Suppose we have the arithmetic functions $a_i : \mathbb{N} \rightarrow \mathbb{R}$ and $b_j : \mathbb{N} \rightarrow \mathbb{R}$ such that $a_n = b_n = 1$ if and only if the index $n$ is a square, and $a_n = b_n = 0$ otherwise. Suppose now that we would like to analyze $\sum_{i \ge 1} {a_i x^i} \sum_{j \ge 1} {b_j x^i}$ without convolutions to begin with. The approach is to extend $a_i$ and $b_j$ with continuous interpolations such that an integral rather than a sum might be evaluated.
That is, we want to create continuous auxiliary functions $a^*_i : \mathbb{R} \rightarrow \mathbb{R}$ and $b^*_j : \mathbb{R} \rightarrow \mathbb{R}$ so that $\forall m \in \mathbb{N}, a^*_m = a_m$ and $b^*_m = b_m$. The hope is that we can now enforce the condition on our $a^*_n, b^*_n$ so that $a^*_i b^*_j + a^*_j b^*_i = 0$ when $i, j \not \in \mathbb{N}$ and $a^*_i b^*_j + a^*_j b^*_i = a_i b_j + a_j b_i$ when $i, j \in \mathbb{N}$ and evaluate the double integral $\int_{i \gt 0} {a^*_i x^i} \int_{j \gt 0} {b^*_j x^j} = \int_{i \gt 0} {\int_{j \gt 0} a^*_i b^*_j x^{i+j}}$ into a target function with support $\mathbb{R} \rightarrow \mathbb{R}$. The goal then, would be to use fourier analysis on this integral to collect coefficients without the usual arithmetic convolution.
Can one prove from the conditions states that no such auxiliary functions exist?
