The Wikipedia page on the Mobius inversion formula gives the following formula in passing: if $$G(x)=\sum_{k=1}^x \alpha(x)F(x/k)$$ for some arithmetic function $\alpha(n)$ possessing a Dirichlet inverse $\alpha^{-1}(n)$, then $$F(x)=\sum_{k=1}^x \alpha^{-1}(k)G(x/k).$$ My question is the following: can we define an operator $*'$ analogous to Dirichlet convolution $*$ defined by the formula $$(f*'g)(x)=\sum_{k=1}^x f(k)g(x/k)$$ and if so what can we do with it? In particular, the formula from Wikipedia states that $(f*'g)*'f^{-1}=g$ where $f^{-1}$ is the Dirichlet inverse of $f$. For what function $f^{-1*}$ do we have $(g*'f)*'f^{-1*}=g$? (Clearly $f*'g \neq g*'f$.)
In general, what else can we say about this convolution? Does there exist a transform analogous to Dirichlet convolution for which $*'$ is the convolution? If so, can we define it and is it useful?
In addition, does anyone know the source of the Wikipedia theorem, and if so does the author discuss these or similar questions?
My attempt to derive a formula to the above question about modified Dirichlet inverses ends up "proving" a false result; if anyone can spot the error, or even suggest how to fix it, that would be greatly appreciated.
Define the function $$G(x)=\sum_{k=1}^x \alpha(k)F(\frac xk).$$ We now consider the sum $$\sum_{k=1}^x \alpha^{-1}(k)G(\frac xk)=\sum_{k=1}^x \alpha^{-1}(k) \sum_{j=1}^{x/k}\alpha(j)F(\frac x{kj})$$ by the definition of $G(x)$. Using Iverson bracket notation, we rewrite $$=\sum_{k=1}^x\alpha^{-1}(k)\sum_{j=1}^{x/k} \alpha(j) \sum_{r=1}^x [r=kj] F(x/r)=\sum_{r=1}^x F(x/r) \sum_{k=1}^x \alpha^{-1}(k)\sum_{j=1}^{x/k} \alpha(j)[j=r/k]\\=\sum_{r=1}^x F(x/r)\sum_{d|r}\alpha(d)\alpha^{-1}(r/d)=\sum_{k=1}^x F(x/k)\epsilon(k)=F(x)$$ which proves the Wikipedia identity. (A similar proof is in fact given on the Wikipedia page.) Proceeding analogously, we define $$G(x)=\sum_{k=1}^x F(k) \alpha(x/k)$$ and consider the sum $$\sum_{k=1}^x G(k) \alpha^{-1}(x/k)=\sum_{k=1}^x \alpha^{-1}(x/k)\sum_{j=1}^k F(j)\alpha(k/j)$$ by the definition of $G(x)$ and rewrite as above $$=\sum_{k=1}^x \alpha^{-1}(x/k)\sum_{j=1}^k\sum_{r=1}^x F(j)\alpha(r)[r=k/j]=\sum_{r=1}^x \alpha(r)\sum_{k=1}^x \alpha^{-1}(x/k) \sum_{j=1}^k F(j)[j=k/r]=\sum_{k=1}^x \alpha^{-1}(x/k)\sum_{r=1}^x \alpha(r) \sum_{j=1}^k F(j)[j=k/r]=\sum_{k=1}^x\alpha^{-1}(x/k)\sum_{d|k}F(k/d)\alpha(d)$$ which by our definition of $*'$ can be written $$(F*'\alpha)*'\alpha^{-1}=(F*\alpha)*'\alpha^{-1}$$ which seems to imply $$F*'\alpha=F*\alpha$$ which is clearly false. What am I doing wrong?
The operator you defined is called a generalized convolution (between some arithmetical function $a(n)$ and some function $F:[1,+\infty)\mapsto\mathbb C$), which is formally defined as follows using the notations in Apostol's Introduction to Analytic Number Theory: $$ a\circ F\triangleq\sum_{n\le x}a(n)F\left(\frac xn\right) $$ Then if we were to introduce another arithmetical function $b(n)$, we have $$ \begin{aligned} a\circ(b\circ F) &=\sum_{n\le x}a(n)\sum_{m\le x/n}b(m)F\left(x\over mn\right) \\ &=\sum_{mn\le x}a(n)b(m)F\left(x\over mn\right) \\ &=\sum_{n\le x}\left[\sum_{d|n}a(d)b\left(\frac nd\right)\right]F\left(x\over n\right) \\ &=(a*b)\circ F \end{aligned} $$ In our specific case, $b(n)$ is the Dirichlet inverse, $a^{-1}(n)$, of $a(n)$, so we have $$ a\circ(a^{-1}\circ F)=(a*a^{-1})\circ F=\sum_{n\le x}[n=1]F\left(\frac xn\right)=F(x) $$