# Convergence of a test function sequence

Let $$\varphi\in\mathcal{D}(\mathbb{R}^d)$$, $$(x_n)\subset\mathbb{R}^d$$ and $$x\in\mathbb{R}^d$$ be such that $$x_n\to x$$ in $$\mathbb{R}^d$$, where $$\mathcal{D}(\mathbb{R}^d)$$ denotes the linear space of smooth compactly supported functions. I need to prove that $$\varphi(\cdot-x_n)\to \varphi(\cdot-x) \quad \text{in } \mathcal{D}(\mathbb{R}^d).$$

My attempt. First, we define $$\psi_n(y)=\varphi(y-x_n),\qquad \psi(y)=\varphi(y-x).$$

First, I will construct a compact such that $$\text{supp}\psi_n$$ and $$\text{supp}\psi$$ are embedded in it.

Since $$\varphi\in\mathcal{D}(\mathbb{R}^d)$$, there is a compact $$K$$ such that $$\text{supp}\varphi\subset K$$, and, given that $$x_n\to x$$ there exists $$R>0$$ such that $$x_n\subset B(x,R)$$ for all $$n$$. Therefore,

$$\psi_n(y)\neq 0 \Leftrightarrow \varphi(y-x_n)\neq0 \Leftrightarrow y\in x_n+\{\xi\in\mathbb{R}^d : \varphi(\xi)\neq0\}.$$

Thus, $$\{y\in\mathbb{R^d} : \psi_n(y)\neq0\} = x_n + \{\xi\in\mathbb{R}^d : \varphi(\xi)\neq0\} \subset B(x,R)+K \Rightarrow \text{supp}\psi_n\subset K',$$ where $$K'=\overline{B(x,R)+K}$$ is a compact (closed and bounded). Analogously, $$\text{supp}\psi\subset K'$$.

All that is left to prove is that, for any multi-index $$\alpha$$, $$\lim_{n\to\infty} \max_{y\in K'} |\partial^\alpha\psi_n(y) - \partial^\alpha\psi(y)| = 0.$$

Given any $$y\in K'$$, since $$x_n\to x$$ and $$\partial^\alpha\varphi(y-\cdot)$$ is continuous, it is easy to deduce that $$|\partial^\alpha\psi_n(y) - \partial^\alpha\psi(y)| = |\partial^\alpha\varphi(y-x_n) - \partial^\alpha\varphi(y-x)| \to 0 \text { as } n\to\infty.$$

But this is not enough to conclude what I need, since the above is a punctual convergence on $$K'$$, and the desired result is a uniform convergence on $$K'$$, and I don't know how to obtain what I want. I would appreciate any help you can give me.

It can be done by a simple application of the mean value theorem: For any $$\alpha\in\mathbb{Z}^n_+$$ $$|\partial^\alpha\phi(y-x)-\partial^\alpha\phi(y-x_n)|\leq\|D\partial^\alpha\phi(u)\||x-x_n|\leq \|D\partial^\alpha\phi\||x-x_n|$$ where $$D$$ is the total derivative and $$u$$ is a point in the line joining $$y-x$$ and $$y-x_n$$, and $$\|\;\|$$ is the norm on $$(\mathbb{R}^n)^*$$ ( we consider $$D\partial^\alpha\phi(u)$$ as a linear functional on $$\mathbb{R}^n$$).

• Thank you very much!! Apr 10, 2021 at 1:43