2
$\begingroup$

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.

$\endgroup$
0

1 Answer 1

1
$\begingroup$

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 $\mathbb{R}^n$).

$\endgroup$
1
  • $\begingroup$ Thank you very much!! $\endgroup$ Apr 10, 2021 at 1:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .