I'm trying to solve the following question:
If $S$ is a distribution with compact support on $\mathbb{R}$, which verifies $\langle S, x^n \rangle=0$ $\forall n\in\mathbb{N}$, then $S$ is the distribution $S=0$.
Thanks in advance for any help with this.
Let me show you the main ideas for a distribution $S$ of order $\leq 1$. Let $K$ be a compact set, whose interior contains the support of $S$. Altogether, you have a $C>0$ such that for all $\varphi\in C_c^\infty(\mathbb{R})$: $$ |S(\varphi)|\leq C\sup_{x\in K, k\in\{0,1\}} |(D^{k}\varphi)(x)| =: Cq_{K,k}(\varphi), $$ i.e. $S$ is continuous with respect to the seminorm $q_{K,k}$. But the polynomials are dense with respect to this seminorm by the Stone-Weierstrass approximation theorem: For $\varphi$ approximate $\varphi'$ with polynomials uniformly on $K$, integrate them, and these integrated polynomials converge to $\varphi$ with respect to the seminorm $q_{K,k}$. Cutting off the polynomials smoothly outside $K$ you can now conclude $S(\varphi)=0$ by continuity.
The general case is handled similarly, as any distribution with compact support has finite order.