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I'm having some troubles in proving these identities relating two good functions (Blue and Red, always reals, always positive, always continuous together with their arguments always reals, positive) $$\color{blue}{B\left( x \right)} = - \int_x^\infty {\left[ {\int_0^\infty \color{blue}{B (x')}\color{red}{R(y)}{\mkern 1mu} dy - \int_0^{x'} \color{red}{R (y)}\color{blue}{B(x' - y)}{\mkern 1mu} dy} \right]} {\mkern 1mu} dx',$$ and $$\color{blue}{B\left( x \right)} = \int_0^x {\left[ {\int_{x - x'}^\infty \color{blue} {B (x')}\color{red}{R(y)}{\mkern 1mu} dy} \right]} {\mkern 1mu} dx'.$$ ...but putting inside whatever shape for Blue and Red... they give the same color (Blue) so probably I'd say... no typos.

I'm getting "colorblinded" :-/

Thanks JD

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    I have introduced the "desirable" (?) colors. Is it what you wanted ? – 2017-02-11
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    The fact that it is integral equations doesn't look important. It looks as if you wanted to show the identity of the 2 RHS, without reference to the LHSides. – 2017-02-11
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    Ha ha ha... You put the wrong colors! B is blue and R is red! come on! ;-) ...but thanks. I didn't know how you did it. – 2017-02-11
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    ...and yes, ultimately it is left to prove that the two RHSs are the same. – 2017-02-11
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    Corrected... I should have guessed...It suffices to write \ color{blue} {what you want be colored blue} with antislash sign attached of course. – 2017-02-11
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    Cool! Thanks! ...still... I can prove that it numerically holds with every couple of different (well-behaving) colors you like, but I don't see the correct passage :-o – 2017-02-12
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    Could you show me an example of how you numerically do for obtaining the same result. I am rather confident that I can see in this way where is the blocking point (maybe a notational misunderstanding). – 2017-02-13

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This is not an answer. It is just for inserting a figure.

The RHS of the second expression amounts to a sweeping of the domain in form of a rectangular trapezoid as represented on the figure below.

I imagine that the first expression corresponds to an integration on a larger domain to which is substracted a superfluous quantity... but all my attempts have been unsuccessful...

May I ask you to check that all the bounds (in particular) are correct in this first expression ?

enter image description here

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    I've double checked, no typos. You can of course prove the equality of the two RHS in a different way. I just give you an example: You choose to express Red and Blue as infinite sum of a "suitable" complete basis function of the space where you want Red and Blue to live (e.g. generalized Gaussians can work well, but even simpler functions are ok). You insert inside the first and the second eqs. your basis, coefficients come out, and you "easily" check that the two equations are effectively the the same. The colors match. Proof done. – 2017-02-14
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    but... but... but... wait a moment: I'm not a mathematician! (I'm indeed a painter!) So what I was asking for, searching for, it was a "direct" proof of the colors equation. In direct way, e.g. the same as you were trying to get by inverting the order of integrals. I've tried quite a few tricks similar to what in your plot (super nice plot!) but the direct solution is still escaping and... I start to think that I might just miss some basic notion, some theory of integral equations? Convolution theorems? I don't know... If you have any idea to solve this game... every help is welcome. Thanks – 2017-02-14