# Questions tagged [hahn-banach-theorem]

The tag has no usage guidance.

20 questions
Filter by
Sorted by
Tagged with
82 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 ...
165 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 ...
160 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 ...
49 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 ...
295 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 ...
123 views

389 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 ...
371 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 ...
433 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 ...
680 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 ...
583 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 ...
547 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 ...
3k 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") ...
754 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 ...
520 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 $V$, ...
469 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 ...
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 ...
### 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 ...