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An additive function $f(n)$ is said to be totally/completely additive if

$$f(ab) = f(a) + f(b)$$

holds for all positive integers $a$ and $b$.

Additionally, if $f$ is a totally/completely additive function, then $f(1) = 0$.

On the other hand, a multiplicative function $g(n)$ is said to be totally/completely multiplicative if

$$g(ab) = g(a)g(b)$$

holds for all positive integers $a$ and $b$.

Additionally, if g is a totally/completely multiplicative function, then $g(1) = 1$.

So now suppose there is a function $h(n)$ such that

$$h(n) = \frac{1}{2}[f(n) + g(n)]$$

where $f$ is totally additive and $g$ is totally multiplicative. (I removed the erroneous assumption on the function $h(n)$, thank you Gerry for pointing that out!)

Then

$$h(1) = \frac{1}{2}(f(1) + g(1)) = \frac{1}{2}$$

Now, let

$$i(n) = h(n + 1)$$

so that

$$i(0) = h(1) = \frac{1}{2}$$

My question is:

How do we then characterize such number-theoretic functions $i(n)$?

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2 
 How do we motivate why anyone should care about such functions? (So far there are 2 votes to close the question, and unless there is a reason given why the average of an additive and multiplicative function is a natural object of interest, I imagine there will be more votes to close.) – KConrad Dec 10 2010 at 10:22
4 
 Huh? How can $h$ be additive or multiplicative if $h(1)=1/2$? – Gerry Myerson Dec 10 2010 at 11:08
Thanks Gerry! :-D (I overlooked that one, I have removed that erroneous assumption on $h(n)$.) – Jose Arnaldo Dris Dec 10 2010 at 15:54
2 
 Arnie Dris, please say something about why such a function would be interesting or useful. – S. Carnahan♦ Dec 10 2010 at 16:41
3 
 Arnie: your analogy with even/odd functions is badly posed. The decomposition of functions (from R to R, say) into sums of even and odd functions is a very natural thing from the viewpoint of symmetry: the reflection operation that sends any function f(x) to the function f(-x) has order 2, hence eigenvalues +/-1, and its eigenspaces for those eigenvalues are the even and odd functions. Your proposal about functions that are a sum of additive and multiplicative functions, on the other hand, is not related to any symmetry that the rest of us can see. That's why the question has been closed. – KConrad Dec 11 2010 at 18:51
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closed as too localized by Denis Serre, Chandan Singh Dalawat, Kevin Buzzard, Pete L. Clark, S. Carnahan♦ Dec 10 2010 at 16:37

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