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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

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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.

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Thank you! I didn't completely understand, what you mean by the ONB of the Hilbert space may not be a basis for the vector space. – Lena Sep 17 '12 at 23:20
do you mean, that if the additional structure of the scalar product is dropped that the basis might be not a basis anymore? – Lena Sep 17 '12 at 23:25
@Lena The orthonormal basis is a linearly independent set, but in infinite dimensional Hilbert Space, it does not span. Remember, something is an algebraic basis if and only if every element can be written as a $\textbf{finite}$ linear combination. You can not write every every element of a Hilbert Space as a finite linear combination of the orthonormal basis. – William Sep 17 '12 at 23:28
@Lena For example, consider $l^1(\mathbb{R})$. The sequence that are $1$ at the $n^\text{th}$ term but zero everywhere else, is an orthonormal basis, but you can not write everything using just finite number of these. – William Sep 17 '12 at 23:29
I see, but it will span the space using limits, i.e. infinite linear combinations right? – Lena Sep 17 '12 at 23:31

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