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I have seen a simple proof that no Banach space over $\mathbb{R}$ can be of countably infinite dimension. However since the space of all square integrable functions on the unit interval forms a Hilbert space, and all Hilbert Spaces are Banach, this space must not be of countable dimension. However we know that each point has a unique decomposition as a sum complex exponentials $e^{2n\pi}$ were $n\in\mathbb{Z}$. Thus these complex exponentials form a basis. But since there is a countable number of such exponentials there must be a contradiction. Where is that contradiction?

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    $\begingroup$ Basis $\ne$ Schauder basis. You’re talking about a countable Schauder basis for the space; it’s not a basis, since you allow infinite sums. $\endgroup$ – Brian M. Scott Sep 27 '12 at 23:10
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    $\begingroup$ Better to say "Hamel basis $\ne$ Schauder basis". A mathematician will use "basis" by itself for his favorite one. $\endgroup$ – GEdgar Jan 5 '16 at 12:45
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The unique decomposition is as an infinite combination (series) of the complex characters, while a linear basis is a set of vectors such that any element of the space can be expressed as a finite combination.

The set $B$ such that any element of the space $V\supseteq B$ can be uniquely expressed as an infinite combination of elements of $B$ is called Schauder basis, as indicated by Brian M. Scott in the comment, as opposed to linear basis, usually called Hamel basis in functional analytic contexts.

For any cardinal number $\kappa$, there is a Banach space with a Schauder basis of cardinality $\kappa$ -- the Hilbert space of Hilbert dimension $\kappa$, for instance, but none of them have linear dimension $\aleph_0$.

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