I used to think that in any Vector space the space spanned by a set of orthogonal
basis vectors contains the basis vectors themselves. But when I consider the vector space $\mathcal{L}^2(\mathbb{R})$ and the Fourier basis which spans this vector space, the same is not true ! I'd like to get clarified on possible mistake in the above argument.
[Math] Does the vector space spanned by a set of orthogonal basis contains the basis vectors themselves always
fourier analysishilbert-spacesvector-spaces
Related Question
- [Math] A vector space with countable and uncountable basis at the same time
- [Math] Does a vector space with dimension 1 have an orthogonal basis
- [Math] Can you construct a basis for an infinite dimensional vector space from a set of vectors that span that space
- [Math] Prove: In a finite dimensional vector space, every spanning set contains a basis.
- Orthogonal projection onto the closure of a subspace spanned by a set of orthonormal vectors
Best Answer
If by "the Fourier basis" you mean the functions $e^{2 \pi i n x}, n \in \mathbb{Z}$, then these functions do not lie in $L^2(\mathbb{R})$ as they are not square-integrable over $\mathbb{R}$, so in particular they can't span that space in any reasonable sense. The functions $e^{2 \pi i n x}$ do span $L^2(S^1)$ (in the Hilbert space sense).
Perhaps you are getting the Fourier transform for periodic functions mixed up with the Fourier transform on $\mathbb{R}$.