A definition of the functional $f\to\sum_{n=1}^\infty\mu(n)f \left( \frac{1}{n} \right) $, involving good functions $f$ and the Möbius function I would like to create a definition, is a conditional definition since the behaviour of arithmetic functions like to the Möbius function $\mu(n)$ depends on the veracity of the Riemann Hypothesis, that is an unsolved problem.
Any case I would like to know how create a definition with mathematical meaning. And as I am saying we are on assumption of the Riemann Hypothesis, because I believe that my definition requires this hypothesis to deduce the absolute convergence of a functional, that is its definition.
The definition is for the functional $\mathcal{F}$, from the set of entire functions (or you can do a restriction to analytic functions on an open subset of the complex plane $A(\Omega)$) to $\mathbb{C}$, defined by the rule $$f\to\sum_{n=1}^\infty\mu(n)f \left( \frac{1}{n} \right) .$$

An example. A particular value of our funtional $\mathcal{F}$, is given by $$\mathcal{F}(\operatorname{erf})=\sum_{n=1}^\infty\mu(n) \operatorname{erf}\left( \frac{1}{n} \right) ,$$ where $\operatorname{erf}$ is the Erf function, and as was said for integeres $n\geq 1$, we are denoting by $\mu(n)$ the Möbius function. Then we find that RHS is absolutely convergent, thus this particular value for the entire function $\operatorname{erf}$ is defined as a complex number.

I presume that our functional $\mathcal{F}$ will be linear (are easy justifications).

Question. How do you create a/this definition, rigurously, for the functional $\mathcal{F}$ that I've evoked, on assumption of the Riemann Hypothesis? Many thanks.

Thus I believe that is required state the domain of our functional as a subset of the linear space of entire functions, (show that is linear) and show that is defined by absolute convergence, using the asymptotic behaviour of means of the Möbius function (I presume that it is the way, and this is the reason for which is neccesary to state that we are on assumption of the Riemann Hypothesis, but if you know how create a definition without this assumption you can add it to your answer).
If there are inaccuracies in my previous words please tell me. My purpose is to know how create this definition in mathematics.
 A: In fact, your functional is independent of the Riemann hypothesis.
First we note that since there are infinitely many primes, we have $\lvert\mu(n)\rvert = 1$ for infinitely many $n$, and since $f\bigl(\frac{1}{n}\bigr) \to f(0)$ by continuity, the series
$$\sum_{n = 1}^{\infty} \mu(n)f\biggl(\frac{1}{n}\biggr)\tag{1}$$
can only converge for functions $f$ with $f(0) = 0$.
If $f$ is holomorphic on a neighbourhood $U$ of $0$ and defined at $\frac{1}{n}$ for all $n \in \mathbb{N}\setminus \{0\}$, with $f(0) = 0$, then on $U$ we can write
$$f(z) = f'(0)\cdot z + z^2g(z)$$
with a holomorphic function $g$. There is an $r > 0$ such that $\overline{D_r(0)} = \{ z\in \mathbb{C} : \lvert z\rvert \leqslant r\} \subset U$, and since $\overline{D_r(0)}$ is compact, $g$ is bounded there, say $\lvert g(z)\rvert \leqslant K$ for $\lvert z\rvert \leqslant r$. Since
$$\sum_{n = 1}^{\infty} \frac{\mu(n)}{n^s}$$
is absolutely convergent if and only if $\operatorname{Re} s > 1$, the series
$$\sum_{n = \lceil 1/r\rceil}^{\infty} \frac{\mu(n)}{n^2}g\biggl(\frac{1}{n}\biggr)$$
is absolutely convergent, and hence $(1)$ is absolutely convergent if and only if
$$\sum_{n = 1}^{\infty} \mu(n) f'(0)\frac{1}{n}$$
is absolutely convergent, that is, if and only if $f'(0) = 0$.
To get an absolutely convergent series $(1)$, we thus have to require $f(0) = f'(0) = 0$. However, the series
$$\sum_{n = 1}^{\infty} \frac{\mu(n)}{n}$$
is conditionally convergent (to $0$), and so the series $(1)$ is (conditionally) convergent for all holomorphic $f$ with $f(0) = 0$.
Since only the values of $f$ at the real points $\frac{1}{n}$, $n\in \mathbb{N}\setminus \{0\}$ are used in $(1)$, we can define the functional $\mathcal{F}$ for continuous functions $g\colon [0,1] \to \mathbb{C}$ with $g(0) = 0$ that are twice continuously differentiable on $[0,r]$ for some $r \in (0,1]$. The corresponding series will then be conditionally convergent if $g'(0) \neq 0$, and absolutely convergent if $g'(0) = 0$, by essentially the same argument as above.
However, for analytic functions whose Maclaurin series converges absolutely at $1$ (in particular, functions that are holomorphic at least on some disk $D_r(0)$ with $r > 1$), we can get a nice alternate form of the functional. Namely, with the Maclaurin expansion
$$f(z) = \sum_{k = 1}^{\infty} a_k z^k$$
we obtain
\begin{align}
\sum_{n = 1}^{\infty} \mu(n)f\biggl(\frac{1}{n}\biggr)
&= \sum_{n = 1}^{\infty} \mu(n)\Biggl(\frac{a_1}{n} + \sum_{k = 2}^{\infty} \frac{a_k}{n^k}\Biggr) \\
&= \sum_{n = 1}^{\infty} \mu(n)\frac{a_1}{n} + \sum_{n = 1}^{\infty}\mu(n)\sum_{k = 2}^{\infty} \frac{a_k}{n^k} \\
&= \sum_{n = 1}^{\infty}\mu(n)\sum_{k = 2}^{\infty} \frac{a_k}{n^k} \\
&= \sum_{k = 2}^{\infty} a_k \sum_{n = 1}^{\infty} \frac{\mu(n)}{n^k} \\
&= \sum_{k = 2}^{\infty} \frac{a_k}{\zeta(k)}
\end{align}
since
$$\sum_{n = 1}^{\infty} \frac{\mu(n)}{n^s} = \frac{1}{\zeta(s)}$$
for $\operatorname{Re} s > 1$, and the change of order of summation is legitimate by absolute convergence:
$$\sum_{k = 2}^{\infty} \sum_{n = 1}^{\infty} \frac{\lvert \mu(n)a_k\rvert}{n^k} \leqslant \sum_{k = 2}^{\infty} \lvert a_k\rvert \sum_{n = 1}^{\infty} \frac{1}{n^k} = \sum_{k = 2}^\infty \zeta(k)\lvert a_k\rvert \leqslant 2\sum_{k = 2}^{\infty} \lvert a_k\rvert < +\infty$$
since we assumed the Maclaurin series converges absolutely at $1$.
Since $\frac{1}{\zeta}$ has a zero at $1$, we can also write
$$\mathcal{F}(f) = \sum_{k = 1}^{\infty} \frac{f^{(k)}(0)}{k!\zeta(k)}$$
for these functions. In that form, we have a natural extension of the functional to holomorphic functions whose Maclaurin series converges absolutely at $1$ even when $f(0) \neq 0$:
$$\tilde{\mathcal{F}}(f) = \sum_{k = 0}^{\infty} \frac{f^{(k)}(0)}{k!\zeta(k)}.$$
