# Challenging proofs

• Mar 24th 2009, 01:41 AM
Ivan
Challenging proofs
Hello!

Can anyone help me solve the problems? It paralyzes my work (Headbang)

It is obvious in the first two problems that $f'(0)=0$ and $f$ is Lipschitz. But I can't go any further.

1) A continuously differentiable function $f$ is defined on $\mathbb{R}$ such that $f(0) = 0$ and $|f'(x)| \leq |f(x)|$ for all $x\in \mathbb{R}$. Show that $f$ is constant.

2) Suppose that for some positive constant $M$, $|f'(x)|\leq M|f(x)|$, $x\in[0,1]$. Show that if $f(0)=0$, then $f(x) =0$ for any $x$, $0\leq x\leq 1$.

3) Show that there exists no real-valued function $f$ such that $f(x)>0$ and $f'(x)=f(f(x))$ for all $x$.

Thank you very much!
• Mar 24th 2009, 02:08 AM
Showcase_22
Quote:

Originally Posted by Ivan
Hello!

Can anyone help me solve the problems? It paralyzes my work (Headbang)

It is obvious in the first two problems that $f'(0)=0$ and $f$ is Lipschitz. But I can't go any further.

1) A continuously differentiable function $f$ is defined on $\mathbb{R}$ such that $f(0) = 0$ and $|f'(x)| \leq |f(x)|$ for all $x\in \mathbb{R}$. Show that $f$ is constant.

Here's my attempt:

$f(x)$ is continuous on $\mathbb{R}$ since it's differentiable. Hence $f(x)$ is continuous on some interval $[0,a]$, and differentiable on $(0,a)$ where $a \in \mathbb{R}$ and $a \neq 0$.

By applying the mean value theorem $\exists \ x_0 \in (0,a) \ s.t \ f'(x_0)=\frac{f(a)-f(0)}{a-0}=\frac{f(a)-f(0)}{a}$.

We know that $|f'(x)| \leq |f(x)| \Rightarrow \ |f'(x_0)| \leq |f(x_0)|$.

Therefore:

$|\frac{f(a)-f(0)}{a}|=|f'(x_0)| \leq |f(x_0)|$

$\Rightarrow |f(a)-f(0)| \leq |a||f(x_0)|$

$\Rightarrow |f(a)-f(0)| \leq |af(x_0)|$

Let $\epsilon>|af(x_0)|$

This gives:

$|f(a)-f(0)| \leq |af(x_0)|<\epsilon$ so $|f(a)-f(0)|< \epsilon$ $\forall a \in \mathbb{R}$ where $a \neq 0$.

Hence $f(a)=f(0)=0$ since $|f(a)| < \epsilon$ as $f(0)=0$.
• Mar 24th 2009, 04:36 AM
Ivan
?
Great, but the $\epsilon$ in your proof isn't arbitrary, is it?
• Mar 24th 2009, 06:16 AM
Showcase_22
I don't think it is , shoot!

hmm, this really is pretty tricky....(Worried)