Is something known about the solution to Fredholm equations of the 2nd type of the following form:
$\displaystyle f(x) = g(x) + \int_{-k}^k f(y) h(x-y) dy$
where $f: [-k, k] \to \mathbb{R}$, $g(x)$ is an even function, and
$\displaystyle h(x - y) = \frac{1}{\sqrt{2\pi}} e^{-(x-y)^2/2}$
Let $$ Tf(x) = \dfrac{1}{\sqrt{2\pi}}\int_{-k}^k f(y) e^{-(x-y)^2/2}\; dy $$ so your integral equation is $(I-T) f = g$. Since $$\dfrac{1}{\sqrt{2\pi}}\int_{-k}^k e^{-(x-y)^2/2}\; dx \le \dfrac{1}{\sqrt{2\pi}}\int_{-k}^k e^{-x^2/2}\; dx < 1$$ (call the middle integral $R$), we have $$ \|Tf\|_\infty \le R \|f\|_\infty$$ Thus $I - T$ is invertible on $C[-k,k]$: there is a unique continuous solution $$ f = (I - T)^{-1} g = \sum_{i=0}^\infty T^i g$$