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For all $x \in \mathbb{R}$, $b(x) =\int_{-\infty}^\infty b(s)a(x,s)ds.$ If it helps, we can assume that $a, b$ are continuous, nonnegative, and $\int_{-\infty}^\infty$ of $a$ or $b$ are both bounded.

Two questions: (1) Is $a$ unique? and (2) to what extent can we solve for $a$?

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    Ah, because $\int_{-\infty}^\infty$ is an inner product. That's a very interesting way to look at this problem.2012-07-14

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The solution for your integral equation is the Dirac Delta function, that is $a(x,s) = \delta(s-x).$ It is known as the sifting propert of the dirac delta function.

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    The dirac delta is a functional and not a function. This is of course an extent to which the equation can be solved by means of distributions.2012-07-14
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Uniqueness is not necessarily guaranteed. For example, when definable, take $a(x,s)=\frac{b(x)}{b(s)}f(s)$ where $f$ is any probability density, that is $\int_{-\infty}^\infty f(s) ds=1$.