[proofplan] The sufficiency direction is straightforward: conditions (1)-(2) place the sequence in a single Fréchet step $\mathcal{D}_K(\Omega)$ and assert convergence there, which lifts to $\mathcal{D}(\Omega)$ by continuity of the inclusion. The necessity direction is the substantive part. For condition (1), we prove the contrapositive: if supports escape every compact set, we construct a continuous seminorm on $\mathcal{D}(\Omega)$ that is unbounded on a subsequence, contradicting convergence. For condition (2), we use the fact that the subspace topology on $\mathcal{D}_K(\Omega)$ inherited from $\mathcal{D}(\Omega)$ coincides with the Fréchet topology, so convergence in $\mathcal{D}(\Omega)$ restricted to a single step gives convergence in all seminorms. [/proofplan] [step:Set up the Fréchet steps of $\mathcal{D}(\Omega)$] Fix a compact exhaustion $K_1 \subset K_2 \subset \cdots$ of $\Omega$ with $K_j \subset \operatorname{int}(K_{j+1})$ and $\bigcup_{j=1}^\infty K_j = \Omega$. For each $j$, the space \begin{align*} \mathcal{D}_{K_j}(\Omega) := \{\varphi \in C^\infty(\Omega) : \operatorname{supp}(\varphi) \subseteq K_j\} \end{align*} carries the Fréchet topology generated by the seminorms \begin{align*} p_{j,N}(\varphi) := \sum_{|\alpha| \leq N} \sup_{x \in K_j} |\partial^\alpha \varphi(x)|, \quad N \in \mathbb{N}_0. \end{align*} The [strict inductive limit topology](/page/Strict%20Inductive%20Limit%20Topology) on $\mathcal{D}(\Omega) = \bigcup_{j=1}^\infty \mathcal{D}_{K_j}(\Omega)$ is the finest locally convex topology making each inclusion $\iota_j: \mathcal{D}_{K_j}(\Omega) \hookrightarrow \mathcal{D}(\Omega)$ continuous. A seminorm $p$ on $\mathcal{D}(\Omega)$ is continuous if and only if $p|_{\mathcal{D}_{K_j}(\Omega)}$ is continuous for every $j$. [/step] [step:Prove sufficiency: conditions (1)-(2) imply convergence in $\mathcal{D}(\Omega)$] Suppose conditions (1) and (2) hold with compact set $K$. Choose $j_0$ such that $K \subseteq K_{j_0}$. Then $\varphi_k, \varphi \in \mathcal{D}_{K_{j_0}}(\Omega)$ for all $k$. Condition (2) states $\sup_{x \in K}|\partial^\alpha \varphi_k(x) - \partial^\alpha \varphi(x)| \to 0$ for every multi-index $\alpha$. Since $\varphi_k - \varphi$ is supported in $K \subseteq K_{j_0}$: \begin{align*} p_{j_0, N}(\varphi_k - \varphi) = \sum_{|\alpha| \leq N} \sup_{x \in K_{j_0}} |\partial^\alpha(\varphi_k - \varphi)(x)| = \sum_{|\alpha| \leq N} \sup_{x \in K} |\partial^\alpha(\varphi_k - \varphi)(x)| \to 0 \end{align*} for every $N \in \mathbb{N}_0$. This is convergence $\varphi_k \to \varphi$ in the Fréchet space $\mathcal{D}_{K_{j_0}}(\Omega)$. Since $\iota_{j_0}$ is continuous, $\varphi_k \to \varphi$ in $\mathcal{D}(\Omega)$. [/step] [step:Prove necessity of condition (1) by constructing a continuous seminorm unbounded on a subsequence] Suppose $\varphi_k \to \varphi$ in $\mathcal{D}(\Omega)$ but condition (1) fails. [claim:Continuous Seminorm Unbounded On Subsequence] There exist a subsequence $\{\varphi_{k_j}\}_{j=1}^\infty$ and a continuous seminorm $p$ on $\mathcal{D}(\Omega)$ with $p(\varphi_{k_j}) \geq j$ for every $j$. [/claim] [proof] Since condition (1) fails, for every $j \in \mathbb{N}$ there exists $k_j$ with $\operatorname{supp}(\varphi_{k_j}) \not\subseteq K_j$. By passing to a subsequence, assume $k_1 < k_2 < \cdots$. For each $j$, pick $x_j \in \operatorname{supp}(\varphi_{k_j}) \setminus K_j$. Since $x_j \in \operatorname{supp}(\varphi_{k_j})$, there exists a multi-index $\alpha_j$ with $\partial^{\alpha_j}\varphi_{k_j}(x_j) \neq 0$. (If all derivatives vanished at $x_j$, smoothness would force $\varphi_{k_j}$ to vanish in a neighbourhood, contradicting $x_j \in \operatorname{supp}(\varphi_{k_j})$.) Set $\delta_j := |\partial^{\alpha_j}\varphi_{k_j}(x_j)| > 0$. Define the seminorm \begin{align*} p(\varphi) := \sum_{j=1}^\infty \frac{j}{\delta_j}\,|\partial^{\alpha_j}\varphi(x_j)|. \end{align*} **Well-definedness and continuity:** Fix $m \in \mathbb{N}$ and $\varphi \in \mathcal{D}_{K_m}(\Omega)$. For $j \geq m$, $x_j \in \Omega \setminus K_j \subseteq \Omega \setminus K_m$ lies outside $\operatorname{supp}(\varphi)$, so $\partial^{\alpha_j}\varphi(x_j) = 0$. Therefore the sum is finite: \begin{align*} p(\varphi) = \sum_{j=1}^{m-1} \frac{j}{\delta_j}\,|\partial^{\alpha_j}\varphi(x_j)| \leq C_m \, p_{m, N_m}(\varphi), \end{align*} where $N_m := \max_{1 \leq j \leq m-1} |\alpha_j|$ and $C_m := \sum_{j=1}^{m-1} j/\delta_j$. Since $p_{m, N_m}$ is continuous on $\mathcal{D}_{K_m}(\Omega)$, so is $p|_{\mathcal{D}_{K_m}(\Omega)}$. As this holds for every $m$, $p$ is continuous on $\mathcal{D}(\Omega)$. **Unboundedness:** The $j$-th term gives $p(\varphi_{k_j}) \geq \frac{j}{\delta_j}|\partial^{\alpha_j}\varphi_{k_j}(x_j)| = j$. [/proof] This contradicts $\varphi_k \to \varphi$ in $\mathcal{D}(\Omega)$: convergence requires $p(\varphi_{k_j} - \varphi) \to 0$ for every continuous seminorm, forcing $p(\varphi_{k_j})$ to remain bounded by $p(\varphi) + p(\varphi_{k_j} - \varphi)$. But $p(\varphi_{k_j}) \geq j \to \infty$, a contradiction. Therefore condition (1) holds. [/step] [step:Prove necessity of condition (2) by reducing to convergence in a single Fréchet step] Condition (1) provides a compact set $K$ with $\operatorname{supp}(\varphi_k) \subseteq K$ for all $k$. Choose $j_0$ with $K \subseteq K_{j_0}$, so $\varphi_k, \varphi \in \mathcal{D}_{K_{j_0}}(\Omega)$ for all $k$. The subspace topology that $\mathcal{D}_{K_{j_0}}(\Omega)$ inherits from $\mathcal{D}(\Omega)$ coincides with its Fréchet topology. (The inclusion $\iota_{j_0}$ is continuous by definition; the reverse inclusion follows from the strict inductive limit property that each $\mathcal{D}_{K_j}(\Omega)$ is closed in $\mathcal{D}_{K_{j+1}}(\Omega)$ with the subspace topology agreeing with the Fréchet topology.) Therefore $\varphi_k \to \varphi$ in $\mathcal{D}(\Omega)$ implies convergence in the Fréchet topology of $\mathcal{D}_{K_{j_0}}(\Omega)$, which means $p_{j_0, N}(\varphi_k - \varphi) \to 0$ for every $N$. Unpacking: $\sup_{x \in K_{j_0}}|\partial^\alpha(\varphi_k - \varphi)(x)| \to 0$ for every $\alpha$. Since $\varphi_k - \varphi$ is supported in $K$, this is $\sup_{x \in K}|\partial^\alpha \varphi_k(x) - \partial^\alpha \varphi(x)| \to 0$ — condition (2). [/step] [step:Confirm the support of the limit] Condition (1) gives $\operatorname{supp}(\varphi_k) \subseteq K$ for all $k$, so $\varphi_k(x) = 0$ for $x \in \Omega \setminus K$. Condition (2) with $\alpha = 0$ gives $\varphi_k \to \varphi$ uniformly. For $x \in \Omega \setminus K$: $\varphi(x) = \lim_{k \to \infty} \varphi_k(x) = 0$. Therefore $\operatorname{supp}(\varphi) \subseteq K$, confirming $\varphi \in \mathcal{D}(\Omega)$. [/step]