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

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?

Answer 1

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.