The function $\phi/2$ is a Green's function for the operator $1-\partial_x^2$ where $\phi=e^{-|x|}$. Thus, formally, the solution to $$ y-y''=f(x)\;\;\;\;\;\;\;(1) $$ on the whole line is $$ y=\frac{1}{2}\phi\ast f,\;\;\;\;\;\;\;(2) $$ where $\ast$ denotes convolution.
My question: how bad can $f$ be for (2) to define a solution to Equation (1)? If $f$ is smooth and bounded for example, it works. How about, for example, if $f$ is
(i) $H^1(\mathbb{R})$?
(ii) $C_b(\mathbb{R})$ (Bounded and continuous)?
(iii) $L^2(\mathbb{R})$?
(iv) ${\mbox{L}}_{1,\,{\mbox{loc}}}(\mathbb{R})$ (locally integrable) with the restriction that $\phi\ast f$ exists?
You need $ \phi *f $ , and $\phi *f'$ or $\phi' *f$ to exist.
Green functions are meaningful in the context of distributions, so the distributional derivative $\phi *f'$ or $\phi' *f$ need to exist.