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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Sign up to join this communityI 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
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 $t\lt 0$.