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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.2012-09-17
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    do you mean, that if the additional structure of the scalar product is dropped that the basis might be not a basis anymore?2012-09-17
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    @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.2012-09-17
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    @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.2012-09-17
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    I see, but it will span the space using limits, i.e. infinite linear combinations right?2012-09-17
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    @Lena Yes. However "infinite linear combination" is not really a correct term. It does not make sense in vector space so speak of infinite sum. This is a topological property involving limits.2012-09-17
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    I see. So I cannot really use this to show that a wavelet basis of $L^2$ is Fourier-transformed to a wavelet basis again.2012-09-17
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    @Lena If your isomorphism is just an ordinary vector space isomorphism and nothing else, then no. However if your isomorphism has additional property such as being a unitary isomorphism (preserve the inner product), then you may be able to say more2012-09-17
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    It preserves the inner product up to a constant factor $\frac{1}{2\Pi}$2012-09-17
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    Then yes. Let $\langle \cdot, \cdot \rangle$ denote the inner product. If $e_1$ and $e_2$ are orthonormal, then $0 = \langle e_1, e_2 \rangle = \frac{1}{2\pi}\langle f(e_1), f(e_2) \rangle$. So $f(e_2)$ and $f(e_2)$ remain orthogonal.2012-09-17
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    yes, but what is with the part about generating system? If I have $x=\sume_i$ for all $x\in L^2$, do I still get $f(x)=\sumf(e_i)$? This way, since f is assumed to be surjective one would know that each element of $L^2$ can be generated as a sum wrt the $f(e_i).$2012-09-17
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    Yes up to dividing out that factor. Remembering that infinite sums are really limits and use the fact that $f$ is continuous (so you can pull limits out).2012-09-17
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    Awesome, got it now! Thank you very much!2012-09-18