A proof relies on a inequality I cant a mange to understand ;
$\sup_{x} \mid x \mid ^{l} \mid g(x-y) \mid \le A_{l}(1+ \mid y \mid )^{l} $ given that $g \in S(\mathbb{R})$ i.e $\sup_{x} \mid x \mid ^{l} \mid g(x) \mid \le A_{l}$
the hint is to consider $\mid x \mid \le 2\mid y\mid$ and $\mid x \mid \ge 2\mid y\mid$
This is the usual issue about comparing translated bump functions.
Unfortunately, the claim seems to be false as stated. Consider the case $x=y$. Then the claim says: $|x|^l |g(0)| \le A_l (1+|x|)^l$ for all $x$. This is absurd, because $A_l$ doesn't control the size of $g(0)$.
To give a specific counterexample, let $l=2$ and $$g(x)=\frac{1}{10^{-100}+e^{x^2}-1}$$ $g$ is a Schwartz function and $|g(x)|\le |x|^{-2}$ (by Taylor expansion of $e^{x^2}$), so $A_2=1$. But $g(0)=10^{100}$...
Here is one possible fix that makes the statement correct:
Define $$A_l = \sup_{x} (1+|x|)^l |g(x)|.$$
This controls $|g(x)|$ also for small $|x|$. Let us prove the claim.
We have for all $x,y$:
$$|x|^l |g(x-y)| \le A_l (1+|x|)^l (1+|x-y|)^{-l} \le A_l (1+|y|)^l,$$
where we used the inequality
$$\frac1{1+|x-y|}\le \frac{1+|y|}{1+|x|}.$$
(Proof: $(1+|y|)(1+|x-y|)=1+|y|+|x-y|+|y||x-y|\ge 1+|y|+|x|-|y|=1+|x|.$)