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

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?

## marked as duplicate by Martin R, Gabriel Romon, Community♦May 17 at 17:45

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 $$0, letting $$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)$$.