Hahn–Banach theorem

From Wikipedia, the free encyclopedia

(Redirected from Hahn-Banach theorem)
Jump to: navigation, search

In mathematics, the Hahn–Banach theorem is a central tool in functional analysis. It allows the extension of bounded linear operators defined on a subspace of some vector space to the whole space, and it also shows that there are "enough" continuous linear functionals defined on every normed vector space to make the study of the dual space interesting. It is named for Hans Hahn and Stefan Banach who proved this theorem independently in the 1920s.

Contents

The most general formulation of the theorem needs some preparation. Given a vector space V over the scalar field K (either the real numbers R or the complex numbers C), a function N : VR is called sublinear if

N(ax + by) ≤ |a| N(x) + |b| N(y)

for all x and y in V and all scalars a and b in K. Every norm on V is sublinear, as is every seminorm, but there are other examples.

The Hahn–Banach theorem states the following:

Let N : VR be sublinear, let φ : UK be a linear functional on a subspace U of V. If φ is dominated by N on U (meaning |φ(x)| ≤ N(x) for all x in U) then there exists a linear extension ψ : VK of φ to all of V (meaning ψ(x) = φ(x) for all x in U) which is dominated by N on all of V.

The extension ψ is in general not uniquely specified by φ and the proof gives no method as to how to find ψ: in the case of an infinite dimensional space V, it depends on Zorn's lemma, one formulation of the axiom of choice.

In fact, the sublinearity condition on N can be slightly relaxed: it suffices to assume that

N(ax + by) ≤ |a| N(x) + |b| N(y)

for all a and b in K with |a| + |b| = 1 (Reed and Simon, 1980).

The Mizar project has completely formalized and automatically checked the proof of the Hahn–Banach theorem in the HAHNBAN file.

The theorem has several important consequences, some of are which also sometimes called "Hahn–Banach theorem":

  • If V is a normed vector space with subspace U (not necessarily closed) and if φ : UK is continuous and linear, then there exists an extension ψ : VK of φ which is also continuous and linear and which has the same norm 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 z is an element of V not in the closure of U, then there exists a continuous linear map ψ : VK with ψ(x) = 0 for all x in U, ψ(z) = 1, and ||ψ|| = ||z||−1.

Another version of Hahn-Banach theorem is known as Hahn-Banach separation theorem.[1][2] It has numerous uses in complex geometry.[3]

Theorem: Let V be a topological vector space over {\Bbb K}={\Bbb R} or {\Bbb C}, and A, B convex, non-empty subsets of V. Assume that A\cap B=\varnothing. Then

(i) If A is open, there exists a continuous linear map \lambda:\; V\mapsto {\Bbb K} and t\in {\Bbb R} such that

 Re \ \lambda(a) < t \leq Re\ \lambda(b)

for all a\in A, b \in B

(ii) If V is locally convex, A is compact, and B closed, there exists a continuous linear map \lambda:\; V\mapsto {\Bbb K} and s, t\in {\Bbb R} such that

 Re \ \lambda(a) < t < s < Re\ \lambda(b)

for all a\in A, b \in B.

As mentioned earlier, the axiom of choice implies the Hahn-Banach theorem. The converse is not true. One way to see that by noting that the ultrafilter lemma, which is strictly weaker than the axiom of choice, can be used to show the Hahn-Banach theorem, although the converse is not the case.

  • Michael Reed and Barry Simon, Functional Analysis, Section III.3. Academic Press, San Diego, 1980. ISBN 0-12-585050-6.

  1. ^ Klaus Thomsen, The Hahn-Banach separation theorem, Aarhus University, Advanced Analysis lecture notes
  2. ^ Gabriel Nagy, Real Analysis lecture notes
  3. ^ R. Harvey and H. B. Lawson, "An intrinsic characterisation of Kahler manifolds," Invent. Math 74 (1983) 169-198.
Advanced Search
Included Web Search Engines


Safe Search

close

Top Matching Results

Occasionally Search.com will highlight specialized results that are based on the context of your query. Examples of specialized results include specific links to news, images, or video.

Top Matching Results may highlight information from other Search.com pages, content from the CNET Network of sites, or third party content. The listings are based purely on relevance. Search.com does not receive payment for listings in this section but our partners that provide this data may get paid for listing these products.

Sponsored Links

This section contains paid listings which have been purchased by companies that want to have their sites appear for specific search terms and related content. These listings are administered, sorted and maintained by a third party and are not endorsed by Search.com.

Search Results

Search.com sends your search query to several search engines at one time and integrates the results into one list which has been sorted by relevance using Search.com's proprietary algorithm. You can customize the list of search engines included in your metasearch from the preferences.

The search engines that are used in your metasearch may allow companies to pay to have their Web sites included within the results. To view the Paid Inclusion policy for a specific search engine, please visit their Web site. Search.com does not accept payment or share revenue with any search engine partner for listings in this section.