§2. Absolutely convergent series
145
and since
If * g(n)1 :-s: L If(p)g(n - p)l,
p
one deduces that the series L If * g(n)1 converges unconditionally and that
L If * g(n)1 :-s: L If(p)g(q)l,
whence (18) in view of (19).
Since the convolution product does not take one out of the set Ll (1:)
where it is defined, one can (and one must ... ) ask oneself whether it is
associative, i.e. whether
(18.20)
for any f, g, hE Ll(1:). To show this, remark first that (16) can be rewritten
as
(18.21)
f * g(s) = L f(q)g(r).
q+r=s
Thus
f * (g* h)(n)
L f(p)g*h(s) = L f(p) L g(q)h(r) =
p+s=n
p+s=n
L L f(p)g(q)h(r)j
p+s=n q+r=s
but the triple series L f(p)g(q)h(r) converges unconditionally, and the expression we have just obtained is, precisely, by the associativity theorem, its
partial sum over the system of all integers (p, q, r) such that p + q + r = n,
a partial sum in which we have grouped together the terms for which q + r
has a given value s. In other words,
(18.22)
f * (g * h)(n) = L f(p)g(q)h(r).
p+q+r=n
A similar calculation provides the same result for (J * g) * hen), but one can
also argue in the reverse direction, and, in (22), group together all the terms
for which p + q has a value given Sj one finds
L her) L f(p)g(q) = L h(r)f * g(s) = (J * g) * hen),
r+s=n
p+q=s
r+s=n
whence associativity.
It is obvious that the sum of two Ll(Z) functions is again in Ll(Z) and
that the convolution product is distributive with respect to addition: