Proof of Mobius inversion formula - the other direction The Mobius inversion formula states that if we define $f$ as $$f(m)=\sum_{d|m}g(d)$$ then $$g(m)=\sum_{d|m}f\left(\frac md\right)\mu(d)$$
We know that the other direction is also true: if we define $g$ as above, then $f$ also satisfies the above.
I've searched for the proof, but failed to find any that does not rely on convolutions. My question is whether there is a proof that relies on manipulating sums that proves the second direction of the theorem. 
 A: We assume 
\begin{align*}
g(m)=\sum_{d|m}f\left(\frac md\right)\mu(d)\tag{1}
\end{align*}
and show the validity of
\begin{align*}
f(m)=\sum_{d|m}g(d)\tag{2}
\end{align*}
It is convenient to use the unit-function $u$ defined as $u(n)=1, n\geq 1$.

We start with the right-hand side of (2) and obtain
  \begin{align*}
\color{blue}{\sum_{m|n}g(m)}&=\sum_{m|n}\left(\sum_{d|m}f\left(\frac {m}{d}\right)\mu(d)\right)u\left(\frac{n}{m}\right)\tag{3}\\
&=\sum_{a\cdot m=n}\left(\sum_{b\cdot d=m}f(b)\mu(d)\right)u(a)\\
&=\sum_{a\cdot b\cdot d=n}f(b)\mu(d)u(a)\\
&=\sum_{b\cdot m=n}f(b)\left(\sum_{a\cdot d=m}\mu(d)u(a)\right)\\
&=\sum_{m|n}f\left(\frac{n}{m}\right)\left(\sum_{d|m}\mu(d)u\left(\frac{m}{d}\right)\right)\\
&=\sum_{m|n}f\left(\frac{n}{m}\right)\sum_{d|m}\mu(d)\\
&=\sum_{m|n}f\left(\frac{n}{m}\right)\left\lfloor\frac{1}{m}\right\rfloor\tag{4}\\
&\,\,\color{blue}{=f(n)}\tag{5}
\end{align*}
and the claim (2) follows.

Comment: 


*

*In (3) we use the identity (1) and multiply for convenience only with $1=u\left(\frac{n}{m}\right)$.

*In (4) we use the identity $\sum_{d|m}\mu(d)=\left\lfloor\frac{1}{m}\right\rfloor=\begin{cases}1&m=1\\0&m>1\end{cases}$.

*In (5) we obtain $f(n)$ since all other summands with $m>1$ in (4) give zero.
Note: We can omit the usage of $u$ in the derivation, but this might reduce readability.
