I need help with this exercise.
Let $f:\;\mathbb{R\rightarrow R}$ be a diferentiable function such that $f(0)=0$ and $|f'(x)|\leq|f(x)|\ \forall x\in\mathbb{R}$. Then $f(x)=0\ \forall x\in\mathbb{R}$.
Idea:
Case 1: $|f'(x)|=|f(x)|$ trivial.
Case 2: $|f'(x)|<|f(x)|$. Okay, we know $f(0)=0$ then $|f'(x)|-|f(x)|<0$. If $x=0$ then $|f'(0)|-|f(0)|<0\Rightarrow|f'(0)|<0$. But here I'm stuck because $|f'(0)|$ never is negative, can someone help me with this exercise?
We shall give a different proof then the one given in the reference by Canis Lupus.
Suppose $0\leq x<1$, then by the mean value theorem we see that \begin{align} |f(x)| =&\ |f(x)-f(0)| = |f'(\xi_1)||x| \leq |f(\xi_1)||x|\\ \leq&\ |f'(\xi_2)||\xi_1||x| \leq |f(\xi_2)||\xi_1||x|\\ \leq&\ \ldots \leq |f(\xi_{n+1})||\xi_{n}|\cdots|\xi_1||x| \end{align} where $0\leq \xi_{n+1} \leq \xi_n\leq x$ for all $n$ which means \begin{align} |f(x)| \leq |f(\xi_{n+1})||x|^{n+1} \leq M|x|^{n+1}\rightarrow 0 \end{align} as $n\rightarrow \infty$. Thus, it follows $f \equiv 0$ on $[0, 1]$.
Next, define $g(x) = f(x+1)$ and observe that $g(0) = f(1) = 0$ and
\begin{align}
|g'(x)| = |f'(x+1)| \leq |f(x+1)| = |g(x)|.
\end{align}
By the above argument we see that $g \equiv 0$ on $[0, 1]$ which means $f(x) \equiv 0$ on $[1, 2]$.
Hence it follows $f \equiv 0$ on $[0, \infty)$.
Note: The proof is exactly the same for $(-\infty, 0]$.
