Show something using the Mean Value Theorem

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.