How is $\{1, x, x^2, x^3, ...\}$ a basis for $L^2(-1, 1)$? I have just read that $\{1, x, x^2, x^3, ...\}$ is a basis for $L^2(-1, 1)$...I don't get this. It seems like that set is a basis for the space of polynomials of degree up to infinity. I don't see how it can be a basis for all square integrable functions?
 A: In the setting of normed linear spaces, a basis spans a dense subspace. To say the monomials constitute a basis of $L^{2}[-1, 1]$ amounts to the assertion that for every function $f$ in $L^{2}[-1, 1]$, there exists a sequence $(p_{n})$ of polynomials such that
$$
\int_{-1}^{1} |f - p_{n}|^{2} \to 0.
$$
The space of continuous functions on $[-1, 1]$ is dense (with respect to the $L^{2}$ norm) in $L^{2}[-1, 1]$ by Lusin's theorem, and the space of polynomials is dense with respect to the sup norm by the Weierstrass approximation theorem, hence dense with respect to the $L^{2}$ norm on $[-1, 1]$. A dense subspace of a dense subspace is a dense subspace, so the monomials are indeed a basis of $L^{2}[-1, 1]$.
If you did mean $L^{2}(-1, 1)$ (open interval), you can use the fact that the space of compactly supported functions in $L^{2}(-1, 1)$ is dense, then use the sketch of the preceding paragraph to see that polynomials are dense in the space of compactly supported $L^{2}$ functions.
A: Maybe I am missing something but 1,x,x^2 is not a basis in L2[-1,1] as the integral of 1*x^2 is not zero, hence 1 and x^2 are not orthogonal to each other.
