Product Rule and Distributions

Question

Suppose $f,g$ are smooth functions and $\phi$ is a test function. the product rule for distributions has that: $$[fg]'\{\phi\}=[fg]\{-\phi'\}=[f'g]\{\phi\}+[fg']\{\phi\} $$ I would like to multiply the derivative $[fg]'\{\phi\}$ by $\frac{1}{g(x)}$. Concretely, suppose everything is defined on $(0,\infty)$ and $g(x)=e^{-x}$. Is is correct that: $$[fg]'\{\phi\}=\int_0^\infty [f(x)'e^{-x}\phi(x) - f(x)e^{-x}\phi(x)]dx$$ and so $$\frac{1}{g(x)}([fg]'\{\phi\})=\int_0^\infty [f(x)'\phi(x) - f(x)\phi(x)]dx$$

Answer 1

Assume that $f,g,\phi$ are supported on $(0,\infty).$ Then, $$ [fg]′\{ϕ\} = -[fg]\{ϕ′\} = -\int_0^\infty f(x)\,g(x)\,\phi'(x)\,dx = \int_0^\infty (fg)'(x)\,\phi(x)\,dx . $$ With $g(x)=e^{-x}$ this becomes $$ \int_0^\infty (fe^{-\bullet})'(x)\,\phi(x)\,dx = \int_0^\infty (f'e^{-\bullet}-fe^{-\bullet})(x)\,\phi(x)\,dx = \int_0^\infty (f'(x)e^{-x}-f(x)e^{-x})\,\phi(x)\,dx . $$