Uniqueness of solution to an integral equation on the half line

Question

The equation in question is $$f(x)=\int_0^\infty f(y)(x+y)e^{-x^2/2-xy}\text{d}y$$ where $f: [0,\infty)\rightarrow[0,\infty)$. It is not hard to see $f(x)=Ce^{-x^2/2}$ solves the equation. However, how would you prove the uniqueness (if it is unique..) of a solution to such an equation?

Answer 1

I'm not sure it helps, but you can write:

$$f(x)=\int_0^\infty f(y) \, (x+y) \, e^{-x^2/2-xy}\mathrm{d}y=e^{-x^2/2}\int_0^\infty f(y) \, (x+y) \, e^{-xy}\mathrm{d}y$$ $$=x \, e^{-x^2/2}\int_0^\infty f(y) \, e^{-xy}\mathrm{d}y + e^{-x^2/2}\int_0^\infty y \, f(y) \, e^{-xy}\mathrm{d}y$$

Then, if F is Laplace transform of f, i.e. $F(x)=\int_0^\infty e^{-x y} \, f(y) \mathrm{d}y$, your equation is: $$ f(x) = x \, e^{-x^2/2} F(x) - e^{-x^2/2} F'(x)$$ Or, $$e^{x^2/2}f(x)=xF(x) - F'(x)$$

Answer 2

it's the only smooth solution: write

$$ f(x) = e^{-x^2/2}\sum_{n=0}^{\infty}{a_n x^n} $$

now some integral/sum switcheroo together with coefficient comparison affords the desired result

EDIT: as pointed out by thomas the above method is unwieldy. new attempt: write

$$ f(x) = C(x)e^{-x^2/2} $$

then by assumption

$$ C(x)e^{-x^2/2} = \int_{0}^{\infty}C(y)(x+y)e^{-\frac{(x+y)^2}{2}}dy \quad (*) $$

the derivative of the right hand side is

$$ \int_{0}^{\infty}C(y)\frac{\partial}{\partial x}[(x+y)e^{-\frac{(x+y)^2}{2}}]dy = \int_{0}^{\infty}C(y)\frac{\partial}{\partial y}[(x+y)e^{-\frac{(x+y)^2}{2}}]dy \\ = C(y)(x+y)e^{-\frac{(x+y)^2}{2}}|_{y=0}^\infty - \int_{0}^{\infty}C'(y)(x+y)e^{-\frac{(x+y)^2}{2}}dy \\ = -C(0)xe^{-\frac{x^2}{2}} - \int_{0}^{\infty}C'(y)(x+y)e^{-\frac{(x+y)^2}{2}}dy $$

now - by $(*)$ - integrating over all $x$'s gives

$$ -C(0) = -C(0)\int_{0}^{\infty}xe^{-\frac{x^2}{2}}dx - \int_{0}^{\infty}C'(y)\int_{0}^{\infty}(x+y)e^{-\frac{(x+y)^2}{2}}dxdy \\ = -C(0)[-e^{-\frac{x^2}{2}}]_{0}^{\infty} - \int_{0}^{\infty}C'(y)[-e^{-\frac{(x+y)^2}{2}}]_{x=0}^{\infty}dy \\ = -C(0) - \int_{0}^{\infty}C'(y)e^{-\frac{y^2}{2}}dy $$

or

$$ \int_{0}^{\infty}C'(y)e^{-\frac{y^2}{2}}dy = 0 $$

while the last equation is fulfilled trivially for $C=constant$, it conceivably has nontrivial solutions. one such can be constructed using alternating smooth bump functions.

i'll leave the thoughts on convergence to you =)

EDIT: wrong, nothing guarantees that such a constructed function would be a solution. it's still preferable to use Taylor expansion or fourier analysis...