The space $C[a,b]$ , space of all real valued continuous functions on $[a,b]$ is an infinite dimensional vector space over the field $\Bbb R$. As every vector space over a field has a basis so definitely $C[a,b]$ has a basis. I want to know a basis of $C[a,b]$. How I can find a basis of it ?
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Fourier series? – BAI Nov 04 '17 at 14:48
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@BAI What do you mean ? – Empty Nov 04 '17 at 14:50
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1A (Hamel) basis cannot be described explicitly. For a Schauder basis you need to specify a norm. – egreg Nov 04 '17 at 14:52
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1@egreg Ok. So if $\sup$ norm is given then what's the Schauder basis ? – Empty Nov 04 '17 at 14:54
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I'm not getting why negative vote ? – Empty Nov 04 '17 at 15:05
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@S717717 I don't know why either. This is a good question. Perhaps it is because it is a bit unclear what you are asking for (Hamel or Shauder basis?) and you have added a strange algebra-precalculus tag. – nullUser Nov 04 '17 at 15:12
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@BAI No, Fourier series do not give a (Schauder) basis for $C([a,b])$. – David C. Ullrich Nov 04 '17 at 15:48
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@DavidC.Ullrich can you say something about why not? At least in $L^2$ topology, I expect trig functions to be a Schauder basis per Fourier series. – ziggurism Nov 04 '17 at 16:01
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@ziggurism Yes of course trig functions are a basis for $L^2$. But not for $C([a,b])$. The Fourier series of a continuous function need not even converge pointwise, much less converge uniformly to $f$. – David C. Ullrich Nov 04 '17 at 23:26
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@DavidC.Ullrich how is the compatible with the Stone-Weierstrass theorem? – ziggurism Nov 04 '17 at 23:36
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@ziggurism S-W, or just W, shows that the trigonometric polynomials are dense. So for every $f\in C([a,b])$ there exists a sequence $P_n$ of trig polys converging uniformly to $f$. That doesn't say anything about a basis; in particular $P_n$ is not the $n$th partial sum of the Fourier series. – David C. Ullrich Nov 04 '17 at 23:45
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@DavidC.Ullrich right, thanks – ziggurism Nov 04 '17 at 23:48
2 Answers
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You mention that every vector space has a basis, which is a fact about Hamel bases, but your question is tagged with shauder-basis. Which one are you asking about? I'll just answer both and highlight the differences.
A Hamel basis is what we usually mean when we just say "basis". I.e. every element can be uniquely written as a finite linear combination of basis elements. Necessarily, any Hamel basis of $C[a,b]$ will be uncountable. See What is a basis for the vector space of continuous functions? for why there is essentially no way to write down a Hamel basis of $C[a,b]$ without invoking the Axiom of Choice.
A Schauder basis is a sequence $(f_n)$ such that for any element $f$ of the space, there is a unique sequence of constants $(\alpha_n)$ such that $f = \lim_n \sum_{i=1}^n \alpha_n f_n$, where the limit is taken with respect to the norm of the space (in this case, the sup norm on $C[a,b]$). The most common Schauder basis for $C[0,1]$ is a certain type of wavelet basis, called a Faber–Schauder system. Describing such a system is a task that I don't attempt here, but I encourage reading the linked wiki.
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If you're looking for a Schauder basis, meaning a basis from which every vector may be written as an infinite sum, then by the Stone-Weierstrass theorem, the polynomials are dense in $C[a,b]$ in the supremum norm. That means every continuous function can be written as an infinite sum of polynomials. For example Bernstein polynomials are a convenient spanning set. But if you don't like polynomials, many other families of functions also serve. The Stone-Weierstrass theorem applies to any family of functions that separates points. Trig functions, exponentials, absolute values, whatever you want. And there is a process for turning these spanning sets into bases.
But if you want an actual algebraic basis in the sense of linear algebra, a Hamel basis, a basis such that every vector is a finite linear combination, well such things exist by the axiom of choice, but this is nonconstructive. No explicit formula for such bases may be given in general, and it is even consistent with ZF set theory that they not exist.
ziggurism
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2The polynomials do span $C[a,b]$ in the Shauder sense, but they are not a basis. – nullUser Nov 04 '17 at 14:59
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@nullUser, I believe this distinction is already addressed in my answer, or do you think it is unclear? – ziggurism Nov 04 '17 at 15:02
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your answer seems to imply that the Bernstein polynomials, trig functions, absolute values are all Shauder bases of $C[a,b]$, when in reality none of them are. For instance, the Bernstein polynomial are called "Bernstein basis polynomials" because the first $n$ of them constitute a (Hamel) basis of the polynomials of degree at most $n$. It is not clear to me that this is what you meant from your post, and if it is what you meant, then you still haven't answered OPs question because OP asked about a basis, not just spanning sets. – nullUser Nov 04 '17 at 15:07
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It is interesting that it is consistent with ZF that no explicit formula for a Hamel basis exists, I did not know this. – nullUser Nov 04 '17 at 15:10
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@nullUser Asaf's answer in the question you linked in your answer shows why there is no Hamel basis in Solovay's model. – ziggurism Nov 04 '17 at 15:18
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@nullUser you're commenting not on the distinction between Hamel and Schauder bases, but rather the distinction between basis and spanning set. You're right that's a weakness in my answer. – ziggurism Nov 04 '17 at 15:22
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"And there is a process for turning these spanning sets into bases." If we're talking about Schauder bases then I don't think this is true. – David C. Ullrich Nov 04 '17 at 15:46
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@DavidC.Ullrich hmm I thought this was something wavelet analysis could do for us, but I confess I do not know the subject very well. – ziggurism Nov 04 '17 at 15:52
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@ziggurism I don't know wavelets much either. I wouldn't be surprised if lots of wavelet systems gave bases for $C([a,b])$. The basis I know is the Faber-Schauder system - see link in one of the answers. – David C. Ullrich Nov 04 '17 at 18:22