I have an infinite orthonormal set in $L^2 (\mathbb{R})$ and want to know under which conditions it is as basis of $L^2 (\mathbb{R})$ and how to prove that.
In the finite dimensional case, the cardinality of the set would just have to be equal to the dimension of the space, but I suppose in the infinite dimensional case (such as $L^2 (\mathbb{R})$) this doesn't hold.
I know the usual conditions for being a basis such as its span being dense in the space or it being a Schauder basis or a total orthonormal set, but do not know how to work with that in the case below:
I take a given orthonormal basis $\lbrace \phi_i \rbrace_{i=1}^{\infty} $ of $L^2 (\mathbb{R})$, say the weighted Hermite Polynomials, and modify it as follows: $ \psi_i(x) = \phi_i( g(x)) * \vert g^\prime (x) \vert^\frac{1}{2} $
This conserves orthonormality, but I do not now whether or under which circumstances this is still a basis. What conditions do I need on g? Maybe just bijectivity?
Thanks for any help or ideas.