Thread: Calculus Problem - Properties of a Function

1. Calculus Problem - Properties of a Function

Hello

This is my first post in this forum

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

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

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

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

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

Thanks for help
Greetings

2. Re: Calculus Problem - Properties of a Function

I think I would be inclined to use the "mean value" theorem. Suppose there exist $\displaystyle x_1$such that $\displaystyle f(x_1)= y_1\ne 0$. Then there exist $\displaystyle x_2$ between 0 and 1 such that $\displaystyle f'(x_2)= \frac{f(x_1)- f(0)}{x_1- 0}= \frac{y_1}{x_1}$