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Does any one know any bounded integral operator?

To put it in another way, I am given the following equation

$ \int_{a}^{b} f(x,y) X(y) dy = b(x) $

where $f(x,y)$ and $b(x)$ are known.

  • What conditions do I need in order to have a unique solution for $X(y)$?
  • How can I find $X(x)$?

Thank you

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    Here is a [reference](http://www.tdx.cat/bitstream/handle/10803/6189/10Jvl10de11.pdf;jsessionid=66561E61887ED93E887D00F4575A1A8C.tdx2?sequence=10).2013-02-19

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I found this useful1 : http://en.wikipedia.org/wiki/Fredholm_integral_equation

Using the mentioned Fourier transform formula, $ X(y) = \mathcal{F}_\omega^{-1}\left[ {\mathcal{F}_x[b(x)](\omega)\over \mathcal{F}_x[f(x,y)](\omega)} \right]=\int_{-\infty}^\infty {\mathcal{F}_x[b(x)](\omega)\over \mathcal{F}_x[f(x,y)](\omega)}e^{2\pi i \omega x} \mathrm{d}\omega $

it should be Lebesgue measurable:

$ \int_{-\infty}^\infty |b(x)| \, dx < \infty $

$ \int_{-\infty}^\infty |f(x,y)| \, dx < \infty $

One necessary condition is $\mathcal{F}_x[f(x,y)](\omega) \neq 0$. The problem with this method is that Fourier transform is not easy to calculate, though I can use DFT.

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    Oh, I see the problem with that ....2013-04-26