TITLE: If two functions are in Schwartz space,then their convolution is also in Schwartz space
QUESTION [9 upvotes]: For the proposition (i), I don't know how to show the inequality in the second line of the proof, can someone help me?

REPLY [2 votes]: Consider the hints from the authors:
When $\left\vert x \right\vert \leq 2\left\vert y \right\vert$, then
\begin{equation*}
 \left\vert x \right\vert^\ell\left\vert g(x-y) \right\vert\leq C_0\left\vert 2y \right\vert^\ell\leq2^\ell C_0 (1+\left\vert y \right\vert)^\ell,
\end{equation*}
where $C_0\geq \sup\left\vert g(x-y) \right\vert$.
When $\left\vert x \right\vert > 2\left\vert y \right\vert$, then
\begin{align*}
 \left\vert x \right\vert^\ell\left\vert g(x-y) \right\vert &= \left\vert (x-y) + y \right\vert^\ell\left\vert g(x-y) \right\vert\\
 &=\left\vert (x-y) \right\vert^\ell \left\vert 1 + \frac{\left\vert y \right\vert}{\left\vert x-y \right\vert} \right\vert^\ell\left\vert g(x-y) \right\vert\\
 &\leq\left\vert (x-y) \right\vert^\ell \left\vert 1 + \frac{\left\vert y \right\vert}{\left\vert x \right\vert - \left\vert y \right\vert} \right\vert^\ell\left\vert g(x-y) \right\vert\\
 &\leq\left\vert (x-y) \right\vert^\ell \left\vert 1 + \frac{\left\vert y \right\vert}{\left\vert 2y \right\vert - \left\vert y \right\vert} \right\vert^\ell\left\vert g(x-y) \right\vert\\
 &=\left\vert (x-y) \right\vert^\ell 2^\ell\left\vert g(x-y) \right\vert\\
 &\leq 2^\ell\left\vert (x-y) \right\vert^\ell C_{\ell}\left\vert x-y \right\vert^{-\ell}\\
  &\leq 2^\ell C_{\ell}(1+\left\vert y\right\vert)^{\ell},
\end{align*}
where $C_\ell\geq \sup \left\vert (x-y) \right\vert^\ell\left\vert g(x-y) \right\vert$.
Now let $A_\ell = 2^\ell\max\{C_0, C_\ell\}$.