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Does the set ${1, x, x^2...}$ form a basis for the vector space of real analytic functions over the real numbers? It seems obvious that they span, but not obvious that they are independent.

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  • $\begingroup$ doesn't the Taylor series prove that they are ? $\endgroup$ – reuns May 5 '16 at 4:02
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    $\begingroup$ No, if it is a basis, then any element can be represented by a finite linear combination of the basis elements. $\endgroup$ – copper.hat May 5 '16 at 4:04
  • $\begingroup$ How could it? Every real analytic function would then be a polynomial. $\endgroup$ – zhw. May 5 '16 at 4:05
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    $\begingroup$ Actually, they are I think clearly linearly independent but do not span. $\endgroup$ – André Nicolas May 5 '16 at 4:09
  • $\begingroup$ Ah, I see. I thought infinite linear combinations were allowed-- I've only done finite dimensional vector spaces thus far. Thanks for clarifying. $\endgroup$ – Vik78 May 5 '16 at 4:15
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Short answer: NO.

Long answer: It depends on your use of the word 'basis'. Remember that when working with infinite dimensional vector spaces you can define 2 types of basis. Schauder's basis and Hamel's basis.

$\textbf{1)}$ Hamel Basis: For any infinite dimensional vector space $V$ a l.i. collection of vectors $ HB = \{ v_1,v_2, \cdots \} $ is named a Hamel basis for $V$ if for every vector $ v \in V $ there is a $\textbf{finite}$ number of elements of $HB$ such that $$ v = \displaystyle \sum_{i=1}^N c_i v_i $$ where of course $c_i$ belongs to the corresponding field.

$ \textbf{2)} $ Schauder Basis: For any infinite dimensional vector space $V$ a l.i. collection of vectors $ SB = \{ v_1,v_2, \cdots \} $ is named a Sachauder basis for $V$ if for every vector $ v \in V $ there is a subset $\{ u_1, u_2, \cdots \} \subseteq SB$ such that $$ v = \sum_{i=1}^{\infty} c_i u_i $$

Is the difference clear? In the first one you can only accept $\textbf{finite}$ linear combinations while in the second one you can have a series. The thing with the Schauder basis it is that they are often very difficult to work with. Banach himself asked if any separable Banach space had a Schauder basis. Years after this question was solved by Enflo. (No). That is a very disappointing thing for the fans of Schauder's basis because existence is one of the first things you want to ensure.

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  • $\begingroup$ Note that the space of real analytic functions does not have a Schauder basis, so your last sentence might be misleading. $\endgroup$ – Christian Feb 11 '18 at 8:13
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Complementing the other answer where the difference between a vector space basis (Hamel basis) and a Schauder basis was introduced, let me mention that in this article P. Domaǹski and D. Vogt show that the space of real analytic functions does not have a Schauder basis.

Note that the Taylor series expansion might lead to a different series at each point but in order for the monomials to form a basis one would need that have the same expansion at every point.

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