The hahn-banach-theorem tag has no wiki summary.

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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 ...