From: "Ashok.R" Subject: Question on the Hahn banach theorem Date: Mon, 27 Sep 1999 16:38:31 -0400 Newsgroups: sci.math Keywords: Reconciling different versions Greetings. I have some confusion relating the geometric form and the extension form of the Hahn Banach theorem. The extension form basically says that given any bounded linear functional in a subspace M of a vector space, X then it is possible to extend the functional to the entire space without blowing up the norm. The geometric form says that given a convex C with a non empty interior set in a normed linear vector space X and a point x not in C, then there is a hyperplane containing x and not containing any interior point of C. Are these theorems one and the same? or do they mean totally different things? Is there a way to relate these two theorems? That is start from the geometric form and show that the extension form is true and vice versa? Any help is greatly appreciated. Thanks. Ashok.R ============================================================================== From: spamless@nil.nil Subject: Re: Question on the Hahn banach theorem Date: 27 Sep 1999 18:10:50 -0400 Newsgroups: sci.math Ashok.R wrote: > Is there a way to relate these two theorems? That is start from the > geometric form and show that the extension form is true and vice versa? Well, for a (semi)-norm, one can consider the convex set {x:||x||<=k} to get a convex set. For a convex set, C, one can consider ||x||=INF{w>0:x/w is in C} (note that as w-->infinity, x is eventually in C if C is a convex neighbourhood of the origin) ============================================================================== From: spamless@Nil.nil Subject: Re: Question on the Hahn banach theorem Date: 28 Sep 1999 10:15:52 -0400 Newsgroups: sci.math spamless@nil.nil wrote: > Well, for a (semi)-norm, one can consider the convex set {x:||x||<=k} to > get a convex set. (for Hahn Banach you only need ||ax||=a||x|| for a positive (useful in some separation theorems) and ||x+y||<=||x||+||y||) > For a convex set, C, one can consider ||x||=INF{w>0:x/w is in C} (note > that as w-->infinity, x is eventually in C if C is a convex neighbourhood > of the origin) If you want a semi-norm (instead of just positive homogeneity), you want a balanced convex set (ax in C whenever x is in C and |a|<=1) Note that the norm (well, it doesn't need a norm ... only homogeneous for positive multipliers) does not have to be the norm defining the topology on the vector space. If it IS the norm which defines the topology, then you get a continuous functional. ============================================================================== From: "David C. Ullrich" Subject: Re: Question on the Hahn banach theorem Date: Tue, 28 Sep 1999 10:35:34 -0500 Newsgroups: sci.math "Ashok.R" wrote: > Greetings. > > I have some confusion relating the geometric form and the extension form > of the Hahn Banach theorem. > > The extension form basically says that given any bounded linear > functional in a subspace M of a vector space, X then it is possible to > extend the functional to the entire space without blowing up the norm. > > The geometric form says that given a convex C with a non empty interior > set in a normed linear vector space X and a point x not in C, then there > is a hyperplane containing x and not containing any interior point of C. > > Are these theorems one and the same? No. > or do they mean totally different > things? No. > Is there a way to relate these two theorems? Yes. Are you going through some book? Does it contain any proofs? If, say, you were looking at Rudin "Functional Analysis" you might note there dozens of "Hahn-Banach" theorems in Chapter 3. They all follow from Theorem 3.2. [deletia -- djr] ============================================================================== From: schectex@math.vanderbilt.edu (Eric Schechter) Subject: Reply to "Question on the Hahn banach theorem" Date: 29 Sep 1999 01:57:40 -0400 Newsgroups: sci.math There are many different theorems in the literature that are called "The Hahn-Banach Theorem." (The use of the word "the" is unfortunate, of course.) Most of them are closely related, in one or both of the following ways: (1) One can be used to prove another, with only slight difficulty. (2) They are equivalent, when considered as weak forms of the Axiom of Choice. In other words: Conventional set theory is ZF + AC; most forms of "the Hahn Banach Theorem" can be proved from each other using just ZF. By the way, most textbooks use AC to prove HB, but it is known that HB is strictly weaker than AC. An introduction to these matters can be found in my book, "Handbook of Analysis and its Foundations." Among other things, my book contains 26 different versions of the Hahn-Banach Theorem, all equivalent in the sense of weak forms of AC, though the proofs of equivalence sometimes involve more than slight difficulty. The book also goes into more detail in explaining what is a weak form of AC. Of course, there are other books on this subject, but mine happens to be the first one that occurs to me. :-) http://www.math.vanderbilt.edu/~schectex/ccc/ ============================================================================== From: Brandsma Subject: Re: Hahn Banach, Axiom of Choice, seperable spaces Date: Thu, 22 Jul 1999 10:46:23 +0200 Newsgroups: sci.math helmut.zeisel@aon.at wrote: > In article <7n4kkp\$t3o\$1@nnrp1.deja.com>, > chri0562@my-deja.com wrote: > > > > > Are there any results in which spaces one can proof > > > the Hahn Banach Theorem without using Axiom of Choice? > > > > > > In particular, > > > can one proof that Hahn Banach holds in seperable spaces > > > without using Axiom of Choice? > > > > Yep. First do it when an element outside a subspace is added, > > induct on the set whose span is dense in the separable space > > and then use density. > > > > Details are far to ghastly to be included here. :) > > Thatīs fine. > > Please, can you point me to any written references? > > Thank you > > Helmut > > Sent via Deja.com http://www.deja.com/ > Share what you know. Learn what you don't. There is some confusion as to what we mean by Hahn Banach, I think. The version in Consequences of the Axiom of Choice (287) is the following: Let V be a separable normed linear space, and p (from V to R) be a subadditive and positively homogeneous map. (p(x+y) <= p(x) + p(y), p(tx) = tp(x) if t>=0) Assume that f is a linear map from a subspace S of V to R with f(x) <= p(x) for x in S. Then there is an extension f~ of f to V such that f(x) <= p(x) on V. This form implies the existence of a non-zero finitely additive measure that is 0 on finite sets (form 222), which is not provable in just ZF (follows from Pincus: the strength of the Hahn-Banach theorem, Bull AMS 78, 776-770 (1972), and Pincus doctoral dissertation) A version of H-B which is provable from ZF alone is the following (0 AM): Let V be a separable (topological) vector space, D a countable dense subset of V, and p as above. Also assume that p satisfies (*) for all epsilon>0 and all x in V there is a y in D such that both p(x-y) and p(y-x) are smaller than epsilon. (so this is satisfied if p = a norm on V, which is what some people here mean by H-B: they use "continuity" of p) Then there is an extension of f as above. ---- I meant the first version of H-B, and others mean (a version of) the second. I hope this clarifies it a bit, Henno Brandsma ============================================================================== From: hrubin@odds.stat.purdue.edu (Herman Rubin) Subject: Re: Hahn Banach, Axiom of Choice, seperable spaces Date: 21 Jul 1999 13:55:45 -0500 Newsgroups: sci.math In article <3795c55a.0@webpc1.vai.co.at>, Zeisel Helmut wrote: >Are there any results in which spaces one can proof >the Hahn Banach Theorem without using Axiom of Choice? >In particular, >can one proof that Hahn Banach holds in seperable spaces >without using Axiom of Choice? It depends on the definitions, and this applies for the full Hahn-Banach Theorem, not just on Banach spaces. But the proof is straightforward. So suppose that one has a linear space X with a linear subspace Y, and a linear functional f defined on Y, and a subadditive positively homogeneous functional g on X with f <= g on Y. Suppose also that there is a countable set {x_i: i = 1, ..., } of elements of X with the property that for all z in x and e > 0, there is a convex "neighborhood" A of 0 such that for all w in A, |g(w)| < e, and there is an i with x_i in z + A. Then f can be extended as usual. The proof consists in adding x_i one at a time; this does not involve choice. But as the x_i have the appropriate denseness property, this functional extends uniquely to all of X. Banach's proof from the Axiom of Choice is just this extended by transfinite induction. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399 hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558