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If $f:\mathbb{R}\rightarrow \mathbb{R}$ is a differentiable function satisfying the conditions $|f'(x)|\leq |f(x)|$ and $f(0)=0$ then show that $f(x)=0$ for all $x\in \mathbb{R}.$ This we have to prove using Mean Value Theorem only.

My approach: For arbitrary $x\neq0$ by MVT $$\frac{f(x)-f(0)}{x-0}=f'(c)$$ for some $c\in(0,x)$ or $(x,0)$ which implies $f(x)=xf'(c).$ Now, differentiating last equation both sides w.r.t $x$ we get $f'(x)=f'(c)$ yielding $$f'(x)=\frac{f(x)}{x}$$ By condition $|f'(x)|\leq |f(x)|$ we get $$\Big|\frac{f(x)}{x}\Big|\leq|f(x)| \implies |f(x)|(1-|x|)\leq0\implies|f(x)|\geq0\quad or\quad (1-|x|)\leq0\\\implies|f(x)|\geq0\quad or\quad x\notin(-1,1)$$ where the possibility $x\notin(-1,1)$ can't hold since $x$ is arbitrary non-zero real number. There fore we are only left with $|f(x)|\geq0$. After it I can't think more. Kindly HELP. Thank you!

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    $c$ depends on $x$, so you can't just say that the derivative of $xf'(c)$ is $f'(c)$. – 2017-01-21
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    yes, I see it, I had some doubt here – 2017-01-21
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    Does the approach I started help anywhere to prove it – 2017-01-21
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    Does "using the Mean Value Theorem only" forbid using theorems about continuous functions? – 2017-01-21
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    @DanielFischer yes, only MVT allowed no other – 2017-01-21

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Fix $0 < \delta < 1$. For any $x \in (0,\delta]$ we apply the MVT to find that that there exists $c \in (0,x)$ such that $$ |f(x)| = |f(x)-f(0)| = |f'(c) (x-0)| = |f'(c)| |x| \le |f(c)| \delta \le \delta \max_{0 \le z \le \delta} |f(z)|. $$ This holds for all such $x$, so $$ \max_{0 \le z \le \delta} |f(z)| \le \delta \max_{0 \le z \le \delta} |f(z)| $$ and hence $$ (1-\delta) \max_{0 \le z \le \delta} |f(z)| \le 0. $$ Since $0 < \delta < 1$ we find that $f(x) =0$ for all $0 \le x \le \delta$.

Now iterate this argument to show that $f=0$ on $[\delta,2\delta]$, $[2\delta, 3\delta]$, etc, and conclude that $f=0$ on $[0,\infty)$. Then apply this result to $g(x) = f(-x)$ to handle $x \le 0$.