Isomorphism on infinite dimensional vector space or Fourier transform of wavelets Is it true that isomorphisms between infinite-dim. vector spaces map basis onto basis, just as in the finite-dim. case?
I am looking for a proof that the Fourier transform of an orthonormal basis on $L^2$ gives a (orthogonal) basis again.
Thank you very much!
Lena
 A: It is true that isomorphisms of vector spaces map basis to basis. But you need to be careful when apply this to orthonormal basis of Hilbert Space. 

If $V$ and $W$ are isomorphic via $f : V \rightarrow W$. Suppose $v_1, ..., v_n$ is a subset of a basis of $V$. In particular $\{v_1, ..., v_n\}$ are linearly independent. If $\{f(v_1), ..., f(v_n)\}$ are not linearly independent, then there exists not all zero coefficient $a_1, ..., a_n$ such that $a_1f(v_1) + ... + a_nf( v_n) = f(a_1v_1 + ... + a_n v_n) = 0$. Hence $a_1 v_1 + ... + a_n v_n \neq 0$ since $\{v_1, ..., v_n\}$ linearly independent. Thus $\text{ker}(f) \neq \{0\}$. $f$ is not injective. So if $\mathcal{B}$ is a basis for $V$, then $f(\mathcal{B}) = \{f(v) : v \in \mathcal{B}\}$ is linearly independent in $W$. If $f(\mathcal{B})$ does not span $W$, then $f$ is not surjective. So $f(\mathcal{B})$ is a spanning linearly independent subset of $W$. 

Before, you apply this result to Hilbert Space, you should be aware that an orthonormal basis may not be a basis for the vector space. For example, every separable Hilbert Space has a countable orthonormal basis, but no infinite dimensional Banach Space has a countable basis. Hilbert spaces are Banach Spaces. So infinite dimensional separable Hilbert spaces have a countable orthonormal basis but uncountable algebraic basis.
You may be aware of some theorem in Hilbert Space theory that asserts that every element of the hilbert space can be written as
$\sum_{i = 1}^\infty a_i e_i$
where $e_i$ come from a fixed orthonormal basis. 
In vector space theory, if $\mathcal{B}$ is a basis, then every element can be written as a finite sum of basis elements $a_1 e_1 + ... + a_ne_n$. Hence the two notion are not necessarily the same. 
