From: israel@math.ubc.ca (Robert Israel) Newsgroups: sci.math Subject: Re: Schauder basis Date: 4 Sep 1996 00:19:37 GMT In article <50i24r\$k9e\$1@mhafn.production.compuserve.com>, Marc Wermann <101536.3500@CompuServe.COM> writes: |> Can anyone tell me about the difference of a so called Schauder |> basis in contrast to a "usual" basis (as understood in linear |> algebra) ?? The difference occurs in infinite-dimensional Banach spaces. In that context, the "usual" basis is called a Hamel basis: it is a collection of vectors {v_i}, i in some index set, such that every vector in the space can be written uniquely as a linear combination of some finite set of v_i's. For a Schauder basis, on the other hand, you allow an infinite "linear combination", where the infinite sum is understood as an infinite series, converging in the norm topology of the Banach space. A Banach space with a Schauder basis must be separable. Hamel bases always exist, but can't be constructed without the Axiom of Choice. Concrete examples of Schauder bases, on the other hand, are known for most of the "natural" separable Banach spaces. Per Enflo found a separable Banach space that doesn't have a Schauder basis. Robert Israel israel@math.ubc.ca Department of Mathematics (604) 822-3629 University of British Columbia fax 822-6074 Vancouver, BC, Canada V6T 1Y4 ============================================================================== From: edgar@math.ohio-state.edu (G. A. Edgar) Newsgroups: sci.math Subject: Re: Basis for continuous functions on Reals Date: Thu, 29 Oct 1998 13:11:16 -0500 In article <3637B35F.2050C97@kentroad.demon.co.uk>, David Chan wrote: > Hi, > > I was wondering if anyone knows if an explicit basis for the vector > space of continuous functions over R has been computed. > As others have said, "explicit" Hamel basis is not possible. For Schauder basis, you need to say what kind of convergence is allowed. For the continuous functions on an interval [a,b] with uniform norm such a computation was done by Schauder himself (and the term "Schauder basis" is named for him). Essentially you can use the indefinite integrals of the Rademacher functions on [a,b] (plus the constant 1). -- Gerald A. Edgar edgar@math.ohio-state.edu ============================================================================== From: ullrich@math.okstate.edu Newsgroups: sci.math Subject: Re: Basis for continuous functions on Reals Date: Fri, 30 Oct 1998 16:09:38 GMT In article , edgar@math.ohio-state.edu (G. A. Edgar) wrote: > In article <3637B35F.2050C97@kentroad.demon.co.uk>, David Chan > wrote: > > > Hi, > > > > I was wondering if anyone knows if an explicit basis for the vector > > space of continuous functions over R has been computed. > > > > As others have said, "explicit" Hamel basis is not possible. > > For Schauder basis, you need to say what kind of convergence > is allowed. For the continuous functions on an interval > [a,b] with uniform norm such a computation was done by Schauder himself > (and the term "Schauder basis" is named for him). Essentially you can > use the indefinite integrals of the Rademacher functions on [a,b] > (plus the constant 1). > -- > Gerald A. Edgar edgar@math.ohio-state.edu Actually it's the integrals of the Haar functions - the Rademacher functions are not nearly enough. (Roughly, the Rademacher functions are to the Haar functions as lacunary trigonometrc series are to general trigonometric series...) David C. Ullrich -----------== Posted via Deja News, The Discussion Network ==---------- http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own ============================================================================== From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Newsgroups: sci.math Subject: Re: Basis for continuous functions on Reals Date: 30 Oct 1998 10:48:39 -0500 In article <3637B35F.2050C97@kentroad.demon.co.uk>, David Chan wrote: >Hi, > >I was wondering if anyone knows if an explicit basis for the vector >space of continuous functions over R has been computed. > >Thanks, >David Chan (dpchan@kentroad.demon.co.uk) Algebraic, also called Hamel, basis, is known to exist (in every vector space) only with the help of the Axiom of Choice or one of its restrictions. There is perhaps, I don't know for sure, a model of set theory in which AC fails so emphatically that this Hamel basis will not exist. If this is the case, forget about any explicit description. For the space of all continuous functions on [0,1], normed by the supremum norm, there is an explicit basis: the binary roof functions. For every n=0, 1, 2, ..., divide [0,1] into 2^n parts, and consider the piecewise linear functions with graphs \_____, /\_____, _/\_____, and so on, until ____/ with breakpoints at the adjacent points of the subdivision. Then take the union of all such collection of these roof functions. A proof that this is a basis can be arranged so that it gives a recursive procedure for finding the coefficients. The expansions will in general be infinite and uniformly convergent. If you mean the normed space of all continuous bounded functions on all of R, this space is not separable, so that there is no sequential basis. If you mean the space of all continuous functions defined on all of R, it is not normable (although it is metrizable). You can manipulate the binary roof functions to obtain a basis the expansions into which will be locally uniformly convergent. If you had still another situation in mind, include the specifics in your question. Cheers, ZVK (Slavek).