Function limiting its own derivative implies implies null function Let $f$ be a continuous differentiable function in $\mathbb{R}$. If $f(0)=0$ and $|f'(x)|\leq|f(x)|,\;\forall x \in \mathbb{R}$, then $f$ is the null function.
Following @DougM's idea, I think I figured it out:
Consider $f$ restricted to the interval $[0,1]$. Then, by Weirstrass theorem there exists $x_1,x_2 \in [0,1]$ such that $f(x_1)\leq f(x)\leq f(x_2),\;\forall x \in [0,1]$. Suppose, WLOG, that $|f(x_1)|\geq|f(x_2)|$. Note that this implies $|f(x_1)|\geq |f(x)|,\; \forall x \in [0,1]$.
If $x_1=0$, then $f(x)=0, \forall x \in [0,1]$.
Suppose $x_1\neq0$. Then, by the mean value theorem, there exists $c \in (0,1)$ such that: 
\begin{align}
f(x_1)-f(0)&=f'(c)(x_1-0)\\
|f(x_1)|&=|f'(c)||x_1|\\
|f(x_1)|&\leq|f'(c)|
\end{align}
But then $|f(c)|\geq|f'(c)|\geq|f(x_1)|$.
As $c\in(0,1)\subset[0,1]$, by the men value theorem there exists $d\in(0,c)\subset[0,1]$ such that:
\begin{align}
f(c)-f(0)&=f'(d)(c-0)\\
|f(c)|&=|f'(d)||c|\\
|f(c)|&<|f'(d)|
\end{align}
But then $|f(d)|\geq|f'(d)|>|f(c)\geq |f(x_1)|$, a contradiction. since $|f(x_1)|\geq |f(x),\; \forall x\in [0,1]$.
Therefore $x_1=0$.
From there you can just follow inductively to prove $f(x)=0$ for all positives, and the analogous proof for the negatives.
Using the MVT twice is kinda weird, wonder if there's a way not to use it like that.
 A: Following Bettybel's hint: let $x_0$ be the supremum of $t\in\mathbb{R}^+$ such that $f(x)=0$ on $[0,t]$.
Assume that $x_0$ is finite, let $I=\left[x_0,x_0+\frac{1}{2}\right]$ and let $M=\max_{x\in I}|f(x)|$. 
For any $x\in I$ we have
$$ f(x) = f(x_0)+\int_{x_0}^{x}f'(\xi)\,d\xi  $$
from which
$$ |f(x)|\leq (x-x_0) M \leq \frac{M}{2} $$
and $M=0$, contradicting $x_0=\sup\left\{t\in\mathbb{R}^+:\forall x\in[0,t],\, f(x)=0\right\}.$
It follows that $f(x)\equiv 0$ on $\mathbb{R}^+$ and by replacing $f(x)$ with $f(-x)$ we also have $f(x)\equiv 0$ on $\mathbb{R}^-$.

As an alternative, we may define $g(x)$ as $f(x)$ on $[0,1]$ and as $f(2-x)$ on $[1,2]$, then apply Wirtinger's inequality to $g(x)$:
$$ \int_{0}^{2}g(x)^2\,dx \leq \frac{4}{\pi^2}\int_{0}^{2}g'(x)^2\,dx \leq \frac{4}{\pi^2}\int_{0}^{2}g(x)^2\,dx $$
implies $g(x)\equiv 0$ on $[0,2]$, hence $f(x)\equiv 0$ on $[0,1]$. By shifting and repeating the same argument, $f(x)\equiv 0$ on $\mathbb{R}$.

We do not need the continuity of $f'$, and indeed $f\in H^{1}_{\text{loc}}$ is enough to prove the claim with the last approach.
A: MVT
if $f(x)$ is continuous in $[a,b]$ and differentiable in $(a,b)$ there exists a $c \in (a,b)$ such that $f'(c) = \frac {f(b) - f(a)}{b-a}$
How can we apply that
$\exists q\in [0,1]: \forall x\in [0,1], |f(q)|\ge |f(x)|$ (extreme value theorem)
i.e. let $|f(q)|$ be the maximal value of $|f(x)|$ when $x$ is in $[0,1]$
$|f'(x)|\le |f(q)|$ for all $x \in [0,1]$
$|f(q)| \ge |f'(q)| \ge |\frac {f(q)}{q}|\\
q|f(q)|\ge |f(q)|$
and since $q<1, |f(q)| = 0.$
Prove by induction.  Use the identical argument so show that if $f(x) = 0$  for all $x \in [0,n]$ then $ \forall x \in [0,n+1], f(x) = 0$
And similarly, if $\forall x \in [-n,0], f(x) = 0$ then $\forall x \in [-n-1,0], f(x) = 0$
