I'm struggling to prove the following statement, which makes intuitive sense:
Let $f: \mathbb{R} \to \mathbb{R}$ be differentiable. Suppose $\forall x \in \mathbb{R} \quad 0 \le f'(x) \le f(x)\,$. If $f$ vanishes at some point, show that $f$ is identically zero.
This is exercise 6 in section 6.7, in Elementary Classical Analysis by Mardsen & Hofmann.
I've managed to do the following:
Let $\omega \in \mathbb{R}$ be the point such that $f(\omega) = 0$ (which exists by hypothesis). Let $x < \omega$. By Cauchy's Mean Value Theorem,
$$
\exists \xi \in (x, \omega)\colon\quad 0 \le f(x) =f(x) – f(\omega) = f'(\xi)(x – \omega)\,.
$$
By hypothesis, we have that $f'(\xi) \ge 0$. On the other hand, $x – \omega < 0$. Thus, $f'(\xi)(x – \omega) \le 0$. By the above proposition, though, $ f'(\xi)(x – \omega) \ge 0\,$. Therefore, $f'(\xi) = 0$ which implies $f(x) = 0$.
With this we have proven that $\forall x \in \mathbb{R}\quad x \le \omega \rightarrow f(x) = 0\,$.
But for $x > \omega$, I haven't been able to prove much. I've got so far:
Let $x > \omega$. By Cauchy's Mean Value Theorem,
$$
\exists \xi_1 \in (\omega, x)\colon\quad 0 \le f(x) = f(x) – f(\omega) = f'(\xi_1)(x – \omega) \le f(\xi_1)(x – \omega)\,.
$$
If we apply this reasoning recursively on $f(\xi_1)$, we get a succession $(\xi_i)_{i\in\mathbb{N}} \subseteq (\omega, x)$ such that $\forall i \in \mathbb{N}\quad \xi_{i+1} \in (\omega, \xi_i)$ and
$$
\forall n \in \mathbb{N}\quad f(x) \le f(\xi_n) \prod_{i = 0}^{n – 1} (\xi_i – \omega)
$$
where $\xi_0 = x$.
Is there a hint?
Best Answer
$(e^{-x}f(x))'=e^{-x}(f'(x)-f(x)) \leq 0$ so $e^{-x}f(x)$ is decreasing. If $f(x_o)=0$ then $e^{-x}f(x)$ is non-negative and $\leq e^{-x_0}f(x_0) =0$ for $x>x_0$ so $f(x)=0$ for $x \geq x_0$. For $x \leq x_0$ you already have a proof.