# prove that the following function is: $f(x) = 0$

let $f: [0,1] \to \mathbb R$ , $f$ is differentiable

$f(0) = 0$

$|f'(x)|\le|f(x)|$ for $x\in [0,1]$

prove that $f(x) =0$ for $x\in [0,1]$

i believe that i need to somehow use the mean value theorem iteratively

any hints?

• What if $f(x) = 1$ for $x \in [0,1]$? – JPLF Jan 4 '14 at 18:01
• @JPLF: $f(0) = 0$ is a given. – Ulrik Jan 4 '14 at 18:02
• Oh ok, I didn't read carefully. – JPLF Jan 4 '14 at 18:02
• So, $f(0) = 0$ and $f'(0) = 0$. Surely, to rise above $0$, $f$ needs a positive derivative. But it can't have a positive derivative as long as it is $0$. – Karolis Juodelė Jan 4 '14 at 18:05
• Wow! Somebody really is posting all the last exercises from the Calculus I homework in the Tel-Aviv University Mathematics Dept ... – DonAntonio Jan 4 '14 at 18:32

Choose $x \in [0,1]$. By the mean-value theorem, $$f(x) = f(x) - f(0) = f'(c)(x-0) = f'(c)x$$

for some $c \in (0,x)$. But $$|f'(c)x| \leq |f(c)||x| = |f(c) - f(0)||x| \leq |f'(d)||x||c| \leq |f'(d)||x|^2$$

for some $d \in (0,c)$. Do you see how to iterate this argument?

More hints:

Note that $f$ is continuous and hence bounded on the compact set $[0,1]$, and by the hypothesis that $|f'| \leq |f|$, $f'$ is also bounded.

Iterate the above to show $|f(x)| \leq M |x|^n$ for some absolute constant $M$ and all $n$.

Conclude that $f(x) = 0$ for all $x \in [0,1)$. To obtain the result at $1$, either use the continuity of $f(x)$, or the following argument: you now know that $f(0.5) = 0$, so repeat the above argument, but center it at $0.5$ instead of $0$. You should get some bound $f(1) \leq M |1 - 0.5|^n$.

• Ah, right. That's not a hypothesis. Let me adjust the proof. – Elchanan Solomon Jan 4 '14 at 18:34
• Now it's fine and rather nice. +1 – DonAntonio Jan 4 '14 at 20:21
• @IsaacSolomon what does it mean to iterate an argument in math or to prove in an iterative way ? I suppose it's different from induction. – GinKin Jan 4 '14 at 21:19
• @GinKin Perhaps a good definition would be that an "iterative argument" is one that reduces to applying the same step over and over. In this case, obtaining the bound $|f(x)| \leq M|x|^n$ for all $n$ requires induction, because I am trying to prove a statement about all natural numbers. However, it is possible to complete the proof without induction, by supposing that $|f(x)| \geq \epsilon > 0$ and obtaining a contradiction by iterating until $M|x|^N < \epsilon$ for some sufficiently large $N$. – Elchanan Solomon Jan 4 '14 at 21:34