Tagged Questions
4
votes
1answer
203 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
95 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 ...
2
votes
2answers
203 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 ...
27
votes
0answers
458 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
258 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
98 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 ...
84
votes
1answer
2k 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 ...
21
votes
4answers
708 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 ...
4
votes
4answers
498 views
Is there any motivation for Zorn's Lemma?
I have been reading Kreszig'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 is ...
30
votes
2answers
1k 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
264 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 ...
8
votes
1answer
629 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 ...
