# If $f(0)=0$ and $0\le f'(x)\le2f(x)$ for all $x\in(0,1)$, prove that $f\equiv0$.

Let $$f: [0,1]\to \Bbb R$$ be a continuous function such that $$f(0)=0$$ and it's differentiable on $$(0,1)$$. In addition, we know that $$0\le f'(x)\le2f(x)$$. Prove that $$f=0$$.

I know that if the function $$f$$ is continuous on $$[a,b]$$, differentiable on $$(a,b)$$ and $$f'(x) = 0$$ on $$(a,b)$$, then $$f$$ must be a constant function on $$[a,b]$$.

But I do not quite know how to use the inequality.

• If $f$ is constant and $f(0)=0$ then $f\equiv 0$ – Nameless Dec 18 '12 at 18:02
• Just like that. But when proving that the function is constant I only use the mean value theorem. When should I use the inequality $0\le f'(x) \le 2f(x)$? – Bilbo Dec 18 '12 at 18:06
• You don't need the MVT or anything. – Nameless Dec 18 '12 at 18:07
• My guess is there is a typo/read again carefully. As stated, it's obvious by first comment. – gnometorule Dec 18 '12 at 18:07

## 2 Answers

It is true even if you ignore the assumption of being constant (which surely must be a typo.)

Note that since $f'(x) \geq 0$, $f$ is non-decreasing. Hence $f(x) \geq 0$ for $x \in [a,b]$. Also, since $f$ is differentiable, it is continuous.

Let $0 \leq a < b \leq 1$, then $f(b)-f(a) \leq \sup_{\xi \in (a,b)} f'(\xi) (b-a) \leq 2 f(b)(b-a)$. Now suppose $f(a) = 0$ and $b-a < \frac{1}{2}$. Then the inequality gives $f(b) \leq 2 f(b) (b-a)$, or $0 \leq f(b)(2(b-a)-1)$. Hence $f(b) = 0$.

Take $a=0,b<\frac{1}{2}$. Then the above shows that $f(b) = 0$ for $b \in (0,\frac{1}{2})$, and continuity shows $f(\frac{1}{2}) = 0$. Now apply the same argument with $a=\frac{1}{2}, b<1$, to get $f(1) = 0$, from which the desired conclusion follows.

• I wish I had known you were writing this - I was about 30 seconds from posting an almost identical proof when yours appeared ;-). – Jason DeVito Dec 18 '12 at 18:34
• I think @richard's proof is the nicest. – copper.hat Dec 18 '12 at 18:37
• @copper.hat: I think your argument looks more natural compared with mine. +1. – 23rd Dec 18 '12 at 18:45
• @Anna: No. The first inequality is just the mean value theorem. It is true here as long as $a<b$. The second comes from the fact that $f'(\xi) \leq 2 f(\xi)$, so $\sup_{\xi \in (a,b)} f'(\xi) \leq \sup_{\xi \in (a,b)} 2 f(\xi) = 2 f(b)$. – copper.hat Dec 18 '12 at 19:25
• Yes, I noticed that just after posting the comment, that's why I deteted it :) Thanks. – Bilbo Dec 18 '12 at 19:31

Since $f(0)=0$ and $f'(x)\ge 0$ on $(0,1)$, $f(x)\ge 0$ on $[0,1]$. Let $g(x)=f(x)e^{-2x}$ on $[0,1]$. Then $g$ is continuous on $[0,1]$ and differentiable on $(0,1)$. On the one hand, $g(x)\ge 0$ on $[0,1]$. On the other hand, since $g(0)=0$ and $g'(x)=e^{-2x}(f'(x)-2f(x))\le 0$ on $(0,1)$, $g(x)\le 0$ on $[0,1]$. The conclusion follows.