suppose $f$ and $g$ are distributions such that $\langle f,\varphi\rangle=0$ if and only if $\langle g,\varphi\rangle=0.$ Show that$\langle f,\varphi\rangle=c\langle g,\varphi\rangle$ for some constant $c$.
I just started with distribution theory with the basic definitions of distribution, e.g. linearity and continuity. I just have no idea how to approach this question. It looks like we have to apply some tricks by linearity? Any hints?
If $f=0$ then clearly $g=0$ and vice versa, so suppose there is some $\psi_0$ such that $f(\psi_0) \neq 0$.
Claim: $g = {g(\psi_0) \over f(\psi_0)} f$.
Choose a test function $\psi$, then note that $\psi = {f(\psi) \over f(\psi_0)} \psi_0 + \psi-{f(\psi) \over f(\psi_0)} \psi_0$ and $\psi-{f(\psi) \over f(\psi_0)} \psi_0 \in \ker f = \ker g$.
Hence $g(\psi) = g({f(\psi) \over f(\psi_0)} \psi_0) = {g(\psi_0) \over f(\psi_0)} f (\psi)$ and so we have the desired result.