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Questions tagged [hahn-banach-theorem]

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Example of space which is weak Hahn-Banach smooth but not Hahn-Banach smooth

A Banach space $X$ is said to be Hahn-Banach smooth if every linear functional on $X$ has a unique norm-preserving extension over $X^{**}$. Weak Hahn-Banach smoothness is what if the above condition ...
Tanmoy Paul's user avatar
0 votes
0 answers
81 views

Is a bounded measurable convex function above its interior lower semi-continuous convex envelope?

Let $E$ be a locally convex topological vector space, let $C$ be a convex set which matches the closure of its relative interior $\mathring C=\{ x\in C : \forall y\in C,\exists z\in C,~x\in\mathopen]y,...
P. Quinton's user avatar
1 vote
1 answer
72 views

Is this notion of being "fully" convex closed under set addition?

While reading through "Linear Operators: General theory" by "Jacob T. Schwartz", reading the corollary to II.10.1 which states that for a compact convex subset $C$ of some ...
P. Quinton's user avatar
6 votes
3 answers
551 views

Hahn-Banach Theorem for convex polytopes and their supporting hyperplanes

A polytope in $\mathbb R^n$ is the convex hull of a nonempty finite set in $\mathbb R^n$. Let $C$ be a polytope in $\mathbb R^n$. We shall say that a hyperplane $H\subseteq \mathbb R^n$ $\bullet$ ...
Taras Banakh's user avatar
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4 votes
0 answers
246 views

A question regarding the Hahn-Banach theorem and Banach limits

Set theorists typically prove the existence of Banach limits (EBL) using the Ultrafilter Theorem or, its equivalent, the Boolean Prime Ideal Theorem (BPI). Analysts, on the other hand, typically prove ...
Philip Ehrlich's user avatar
2 votes
0 answers
168 views

On weak Hahn-Banach smoothness

Let us recall Phelp's property-$U$: A subspace $Y\subset X$ is said to have property-$U$ if every $y^*\in Y^*$ has unique norm preserving extension over $X$. $Y$ is weak Hahn-Banach smooth if $y^*$ ...
Tanmoy Paul's user avatar
1 vote
1 answer
203 views

Hyperplane separation of a concave functional and a point, in domain theory

Problem: Let $D$ be an $\omega$-BC domain, and $[D\to[0,\infty]]$ be the space of Scott-continuous nonnegative functions on $D$, equipped with the obvious ordering and the Scott-topology. EDIT: I don'...
Alex Appel's user avatar
13 votes
0 answers
490 views

Does Hahn-Banach for $\ell^\infty$ imply the existence of a non-measurable set?

Working over ZF but without the Axiom of Choice (AC), assume that the Hahn–Banach Theorem holds for $\ell^\infty$. Does it follow that there exists a set of real numbers that is not Lebesgue ...
Timothy Chow's user avatar
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5 votes
1 answer
500 views

How to apply Hahn-Banach to the convex hull?

I am trying to understand the proof of Lemma 4.1.2 in Michel Talagrand's publication from 1995 on concentration inequalities (see below for the precise question statement): A bit of context: ...
Maximilian Janisch's user avatar
6 votes
1 answer
692 views

Do multiplicative Banach limits exist?

Let $(D, \succeq)$ be a directed set, and let $B$ be the space of real-valued bounded functions on $D$. A Banach limit $\ell$ on $D$ is a linear functional that satisfies $$\sup_{d \in D} \inf_{c \...
aduh's user avatar
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2 votes
1 answer
122 views

When does the map from a normed vector cone to its double dual preserve norms?

If $V$ is a normed vector space then the natural map from $V$ to its double dual $V''$ is norm-preserving as follows from Hahn-Banach theorem. This is well-known. Now assume that P is just a vector ...
Sergey Slavnov's user avatar
2 votes
1 answer
982 views

Are closed convex subsets of a Banach space weakly closed without the axiom of choice?

It is a well-known fact that closed convex sets in Banach spaces are weakly closed. The common proof is based on the Hahn-Banach theorem that uses the axiom of choice. Is there any proof of this fact ...
Jan's user avatar
  • 23
0 votes
1 answer
208 views

Can a hyperplane be contained in a subspace?

Suppose $Y$ is a subspace of a normed linear space $X$ and let $y\in S_Y, y^*\in S_{Y^*}$ such that $y^*(y)=1$, where $S_Y$ denotes the closed unit sphere in $Y$. My question is the following: Is it ...
Anupam's user avatar
  • 585
5 votes
1 answer
309 views

Hahn-Banach smoothness of $Y^{**}$ in $X^{**}$

A subspace $Y$ of a Banach space is said to be Hahn-Banach smooth if every $f\in Y^*$ has unique norm preserving extension to whole $X$. This notion is related to many other geometric properties of ...
Tanmoy Paul's user avatar
3 votes
1 answer
98 views

Separation of Infinite-Dimensional Salient Convex Cones

Let X be the set of all summable sequences of reals endowed with the $l^1$ norm. That is, two elements of x are $a=(a_1,a_2,....)$ and $b=(b_1,b_2,...)$ and $d(a,b) = \sum_n |a_n-b_n|$. In this set ...
Tychonoff's user avatar
20 votes
0 answers
460 views

Hahn-Banach and the "Axiom of Probabilistic Choice"

Stipulate that the Axiom of Probabilistic Choice (APC) says that for every collection $\{ A_i : i \in I \}$ of non-empty sets, there is a function on $I$ that assigns to $i$ a finitely-additive ...
Alexander Pruss's user avatar
2 votes
1 answer
159 views

induced map on state spaces

A $*$-homomorphism $f:A\to B$ between C*-algebras is called non-degenerate if $f(A)B=B$. I guess that I can prove that a non-degenerate *-homomorphism always induces a map on state spaces $f^\ast:S(B)...
Philip's user avatar
  • 23
8 votes
1 answer
975 views

Equivalence of the Banach–Tarski paradox

I am working on the Banach–Tarski paradox and the fact that the Hahn–Banach theorem implies that paradox. The proof involves the equivalence of the Hahn–Banach theorem and the fact that for every ...
bdt's user avatar
  • 81
33 votes
4 answers
2k views

Hahn-Banach theorem with convex majorant

At least 99% of books on functional analysis state and prove the Hahn-Banach theorem in the following form: Let $p:X\to \mathbb R$ be sublinear on a real vector space, $L$ a subspace of $X$, and $f:L\...
Jochen Wengenroth's user avatar
4 votes
2 answers
519 views

Maximal cones and lexicographic orderings

Let $V$ be a real vector space. It is well known that, given a totally ordered basis of $V$ (say $(b_i)_{i\in I}$ where $I,<$ is totally ordered), $V$ is totally ordered by the lexicographic ...
Duchamp Gérard H. E.'s user avatar
5 votes
1 answer
519 views

Hahn Banach type extension of a Lipschitz map

The problem that I posted was a much generalized form of what I had in my mind. All I want to know the literature of Hahn-Banach type extension of Lipschitz map. I know only about the result by ...
Tanmoy Paul's user avatar
2 votes
1 answer
583 views

A question regarding the Hahn-Banach theorem

Wikipedia states that, in $ZF$, the Axiom of Choice ($AC$) implies the Hahn-Banach theorem, but that the Hahn-Banach theorem does not imply $AC$. It also states that in $ZF$, the Hahn-Banach theorem ...
Thomas Benjamin's user avatar
10 votes
0 answers
759 views

Full conditional probabilities and versions of AC?

A probability is a finitely additive measure on a boolean algebra with total measure $1$. A function $P:\scr B \times (\scr B - \{ 0 \})$ is a full conditional probability on $\scr B$ (for a boolean ...
15 votes
1 answer
778 views

Does ZF imply a weak version of Hahn-Banach?

I have encountered this when I was thinking about differentiability in Banach spaces. There, for $x\in X$ we usually need functionals $u\in X^*$ such that $|u|=1$ and $u(x)=|x|$. This is a simple ...
Mohammad Safdari's user avatar
4 votes
1 answer
767 views

binary intersection property in finite dimension

I have recently discovered a survey article called "The Hahn-Banach Theorem: The Life and Times" by Narici and Beckenstein and I have read it with great amusement: I was particularly fascinated by ...
Delio Mugnolo's user avatar
10 votes
2 answers
5k views

Direct proof of the separation theorem of Hahn-Banach

The "extension" (or "analytic") form of the theorem of Hahn-Banach has a natural and yet elegant proof. In just any textbook I have ever seen, it is proved first; the "separation" (or "geometric") ...
Delio Mugnolo's user avatar
17 votes
1 answer
911 views

$(1+\epsilon)$-injective Banach spaces, complex scalars

It is well known that a real Banach space which is $(1+\epsilon)$-injective for every $\epsilon >0$ is already 1-injective (Lindenstrauss Memoirs AMS, 1964, download here). Using common ...
Fred Dashiell's user avatar
4 votes
1 answer
807 views

Hahn-Banach restricted to a pre-dual

If $V$ is a locally convex topological space, the Hahn-Banach theorem shows that a continuous linear functional on a closed subspace can be extended to a continuous linear functional on all of $V$, ...
Igor Khavkine's user avatar
3 votes
1 answer
822 views

Extension of equivalent norms

Let $(X,||\cdot||_1)$ be a normed space and $Y$ a linear subspace of $X$. Let $||\cdot||_2$ be a norm on $X$ which is equivalent to $||\cdot||_1$ on $Y$. Does there exist a norm on $X$ that coincides ...
Max's user avatar
  • 33
4 votes
0 answers
94 views

Algebraic conditions of separability

Let $X$ be a real vector space (without any norm), and $Y$ be a convex subset of $X$, $0\notin Y$. The goal is to find a hyperplane $L$ passing through 0 such that $Y$ lies in a closed halfspace ...
Fedor Petrov's user avatar
4 votes
0 answers
239 views

When separation in $L^1$ is possible?

Let $A$, $B$ be disjoint convex closed subsets of the Banach space $L^1[0,1]$. Assume additionally that $A$ is bounded and $A$, $B$ are closed under convergence in measure. Then there exists a closed ...
Fedor Petrov's user avatar
23 votes
5 answers
6k views

Hahn-Banach without Choice

The standard proof of the Hahn-Banach theorem makes use of Zorn's lemma. I hear that, however, Hahn-Banach is strictly weaker than Choice. A quick search leads to many sources stating that Hahn-Banach ...
Mark Kim-Mulgrew's user avatar
66 votes
7 answers
10k views

Why is the Hahn-Banach theorem so important?

Every time I hear it mentioned it is praised in the highest possible terms, and I remember one of my old lecturers saying that it is one of the 3 most important theorems in analysis. Yet the only ...
teil's user avatar
  • 4,341
7 votes
2 answers
2k views

What is the "Krein-Milman theorem for cones"?

Update: The question is completely answered. I had overlooked a reduction to the self-adjoint case, and the latter can be proved using a Hahn-Banach separation theorem. Thanks to Matthew Daws for ...
Jonas Meyer's user avatar
  • 7,329
45 votes
7 answers
9k views

What's an example of a space that needs the Hahn-Banach Theorem?

The Hahn-Banach theorem is rightly seen as one of the Big Theorems in functional analysis. Indeed, it can be said to be where functional analysis really starts. But as it's one of those "there ...
Andrew Stacey's user avatar