = e^x f(x)$?

Ricardo Andrade
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STrick
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    these problems are either trivial or extremely elaborate. Why do you want to know? – Will Jagy Dec 31 '13 at 21:22
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    my answer to my own question gives a pretty good idea of the intricacy http://mathoverflow.net/questions/45608/formal-power-series-convergence – Will Jagy Dec 31 '13 at 21:45
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    Related question? http://mathoverflow.net/questions/49724/the-functional-equation-ffx-xfx2 – Timothy Chow Dec 31 '13 at 23:11
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    Oh, well. Lots of references at http://zakuski.utsa.edu/~jagy/other.html – Will Jagy Dec 31 '13 at 23:42
  • One obvious observation would be that $f(0)=f(f(0))$. Since $f$ is monotonically increasing on $\mathbb{R}$, this would seem to imply that $f(0)=0$, otherwise we'd have to have $f(x)=$ constant for all x in between 0 and $f(0)$, which would break the basic property of $f$ - Or am I mistaken ? – Lucian Jan 01 '14 at 07:45
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    Let $c=f(0)$. Then $c = f(f(c)) = e^c f(c) = e^c c$, so $c=0$. If a power series expansion exists at zero, then it is unique, and has the form $x + x^2 + \frac{1}{2}x^3 - \frac{7}{3}x^4 + \frac{71}{8}x^5 + \cdots$ [Edit 2016-5-17: this is not quite correct]. It seems unlikely that this power series has infinite radius of convergence. – S. Carnahan Jan 01 '14 at 13:13
  • But why would you need infinite radius of convergence? The singularity might not be on the real axis. If the power series yields a local solution in a neighborhood of zero, you might actually use the functional equation to extend the solution further. – Michael Renardy Jan 01 '14 at 15:00
  • @S.Carnahan: $\ln(1-x)=-\displaystyle\sum_1^\infty\frac{x^n}n$ is also a power series that does not have an infinite radius of convergence. That doesn't mean that the natural logarithm is undefined outside of $[-1,+1]$. – Lucian Jan 02 '14 at 01:35
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    If we use the recursion relation on coefficients, a naive estimate for the growth rate is roughly factorial. Barring some nice cancellation, I suspect the radius of convergence is zero. – S. Carnahan Jan 02 '14 at 02:42
  • If $f$ were monotonic then $f$ is invertible, it would then imply that $f(x) = e^{f^{-1}(x)}x$ that might help as well. – Daniel Parry Jan 03 '14 at 07:53
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    @S.Carnahan I actually get something different here, namely $x+x^2-\frac12x^3+\frac53x^4-\frac{53}8x^5+\frac{159}5x^6-\frac{25075}{144}x^7+...$ – მამუკა ჯიბლაძე May 16 '16 at 11:26
  • @მამუკაჯიბლაძე Thank you for pointing this out. SAGE agrees with you. – S. Carnahan May 16 '16 at 16:09

1 Answers1

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By monotonically increasing, do you mean strictly increasing or non-decreasing?

If you mean non-decreasing, then there is the trivial case $f(x)=0$. However, there may well be other cases that are harder to find.

If you mean strictly increasing, then there are no continuous solutions (there might still be discontinuous ones):

As proven by Carnahan, $f(0)=0$.

Now, consider the following limit: $$\alpha=\lim_{x \to -\infty}{f(x)}$$ We shall consider three cases, $\alpha=0$, $-\infty<\alpha<0$, and $\alpha=-\infty$.

In the first case, one of three things occurs: it approaches from above, from below, or from equal to zero. All three of these contradict that $f$ is strictly increasing.

In the second case, we have: $$f(\alpha)=\lim_{x \to -\infty}{f(f(x))}=\lim_{x \to -\infty}{{e^x}{f(x)}}=\lim_{x \to -\infty}{e^x} \times \lim_{x \to -\infty}{f(x)}=0 \times \alpha=0$$ So, $f(\alpha)=0=f(0)$ and since $\alpha<0$, we get a contradiction.

In the third case, $$-\infty=\lim_{x \to -\infty}{f(f(x))}=\lim_{x \to -\infty}{{e^x}{f(x)}}=\lim_{x \to -\infty}\frac{f(x)} {e^x}$$

Which means that it goes down to negative infinity at a greater than exponential rate. Let $k=f(-1)$

Firstly, $k<0$ because $k=f(-1)<f(0)=0$.

Secondly, $f(k)=f(f(-1))={e^{-1}}{f(-1)}=\frac{k}{e}$

Thus, $f(k)>k$

Also, since it descends to negative infinity faster than exponentially, there must exist a finite $N$ such that $n<N \implies f(n)<n$

Thus, by assumed continuity of $f$, there must be a $N<p<k$ such that $f(p)=p$.

But, $p=f(p)=f(f(p))={e^p}{f(p)}=p{e^p}$

So, $p=p{e^p}$ which only has $p=0$ as a solution.

But, $p<k<0$. A contradiction.

Dylan Pizzo
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