# Calculus Problem - Properties of a Function

1. Jan 30, 2013

### Joschua_S

Hello

I have a mathematical problem that I could not solve. Could you please give me some hints how to solve it?

Let $f: [0,1] \rightarrow \mathbb{R}$ be a continous and on $(0,1)$ a differentiable function with following properties:

a) $f(0) = 0$
b) there exists a $M>0$ with $|f'(x)| \leq M |f(x)|$ for all $x \in (0,1)$

Now the problem is: Show that $f(x) = 0$ is true for all $x \in [0,1]$

There is a hint given but it doesn't help me The hint is: Consider the set $D = \{ x \in [0,1]: ~ f(t) =0$ for $t \in [0,x] \}$ and show that the the supremum of this set is $1$.

Thanks for help
Greetings

2. Jan 30, 2013

### jbunniii

What if we look at the interval $I = [0, \frac{1}{2M}]$ or $I = [0, 1]$, whichever is smaller. This ensures that any point in $I$ will be no larger than $\frac{1}{2M}$.

As $f$ is continuous on $I$, so is $|f|$, so $|f|$ achieves a maximum at some point $x \in I$. If $x = 0$ then we're done. Otherwise, we can apply the mean value theorem to $f$ on $[0,x]$ and, with a bit of work, derive a contradiction.

3. Jan 30, 2013

### jbunniii

P.S. The above will show that under the given hypotheses, we will have $f(x) = 0$ for all $x \in I$. If $I = [0,1]$ then we're done. Otherwise, this shows that $f(x) = 0$ for all $0 \leq x \leq \frac{1}{2M}$. In that case, you can repeat the argument for $[\frac{1}{2M}, \frac{2}{2M}]$ and so forth until you have covered all of $[0,1]$.