Let $f:[0,1] \to \Bbb{R}$ a differentiable function and $M>0$ such that $f'(x) \geq M,\forall x \in [0,1].$
Prove that exists an interval $I$ with length $\frac{1}{4}$ such that $|f(x)| \geq \frac{M}{4}$.
For the solution i assumed that $f(0)=0$ and applying the MVT, i proved that $\forall x \in [0,1]$ we have $|f(x)| \geq Mx$ whick gives me the solution
I have a litle difficulty in the case where $f(0) \neq 0$
Can someone provide me a hint,or a different direction without the MVT?
Thank you in advance.
Best Answer
Note that $f$ is increasing. Let $S = \{x \in [0,1]\,:\, |f(x)|< \frac{M}{4}\}$. Since $f$ is increasing, this will be an interval $(a,b)\subseteq [0,1]$. If no interval exists of length $\frac{1}{4}$ where $|f(x)|\geq \frac{M}{4}$ then $b-a > \frac{1}{2}$. Now use MVT to reach a contradiction.