Here's the proof given by my professor. Hopefully it meets your standards.
Recall the following definitions concerning distributions with compact support:
Definitions:
- If $X \subset \mathbb{R}^d$ is open then
$$
\mathcal{E}'(X) := \{u \in \mathcal{D}'(X) : \operatorname{supp} u \text{ is a compact subset of } X\}
$$
- If $u \in \mathcal{E}'(X)$ and $\psi \in C^{\infty}(X)$ then
$$
\langle u, \psi \rangle := \langle u, \rho \psi\rangle
$$
where $\rho \in C^{\infty}_C(X)$ with $\rho = 1$ on a neighborhood of $\operatorname{supp} u$.
Remark: Definition 2. is independent of the choice of $\rho$ and coincides with the usual definition when $\psi \in C^{\infty}_C(X)$.
Lemma: If $u \in \mathcal{E}'(X)$ and $V$ is a neighborhood of $\operatorname{supp} u$ then there exists constants $C$, $k$ such that
$$
|\langle u,\psi \rangle |\leq C\sum_{|\alpha|\leq k}\sup_V |\partial^{\alpha}\psi|\quad\quad\forall\psi\in C^{\infty}(X)
$$
Proof: Fix $\rho \in C^{\infty}_C(X)$ such that $\rho = 1$ on a neighborhood of $\operatorname{supp} u$ and $\operatorname{supp} \rho \subset V$. Let $K := \operatorname{supp} \rho$. Then there are constants $C$, $k$ such that
$$
|\langle u,\varphi \rangle |\leq C\sum_{|\alpha|\leq k}\sup |\partial^{\alpha}\varphi|\quad\quad\forall\varphi\in C^{\infty}_C(X), \operatorname{supp} \varphi \subset K
$$
Let's take an arbitrary $\psi \in C^{\infty}(X)$ and apply this last inequality with $\varphi = \rho \psi$. We obtain
\begin{align}
|\langle u,\psi \rangle| &= |\langle u,\rho\psi \rangle| \\
&\leq C\sum_{|\alpha|\leq k}\sup |\partial^{\alpha}(\rho\psi)| \\
&= C\sum_{|\alpha|\leq k}\sup \left|\sum_{\substack{\beta,\gamma \\ \beta+\gamma=\alpha}}\frac{\alpha!}{\beta!\, \gamma!}\partial^{\beta}\rho\,\partial^{\gamma}\psi\right| \\
&\leq C\sum_{|\alpha|\leq k} \sum_{\substack{\beta,\gamma \\ \beta+\gamma=\alpha}} \frac{\alpha!}{\beta!\, \gamma!} \sup_V |\partial^{\beta}\rho|\sup_V|\partial^{\gamma}\psi| \\
&\leq C' \sum_{|\alpha|\leq k}\sup_V|\partial^{\gamma}\psi|
\end{align}
Theorem: Let $u \in \mathcal{D}'(X)$ with $\operatorname{supp}u=\{0\}$. Then $\displaystyle u = \sum_{|\alpha|\leq k} c_{\alpha} \delta^{(\alpha)}$ where $k \in \Bbb{Z}^+$, $c_{\alpha} \in \Bbb{C}$ and $\delta^{(\alpha)}=\partial^{\alpha} \delta$.
Proof: Let's fix $r>0$ such that $\{x : |x|\leq r\} \subset X$. By the lemma there exists constants $C$, $k$ such that
$$
|\langle u,\psi \rangle |\leq C\sum_{|\alpha|\leq k}\,\sup_{|x|
If $\chi \in C^{\infty}(X)$ and if $\partial^{\alpha}\chi(0)=0$ for all $|\alpha|\leq k$ then $\langle u, \chi \rangle = 0$
Let's fix $\rho \in C^{\infty}_C(\mathbb{R}^d)$ with $\operatorname{supp} \rho \subset \{x : |x| < r\}$ and $\rho = 1$ on $\{x : |x| \leq r/2\}$. Then for all $\epsilon > 0$,
\begin{align}
|\langle u,\chi \rangle| &= |\langle u, \rho(x/\epsilon) \chi(x)\rangle| \\
&\leq C\sum_{|\alpha|\leq k}\,\sup_{|x|
But the function $\partial^{\gamma}\chi$ satisfies $\partial^{\eta}(\partial^{\gamma}\chi)(0)=0$ for all multi index $\eta$ with $|\eta|\leq k-|\gamma|$. Hence, by Taylor's theorem applied to $\partial^{\gamma}\chi$ at $0$, we have
$$
\partial^{\gamma}\chi (x) = o(|x|^{k-|\gamma|}) \quad \quad (x \to 0)
$$
So
$$
\sup_{|x|
Substituting this information in $(\star)$, it follows that
$$
|\langle u,\chi \rangle| = o(1)\quad\quad (\epsilon \to 0)
$$
Since $|\langle u,\chi \rangle|$ does not depend on $\epsilon$, it follows that $\langle u,\chi \rangle = 0$.
$\displaystyle u = \sum_{|\alpha| \leq k} c_{\alpha} \delta^{(\alpha)}$ for some $c_{\alpha} \in \Bbb{C}$
Let's take $\varphi \in C^{\infty}_C(X)$. Let $\displaystyle \chi(x) := \varphi(x) - \sum_{|\alpha| \leq k}\frac{\partial^{\alpha}\varphi(0)}{\alpha!}x^{\alpha}$. Then $\chi \in C^{\infty}(X)$ and $\partial^{\alpha}\chi(0)=0$ for all $|\alpha| \leq k$. By the first part of the proof, $\langle u,\chi \rangle=0$. Hence
\begin{align}
\langle u,\varphi \rangle &= \left\langle u, \sum_{|\alpha| \leq k}\frac{\partial^{\alpha}\varphi(0)}{\alpha!}x^{\alpha}\right\rangle \\
&= \sum_{|\alpha| \leq k} \partial^{\alpha}\varphi(0) \langle u,x^{\alpha}/\alpha! \rangle \\
&= \sum_{|\alpha| \leq k} \langle u,x^{\alpha}/\alpha! \rangle (-1)^{|\alpha|} \langle \delta^{(\alpha)}, \varphi \rangle
\end{align}
and it follows that
$$
u = \sum_{|\alpha| \leq k} c_{\alpha} \delta^{(\alpha)}
$$
where $c_{\alpha} = (-1)^{|\alpha|}\langle u,x^{\alpha}/\alpha! \rangle$.
