Prove f(x) =0 for all x belonging to [0,1]

Question

Suppose f : [0, 1] → R is differentiable and f (0) = 0. Suppose $|f′(x)| ≤ |f(x)|^2$, $∀x ∈ [0, 1]$.

Prove that $f(x) = 0, ∀x ∈ [0,1]$.

Since f is differentiable, then $f$ is continuous, then if such a function is continuous, then for any given $\varepsilon > 0$ and $x \in [a, b]$, there exists a $\delta > 0$ such that $|x-y| < \delta \implies |f(x) - f(y)|< \varepsilon$...I am not too sure how to continue from here.

Answer 1

Here is a hint:

Note that $f(x) = f(0) + f'(\xi) x$ for some $0<\xi < x$. Hence $|f(x)| \le |f'(\xi)| \le |f(\xi)|^2$.

By continuity there is some $\delta>0$ such that $|f(x)| \le {1 \over 2}$ for $x\in [0,\delta]$.

Then we have $|f(x)| \le {1 \over 2^2}$ for $x \in [0,\delta]$. Repeating shows that $f(x) = 0 $ for $x \in [0,\delta]$.

Now use uniform continuity to show that this holds for the entire interval.