# Questions tagged [hahn-banach-theorem]

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32
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### 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 ...

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### 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 $H\subseteq \mathbb R^n$
$\bullet$ ...

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### 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 ...

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### 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 $y^*$ ...

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### 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'...

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### 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 ...

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### 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: ...

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### 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 \...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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)...

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### 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 ...

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### 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\...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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") ...

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### $(1+\epsilon)$-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 ...

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### 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 $V$, ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...

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### 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 ...