# Can we solve for $a$ in $b(x) =\int_{-\infty}^\infty b(s)a(x,s)ds$

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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This looks like a special case of Fredholm Integral Equations of the Second Kind, with f(x) = 0, $\lambda$ = 1, and where $\phi$ is known instead of the Kernel. That might be useful. – GMB Jul 14 '12 at 20:15
A corresponding matrix problem: Given an eigenvector $b$ for an $n \times n$ matrix $A$, solve for the matrix. Answer: no, $A$ is not unique. – GEdgar Jul 14 '12 at 21:17
This sounds helpful, but I'm not sure what you mean: in what sense is that a corresopnding matrix problem? Thanks for your help. – GMB Jul 14 '12 at 21:38
Ah, because $\int_{-\infty}^\infty$ is an inner product. That's a very interesting way to look at this problem. – GMB Jul 14 '12 at 22:10

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.
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 $\int_{-\infty}^\infty f(s) ds=1$.