I'm currently reading the paper available at https://arxiv.org/abs/1503.05724.
In this paper, the authors define a function iteration as shown here:
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They also discuss Abel's functional equation:
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Given that $f(x) = e^x$ and $\beta = 1$, the equation $$\psi(f(x)) = \psi(x) + \beta$$ simplifies to $$\psi(e^x) = \psi(x) + 1.$$ If we substitute $\log^{(k)}(x) + k$ for $\psi(x)$, we obtain the equation $$\log^{(k)}(e^x) + k = \log^{(k)}(x) + k + 1.$$
However, this doesn't seem to hold true for $k \in \mathbb{N}$, where $0 \le \log^{(k)}(x) < 1$.
For instance, if $k = 1$, which belongs to $\mathbb{N}$, we get $x = \log(x) + 1$, but this only holds true when $x = 1$.
Could someone clarify what I might be missing or misunderstanding in this scenario?
It is known and easy to see that Abel's functional equation $$\tag1 \psi(f(x))=\psi(x)+1 $$ is equivalent to $$\tag2 f(h(x))=h(x+1) $$ when $\psi$ is invertible. In fact, $h=\psi^{-1}\,.$ The question how to solve $\exp(h(x))=h(x+1)$ is handled for example in this paper.
Now to your arxiv.org/abs/1503.05724:
They take $f(x)=e^x\,.$ When we interpret $\log^{(k)}(x)$ naively as $$ \log\circ\log\dots\circ\log\,,\text{ ($k$ times) } $$ then their function $$\tag3 \psi(x)=\log^{(k)}(x)+k $$ does not solve (1). Not even for $k=1\,:$ $$ \log(e^x)+1=x+1\not=\log x+2\,. $$ It seems to me that (3) is a failed attempt to find a just a notation that is supposed to reflect the fact that every invertible solution $\psi$ of (1) with $f(x)=e^x$ must satisfy $$\tag4 \exp^{(k)}(x)=\psi^{-1}(\psi(x)+k)\, $$ where $\exp^{(k)}$ is the usual $k$-fold iterate of $\exp\,.$