# Question on “Absolute value ”

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

• $t=|t|$ for $t\geq 0$. – Jack Aug 20 '13 at 20:43
• but we have $t\in \mathbb{R}$ – Vrouvrou Aug 20 '13 at 20:44
• That's because you're bounding $|F(t)|$ rather than $F(t)$. – Jonathan Y. Aug 20 '13 at 20:48
• @JonathanY. i don't understand – Vrouvrou Aug 20 '13 at 20:50
• 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 $t<0$). – Jonathan Y. Aug 20 '13 at 20:55

## 1 Answer

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 $t\lt 0$.