# Note on Generalized Convolution in Number Theory

In Tom Apostol’s famous Introduction to Analytic Number Theory there is a section about generalized convolutions which sometimes causes confusion (see Apostol, Chapter 2, p.39 and also my Advanced Number Theory Note #3). In this short note I want to quickly try to clarify this bit.

Let $F$ denote a real or complex-valued function defined on the positive real axis such that $F(x) = 0$ for $0 < x < 1$.

$\alpha$ is any arithmetical function.

Generalized convolution is then defined by

$(\alpha \circ F)(x)= \sum_{n \leq x} \alpha(n)F(\frac{x}{n})$

Apostol says:

“If $F(x) = 0$ for all nonintegral values of $x$, the restriction of $F$ to integers is an arithmetical function and we find that $(\alpha \circ F)(m) =(\alpha \ast F)(m)$.”

Here $(\alpha \ast \beta)(n) = \sum_{d | n} \alpha(n)\beta(\frac{n}{d})$.

The question that sometimes arises in some form or other in relation to this is: why does $F(x)$ have to be $0$ for all nonintegral values of $x$ for this to work? Is the point that we want $F(\frac{x}{n})$ to be $0$ when $\frac{x}{n}$ is not an integer?

The answer is yes. For example, to isolate $F$, take $\alpha$ to be the unit function $u$. Then the generalized convolution for integer $m \geq 1$ is

$(u \circ F)(m) = \sum_{n \leq m} F(\frac{m}{n})$

If $F(\frac{m}{n}) = 0$ whenever $\frac{m}{n}$ is not an integer, the above sum reduces to

$\sum_{n | m} F(\frac{m}{n}) = (u \ast F)(m)$

I think Apostol could have simply said “If $F(x) = 0$ for all nonintegral values of $x$, we find that $(\alpha \circ F)(m) = (\alpha \ast F)(m)$ for all integers $m \geq 1$…”