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I am trying to solve the equation:

$\phi(x)=\int_{-\infty}^\infty K(x, t)\phi(t)dt$

for $K$ given $\phi$. This closely resembles a Fredholm Integral of the Second Kind, which has the form:

$\phi(x)=f(x)+\lambda \int_{-\infty}^\infty K(x, t)\phi(t)dt$

with $\phi$ unknown and $\lambda$, $K$, $f$ known. Is Fredholm Theory helpful, even though I'm solving for a different term in the expression than usual?

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The finite-dimensional analog of this problem would be to find a matrix $K$ given only that one particular vector is an eigenvector with eigenvalue 1. That problem is obviously very underdetermined (as soon as the dimension exceeds 1). In the infinite-dimensional case of integral kernels, the problem is even "more" underdetermined. That's why Robert Israel's answer gives lots of $K$'s that work. –  Andreas Blass Jul 15 '12 at 18:10
    
Seconding @Andreas Blass' comment, and noting how simple Bob Israel's solution is, I might go so far as to speculate that there's some misunderstanding about the context of the question... and perhaps upon closer examination the question should be somewhat different than as posed. –  paul garrett Jul 29 '12 at 18:27

1 Answer 1

One family of solutions is $K(x,t) = \phi(x) h(t)$ where $h(t)$ is any function such that $\int_{-\infty}^\infty h(t) \phi(t)\ dt = 1$.

Somewhat more generally, if $\phi(x) = \sum_{n=1}^N \phi_n(x)$ take $K(x,t) = \sum_{n=1}^N \phi_n(x) h_n(t)$ where $\int_{-\infty}^\infty h_n(t) \phi(t)\ dt = 1$ for each $n$.

You could also add any linear combination of functions $u(x) v(t)$ where $\int_{-\infty}^\infty v(t) \phi(t)\ dt = 0$.

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