TITLE: Why it is true for rapid decreasing function $g$ that: $\sup_{x\in\mathbb{R}}|x|^{l\geq 0}|g^{(k\geq 0)}(x-y)|\leq A_{l,k}(1+|y|)^{l}$
QUESTION [5 upvotes]: If $g$ is of rapid decrease, that is $\displaystyle\sup_{x\in\mathbb{R}}|x|^{l\geq 0}|g^{(k\geq 0)}(x)|<\infty$, then we have: $$\displaystyle\sup_{x\in\mathbb{R}}|x|^{l\geq 0}|g^{(k\geq 0)}(x-y)|\leq A_{l,k}(1+|y|)^{l}$$
where $A_{l,k}\geq 0$ is a constant dependent on $l,k$.
How can I induce the above inequality?

REPLY [2 votes]: Hint. You may observe that
$$
\begin{align}
(1+|x|^2)^{l}|g^{(k)}(x-y)|&=(1+|x-y|^2)^{l}|g^{(k)}(x-y)|\frac{(1+|x|^2)^{l}}{(1+|x-y|^2)^{l}}\\
&\leq a_{l,k}\:(1+|y|^2)^{l}
\end{align}
$$ where we have used Peetre's inequality, whose proof may be found here.