I have a small question please :
If $F(t)=\int_0^t f(s)ds$ and $f$ is bounded,
why does there exist $M>0$, $|F(t)|\leq M|t|$ ?
I don't understand why we have $|t|$ and not $t$.
Please
$\begingroup$
$\endgroup$
6
-
1$\begingroup$ $t=|t|$ for $t\geq 0$. $\endgroup$– user9464Aug 20, 2013 at 20:43
-
1$\begingroup$ but we have $t\in \mathbb{R}$ $\endgroup$– VrouvrouAug 20, 2013 at 20:44
-
$\begingroup$ That's because you're bounding $|F(t)|$ rather than $F(t)$. $\endgroup$– Jonathan Y.Aug 20, 2013 at 20:48
-
1$\begingroup$ @JonathanY. i don't understand $\endgroup$– VrouvrouAug 20, 2013 at 20:50
-
1$\begingroup$ We've introduced the absolute value by examining $|F(t)|$. It should come as no surprise that it wouldn't vanish (and, what's more, we can't bound a positive quantity from above by a negative number, such as $Mt$ for $t<0$). $\endgroup$– Jonathan Y.Aug 20, 2013 at 20:55
|
Show 1 more comment
1 Answer
$\begingroup$
$\endgroup$
Let $M$ be a bound for $f$, that means $|f(s)|\le M$ for all $s\in \Bbb R$, then we get $$\begin{align} |F(t)|&=&\left|\int_0^tf(s)ds\right|\le \int_0^t|f(s)|ds\\&\le& \int_0^tMds=Mt=M|t| \end{align}$$if $t\ge 0$ and$$\begin{align}|F(t)|&=&\left|\int_0^tf(s)ds\right|=\left|-\int_t^0f(s)ds\right|=\left|\int_t^0f(s)ds\right|\\&\le&\int_t^0|f(s)|ds\le\int_t^0Mds=M(-t)=M|t|\end{align}$$if $t\lt 0$.
