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
Now for any fixed
so
whence
and by the triangle inequality
hence
so
so indeed
#state/tidy | #lang/en | #SemBr
Footnotes
-
There is no need to take a normed quotient here,
is already a full norm due to properties of the counting measure. ↩