I've got an exercise about differentiability, mean value theorem and suprema.
To be honest I don't understand the structure of this question. Maybe you guys are so kind to help me out :)
Let $f: [0,1] \rightarrow \mathbb{R}$ be differentiable with $f(0) = 0$, and satisfying
$$|f'(x)|\le M|f(x)|, x\in[0,1] $$ for some $M>0$
a.) Use the Mean Value Theorem to show that for all $x \le x_0 \in[0,1], y\in[0,x_0]:$ $$ |f(x)|\le x_0 \text{ sup} |f'(y)|\le M x_0 \text{ sup} |f(y)|$$ b.) Use the previous part to show that f is the zero-function on [0,1].
(Hint: What happens if we choose $x_0$ such that M$x_0<1$?)
Known definitions, theorems:
- Mean Value Theorem (There is a c for which f'(c) equals f(b)-f(a)/(b-a) if [a,b] is the domain, and f is continuous on [a,b], differentiable on (a,b)
- Differentiable function means that that for all points c $\in$ A, the limit of f(x)-f(c)/(x-c) exists.
- Interior Extremum Theorem (Intermediate Value theorem). States that if f attains a maximum or minimum value on a open interval, then at some point c $\in$(a,b), f'(c)=0
- Darboux Theorem: If f is differentiable on an interval [a,b] and if $\alpha$ satisfies $f'(a) < \alpha < f'(b)$, then there exists a point c$\in$(a,b) where $ f'(c) = \alpha $.