Riemann Rearrangement Theorem for Convolution

Question

I was recently studying the Riemann rearrangement theorem.
The theorem describes that:
There exist two function, such as $F:=f_1 + f_2 + f_3 + \cdots$, and $G:=g_1 + g_2 + g_3 + \cdots$.
Assume that the set of this two functions $\left\{f_i\right\}_{i=1}^{\infty}$ has a one-to-one mapping with $\left\{g_i\right\}_{i=1}^{\infty}$. $F$ and $G$ is conditionally convergent.
Then we got $F \neq G$.

However, I am wondering that:
If we replace + with * (convolution operator),
Could we still get the result $F \neq G$ ?

Any ideas or counter-example are highly appreciated.

Answer 1

Here is my proof about your problem (used the commutative property and mathematical induction):
The proof defines a sequence $\left\{G_n\right\}_{n=1}^{\infty}$ as follows:

$G_1$: There exists an $i_1$ such that $g_{i_1} = f_1$

$G = g_1 * g_2 * g_3 \ldots * g_{i-1} * g_{i_1} * g_{i+1}$

$= g_{i_1} * g_1 * g_2 * \ldots * g_{i-1} * g_{i+1}$

$= f_1 * g_1 * g_2 * \ldots * g_{j-1} * g_{i+1}$

$\triangleq G_1$

$G_2$: There exists an $i_2 \neq i_1$ such that $g_{i_2} = f_2$

$G = g_{i_1} * g_{i_2} * g_1 * g_2 \ldots * g_{i_1-1} * g_{i_1+1} * \cdots * g_{i_2-1} * g_{i_2+1} * \cdots$

$= f_1 * f_2 * g_1 * g_2 \cdots * g_{i_1-1} * g_{i_1+1}* \cdots * g_{i_2-1} * g_{i_2+1} * \cdots$

$\triangleq G_2$

$G_n$: According to the above proof, we can get $G_n\rightarrow G = G_n$

\begin{align*} \lim _{n \rightarrow \infty} G_n=f_1 * f_2 * \cdots=F \\ \lim _{n \rightarrow \infty} G_n=\lim _{n \rightarrow \infty} G=G \end{align*}

Finally, it is concluded that $F=G$