Talk:Schauder basis

Complete vs. Total

The article currently states that "Every complete set of vectors is total, but the converse need not hold in an infinite-dimensional space." Total is defined, but complete is not. In what sense is complete being used here? Heil, in his "Basis Theory Primer," defines both complete and total as the linear span being dense in the space, or the closed linear span equalling the full space. Nym6433 (talk) 23:09, 29 November 2009 (UTC)

Me too, I wanted to ask about this. What is a complete set? Maybe we should delete this statement... --Txebixev (talk) 10:41, 11 October 2012 (UTC)

why countable?

Why does this article require Schauder bases to be countable? This probably makes it impossible for a nonseparable space to have a Schauder basis, whereas for example, $\ell ^{p}(X)$ has a natural basis with the same cardinality as X.

OK, actually, I think I know the reason: if X is uncountable, then modifications must be made to the definition of infinite sum. Although it is true that the space of finite linear combinations of basis elements is dense in $\ell ^{p}(X)$, the limit points cannot all be limits of sequences of finite partial sums. Rather, we would need to define summation of uncountably many terms as the limit of the uncountable net of finite partial sums. This in itself doesn't really entail a modification of the definition though. We would just have to remove the word "sequence" from the article. You're still taking the limit over all partial sums, you just can't use the word "sequence" to describe such a limit.

So maybe the only reason that Schauder doesn't allow nonseparable spaces in his definition is that he thinks that people who deal with nonseparable spaces are miscreants and deviants? -lethe ^{talk +} 19:11, 14 April 2006 (UTC)

I don't know the reason - but the definition does look that way really. Note that the famous "basis problem", posed maybe by Schauder himself (I don't know) and solved by Per Enflo, was whether evry countable space has a (Schauder) basis. anon 18:14, 4 June 2007 User:68.34.29.30

Tradition. Schauder and Mazur worked in the 1930's. It took 40 years for Enflo to come up with the counterexample -- so its not easy. Also, every seperable space embeds isometrically into C[0,1], and so Enflo's result has the weird corrollary that C[0,1] has a closed subspace that does not have a basis. So things seem weird enough when its just countable.linas (talk) 14:57, 26 October 2008 (UTC)

Just saying "limit of the net of partial sums" would give a different concept. Because even for a standard countable Schauder basis we're not necessarily talking about that net! In general the ordering of the basis elements is crucial; the specific partial sums wrt that ordering converge as desired, but the net of arbitrary finite sums does not. 76.242.183.215 (talk) 23:38, 23 November 2014 (UTC) David C. Ullrich

Separable spaces without bases

Can someone add an example or at least give a heuristic justification for how such spaces can exist? 203.167.251.186 (talk) 20:31, 21 January 2009 (UTC)

There is already a piece of answer in this page: It took 40 years for Enflo to come up with the counterexample -- so its not easy. Actually Enflo gave a counterexample to the approximation problem, not directly to the basis problem. —Preceding unsigned comment added by Bdmy (talk • contribs) 22:29, 21 January 2009 (UTC) The results in these directions are often proved by random methods, by selecting for example a random 2n-dimensional subspace F of R⁴ⁿ, equipped with a $\ell ^{p}$-norm for p ≠ 2; then there is a high probability that the normed space F behaves badly, for example the rank-n projections on F have large norm, tending to infinity with n; try to look in the rather recent Handbook of Banach space theory. There are papers there by S. Szarek or N. Tomczak that might contain answers to your question. But I don't think that there is a simple explanation. --Bdmy (talk) 22:52, 21 January 2009 (UTC)

thank you for your answer, that sounds esoteric indeed! I will look this reference up. —Preceding unsigned comment added by 203.167.251.186 (talk) 23:43, 21 January 2009 (UTC)

Schauder basis versus topological basis

"It follows from the Banach–Steinhaus theorem that the linear mappings {P_n} defined by

$v=\sum _{k=0}^{\infty }\alpha _{k}b_{k}\ \ {\overset {\textstyle P_{n}}{\longrightarrow }}\ \ P_{n}(v)=\sum _{k=0}^{n}\alpha _{k}b_{k}$

are uniformly bounded by some constant C."

No, this is incorrect, since at this moment we do not know yet that $P_{n}$ are continuous.

Looking at a survey

C. W. McARTHUR, "DEVELOPMENTS IN SCHAUDER BASIS THEORY", BULLETIN OF THE AMERICAN MATHEMATICAL SOCIETY Volume 78, Number 6, November 1972 [1]

I see on page 878 that the notion called "Schauder basis" in our article is called "(topological) basis" by McArthur. A Schauder basis, according to McArthur, is a special case of a biorthogonal system. The distinction is that, according to McArthur, coefficient functionals $v\mapsto \alpha _{n}$ are continuous by definition for a Schauder basis, but not for a (topological) basis.

"Banach [6, p. 111] showed that the coefficient functionals of a basis for a Banach space are necessarily continuous, i.e., a basis for a Banach space is a Schauder basis." (McArthur, p. 878)

[6]. S. Banach, Théorie des opérations linéaires, Monografie Mat., PWN, Warsaw, 1932; reprint, Chelsea, New York, 1955. MR 17, 175.

Thus, restricing ourselves to Banach spaces, we may treat these two notions of basis as a single notion. And still, the phrase "It follows from the Banach–Steinhaus theorem..." is incorrect; an additional argument (provided by Banach) is needed, and is far not self-evident. Or else we should replace the definition.

Boris Tsirelson (talk) 07:17, 23 October 2014 (UTC)

I was about to change that, decided to look her first. Yes, it's wrong as stated. 76.242.183.215 (talk) 23:40, 23 November 2014 (UTC) David C. Ullrich.

Yes, please! Boris Tsirelson (talk) 09:25, 24 November 2014 (UTC)