If $f(0) = 0$ and $|f'(x)|\leq |f(x)|$ for all $x\in\mathbb{R}$ then $f\equiv 0$ Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous and differentiable function in all $\mathbb{R}$. If $f(0)=0$ and $|f'(x)|\leq |f(x)|$ for all  $x\in\mathbb{R}$, then $f\equiv 0$.
I've been trying to prove this using the Mean Value Theorem, but I can't get to the result. Can someone help?

Intuitively, the solution to $|f'| \leq |f|$ with $f(0) = c$ can not grow out of the region bounded by the solutions to $f' = +f$ and $f' = -f$ with the same initial condition $f(0) = c$. The boundary solutions are $f_\pm(x) = c e^{\pm x} = 0$ with $c = 0$ we have $f_\pm(x) \equiv 0$.
It's so obvious intuitively, I think, but was not at all as easy to prove as I thought, but finally I came up with a proof. But first a lemma:
Lemma
Let $h : \mathbb [0, \infty) \to \mathbb R$ be differentiable and satisfy 


*

*$h(x_0) > 0$,

*$h'(x) > 0$ when $x > x_0$ and $h(x) > 0$.


Then $h(x) > 0$ for all $x \geq x_0$.
Proof of lemma
Assume that $h(a) \leq 0$ for some $a > x_0$. Since $h$ is continuous, by the intermediate value theorem, $h$ takes the value $0$ in at least one point between $x_0$ and $a$. Let $x_1 = \inf \{ t \in (x_0, a) \mid h(t) = 0 \}$. Since $h$ is continuous and $h(x_0) > 0$ we have $x_1 > x_0$ and $h(x) > 0$ when $x_0 < x < x_1$. Then $h(x_1) - h(x_0) < 0$ and by the mean value theorem there exists some $\xi \in (x_0, x_1)$ such that $h'(\xi) = (h(x_1) - h(x_0))/(x_1 - x_0) < 0$. But this contradicts that $h'(x) > 0$ when $x > x_0$ and $h(x) > 0$. Thus $h(x) > 0$ for $x \geq x_0$.
Proof of statement in question
Take $\lambda>0$.
Let $g(x) = \lambda e^x$. Then $g-f$ satisfies the conditions of the lemma with $x_0=0$. Thus, for all $x > 0$ we have $(g-f)(x) > 0$, i.e. $f(x) < \lambda e^ x$.
In the same way, taking $g(x) = -\lambda e^x$ we have $f-g$ satisfying the conditions of the lemma so $f(x) > -\lambda e^x$ for $x > 0$.
Thus, for $x > 0$ we have $-\lambda e^x < f(x) < \lambda e^x$. Since $\lambda>0$ was arbitrary we must have $f(x) \equiv 0$ for $x > 0$.
Reversing the function, i.e. letting $f(x) \to f(-x)$ in the above, we also get that $f(x) \equiv 0$ for $x < 0$.
