infinity right-parenthesis comma upper X right-parenthesis"/> with the space of bounded functions from to , in which case, we write and equip such space with the supremum norm,
where . Alternatively, we can consider a subspace of formed by all functions such that
The next result shows that is a Banach space with respect to a special norm. This result, whose proof follows ideas similar to those of [124, 220], will be largely used in. Chapters 5 and 8
Proposition 1.9:The space , equipped with the norm
is a Banach space.
Proof. Let be the linear mapping defined by
for all and .
Claim. is an isometric isomorphism. Indeed, is an isometry because
for all . Moreover, if , then defined by
is such that and , since
Therefore, is onto and the Claim is proved.
Once is an isometric isomorphism and is a Banach space, we conclude that is also a Banach space.
