I'm reading Functional Analysis by Rudin, and have trouble understanding a part of the proof of theorem 6.33, in page 174. This theorem states an one-to-one relationship between a linear continuous mapping $L:\phi\mapsto u*\phi$ from $\mathscr{D}$ to $\mathscr{C}^{\infty}$ and a distribution $u\in \mathscr{D}'$.
At the end of proof, it tries to confirm the uniqueness of $u$. It says
The uniquess of $u$ is obvious, for if $u\in\mathscr{D'}$ and $u*\phi=0$ for every $\phi\in\mathscr{D}$, then $u(\check{\phi})=(u*\phi)(0)=0$ for every $\phi\in \mathscr{D}$; hence $u=0$.
I don't quite understand why this is obvious...Actually I don't see why the "for" part implies the uniqueness.
The key is that linear functions on vector spaces are one-to-one iff they only map $0$ to $0$, or said another way, iff the kernel contains exactly 0.
For the forward direction, suppose $L:X\to Y$ is not one-to-one. Then $\exists$ $x \ne y$ such that $L(x) = L(y)$. By linearity, this implies $L(x) - L(y) = L(x - y) = 0.$ Therefore the kernel of $L$ contains $x - y \ne 0.$
For the other direction, if the kernel contains a vector $x \ne 0$ such that $L(x) = 0$, then trivially $L(x) = 0$ and $L(0) = 0$, so the mapping is not one-to-one.