Translation is continuous

Question

Let $\mathcal D$ be the space of 'test-functions'. Those are infinitely differentiable functions with compact support.

Define the following convergence on $\mathcal D$. $(\phi_j) \to \phi$ in $\mathcal D$ if:

1. There exists a compactum $K$ that contains all the supports of $(\phi_j),j=1,2,\dots$ and $\phi$.

2. For every multi-index $\alpha$ $(D^\alpha \phi_j)$ converges uniformly to $D^\alpha \phi$ on $K$.

Let now be $\phi \in D$ and $(v_n) \to 0$ in $\mathbb{R}$ and define $\phi_n(x)=\phi(x-v_n)$. I want to prove that $(\phi_n) \to \phi$ with respect to the convergence in $\mathcal D$. It seems quite obvious but I can't prove why. I started like this:

$|D^\alpha \phi_n - D^\alpha \phi | = |D^\alpha(\phi_n - \phi)|$ is bounded because $\phi_n - \phi$ is infinitely differentiable. Also $\phi_n - \phi \to 0$ pointwise and so $|D^\alpha(\phi_n - \phi)| \to 0$ put I can't see how uniform convergence follows.

Answer 1

Let $R$ such that the support of $\phi$ is contained in $[-R,R]$ and $\lVert v\rVert_\infty:=\sup_n|v_n|$. Then for each $n$, $\operatorname{supp}\phi_n\subset [-R-\lVert v\rVert_\infty,R+\lVert v\rVert_\infty]$ so 1. is fulfilled.

We have $$|\phi_n(x)-\phi(x)|\leqslant \int_{x-|v_n|}^{x+|v_n|}|\phi'(t)|\mathrm dt\leqslant 2|v_n|\sup_{t\in\mathbb R}|\phi'(t)|.$$ The same argument applies for the other derivatives.