I will preface this by saying that I know enough complex analysis to understand analytic continuation and have played around with operator theory. I knew no functional analysis before considering this problem and have since learnt some.
Let $f$ and $g$ be functions and let $A$ be an operator which acts on functions of $x$. Given that $f$ converges in a subset $R_f$ of $\mathbb{C}$ and has an analytic continuation $F$ in another subset $R_F$, when (if at all) does the following hold?
$$f(A)g(x) = F(A)g(x)$$
My attempt to make progress involved supposing that $g$ could be written in the appropriate form for the application of the spectral theorem, i.e. as a sum of eigenfunctions of $A$. I then replaced $f(A)$ with $f(\lambda)$ where $\lambda$ is the eigenvalue corresponding to a particular eigenfuction. If $\lambda$ can be restricted to $R_f$, that is to say that $\lambda \in R_f$ for all eigenfunctions required to construct $g$, then in the sense of analytic continuation this relation holds: $f(\lambda) = F(\lambda)$. This seems to suggest that the equation holds (under the conditions outlined).
I am unsure how to proceed further and whether my reasoning makes sense. Does this question have a well-defined solution, and if so, how should I approach it?
Edit: I'm sorry for the lack of clarity in my original post, it's due to a lack of knowledge about functional analysis and likely a poorly formulated question. I don't know much at all about (un)bounded operators. I will demonstrate what I mean through an example. It will require 2 main results that stem from the theory of exponentiating certain differential operators. Those results can be found here: https://math.stackexchange.com/questions/719487/exponential-of-a-function-times-derivative#:~:text=This%20can%20be%20easily%20verified,%E2%88%82mxn%3Dn!
$$\mathrm{e}^{a \frac{\mathrm{d}} {\mathrm{d} x} } f(x) = f(x+a)$$ $$\lambda^{x\frac{\mathrm{d}}{\mathrm{d}x}} f(x) = f(\lambda x)$$ My example uses the following functional equation for the gamma function. $$z\Gamma(z)=\Gamma(z+1)$$ Which meromorphically continues $\Gamma$ from the right half-plane $\Re(z)>0$ to $\Re(z)>-1$ with a pole at $z=0$. Replacing $z$ with a linear operator $A$ maintains commutativity of the $LHS$ since the term inside the gamma function carrying $A$ commutes with $A$ (it can be expanded as a power series in $A$), so that $A\Gamma(A)=\Gamma(A)A$.
Applying the $LHS$ to $f$, first with $A=\frac{\mathrm{d}}{\mathrm{d}s}$. $$I=\frac{\mathrm{d}}{\mathrm{d}s} \int_0^\infty t^{\frac{\mathrm{d}}{\mathrm{d}s}} \mathrm{e}^{-t} \frac{\mathrm{d}t}{t} f(s) = \int_0^\infty f'(s+\log t) \mathrm{e}^{-t} \frac{\mathrm{d}t}{t}$$ Applying the $RHS$ gives: $$\int_0^\infty t^{\frac{\mathrm{d}}{\mathrm{d}s}} \mathrm{e}^{-t} \mathrm{d}t\, f(s) = \int_0^\infty f(s+\log t) \mathrm{e}^{-t} \mathrm{d}t = -f(s+\log t)\mathrm{e}^{-t}\bigg|_0^\infty + I$$ Partial integration reveals the integrals to be the same, excepting the term $f(s+\log t)\mathrm{e}^{-t} \bigg|_0^\infty$. This term vanishes when $f(s+\log t) = \mathcal{o}(\mathrm{e}^t)$ and when $f(-\infty+is_y)=0$ (or they are negatives of each another). If these conditions are not met the result is either constant or divergent.
A similar result is found when repeated with $A=s\frac{\mathrm{d}}{\mathrm{d}s}$. $LHS$: $$J = s\frac{\mathrm{d}}{\mathrm{d}s} \int_0^\infty t^{s\frac{\mathrm{d}}{\mathrm{d}s}} \mathrm{e}^{-t} \frac{\mathrm{d}t}{t} f(s) = s\int_0^\infty f'(st) \mathrm{e}^{-t} \mathrm{d}t$$ $RHS$: $$\int_0^\infty t^{s\frac{\mathrm{d}}{\mathrm{d}s}} \mathrm{e}^{-t} \mathrm{d}t\, f(s) = \int_0^\infty f(st) \mathrm{e}^{-t} \mathrm{d}t = -f(st)\mathrm{e}^{-t}\bigg|_0^\infty + J$$ The leftover term $f(st)\mathrm{e}^{-t}\bigg|_0^\infty$ vanishes when $|f(st)|=\mathcal{o}(\mathrm{e}^t)$ and $f(0) = 0$ (or if they are negatives of each other)
The results are the same except for the extraneous constants. These constants seem to come from the way the functional equation itself is derived, by integrating by parts. I wondered if other functions continued by a functional equation give a similar result with residual constants or functions. The Jacobi theta gives half iterated operators which I decided not to try working with. The other function I considered was the Riemann Zeta that gave, according to my work, infinities for even nice functions $f$.
