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Let $\ell^2 = \ell^2(\mathbb{Z})$ denote the Hilbert space of square summable complex sequences on $\mathbb{Z}$ and suppose that $\sigma:\mathbb{Z} \to \mathbb{Z}$ is a function such that the linear operator $C_\sigma: \ell^2 \to \ell^2$ defined by $C_\sigma(\{\alpha_n\}) = \{\alpha_{\sigma(n)}\}$ is bounded. In other words, let $C_\sigma$ be the composition operator induced by $\sigma$.

Now let $F$ denote the isomorphism of $L^2(-\pi,\pi) \to \ell^2$ given by Fourier series. Then every operator $T:L^2(-\pi,\pi) \to L^2(-\pi,\pi)$ naturally induces an operator $\hat{T}: \ell^2 \to \ell^2$ given by $\hat{T} = FTF^{-1}$. My question is now the following:

Given $\sigma: \mathbb{Z} \to \mathbb{Z}$, is there a practical way of recovering the operator $T$ such that $\hat{T} = C_\sigma$? Is there a characterization of the operators $T$ such that $\hat{T} = C_\sigma$ for some $\sigma$?

I have tried simply plugging into the definition of Fourier series and working from there, but this has not yielded much help. Oh also, if someone can think of a better title, please edit.

Thanks in advance.

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What do you mean here by 'practical'? I mean, obviously we have $T(f) = (F^{-1}\circ C_\sigma\circ F)(f)$. – William Apr 10 '12 at 22:21
Sure. I meant some other useful representation say as an integral operator or a composition operator induced by $\phi:(-\pi,\pi) \to (-\pi, \pi)$... I understand the question is a bit vague. – user12014 Apr 10 '12 at 22:45
up vote 1 down vote accepted

Unless I'm missing something, your operator $T$ is the map $e^{int}\mapsto e^{i\sigma^{-1}(n)t}$.

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