Prove that $\forall x \in \mathbb{R}, f(x)=0$

Question

Suppose $f\colon\mathbb{R}\to\mathbb{R}$ is differentiable function and satisfies $$\forall x \in \mathbb{R}, \vert f'(x)\vert \leq \vert f(x)\vert, \quad f(0)=0.$$

Prove or disprove $f(x)=0$

How can I approach this problem?

Answer 1

Let $\varphi(n)$ be a "formula" on $n\in\mathbb{N}$ defined by $\varphi(n):$ For any $x\in[0,n]$, $f(x)=0$.

We prove that $\varphi(n)$ is true for any positive integer $n$ by induction.

Let $x\in[0,1]$. Let $a=0$. By applying mean-value theorem repeatedly, we obtain a sequence $\langle x_{n}\mid n\in\mathbb{N}\rangle$ such that $x>x_{1}>x_{2}>\ldots>0$ and \begin{eqnarray*} \left|f(x)\right| & = & \left|f(x)-f(a)\right|\\ & = & \left|f'(x_{1})(x-a)\right|\\ & \leq & x\left|f(x_{1})\right|\\ & = & x|f(x_{1})-f(a)|\\ & = & x(x_{1}-a)|f'(x_{2})|\\ & \leq & xx_{1}|f(x_{2})|\\ & \leq & \cdots\\ & \leq & xx_{1}x_{2}\ldots x_{k-1}|f(x_{k})|\\ & \leq & x(x_{1})^{k-1}M, \end{eqnarray*} where $M:=\sup_{t\in[0,1]}|f(t)|<\infty$. Since $0<x_{1}<1$, letting $k\rightarrow\infty$, it follows that $f(x)=0$. Hence $\varphi(1)$ is true.

Assume that $\varphi(n)$ is true. Let $x\in(n,n+1]$ be arbitrary. Denote $b=n$. Observing that $f(x)=f(x)-f(b)$ and repeating the above argument, with $a$ replaced by $b$, we can prove that $f(x)=0$. By mathematical induction, $\varphi(n)$ is true for all positive integers.

Therefore $f(x)=0$ for all $x\in[0,\infty)$. We can argue similarly that $f(x)=0$ for all $x\in(-\infty,0)$.