Let $\mathbb{B}^n$ denote the open unit ball in $\mathbb{R}^n$. I am reading a paper which claims that functions $a \in C^\infty(\overline{\mathbb{B}^n})$ such that $a|_{\partial{\overline{\mathbb{B}^n}}} = 0$ belong to the symbol class $S^{-1}(\mathbb{R}^n)$. The paper does not precisely define what $C^\infty(\overline{\mathbb{B}^n})$ is, but I think it's meant to be $$a\in C^\infty(\overline{\mathbb{B}^n}) \iff a \in C^\infty(\mathbb{B}^n) \text{ and all its partial derivatives have continuous extensions to $\overline{\mathbb{B}^n}$}.$$
On the other hand, the class $S^{-1}(\mathbb{R}^n)$ is precisely defined to be $$a \in S^{-1}(\mathbb{R}^n) \iff a \in C^\infty(\mathbb{R}^n) \text{ and } |\partial^\alpha a(x) | \le C_{\alpha}\langle x\rangle^{-1-|\alpha|} \text{ for all multi-indices $\alpha$}. $$ Above, $\langle x \rangle = (1 + |x|^2)^{\frac{1}{2}}.$
To prove that functions $a \in C^\infty(\overline{\mathbb{B}^n})$, $a|_{\partial{\overline{\mathbb{B}^n}}} = 0$, belong to $S^{-1}(\mathbb{R}^n)$, one first must explain how $C^\infty(\overline{\mathbb{B}^n})$ is identified as a subset of $C^\infty(\mathbb{R})$. I think the author means to do it just using $a|_{\mathbb{B}^n}$, via the map
$$\mathbb{R}^n \ni x \mapsto \varphi(x) = \frac{x}{\langle x \rangle} \in \mathbb{B}^n , \quad C^\infty(\overline{\mathbb{B}^n}) \ni a \mapsto a(\varphi(\cdot)) \in C^\infty(\mathbb{R}^n),$$
or something similar.
Now the goal is to show that $a \in C^\infty(\overline{\mathbb{B}^n})$, $a|_{\partial{\overline{\mathbb{B}^n}}} = 0$, implies $ \partial_x^{\alpha}a(\varphi(x)) \le C_\alpha \langle x \rangle^{-1-|\alpha|}$ for all $\alpha$.
I am having trouble even showing the base case that $\langle x \rangle a(\varphi(x))$ is bounded on $\mathbb{R}^n.$ I wonder if a higher dimensional version of Taylor's formula would be useful but I'm unsure how to proceed.