3

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ diff. $f(0)=0$ with $\left| f'(x)\right| \leq \left| f(x)\right|$ for $x\in\mathbb{R}$. Show that $f$ is identically zero.

My proof-trying. Let $x\in\mathbb{R^{>0}}$.

Consider

$f:\left[ 0,x\right] \rightarrow \mathbb{R}$ ; continuous,

$f:\left( 0,x\right) \rightarrow \mathbb{R}$; differentiable.

By the mean value theorem, there exist a $c$ in $(0,x)$ such that

$\dfrac {f\left( x\right) -f(0)} {x-0}=\dfrac {f(x)} {x}=f'(c)$. Hence, $f'(c) x=f(x)$.

So, what should I do?

  • Take any $\xi\in(0,x)$, and consider the two cases $f(\xi)<\xi f(x)/x$ and $f(\xi)>\xi f(x)/x$. Show, using the MVT on $(\xi,x)$ or $(0,\xi)$, that these lead to a contradiction. Therefore $f(\xi)=\xi f(x)/x$ for all $\xi\in(0,x)$. But then $f'(0)\ne 0$. – TonyK Apr 04 '17 at 00:22

1 Answers1

2

We first consider $f$ in the interval $[0,1]$: Fix some $x_0 \in [0,1]$ and set $$M_0:=\sup_{x \in[0,x_0]} |f(x)|, \quad M_1:=\sup_{x \in [0,x_0]} |f'(x)|. $$ For any $x \in [0,x_0]$ we then have $$|f(x)| \leq M_1 x_0 \leq M_0 x_0, $$ and thus it follows that $f(x) \equiv 0$ on the inverval $[0,1]$. By induction you can then show that $f \equiv 0$ on $[0,\infty)$. Then, you can work out the negative numbers as well.

Note: These ideas were taken from Exercise 26 on pp. 119 of baby Rudin.

user1337
  • 24,381