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?.
Nonincreasing derivative implies nondecreasing function
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calculus
real-analysis
analysis
1 Answers
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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.
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0@TaoHong洪涛 `mean value theory can not get it`... ?? – 2012-03-30