real analysis - Show something using the Mean Value Theorem

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 $f(c)=\alpha$

Answer

Let f:[0,1]→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 $x \le x_0 \in[0,1], y\in[0,x_0]:$
$$ |f(x)|\le x_0 \sup |f'(y)|\le M x_0 \sup |f(y)|$$
b.) Use the previous part to show that f is the zero-function on [0,1].

Let's begin with part a. So fix an $x_0$ in $(0,1)$. Suppose that there is an $x$ in $[0,x_0]$ such that $|f(x)| > x_0 \sup|f'(y)|$. Then in particular, $\dfrac{|f(x) - f(0)|}{x-0} = \dfrac{|f(x)|}{x} > \dfrac{x_0}{x}\sup f'(y) \geq \sup f'(y)$ as $x_0 \geq x$.

But then by the mean value theorem, there must exist a $c$ such that $f'(c) = \dfrac{f(x) - f(0)}{x - 0}$. This is a contradiction, as then $f'(c) > \sup f'(y)$.

The second inequality is much easier, relying just on using $|f(x)|\le x_0 \sup |f'(y)|\le M x_0 \sup |f(y)|$ and interpreting sup.

And then you can follow the hint for part b, and the answer falls out as $|f(x)| < \sup |f(y)|$ is nonsense.