I'm studying some functional analysis, and currently focusing on the theory of distributions. I came across the following exercise (ex. 21.2 in F. Treves, "Topological vector spaces, distributions and kernels") and I wanted to know if my approach to the solution is correct, as I'm still not familiar with these concepts. I would also appreciate any kind of alternative approach.
Let $\{x_j\}$ be a sequence of points in $\Omega \subset \mathbb{R}^d$ (open) which does not have any accumulation point in $\Omega$. I need to prove that for any sequence of complex numbers $\{a_j\}$, the functional on $\mathcal{D}(\Omega)$ defined by $$ \varphi \mapsto \sum_{j = 1}^\infty a_j D^\mathbb{j} \varphi(x_j)$$ is a distribution in $\Omega$ (that is, an element of the dual $\mathcal{D}'(\Omega)$). The differential operator is defined as usual: $$ D^\mathbb\alpha = \left( \frac{\partial}{\partial x_1} \right)^{\alpha_1} \dotsb \left( \frac{\partial}{\partial x_d} \right)^{\alpha_d}$$ for the multi-index $\mathbb{\alpha} = (\alpha_1, \dotsc, \alpha_d) \in \mathbb{N}^d$, and $\mathbb{j} = (j,\dotsc, j) \in \mathbb{N}^d$. The set of test functions $\mathcal{D}(\Omega)$ is the set of functions $C^\infty(\mathbb{R}^d, \mathbb{C})$ whose support is a compact subset of $\Omega$.
My solution: We know that the continuity of a map $\mathcal{D}(\Omega) \to Y$, where $Y$ is a locally convex topological vector space, is equivalent to sequential continuity in $0$. We thus consider a sequence $\{\varphi_n\}$ in $\mathcal{D}(\Omega)$ such that $\varphi_n \to 0$ (the topology on $\mathcal{D}(\Omega)$ is the canonical LF topology) and show that its image under the linear functional converges to $0$ in the locally convex TVS $\mathbb{C}$.
By a characterization of convergence in the space of test functions $\mathcal{D}(\Omega)$, there exists a compact $K \subset \Omega$ such that $\text{supp}(\varphi_n) \subset \Omega$ for every $n \in \mathbb{N}$ and $D^\alpha \varphi_n \to D^\alpha \varphi$ uniformly on $\Omega$ for every $\alpha \in \mathbb{N}^d$.
Now, since the support of the $\varphi_n$ is compact and $\{x_j\}$ does not have any accumulation point, we can extract a finite number of points $\{x_{j_1}, \dotsc, x_{j_r}\} = K \cap \{x_1, x_2, \dotsc\}$ where $\varphi_n$ does not vanish for every $n \in \mathbb{N}$. We have $$ \left| \sum_{j = 1}^\infty a_j D^\mathbb{j} \varphi_n(x_j) \right| \leq \sum_{i = 1}^r \left| a_{j_i} D^\mathbb{j_i} \varphi_n(x_{j_i}) \right| \leq \max_{i=1, \dotsc, r} |a_{j_i}| \sum_{i = 1}^r \left| D^\mathbb{j_i} \varphi_n(x_{j_i}) \right|. $$ Let us denote $A = \max_{i = 1,\dotsc, r} |a_{j_i}|$. Let $\epsilon > 0$. By the convergence $D^\mathbb{j_i} \varphi_n \to 0$ for every $i = 1, \dotsc, r$ (I think that pointwise convergence should even suffice here), we can find $n_i \in \mathbb{N}$ such that for every $n \geq n_i$ $$ \left| D^\mathbb{j_i} \varphi_n(x_{j_i}) \right| < \frac{\epsilon}{r A}.$$ Letting $n_0 = \max_{i = 1,\dotsc,r} n_i$ we have, for $n \geq n_0$, $$ \left| \sum_{j = 1}^\infty a_j D^\mathbb{j} \varphi_n(x_j) \right| < \epsilon, $$ so that the series converges to $0$ (in $\mathbb{C}$), concluding the proof.