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We know that every vector space has a Hamel basis and also every normed space need not have a Schauder basis. As the normed space $l^\infty$ is not Separable so can't have the Schauder basis, but on the other side $l^\infty$ is also a vector space so what will be the Hamel basis of $l^\infty$?

Martin Sleziak
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Prince Khan
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2 Answers2

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An indication that you cannot expect explicit example is that this cannot be proven in ZF.

In an infinite-dimensional normed space, existence of Hamel basis implies existence of discontinuous linear functional. And it is consistent with ZF that there are no discontinuous linear functionals on $\ell_\infty$, see here: On every infinite-dimensional Banach space there exists a discontinuous linear functional. (Perhaps there is a more straightforward argument, but this is is what I was able to come up with.)

However, if you only want proof that $\ell_\infty$ has Hamel basis, it is as standard application of Zorn's lemma. In fact, every linearly independent set is contained in a basis. (And for this proof there is nothing special about $\ell_\infty$ - it works for any vector space exactly in the same way.)

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Martin Sleziak
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    Does the Schauder basis of a finite dimensional normed space X is the same as the Hamel basis of X?. For example, the Schauder basis of $R^{n}$? – Prince Khan Dec 09 '16 at 08:25
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    If Schauder basis is defined for finite-dimensional spaces at all, them it will be the same as Hamel basis. (It depends on whether in definition of Schauder basis you allow it to be finite countable or you require infinite countable set/sequence. I am not sure, but I'd guess both possibilities appear in literature.) If you are interested in differences between Hamel and Schauder basis, this might be useful: What is the difference between a Hamel basis and a Schauder basis? – Martin Sleziak Dec 09 '16 at 08:48
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    thanks that was really an informative link to me. How can we define Schauder basis for a finite dimensional normed space as the schauder basis must include the infinite linear combinations? – Prince Khan Dec 09 '16 at 17:58
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    You can define it by cases and say that for finite-dimensional you define Schauder base as the usual base. This seems to be how they do it in Banach Space Theory by Fabian, Habala et al., see the remarks right after Definition 4.6. Another possibility would be to formulate the definition in such way that we say that it is either finite or countable sequence or that Schauder basis is a countable set. (Countable sets include finite sets.) My impression is that when studying Schauder basis, fin.-dim. case is not that interesting. – Martin Sleziak Dec 10 '16 at 07:25
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Even for separable infinite-dimensional Banach spaces a Hamel basis is not something we can exhibit concretely. It lives out in Axiom-of-Choice land. Even more so for $\ell^\infty$.

Robert Israel
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    So what we can say about the space $l^\infty$? – Prince Khan Dec 09 '16 at 06:18
  • Why "even more so for $\ell^\infty$"? – Vim Dec 09 '16 at 06:18
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    @PrinceKhan it has an uncountable Hamel basis, which doesn't have a good representation. – Vim Dec 09 '16 at 06:20
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    @Vim how can we prove that $(e_{n})=((0,0,0,...0,1,0,0,...))$ is not the basis of the space $l^\infty$? – Prince Khan Dec 09 '16 at 06:22
  • I would like to go to Axiom-of-Choice land... – copper.hat Dec 09 '16 at 06:23
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    @PrinceKhan consider $(1,1,1,\cdots)\in \ell^\infty$ which cannot be finitely spanned by $(e_n)_n$. – Vim Dec 09 '16 at 06:24
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    @Vim thats so nice of you.... thanks to all of you – Prince Khan Dec 09 '16 at 06:25
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    @PrinceKhan this has nothing to do with convergence. The only reason is that $(1,1,\cdots)$ has infinitely many $1$s and you cannot use only finitely many $e_n$ to add up to it. – Vim Dec 09 '16 at 06:54
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    @Vim check please whether I am right? The sequence $(1,1,1,...) \in l^\infty$ and as the series $\sum || e_{n} ||=1+1+1+1...)$.i.e the series $\sum e_{n}$ is not absolutely convergent so $\sum e_{n}$ is not convergent as the space $l^\infty$ is Banach space. Therefore we can say that there does not any sequence of scalars $(x_{n})$ such that $(1,1,1,...)=\sum x_{n}e_{n}$ – Prince Khan Dec 09 '16 at 06:57
  • @Vim! Thus we can say that $(e_{n})$ is not the Schauder basis of $l^\infty$. – Prince Khan Dec 09 '16 at 07:00
  • @PrinceKhan It is. A Schauder basis is not the same as a Hamel basis. – Vim Dec 09 '16 at 07:12
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    @Vim! I am talking about Schauder basis, The space $l^\infty$ is not separable so it has no Schauder basis. Therefore $(e_{n})$ should not be the Schauder basis of $l^\infty$. Am I right? – Prince Khan Dec 09 '16 at 07:24
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    @PrinceKhan Oh sure.. sorry I made a mistake. You are right it is not a Schaefer basis either. – Vim Dec 09 '16 at 07:25
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    But still there is a bit confusion @Vim. Note that, $(e_{n})$ is a Schauder basis of a normed space X iff there is a unique sequence of scalars $(x_{n})$ such that every $x \in X$ can be written as $x=\sum x_{n}e_{n}$. Now note that $(1,1,1....)=\sum 1\times e_{n})$ satisfies above definition, therefore we should have to prove in a different way that $e_{n})$ is not a Schauder basis of $l^\infty$. (In short I am still failed to show that $(e_{n})$ is not the Schauder basis – Prince Khan Dec 09 '16 at 07:32
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    $(1,1,...)$ is not $\sum e_n$. This might be true in a pointwise sense. We would need it however in the $\ell^\infty$ sense where this is wrong. – Lukas Betz Dec 09 '16 at 07:36
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    @PrinceKhan well to prove it's not a Schauder basis $(1,1,\cdots)$ already suffices. Note that for any $n$, $|\sum_{k=1}^n c_ke_k-(1,1,\cdots)|_\infty\ge1> 0.$ So it's impossible to approach $(1,1,\cdots)$ using elements finitely spanned by ${e_n}$. – Vim Dec 09 '16 at 07:39
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    Yes I got it now completely... Thanks to both of you @Vim and LeBtz – Prince Khan Dec 09 '16 at 07:41