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I've been searching all over the Internet for this but without finding a satisfying answer. This might be a dumb question, but I would like to know the answer anyway.

Is there a set of continuous functions which when combined linearly (or not maybe) span all the functions space ? Could we decompose a log, a sine or an exponent to simpler components ? And if not why ?

I know that Fourier analysis is a powerful tool for functions decomposition, but I wanted to know If we could go further and decompose even trigonometric functions. I wondered if there was is theory about this ?

Thanks for any answer.

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  • $\begingroup$ Yes. In fact, every vector space has a Hamel Basis. For the space in question, see this. $\endgroup$ – David Mitra Oct 18 '14 at 21:55
  • $\begingroup$ Expanding upon what David said, the set of all continuous maps that take on real (or complex) values forms a vector space, and using the Axiom of Choice shows that this vector space has a basis. It will be an extremely large basis, however, certainly uncountable. $\endgroup$ – Hayden Oct 18 '14 at 21:57
  • $\begingroup$ Well, every vector space has a basis (which is implied, and in fact equivalent, to the axiom of choice), so yes, a basis exists. As for constructing such a thing... well, it'd be uncountable, so that could be tricky. $\endgroup$ – Milo Brandt Oct 18 '14 at 21:58
  • $\begingroup$ @DavidMitra thanks for the link, very interesting ! $\endgroup$ – vphenix Oct 18 '14 at 23:47
  • $\begingroup$ @Hayden And if you didn't have the axiom of choice, then the answer at math.stackexchange.com/a/151186/26369 would explain why the continuous functions might not have a Hamel basis. $\endgroup$ – Mark S. Oct 28 '16 at 17:43
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For a very large set of functions, the set of complex exponentials (or the Fourier Basis) is a basis. This set includes the set of all those functions with a finite L2 norm.

We can construct a simple basis for all functions as follows. Let $f_r(x)=\left\{ \begin{array}{cl} 1 & \text{ if } x=r \\ 0 & \text{ otherwise} \end{array} \right.$. Then any function can be made of a linear combination of these very simple functions. However, this basis is not very useful or interesting.

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  • $\begingroup$ I would argue that none of those collections are bases with respect to the definition for a vector space, since not every function is a finite linear combination of functions in those collections. $\endgroup$ – Hayden Oct 18 '14 at 22:02
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    $\begingroup$ @Hayden I think that in analysis, one uses a slightly different notion of basis. Analysts often do not deal with "naked" vector spaces, but topological vector spaces. So one might call these "topological" bases. $\endgroup$ – Baby Dragon Oct 18 '14 at 22:37
  • $\begingroup$ @BabyDragon I agree, but I think that it is important to make the distinction (and make clear which is being discussed), especially because of the fact that the proofs of the existence of such algebraic bases versus these "topological" bases are of very different flavor. $\endgroup$ – Hayden Oct 19 '14 at 1:01

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