Assume $|f'(x)|\le |f(x)|$ for all $x\in(0,1).$ Prove that $f(x)=0$ for all $x\in[0,1]$.
I thought maybe I could use the limit but since there is no equation it won't work.
Suppose $f : [0,1]\mapsto R$ is continuous, $f$ is differentiable on $(0,1) $and $f(0)=0$.
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calculus
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0Thank you Zachary Selk for editing not use to the text conversion yet. :) – 2017-01-09
2 Answers
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Pick $x$ such that $|f(x)|$ reaches the maximum, then $$ |f(x)|=\biggl|\,\int \limits_{0}^x f'(t)\, dt\,\biggr|\leq\int \limits_0^x |f'(t)|dt\leq \int\limits_{0}^x |f(t)|dt\leq x |f(x)|$$
Notice that if $|f(x)|\neq 0$ the last inequality is strict by continuity.
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The mean value theorem says that, for $x\in(0,1]$, $$ \frac{f(x)-f(0)}{x-0}=f'(\xi_x) $$ with $\xi_x\in(0,x)$. Then $$ |f(x)|=x|f'(\xi_x)|\le x|f(\xi_x)| $$ Consider $y\in(0,1)$; then $|f|$ attains a maximum at $z\in(0,y]$. Then, if $|f(z)|>0$, $$ |f(z)|=z|f'(\xi_z)|\le z|f(\xi_z)|\le z|f(z)|<|f(z)| $$ a contradiction. Therefore the maximum of $|f|$ over each interval $[0,y]$ is zero. By continuity, also $f(1)=0$.