This problem is from Exercise 1.13.22, Tao's An Epsilon of Room.
Let $\lambda$ be a distribution whose domain is the set of compactly supported smooth functions on $\mathbb{R}$.
(i) Show that if $\lambda$ is a distribution and $n\ge 1$ is an integer, then $\lambda x^n=0$ iff $\lambda$ is a linear combination of $\delta$ and its first $n-1$ derivatives $\delta', \delta'', ...,\delta^{(n-1)}$.
(ii) Show that a distribution $\lambda$ is supported on ${0}$ iff it is a linear combination of $\delta$ and finitely many of its derivatives.
My attempt: For both (i) and (ii), the 'if' parts are easy, and I already did them easily. The problems are the 'only if' parts.
For the 'only if' part of (i), I let $a_i = (-1)^i \dfrac{\lambda(x^i)}{i!}$, and tried to prove that $\lambda= \sum_{i=0} ^{n-1}a_i \delta^{(i)}$, but I don't know how to proceed further.
For the 'only if' part of (ii), I don't have any idea to proceed.
Does anyone have ideas?
Any hints or advices will help a lot! Thanks!
Given any smooth $f$, you can write, using the Taylor polynomial, $$ f(x)=\sum_{k=0}^{n-1}\frac{f^{(k)}(0)}{k!}\,x^k\ + \frac{f^{(n)}(\xi(x))}{n!}\,x^{n}. $$ Now, when you apply $\lambda$ to the last term, you get (by definition of multiplication) $$ \left\langle \frac{f^{(n)}(\xi(x))}{n!}\,x^{n},\lambda \right\rangle =\left\langle \frac{f^{(n)}(\xi(x))}{n!},\lambda x^{n} \right\rangle=0. $$ Thus $$ \langle f,\lambda\rangle=\sum_{k=0}^{n-1}\frac{\langle x^k,\lambda\rangle}{k!}\,(-1)^k\langle f,\delta^{(k)}\rangle, $$ and $$ \lambda = \sum_{k=0}^{n-1}\frac{\langle x^k,\lambda\rangle}{k!}\,(-1)^k \delta^{(k)}. $$
The second question looks less trivial. Here is a proof.