Let $f : [0, 1] → ℝ$ be continuous over $[0, 1]$, differentiable over $(0, 1)$, $f(0) = 0$, and $0 ≤ f'(x) ≤ 2f(x)$. Show that $f$ is identically $0$.
Let $g(x) = e^{-2x} f(x)$. Then $g(0) = f(0) = 0$ and $g'(x) = e^{-2x} [f'(x) - 2f(x)] ≤ 0$ for all $x$ in $(0, 1]$, since $0 ≤ f'(x) ≤ 2f(x)$.
Since $g$ is continuous on $[0, 1]$ and differentiable on $(0, 1)$, it follows that $g$ is non-increasing on $[0, 1]$. That is, $g(x) ≤ g(0) = 0$ for all $x$ in $[0, 1]$. But $g(x) = e^{-2x} f(x) ≥ 0$ for all $x$ in $[0, 1]$, since $e^{-2x} > 0$ and $f(x) ≥ 0$ for all $x$ in $[0, 1]$.
Therefore, $g(x) = 0$ for all $x$ in $[0, 1]$. In particular, $g(1) = 0$, which implies that $f(1) = 0$. Hence, $f$ is identically $0$ on $[0, 1]$ by the following argument.
Suppose for the sake of contradiction that there exists a point c in $(0, 1]$ such that $f(c) > 0$. Since $f$ is continuous on $[0, 1]$, it follows that there exists a point $a$ in $[0, c]$ such that $f(a) = 0$ and $a$ is the largest such point. Then, by the mean value theorem, there exists a point $b$ in $(a, c)$ such that $f'(b) = \frac {f(c)}{(c-a)} > 0$, which contradicts the fact that $0 ≤ f'(x) ≤ 2f(x)$ for all $x$ in $(0, 1)$.
Therefore, $f$ is identically $0$ on $[0, 1]$.
Please cheak this.. either give another solution...
Your solution is correct. Besides, I don't know why you need this step "Suppose for the sake of contradiction....".
Another solution is to apply the Gronwall's inequality, from this inequality you get $f(x) \le 0$. From the assumption that $2f(x) \ge f'(x) \ge 0$ you deduce that $f(x) = 0$ for all $x \in [0,1]$.