I'm having trouble seeing this bound I've seen on a proof.
Let $f$ be a polynomial, and $F$ the polynomial obtained from $f$ by replacing each coefficient by its absolute value. Then:
$$\bigg{|}\int_0^t e^{t-u}f(u)du\bigg{|}\leq |t|e^{|t|}F(|t|)$$
Thanks in advance for any help.
$u \mapsto e^{t-u}$ is decrasing, so :
$$\bigg{|}\int_0^t e^{t-u}f(u)du\bigg{|}\leq \int_0^t |e^{t-u}||f(u)|du \leq e^{|t|}\int_0^t F(|u|)du$$
But $F'(|u|) \ge 0$ because it is the derivative of a polynomial with positive coefficients (fill with details, you have to consider the cases $t>0$ and $t<0$), so $F(|u|)$ is increasing and so $\int_0^t F(|u|)du\leq tF(|t|).$
Finally : $\bigg{|}\int_0^t e^{t-u}f(u)du\bigg{|}\leq |t|e^{|t|}F(|t|)$