Let $f:[0,1] \to \mathbb R$ differentiable s.t $f(0)=0$. $\forall x \in [0,1]$: $|f^{'}(x)| \le |f(x)|$. Prove $f(x) =0$, $\forall x \in [0,1]$.

Question

Let $f:[0,1] \to \mathbb R$ differentiable s.t $f(0)=0$. $\forall x \in [0,1]$: $|f^{'}(x)| \le |f(x)|$. Prove $f(x) =0$, $\forall x \in [0,1]$.

I've tried proving this with Mean Value Theorem but I think one of my implementations isn't good enough. I'd love some help!

First of all we have $f(x)$ is differentiable $\implies$ $f(x)$ is continuous in $[0,1]$.

From MVT we have: exists some $c_1 \in (0,1)$ s.t $|f(1)| = {f(1)-f(0) \over 1-0} = f^{'}(c_1)$.

From the datum we can deduce: $0 \le f(1) = |f^{'}(c_1)| \le |f(c_1)|$.

Now take some $b \in (0,1)$. Let us examine the interval $[0,b]$. From MVT we have some $c_2 \in (0,b)$ s.t:

$|f(b)| \lt {f(b) \over b} = {f(b)-f(0) \over b-0} = f^{'}(c_2) \implies 0 \le f(b) \lt |f^{'}(c_2)| \le |f(c_2)|$.

This is the tricky part:

Since we took some random $b \in (0,1)$ we have $\forall x \in (0,1]$ we have $0 \le |f(x)| \le f(0) = 0$ thus implementing $f(x) = 0, \forall x \in [0,1]$ as needed.

I understand the last conclusion is based on a limit type of argument which I have not provided.

Answer 1

The "tricky part" does not make sense. Just because you chose a random $b \in (0, 1)$, you don't automatically get $|f(x)| \leq f(0) = 0$. In fact, none of the results regarding $c_1, c_2$ shown before have any relation to the last few lines so they don't back up $|f(x)| \leq f(0) = 0$ either.

Instead, note that $f$ is continuous on the compact interval $[0, 1]$ and hence by EVT its absolute value $|f|$ achieves a maximum at some $0 \leq x_m \leq 1$. If $x_m = 0$, you are done. Otherwise by MVT there is some $0 < c < x_m$ s.t. $f'(c) = \frac{f(x_m)}{x_m}$. But then $$ |f(x_m)| = x_m |f'(c)| \leq x_m |f(c)| \leq x_m |f(x_m)| $$ So $|f(x_m)| = 0$.

Answer 2

I want to present an alternative solution, in which the mean value theorem is used differently.

Define $g(x) = f(x) f(x) \mathrm{e}^{-2x}$. Then $$ g'(x) = (2f(x) f'(x) - 2f(x) f(x)) \mathrm{e}^{-2x}. $$ Since $|f'(x)| \leq |f(x)|$, we have $$ -f(x) f(x) \leq f(x) f'(x) \leq f(x) f(x), $$ which implies that $g'(x) \leq 0$.

Let $x$ be larger than $0$. On one hand, $g(x) \geq 0 = g(0)$, by definition; on the other hand, for $x > 0$, $$ g(x) - g(0) = g'(\xi) (x - 0) \leq 0 $$ for some $\xi$ in $(0, x)$. Hence $g(x)$ must be $0$, which means that $f(x)$ must be $0$.