space
For
where in the case of
Thus
Proof of Banach space
This all follows from the general case of a Lebesgue space, however we will explicitly show completeness for
. Let
be a Cauchy sequence in , i.e. for every there exists some such that for all we have Now for any fixed
so
is a Cauchy sequence in convergent to some . It remains to show that and . For any fixed whence
and by the triangle inequality
hence
so
. Furthermore so indeed
.
Footnotes
-
There is no need to take a normed quotient here,
is already a full norm due to properties of the counting measure. ↩