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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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I suppose complete uniqueness is out of the question: Let $X$ be a solution, $X'$ a function that differs from $X$ only in a zero set of points. Then $X'$ is also a solution. For every reasonable integral I can think of, finite sets will have measure zero. Hence, the result will not be unique, since you can change every solution in a finite numbers of points and get another solution. –  Johannes Kloos Mar 10 '12 at 15:36
    
Yeah, you are right. Let's put it in this way: Give one solution which holds for any arbitrary f(.,.) and b(.)? What condition(s) are needed to guarantee the uniqueness? –  Daniel Oct 31 '12 at 15:44
    
Do you want absolute uniqueness or uniqueness up to some equivalence relation (e.g., equivalent almost everywhere)? –  Johannes Kloos Oct 31 '12 at 15:55
    
Any kind of information could be useful. Do you have any ideas about such equivalence relation? –  Daniel Oct 31 '12 at 15:59
    
Try "equivalent almost everywhere": Two functions $f, g$ are equivalent almost everywhere if the set $\{x | f(x) \neq g(x) \}$ has measure zero. This is still a very weak condition here, for example: Choose $f(x,y) = 1$ and $b(x) = 0$. Then you want to solve the equation $\int_a^b X(y) \operatorname{d}y = 0$, which does have infinitely many solutions, e.g., all functions of the form $X(y) = c \cdot \left(y - \frac{a+b}{2}\right)$. I am no expert in higher analysis, but I doubt there is any integral operator that will give you unique solutions except when adding some other stronger constraints. –  Johannes Kloos Oct 31 '12 at 16:08

1 Answer 1

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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You can do this only if $f(x,y)$ is of the form $f(x-y)$ which is known as the difference kernel. –  Mhenni Benghorbal Feb 19 '13 at 19:33
    
Oh, I see the problem with that .... –  Daniel Apr 26 '13 at 20:21

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