[Home]History of Hahn-Banach theorem

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Revision 4 . . (edit) December 1, 2001 1:33 am by (logged).149.1.xxx
Revision 3 . . September 30, 2001 4:29 am by AxelBoldt [important consequences]
Revision 2 . . September 29, 2001 11:03 pm by Simon J Kissane [*s/&psi:/ψ:/]
  

Difference (from prior major revision) (minor diff, author diff)

Changed: 1c1
The Hahn-Banach theorem is a central tool in functional analysis; it shows that there are "enough" continuous linear functionals defined on every normed vector space to make the study of the dual space interesting.
The Hahn-Banach theorem is a central tool in functional analysis; it shows that there are "enough" continuous linear functionals defined on every normed vector space to make the study of the dual space interesting ( though that is a matter of opinion ).

Changed: 5c5
Now let U be a subspace of V and let φ : U -> K be a [linear function]? such that |φ(x)| ≤ N(x) for all x in U. Then the Hahn-Banach theorem states that there exists a linear map ψ : V -> K which extends φ (meaning ψ(x) = φ(x) for all x in U) and which is dominated by N on all of V (meaning |ψ(x)| ≤ N(x) for all x in V.
Now let U be a subspace of V and let φ : U -> K be a [linear function]? such that |φ(x)| ≤ N(x) for all x in U. Then the Hahn-Banach theorem states that there exists a linear map ψ : V -> K which extends φ (meaning ψ(x) = φ(x) for all x in U) and which is dominated by N on all of V (meaning |ψ(x)| ≤ N(x) for all x in V).

Added: 7a8,11

Several important consequences of the theorem are also sometimes called "Hahn-Banach theorem":
* If V is a normed vector space with subspace U (not necessarily closed) and if φ : U -> K is continuous and linear, then there exists an extension ψ : V -> K of φ which is also continuous and linear and which has the same norm as as φ (see Banach space for a discussion of the norm of a linear map).
* If V is a normed vector space with subspace U (not necessarily closed) and if x0 is an element of V not in U, then there exists a continuous linear map ψ : V -> K with ψ(x) = 0 for all x in U, ψ(x0) = 1, and ||ψ|| = ||x0||-1.

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