# Nonincreasing derivative implies nondecreasing function

Suppose $f\colon[0,\infty)\to[0,\infty)$ is differentiable with continuous derivative such that its derivative $f'$ is nonincreasing. Does this imply that $f$ is non-decreasing?.

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Yes. If $f'(x)\lt0$ for some $x\geqslant0$, then $f'(y)\leqslant -c$ for every $y\geqslant x$, with $c\gt0$. The mean value theorem yields $f(y)\leqslant f(x)+cx-cy$ for every $y\geqslant x$. Since $f(x)+cx-cy\to-\infty$ when $y\to+\infty$, this contradicts the assumption that $f\geqslant0$ everywhere. Thus, $f'(x)\geqslant0$ for every $x\geqslant0$, in particular $f$ is nondecreasing.
The continuity of $f'$ is not needed. – Did Mar 30 '12 at 12:40
I think I see what you mean: you use that $f$ is differentiable at $y$, so by considering the difference quotient, $\frac{f(y)-f(x)}{y-x}=f'(y)\le-c$. – bbm Mar 30 '12 at 14:20
@Didier how do you get $f(y)\leqslant f(x)+cx-cy$,mean value theory can not get it.the defination of quotient also can not get it.because $x$ in$\frac{f(y)-f(x)}{y-x}=f'(y)\le-c$is not the first given $x$. – noname1014 Mar 30 '12 at 15:34
@TaoHong洪涛 mean value theory can not get it... ?? – Did Mar 30 '12 at 16:06