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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

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    $\begingroup$ $t=|t|$ for $t\geq 0$. $\endgroup$
    – user9464
    Aug 20, 2013 at 20:43
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    $\begingroup$ but we have $t\in \mathbb{R}$ $\endgroup$
    – Vrouvrou
    Aug 20, 2013 at 20:44
  • $\begingroup$ That's because you're bounding $|F(t)|$ rather than $F(t)$. $\endgroup$ Aug 20, 2013 at 20:48
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    $\begingroup$ @JonathanY. i don't understand $\endgroup$
    – Vrouvrou
    Aug 20, 2013 at 20:50
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    $\begingroup$ 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 $t<0$). $\endgroup$ Aug 20, 2013 at 20:55

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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 $t\lt 0$.

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