This question is more about a curious identity I have come across, than to do with explicit research. The question is somewhat advanced so I'm posting it here rather than on math stackexchange. It comes across as a sort of juvenile identity, but the work put in to show the result is rather tenuous.
For starters I'll give the motivation for the question. Consider the unit disk $\mathbb{D}$ and holomorphic functions $\phi$ taking $\mathbb{D}$ to itself and fixing $0$, where additionally $\lim_{n\to\infty} \phi^{\circ n}(\xi) = 0$ for all $\xi \in \mathbb{D}$. It is well known, that if $\phi(z) \neq \lambda z$ where $|\lambda| = 1$, the limit
$$\lim_{n\to\infty} \frac{\phi^{\circ n}(\xi)}{\phi'(0)^n} = \Psi(\xi)$$
constructs a holomorphic function for $\xi$ in $\mathbb{D}$. This function produces the solution to the linearizing of $\phi$, so that
$$\Psi(\phi(\xi)) = \phi'(0)\Psi(\xi)$$
Now I've seen this idea phrased slightly differently--in two different manners to be exact. Which leads me to the following. Letting $S$ be the set of all possible such $\phi$, and define the linear operator $C_\phi : S \to S$ such that
$$C_\phi f = f \circ \phi$$,
and now the construction of the Schroder function can be phrased slightly different. We will skip a small step (proving this is valid is a little space heavy), but the following defines an operator on $S$, for $f \in S$:
$$\Psi f = \lim_{n\to\infty}\frac{C_\phi^n f}{\phi'(0)^n}$$
Now let us go ahead denote $\Psi$ as $\Psi_{C_\phi}$, so that now it is an operator which takes an operator as one of its arguments and a function in $S$ as its other argument. (No it is not linear in the operator argument, just a non-linear operator.) So that $\Psi_{C_\phi}$ is the solution to the operator equation
$$\Psi_{C_\phi} C_\phi f = \phi'(0) \Psi_{C_\phi} f$$
Now the second valuable point is the following
$$\lim_{n\to\infty} \frac{C_\phi^{n+1} f}{C_\phi^n f} = \phi'(0)$$
Which leads me to the question that is sort of a reference request. Firstly, if $E : S \to S$, and
$$\lim_{n\to\infty} \frac{E^{n+1}f}{E^n f} = \lambda$$
where convergence is geometric, so that for some $0 <r <1$
$$|\frac{E^{n+1}f}{E^nf} - \lambda| < Cr^n$$
then instantly
$$\Psi_E f = \lim_{n\to\infty} \frac{E^n f}{\lambda^n}$$
defines a valid operator, because the following product converges absolutely.
$$E^0 f \prod_{i=0}^n\frac{E^{i+1}f}{E^{i} f \cdot \lambda} = \frac{E^1f}{E^0 f \cdot \lambda} \cdot \frac{E^2f}{E^1 f \cdot \lambda} \cdot \frac{E^{3}f}{E^2 f \cdot \lambda} ... = \frac{E^{n+1} f}{\lambda^{n+1}}$$
Wherein this new operator $\Psi_E$ satisfies the linearization
$$\Psi_E E f = \lambda \Psi_E f$$.
Now my question is easier to ask, now that all of this has been cleared up. The notion I was trying to define by the above is "linearizing an operator". Very much like linearizing a function from complex dynamics, but now we are linearizing an operator. Does this exist in mathematics already? Does anyone have a reference to something similar or a landmark where things like this have been studied?
Secondly, are there simpler ways of going about and showing results like this? I am noting fervently that for arbitrary $E$ proving that $\lim_{n\to\infty} \frac{E^{n+1}f}{E^n f}$ converges geometrically is really quite difficult. The goal at hand is to linearize an arbitrary operator similarly to how the Schroder function linearizes the operator $C_\phi$, but this one step blocks the path.
Can we relax the condition $\lim_{n\to\infty} \frac{E^{n+1}f}{E^n f}$ converges geometrically somehow?
I hope this question isn't too broad and too abstract, this is just something that's been bugging me for awhile and I haven't gotten much feedback from peers about this.