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I know coefficients of some function in basis $p_j,j=1...K$ where

$p_{j}(x)=\sum_{s\in Z}a_{s,j}\exp\left(-2\pi i(j+sK)x\right)$

With respect to inner product $(f,g)=\int_{0}^{1}f(x)\overline{g(x)}dx$ this basis is orthogonal. What kind of basis is it? Can I somehow use Fourier analysis framework for this basis?

One more question about this basis. How to solve this integral equation? $$a_{j}=\int_{-\infty}^{\infty}g(x)p_{j}(x)dx$$ $g-$unknown function. Could you give me some hint, or some links?

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If $a_{s,j}$ is in $\ell_2$ for any fixed $j$, $p_j$ forms an orthogonal basis for some finite dimensional sub-space of $L^2(\mathbb{T})$. If $f(x) = \sum_{j=0}^K b_j p_j(x)$, then its fourier coefficients are just given by $\hat{f}(\xi) = a_{s,j} b_j$ where $j = \xi \mod K$ and $s = \lfloor \xi / K \rfloor$, so surely normal Fourier analytic techniques apply.

I don't quite understand your question about "What kind of basis is it?" Can you clarify?

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  • $\begingroup$ I am wondering if it is possible to use basis $p_j$ instead of$\exp(2\pi ij)$ to define Fourier transform, inverse fourier transform and so on. $\endgroup$
    – vilvarin
    Commented May 25, 2010 at 14:31
  • $\begingroup$ It is a basis of a finite dimensional sub-space, so not all functions will be in the span of $p_j$. Recall that the Fourier coefficients can be defined as $\hat{f}(\xi) = \langle f(\cdot), \exp(2\pi i \xi \cdot)\rangle$. A general representation formula is that, for some Hilbert space $H$ with orthonormal bases $e_1,e_2, \ldots$ you can write $f = \sum \lambda_j e_j$ where $\lambda_j = \langle f,e_j\rangle$ (this essentially captures the Fourier transform and its inverse, completely analogous to the finite dimension inner product space case)... $\endgroup$
    – Willie Wong
    Commented May 25, 2010 at 16:20
  • $\begingroup$ ... so consider $f(x) = 1$, the constant function on $[0,1]$. Compute $\langle f, p_j\rangle = \delta_{jK} a_{-1,0}$ (in your notation where numbering of $j = 1... K$. But obviously $a_{-1,0} p_0 \neq f$ unless all other $a_{s,0} = 0$. But if you have some a priori knowledge that all the functions you care about lies in the span of the $p_j$s, then you don't even need Fourier analysis: you are just working with finite dimensional linear algebra on some inner product space. $\endgroup$
    – Willie Wong
    Commented May 25, 2010 at 16:24
  • $\begingroup$ .. I meant to write that $\langle f,p_j\rangle = \delta_{jk}a_{-1,K}$ and $a_{-1,K}p_K \neq f$ unless all other $a_{s,K} = 0$. Sorry for the typo. $\endgroup$
    – Willie Wong
    Commented May 25, 2010 at 16:25
  • $\begingroup$ thank you very much. The last question I have is if Fourier inversion theorem is valid for this basis. It is not true in general case, isn't it? Probably i need to check this $\endgroup$
    – vilvarin
    Commented May 25, 2010 at 19:14

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