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Let $(\varphi_n)_{n \in \mathbb{N}} \subset \mathcal{D}(\mathbb{R}^{d})$ such that $\langle T, \varphi_n \rangle \rightarrow 0$ for all distributions $T$ of finite order. Prove that $\varphi_n \rightarrow 0$ in $\mathcal{D}(\mathbb{R}^{d})$.

My attempt: We have to prove that there exists a compact $K \subset \mathbb{R}^{d}$ such that

  • $\operatorname{supp}(\varphi_n) \subset K$ for all $n \in \mathbb{N}$ and
  • $|D^\alpha \varphi_n(x)|\rightarrow 0$ uniformly in $K$ for all $\alpha\in\mathbb{N}^{d}$.

Let $T \in \mathcal{D}'(\mathbb{R}^{d})$ of finite order, say $k$. Then, there exists a compact $K_k$ and $C_{K_k}>0$ such that $|\langle T, \varphi \rangle| \leq C_{K_k}\max_{|\alpha|\leq k}\sup_{x \in K}|D^\alpha\varphi (x)|$ for all $\varphi \in C_{0}^{\infty}(K_{k})$.

But I don't know how to proceed.

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The desired result is true.


Compactly supported distributions on $\mathbb R^d$ have finite order, so $\langle T,\varphi_n\rangle\to 0$ for every $T\in\mathcal E'(\mathbb R^d)$. And as in the post linked by PhoemueX, the fact that $\mathcal E(\mathbb R^d)$ is a Montel space implies that the sequence $(\varphi_n)_{n\in\mathbb N}$ converges to $0$ in $\mathcal E(\mathbb R^d)$.

Convergence of $(\varphi_n)_{n\in\mathbb N}$ to $0$ in $\mathcal E(\mathbb R^d)$ means that for every multiindex $\alpha\in\mathbb N^d$, the sequence $(D^\alpha\varphi_n)_{n\in\mathbb N}$ converges to $0$ uniformly in every compact subset of $\mathbb R^d$. So all that's left to show is that there is a compact subset of $\mathbb R^d$ which contains the support of every $\varphi_n$.

Let's proceed by contradiction and suppose that no compact subset of $\mathbb R^d$ contains the support of every $\varphi_n$. For every $m\in\mathbb N$, since the compact set $$K_m=[-m,m]^d\cup\bigcup_{k=0}^m\operatorname{supp}\varphi_k$$ fails to contain the support of every $\varphi_n$, there is an integer $n>m$ and a point $x$ outside of $K_m$ such that $\varphi_n(x)\neq 0$. So by induction, we are able to construct a strictly increasing sequence $(n_k)_{k\in\mathbb N}$ of elements of $\mathbb N$ as well as a sequence $(x_k)_{k\in\mathbb N}$ of elements of $\mathbb R^d$ such that for every $k\in\mathbb N$, $$x_{k+1}\notin K_{n_k}\quad\text{and}\quad\varphi_{n_k}(x_k)\neq 0.$$ Then, $$T=\sum_{k\in\mathbb N}\frac 1{\varphi_{n_k}(x_k)}\delta_{x_k}$$ is a well-defined distribution of order $0$ on $\mathbb R^d$, but the sequence $(\langle T,\varphi_n\rangle)_{n\in\mathbb N}$ takes the value $1$ infinitely many times, which contradicts the fact that $(\varphi_n)_{n\in\mathbb N}$ converges to $0$ when evaluated against distributions of finite order.

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I don't believe the statement is true. In dimension 1, if $\varphi_n(x)=\frac{x^n}{n!}\theta(x)$ where $\theta$ is some adequate bump function, for any fixed integer $k$ the sequence $(\varphi^{(k)}_n)_n$ goes uniformly to $0$ as $n\rightarrow +\infty$ (which implies convergence towards $0$ when applying any finite order distribution ) but the uniform norm of $\varphi_n^{(n)}$ is at least $1$ (which forbids convergence to $0$ in $\mathcal{D}(\mathbb{R})$.

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