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Does there exist an inner product, such that the all of the monomials $1,x,x^2,\ldots,x^n,\ldots$ (viewed as real valued functions) form an orthogonal (or even orthonormal) set? And what about a finite subset: $1,x,x^2, \ldots,x^N$?

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Clearly not a $\mathbb L^2(\mu)$ space for some measure $\mu$ on $\mathbb R$ since it would require that $\langle x,x \rangle = \langle 1,x^2 \rangle$ is both zero and non-zero. –  roger Mar 28 '13 at 21:13

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Of course. You can trivially define such an inner product on the vector space of polynomials: if $p = \sum_n a_n x^n$ and $q = \sum_n b_n x^n$, set $$\langle p,q \rangle = \sum_n a_n b_n.$$ This is clearly bilinear, symmetric, and positive definite.

Similarly, given any vector space $V$ and any linearly independent set $E \subset V$, there is an inner product on $V$ such that $E$ is orthonormal. Extend $E$ to a basis $F$; then every vector $x \in V$ has a unique representation $x = \sum_{e \in F} a_e e$, where only finitely many $a_e$ are nonzero. If $x = \sum_{e \in E} a_e e$, $y = \sum_{e \in E} b_e e$, let $\langle x, y \rangle = \sum_{e \in E} a_e b_e$.

If you're looking at complex vector spaces, then you want $\langle x,y \rangle = \sum a_e \bar{b}_e$.

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Could you please extend this to functions that are analytic near 0? (that is, they admit power series representation of the $\sum a_n x^n$) –  user1337 Mar 28 '13 at 21:28
    
@MichaelTouitou: For this case I'm not aware of an explicit formula, but the abstract argument still goes through. –  Nate Eldredge Mar 28 '13 at 21:51

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