3
votes
1answer
74 views

Does existence of a non-continuous linear functional depend on Axiom of Choice?

Well, it is easy to construct a non-continuous linear functional on an arbitrary infinite-dimensional vector space (assuming Choice, and taking a basis etc.). I think it is intuitive to say that: ...
0
votes
2answers
51 views

Is this a basis for the dual space?

There is an example on Wikipedia that I don't understand and I'd appreciate some help. They define $\mathbb R^\infty$ to be the space of all sequences that are zero except for finitely many indexes. ...
2
votes
0answers
186 views

Hahn-Banach via Hamel Basis

my question for tonight: Is there a proof for Hahn-Banach using a Hamel Basis? I know, the proof for existence of Hamel Bases uses already Axiom of Choice, but I'd like to apply this without refering ...
4
votes
2answers
72 views

Does the dualizing process on vector spaces necessarily terminate?

It's well-known (assuming the axiom of choice) that the inclusion $\ell^1 \subset (\ell^1)^{**}$ is proper as a simple corollary of the Hahn-Banach theorem. But is this the end of the dualizing ...
5
votes
1answer
501 views

On every infinite-dimensional Banach space there exists a discontinuous linear functional.

On every infinite-dimensional Banach space there exists a discontinuous linear functional. Assuming the axiom of choice, every vector space has a basis. With an infinite basis, I can define on a ...
1
vote
1answer
157 views

Is there a clever way to avoid choice in Riesz Representation Theorem?

Rudin RCA p.43 Riesz Representation for LCH: Let $X$ be a locally compact Hausdorff space, and let $\Lambda$ be a positive linear functional on $C_c(X)$. Then there exists a $\sigma$-algebra ...
3
votes
2answers
669 views

Every Hilbert space has an orthonomal basis - using Zorn's Lemma

The problem is to prove that every Hilbert space has a orthonormal basis. We are given Zorn's Lemma, which is taken as an axiom of set theory: Lemma If X is a nonempty partially ordered set with the ...
49
votes
2answers
1k views

Explicit norm on $\mathcal{C}^0(\mathbb{R},\mathbb{R})$

Do you know an explicit norm on $\mathcal{C}^0(\mathbb{R},\mathbb{R})$? Using the axiom of choice, every vector space admits a norm but have you an explicit formula on ...
10
votes
1answer
505 views

A Hamel basis for $l^{\,p}$?

I am looking for an explicit example for a Hamel basis for $l^{\,p}$?. As we know that for a Banach space a Hamel basis has either finite or uncountably infinite cardinality and for such a basis one ...
4
votes
1answer
177 views

Is “$K$ convex + absorbing $\not\Rightarrow$ $0\in \mathrm{Int }\, K$” dependent on AC?

I have encountered the following problem in Dirk Werner's "Funktionalanalysis" (English translation by me): Definition: A convex set $K\subset X$ is called absorbing, if given $x\in X$ there exists ...
103
votes
1answer
3k views

Does the open mapping theorem imply the Baire category theorem?

A nice observation by C.E. Blair1, 2, 3 shows that the Baire category theorem for complete metric spaces is equivalent to the axiom of (countable) dependent choice. On the other hand, the three ...
23
votes
4answers
950 views

Is Banach-Alaoglu equivalent to AC?

The Banach-Alaoglu theorem is well-known. It states that the closed unit ball in the dual space of a normed space is $\text{wk}^*$-compact. The proof relies heavily on Tychonoff's theorem. As I have ...
7
votes
4answers
731 views

Is there any motivation for Zorn's Lemma?

I have been reading Kreyszig's book on functional analysis, where it uses Zorn's lemma to prove the Hahn Banach theorem. However I don't quite get what Zorn's lemma is saying. I understand that it ...
33
votes
2answers
2k views

Is there an explicit isomorphism between $L^\infty[0,1]$ and $\ell^\infty$?

Is there an explicit isomorphism between $L^\infty[0,1]$ and $\ell^\infty$? In some sense, this is a follow-up to my answer to this question where the non-isomorphism between the spaces $L^r$ ...
4
votes
1answer
341 views

$\ell^1$ vs. continuous dual of $\ell^{\infty}$ in ZF+AD

Let the base field be the real numbers or the complex numbers (I don't think it will matter). Let $(\ell^{\infty})'$ be the continuous dual of the Banach space $\ell^{\infty}$. Let $\: f : \ell^1 ...
9
votes
1answer
827 views

Nonnegative linear functionals over $l^\infty$

My purpose is a clarification of the role of the axiom of choice in constructing limits for bounded sequences. Namely, we want a linear functional of norm 1 defined on the space of all bounded complex ...