proof of Hahn-Banach theorem
Consider the family of all possible extensions of , i.e. the set of all pairings where is a vector subspace of containing and is a linear map such that for all and for all . is naturally endowed with an partial order relation: given we say that iff is an extension of that is and for all . We want to apply Zorn’s Lemma to so we are going to prove that every chain in has an upper bound.
Let be the elements of a chain in . Define . Clearly is a vector subspace of and contains . Define by “merging” all ’s as follows. Given there exists such that : define . This is a good definition since if both and contain then in fact either or . Notice that the map is linear, in fact given any two vectors there exists such that and hence . The so constructed pair is hence an upper bound for the chain because is an extension of every .
We notice that given any it holds
in particular we find that and for all it holds
Define as follows:
Clearly is a linear functional. We have
and by letting by the previous estimates on we obtain
which together give
So we have proved that and which is a contradiction.
|Title||proof of Hahn-Banach theorem|
|Date of creation||2013-03-22 13:31:58|
|Last modified on||2013-03-22 13:31:58|
|Last modified by||paolini (1187)|