# 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 $\displaystyle f'(0)=0$ and $\displaystyle f$ is Lipschitz. But I can't go any further.

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

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

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

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 $\displaystyle f'(0)=0$ and $\displaystyle f$ is Lipschitz. But I can't go any further.

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

Here's my attempt:

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

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

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

Therefore:

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

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

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

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

This gives:

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

Hence $\displaystyle f(a)=f(0)=0$ since $\displaystyle |f(a)| < \epsilon$ as $\displaystyle f(0)=0$.
• Mar 24th 2009, 04:36 AM
Ivan
?
Great, but the $\displaystyle \epsilon$ 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)