# Do monomials form a basis for the vector space of real analytic functions?

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.

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

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.

• Note that the space of real analytic functions does not have a Schauder basis, so your last sentence might be misleading. – Christian Feb 11 '18 at 8:13

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.