Integral transforms - convergence

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    Convergence Integral
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SUMMARY

The discussion centers on the convergence of the sequence (f_n) defined by the integral transform f_n(x) = ∫_{ℝ} f_{n-1}(y) K(y,x) dy. The key condition for convergence is that the kernel function K(y,x) must satisfy |K(y,x)| < 1, alongside specific conditions imposed on the initial function f_0(ℝ). The mention of L^1 spaces indicates that the functions involved should be integrable to ensure convergence.

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Pere Callahan
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Hi,

I have a question which is most probably standard, but I don't have access to a textbook right now...

Given a function [itex]K(y,x):\mathbb{R}^2\to\mathbb{R}[/itex] and a funtion [itex]f_0:\mathbb{R}\to\mathbb{R}[/itex], under what circumstances does the sequence [itex](f_n)_{n\geq0}[/itex] with
[tex] f_n(x) = \int_{\mathbb{R}}{dy\, f_{n-1}(y)K(y,x)}[/tex]
converge?

I have a feeling [itex]|K(x,y)|<1[/itex] could be sufficient (maybe if one assumes that f_0 tends to zero fast enough...)

Thanks

Pere
 
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You will have to impose conditions on ##f_0##, yes, since otherwise we could set ##K=\frac{1}{2}## and chose a function ##f_0## which hasn't a finite integral.

In general I would look at ##L^1## spaces.
 

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