My question is on Mobius inversion formula convergence/properties when used with infinite sums of function.

Lets consider (on $\mathbb{R}^{+}$): 

$$S(x)= \sum\limits_{n=1}^{\infty} f(nx)$$

I call $p_i$ the $i^{nth}$ prime number and I define operators $J_{p_i}$ such that:

$$J_{p_i} (g(x))= g(x) - g(p_i x) $$

So if I apply successively $J_{p_1}$, $J_{p_2}$... to $S(x)$ I obtain:

$$J_{p_j} ... J_{p_1} (S(x)) = S(x) - S(p_1 x) - S(p_2 x)-... + S(p_1 p_2 x)+... + (-1)^j S(p_1 p_2 ... p_jx) = \sum\limits_{n=1}^{p_j} \mu_n S(nx) + \sum\limits_{n=p_j+1}^{p_1 ... p_j} r_n \mu_n S(nx)$$

Where $\mu_n$ is the Mobius function and $r_n=1$ if n has only primes $p_1...p_j$ in its prime number decomposition.

On the other side, using definition of $S(x)$ we see that applying $J_{p_i}$ operators removes the terms in the sum defining $S(x)$ (multiples of $p_1$ are removed then multiple of $p_2$ etc...) We see the inversion in progress as we apply the $J_p$ operators:

$$J_{p_j} ... J_{p_1} (S(x)) =  f(x) + \sum\limits_{n=p_j+1}^{\infty} a_n f(nx)$$
 
If we continue to infinity we obtain only $f(x)$ and this is due to the Mobius inversion property (Finally we have the inverison formula if everything converges nicely $\sum\limits_{n=1}^{\infty} \mu_n S(nx) =  f(x)$). 

My question is on the behavior of $J_{p_j} ... J_{p_1} (S(x))$ as $j \to \infty$.
 
For a "good" $f(x)$ can we have $J_{p_j} ... J_{p_1} (S(x))$ converging uniformly to $f(x)$ for $j\to \infty$ ?

For example suppose that we chose  $f(x)= x^2 e^{-x^2} - \alpha (x \alpha)^2 e^{-(\alpha x)^2})$, as here $\int\limits_{0}^{\infty} f(x) dx =0$ and $f(x)$ has a "nice" Fourier transform by Poisson formula we have  $S(x) \to 0$ for $x \to 0$.

In this case we have simple convergence for all $x$ when $j \to \infty$:

$J_{p_j} ... J_{p_1} (S(x)) \to f(x)$

And for each function $J_{p_j} ... J_{p_1} (S(0))=0$ with also $f(0)=0$, but what is really happening in zero?
Can I bound $J_{p_j} ... J_{p_1} (S(x)) - f(x)$ on $\mathbb{R}^{+}$ ?

Any reference on this subject ?