[proofplan]
We show that a distribution supported at a single point must be a finite linear combination of the Dirac delta and its derivatives. The argument proceeds in three stages: first, invoke the characterisation of distributions to obtain a finite-order seminorm bound on $T$; second, use a cutoff-and-rescaling argument to show that $T$ annihilates every test function whose Taylor expansion vanishes to order $N$ at $x_0$; third, decompose an arbitrary test function via Taylor's theorem into a polynomial part (which $T$ acts on explicitly) and a remainder (which $T$ annihilates), thereby identifying $T$ with a linear combination of $\partial^\alpha \delta_{x_0}$.
[/proofplan]
[step:Establish a finite-order seminorm bound on $T$]
Since $\mathrm{supp}(T) = \{x_0\}$, the distribution $T$ acts only on test functions supported near $x_0$.
Fix $r > 0$ small enough that $\overline{B}(x_0, r) \subset \Omega$, and set $K := \overline{B}(x_0, r)$.
By the [Characterisation of Distributions](/theorems/449) applied to $T$ and the compact set $K$, there exist $N \in \mathbb{N}_0$ and $C > 0$ such that
\begin{align*}
|T(\varphi)| &\leq C \sum_{|\alpha| \leq N} \sup_{K} |\partial^\alpha \varphi|
\end{align*}
for all $\varphi \in \mathcal{D}(B(x_0, r))$.
[guided]
Every distribution, when restricted to test functions supported in a fixed compact set, is controlled by finitely many derivatives.
We need this finite-order bound because the rest of the proof hinges on Taylor expansions of order $N$.
The [Characterisation of Distributions](/theorems/449) provides exactly this: for any compact $K \subset \Omega$, the action of $T$ on $\mathcal{D}(K)$ is bounded by a finite combination of suprema of derivatives.
We apply it with $K = \overline{B}(x_0, r) \subset \Omega$ (which is compact since $r$ is small enough).
This yields $N \in \mathbb{N}_0$ and $C > 0$ such that
\begin{align*}
|T(\varphi)| &\leq C \sum_{|\alpha| \leq N} \sup_{K} |\partial^\alpha \varphi|
\end{align*}
for all $\varphi \in \mathcal{D}(B(x_0, r))$.
The integer $N$ is the order of $T$ at $x_0$ and will determine the number of delta derivatives appearing in the final representation.
[/guided]
[/step]
[step:Show that $T$ annihilates test functions vanishing to order $N$ at $x_0$]
Let $\varphi \in \mathcal{D}(B(x_0, r))$ satisfy $\partial^\alpha \varphi(x_0) = 0$ for all $|\alpha| \leq N$.
By Taylor's theorem with integral remainder, $\varphi(x) = \sum_{|\beta| = N+1} R_\beta(x)(x - x_0)^\beta$ where each $R_\beta \in C^\infty$.
Fix a cutoff $\chi \in \mathcal{D}(B(0, 2))$ with $\chi = 1$ on $B(0, 1)$ and $0 \leq \chi \leq 1$.
Define $\chi_\varepsilon(x) := \chi((x - x_0)/\varepsilon)$ and set $\varphi_\varepsilon := (1 - \chi_\varepsilon)\varphi$.
[claim:$T(\chi_\varepsilon \varphi) \to 0$ as $\varepsilon \to 0$]
The seminorm bound and the Taylor vanishing of $\varphi$ at $x_0$ imply $T(\chi_\varepsilon \varphi) \to 0$.
[/claim]
[proof]
By the Leibniz rule, $\partial^\alpha(\chi_\varepsilon \varphi)$ is a sum of terms $\binom{\alpha}{\gamma}(\partial^\gamma \chi_\varepsilon)(\partial^{\alpha - \gamma}\varphi)$.
The chain rule gives $|\partial^\gamma \chi_\varepsilon(x)| \leq \varepsilon^{-|\gamma|} \|\partial^\gamma \chi\|_{L^\infty}$, and this factor is supported on $B(x_0, 2\varepsilon) \setminus B(x_0, \varepsilon)$.
On this annular region, $|x - x_0| \leq 2\varepsilon$, so the Taylor vanishing $\partial^\alpha \varphi(x_0) = 0$ for $|\alpha| \leq N$ gives $|\partial^{\alpha - \gamma}\varphi(x)| \leq C_{\alpha,\gamma} |x - x_0|^{N + 1 - |\alpha - \gamma|}$ for $|\alpha - \gamma| \leq N$.
The product satisfies $|\partial^\gamma \chi_\varepsilon \cdot \partial^{\alpha - \gamma}\varphi| \leq C \varepsilon^{-|\gamma|} \varepsilon^{N + 1 - |\alpha - \gamma|} = C \varepsilon^{N + 1 - |\alpha|}$.
For $|\alpha| \leq N$, this tends to zero as $\varepsilon \to 0$.
Summing over $\gamma$ and then over $\alpha$, each term in $\sum_{|\alpha| \leq N} \sup_K |\partial^\alpha(\chi_\varepsilon \varphi)|$ tends to zero.
The seminorm bound then gives $|T(\chi_\varepsilon \varphi)| \to 0$.
[/proof]
Since $T(\varphi) = T(\chi_\varepsilon \varphi) + T(\varphi_\varepsilon)$ and $\mathrm{supp}(\varphi_\varepsilon) \cap \{x_0\} = \varnothing$ for each $\varepsilon > 0$ (because $\chi_\varepsilon = 1$ near $x_0$), the distribution $T$ vanishes on $\varphi_\varepsilon$ (its support is $\{x_0\}$ and $\varphi_\varepsilon$ is supported away from $x_0$).
Taking $\varepsilon \to 0$: $T(\varphi) = \lim_{\varepsilon \to 0} T(\chi_\varepsilon \varphi) = 0$.
[guided]
The goal is to show $T(\varphi) = 0$ whenever $\varphi$ vanishes to order $N$ at $x_0$.
The difficulty is that $T$ is a distribution, not a pointwise object, so we cannot simply evaluate at $x_0$.
The strategy is to split $\varphi = \chi_\varepsilon \varphi + (1 - \chi_\varepsilon)\varphi$, where $\chi_\varepsilon$ is a smooth cutoff equal to $1$ on $B(x_0, \varepsilon)$.
The second piece $(1 - \chi_\varepsilon)\varphi$ is supported away from $x_0$.
Since $\mathrm{supp}(T) = \{x_0\}$, the distribution $T$ vanishes on any test function whose support does not contain $x_0$.
So $T((1 - \chi_\varepsilon)\varphi) = 0$ for each $\varepsilon > 0$.
For the first piece $\chi_\varepsilon \varphi$, we use the seminorm bound from the previous step.
By the Leibniz rule, $\partial^\alpha(\chi_\varepsilon \varphi) = \sum_{\gamma \leq \alpha} \binom{\alpha}{\gamma}(\partial^\gamma \chi_\varepsilon)(\partial^{\alpha-\gamma}\varphi)$.
The derivatives of $\chi_\varepsilon$ grow as $\varepsilon^{-|\gamma|}$ (chain rule: $\partial^\gamma \chi_\varepsilon(x) = \varepsilon^{-|\gamma|}(\partial^\gamma \chi)((x - x_0)/\varepsilon)$), but the Taylor vanishing of $\varphi$ provides compensating decay.
Since $\partial^{\alpha - \gamma}\varphi(x_0) = 0$ for $|\alpha - \gamma| \leq N$ (because all derivatives up to order $N$ vanish at $x_0$), Taylor's theorem gives $|\partial^{\alpha - \gamma}\varphi(x)| \leq C|x - x_0|^{N + 1 - |\alpha - \gamma|}$ near $x_0$.
On the support of $\partial^\gamma \chi_\varepsilon$ (which lies in $B(x_0, 2\varepsilon)$), we have $|x - x_0| \leq 2\varepsilon$, so the product is bounded by $C\varepsilon^{-|\gamma|}\varepsilon^{N + 1 - |\alpha - \gamma|} = C\varepsilon^{N + 1 - |\alpha|}$.
For $|\alpha| \leq N$ this power is at least $1$, so each term vanishes as $\varepsilon \to 0$.
The seminorm bound then gives $|T(\chi_\varepsilon \varphi)| \leq C \sum_{|\alpha| \leq N} \sup |\partial^\alpha(\chi_\varepsilon \varphi)| \to 0$.
Therefore $T(\varphi) = \lim_{\varepsilon \to 0}[T(\chi_\varepsilon \varphi) + T((1 - \chi_\varepsilon)\varphi)] = 0 + 0 = 0$.
[/guided]
[/step]
[step:Decompose an arbitrary test function via Taylor's theorem and identify $T$]
Let $\varphi \in \mathcal{D}(B(x_0, r))$ be arbitrary.
By Taylor's theorem,
\begin{align*}
\varphi(x) &= \sum_{|\alpha| \leq N} \frac{\partial^\alpha \varphi(x_0)}{\alpha!}(x - x_0)^\alpha + \psi(x),
\end{align*}
where $\psi \in C^\infty$ satisfies $\partial^\alpha \psi(x_0) = 0$ for all $|\alpha| \leq N$.
By the previous step, $T(\psi) = 0$.
Fix $\eta \in \mathcal{D}(B(x_0, r))$ with $\eta = 1$ in a neighbourhood of $x_0$.
The monomial $(x - x_0)^\alpha$ does not have compact support, but $\eta \cdot (x - x_0)^\alpha \in \mathcal{D}(B(x_0, r))$ and agrees with $(x - x_0)^\alpha$ near $x_0$.
Since $T$ is supported at $\{x_0\}$, the value $T(\eta \cdot (x - x_0)^\alpha)$ is independent of the choice of $\eta$ (any two such cutoffs agree near $x_0$, and their difference is supported away from $x_0$, where $T$ vanishes).
By linearity of $T$:
\begin{align*}
T(\varphi) &= \sum_{|\alpha| \leq N} \frac{\partial^\alpha \varphi(x_0)}{\alpha!} T(\eta \cdot (x - x_0)^\alpha).
\end{align*}
Define $c_\alpha := \frac{(-1)^{|\alpha|}}{\alpha!} T(\eta \cdot (x - x_0)^\alpha)$.
Recall that $\partial^\alpha \delta_{x_0}$ acts by $(\partial^\alpha \delta_{x_0})(\varphi) = (-1)^{|\alpha|} \partial^\alpha \varphi(x_0)$.
Therefore
\begin{align*}
T(\varphi) &= \sum_{|\alpha| \leq N} c_\alpha \cdot (-1)^{|\alpha|} \partial^\alpha \varphi(x_0) = \sum_{|\alpha| \leq N} c_\alpha (\partial^\alpha \delta_{x_0})(\varphi).
\end{align*}
Since this holds for every $\varphi \in \mathcal{D}(\Omega)$, we conclude $T = \sum_{|\alpha| \leq N} c_\alpha \, \partial^\alpha \delta_{x_0}$.
[/step]
