# $\langle T,\varphi_n\rangle\rightarrow 0$ for all distributions $T$ of finite order $\implies\varphi_n\rightarrow 0$ in $\mathcal D(\mathbb{R}^{d})$.

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.

• The argument in the answer here might be helpful: math.stackexchange.com/questions/2106333/… Mar 26 at 20:20
• @PhoemueX Wouldn't it be possible to give direct proof?
– Math
Mar 26 at 20:54

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.
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 $$\mathcal{D}(\mathbb{R})$$.