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I have a definition of a Schauder basis but I’m unsure of it.

The definition I have is

A sequence $\{e_k : k \in \mathbb{N} \}$ in a normed space $(V, \| \cdot \| )$ is a Schauder basis if

  1. $\sum_{k=1}^{\infty } \alpha _k e_k =0 $ implies $\alpha _k =0 $ for all $k$.

  2. every $x \in V$ can be written in the form $x=\sum_{k=1}^{\infty } \alpha _k e_k $(i.e. $ \lim_{n \to \infty }\| (\sum_{k=1}^{n} \alpha _k e_k ) -x\|=0$

This definition to me seems to mean that the Schauder basis is countable.

However a theorem is ‘an infinite dimensional Hilbert space is separable if and only if it has a countable orthonormal basis’ .

This theorem seems to contradict that a Schauder basis has to be countable.

What’s the deal here?

Anonmath101
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2 Answers2

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Yes, Schauder bases are countable. At least this is how it's defined in Linear analysis by Bollobás (p. 37) and Topics in Banach Space Theory by Albiac and Kalton (Defn 1.1.1). For a separable Hilbert space, the orthonormal basis of a Hilbert space can be taken to be a Schauder basis.

Note that if a normed space admits a Schauder basis then it is necessarily separable. Indeed, the $\mathbb{Q}$-span of the Schauder basis is a countable dense set. Thus, a non-separable Hilbert space (i.e. one whose orthonormal basis is of uncountable cardinality) does not admit a Schauder basis.

Jose Avilez
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  • It was my (obviously incorrect) understanding that an orthonormal basis was a Schauder basis where the $e_k$ are orthonormal. What is an orthonormal basis? – Anonmath101 May 09 '22 at 22:44
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    @Ben In a Hilbert space $H$, an orthonormal basis is a set $(e_\alpha)$ such that every element in $x$ can be written as an infinite linear combination of elements in $(e_\alpha)$. Note that you can represent this as $x = \sum_{\alpha} \left\langle x, e_\alpha \right\rangle e_\alpha$ and only countably many of the summands will be non-zero. However, you may need uncountably many $e_\alpha$ to represent all the vectors in $H$. – Jose Avilez May 09 '22 at 23:21
  • I see, so basis in this sense is writing elements of the Hilbert space as a countable linear combination of elements $e_a $ – Anonmath101 May 09 '22 at 23:51
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Perhaps to clarify some accidental "glosses" here:

In some contexts, a Schauder basis $B=\{v_\alpha:\alpha\in A\}$ for a vector space $V$ is a (finitely linearly independent) subset of $V$ such that the finite linear combinations of $B$ are dense in $V$.

There is no compulsion that the index set be countable, although if $V$ is "separable" in the sense of having a countable, dense subset, this will be the case.

Yes, since a sum of positive real numbers can be finite only if the index set is countable, in a Hilbert space the expressions of elements as infinite sums of linear combinations of basis elements can only involve countably many. :)

But that does not preclude the existence of Hilbert spaces without countable (Hilbert-space) bases. :)

paul garrett
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  • How does this compare with the other answer here that says a Schauder basis is countable. Do different texts define it slightly differently? – Anonmath101 May 10 '22 at 16:39
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    @Ben, I don't know what to say about anyone's apparent requirement that a Schauder basis be countable... in part because there do exist Hilbert spaces requiring uncountable Hilbert-space bases (though these tend _not_ to arise in practice). "Defining" words in a fashion that makes it impossible to say certain things makes me uneasy. E.g., declaring that "group" means "finite group"... while, hm, there _are_ groups that are not finite? :) – paul garrett May 10 '22 at 16:52
  • And any of these Hilbert spaces with uncountable bases are not separable right? Due to the theorem above I posted – Anonmath101 May 10 '22 at 17:12
  • Right, Hilbert spaces with uncountable Hilbert space bases (such as $L^2$ of an uncountable set with counting measure) are not separable. They do not have dense countable subsets. – paul garrett May 10 '22 at 17:13
  • What is an orthonormal basis in this context? Just a Schauder basis where $E=\{e_a \} \subset V $ are orthonormal ? – Anonmath101 May 10 '22 at 17:33
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    That's what I'd think, but apparently some sources have other opinions. Bottom line: to know with certainty what anyone means by "Schauder-whatever basis" you have to ask them. Some risk that they'll be disdainful that you don't know their secret definition, but that can't be avoided. :) – paul garrett May 10 '22 at 17:36
  • I see. Thank you. That’s maths unfortunately sometimes – Anonmath101 May 10 '22 at 18:40
  • Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/136216/discussion-between-ben-and-paul-garrett). – Anonmath101 May 11 '22 at 12:38