Let $\mathcal S$ be the Schwartz space on $\mathbb R$, the set of functions $f\in\cal C^{\infty}$ on $\mathbb R$ of rapid decrease, i.e., such that for every $k,l\ge 0$ $$ \sup |x|^k|f^{(l)}(x)|<\infty. $$ Given a fixed $x\in\mathbb{R}$ and $f,g\in\cal{S}$, show that $F(t)=f(x-t)g(t)$ is also of rapid decrease.
I was able to show that $fg\in\cal S$. Since $|x|^k(\frac{d}{dx})^l[f(x)g(x)] = |x|^k|\sum_{i,j} A_{i,j}f^{(i)}(x)g^{(j)}(x)|\le \sum_{i,j}A_{i,j}|x|^k|f^{(i)}(x)||g^{(j)}(x)|$ and $|x|^k|f^{(i)}(x)|\le M_i$ and $|g^{(j)}(x)|\le P_j$ and the sum is finite, we can take $M=\max_{i,j}(M_i, P_j)$ then we will have a upper bound for $\sum_{i,j}A_{i,j}|x|^k|f^{(i)}(x)||g^{(j)}(x)|$, therefore $fg$ is of rapid decrease.
But I'm struggling to show that $F(t) = f(x-t)$ is of rapid decrease to conclude the argument.
Any help will be much appreciated. Thanks in advance.
Hints: We have to show that $\sup |t|^{l}|f^{(l)}(t)|<\infty$.
$|t|^{l} =|x-(x-t)|^{l}\leq 2^{l} (|x|^{l}+|x-t|^{l})$. Can you check that $\sup|f^{(l)}(t)|<\infty$?