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.

  • 5
    $\begingroup$ If $f$ is constant and $f(0)=0$ then $f\equiv 0$ $\endgroup$ – Nameless Dec 18 '12 at 18:02
  • $\begingroup$ Just like that. But when proving that the function is constant I only use the mean value theorem. When should I use the inequality $0\le f'(x) \le 2f(x)$? $\endgroup$ – Bilbo Dec 18 '12 at 18:06
  • $\begingroup$ You don't need the MVT or anything. $\endgroup$ – Nameless Dec 18 '12 at 18:07
  • $\begingroup$ My guess is there is a typo/read again carefully. As stated, it's obvious by first comment. $\endgroup$ – gnometorule Dec 18 '12 at 18:07

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.

  • $\begingroup$ I wish I had known you were writing this - I was about 30 seconds from posting an almost identical proof when yours appeared ;-). $\endgroup$ – Jason DeVito Dec 18 '12 at 18:34
  • $\begingroup$ I think @richard's proof is the nicest. $\endgroup$ – copper.hat Dec 18 '12 at 18:37
  • $\begingroup$ @copper.hat: I think your argument looks more natural compared with mine. +1. $\endgroup$ – 23rd Dec 18 '12 at 18:45
  • $\begingroup$ @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 $\sup_{\xi \in (a,b)} f'(\xi) \leq \sup_{\xi \in (a,b)} 2 f(\xi) = 2 f(b)$. $\endgroup$ – copper.hat Dec 18 '12 at 19:25
  • $\begingroup$ Yes, I noticed that just after posting the comment, that's why I deteted it :) Thanks. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.