Question

Suppose $f: [0, 1] \rightarrow \mathbb{R}$ is differentiable with $f(0) = 0$. If $|f'(x)| \leq f(x)$ for all $x \in [0, 1]$, then show that $f(x) = 0$ for all $x$.

Answer 1

The problem asks to show that $f(x) = 0$ for all $x \in [0, 1]$. The proof above demonstrates this conclusion.

Answer 2

1. Given $|f'(x)| \leq f(x)$ for all $x \in [0, 1]$. Since $|f'(x)| \geq 0$, this implies $f(x) \geq 0$ for all $x \in [0, 1]$.
2. The inequality $|f'(x)| \leq f(x)$ is thus equivalent to $-f(x) \leq f'(x) \leq f(x)$.
3. Consider the inequality $f'(x) \leq f(x)$. Rearranging gives $f'(x) - f(x) \leq 0$.
4. Multiply by $e^{-x}$ (which is positive for all $x$): $e^{-x}f'(x) - e^{-x}f(x) \leq 0$.
5. The left side is the derivative of $e^{-x}f(x)$: $\frac{d}{dx}(e^{-x}f(x)) \leq 0$.
6. Let $g(x) = e^{-x}f(x)$. Then $g'(x) \leq 0$, which means $g(x)$ is a non-increasing function on $[0, 1]$.
7. Therefore, for any $x \in [0, 1]$, $g(x) \leq g(0)$.
8. Given $f(0) = 0$, we have $g(0) = e^{-0}f(0) = 1 \cdot 0 = 0$.
9. So, $g(x) \leq 0$, which implies $e^{-x}f(x) \leq 0$.
10. Since $e^{-x} > 0$, we must have $f(x) \leq 0$ for all $x \in [0, 1]$.
11. Combining the results from step 1 ($f(x) \geq 0$) and step 10 ($f(x) \leq 0$), we conclude that $f(x) = 0$ for all $x \in [0, 1]$.