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Assume $f(x)$ is a real-function defined on $[0,+\infty)$ and satisfies the followings:

  1. $f'(x) \geq 0$

  2. $f(0)=0$

  3. $f'(x) \leq f(x)$

Should we always have $f(x) \equiv 0$ ? Thanks for any solution.

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  • $\begingroup$ Counterexample: $f(x)=a \chi_{(0,\infty)}$ with $a>0$ it respects the conditions and is not the zero function (although it's not continuous) $\endgroup$
    – b00n heT
    Jan 23, 2014 at 13:00
  • $\begingroup$ @b00nheT But in your example, $f(x)$ isn't continuous at $x=0$ and hence $f'(0)$ doesn't exist. $\endgroup$ Jan 23, 2014 at 13:04

3 Answers 3

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As $f(0)=0$ and for all $x\geq0$, $f'(x)\geq0$, we have $f(x)\geq0, \forall x\geq0$.

Define $$g(x)=\mathrm e^{-x}f(x)$$ and compute $$g'(x)=\mathrm e^{-x}\left(f'(x)-f(x)\right)\leq 0$$ as $g(0)=0$ we have $g(x)\leq 0$ for all $x\geq0$. Therefore we have $$f(x)=\mathrm e^xg(x)\leq 0,\quad \forall x\geq0,$$ which proves $f\equiv0$.

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  • 1
    $\begingroup$ This demonstration is based on the idea of Gronwall's inequality. $\endgroup$
    – Tom-Tom
    Jan 23, 2014 at 13:42
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    $\begingroup$ Pretty idea! Thank you Rossetto! $\endgroup$ Jan 23, 2014 at 13:45
  • $\begingroup$ I found a similar older question $\endgroup$
    – Tom-Tom
    Jan 23, 2014 at 14:42
  • $\begingroup$ It used the similar trick. $\endgroup$ Jan 23, 2014 at 14:44
  • $\begingroup$ Nice!${}$${}$${}$ $\endgroup$ Jan 23, 2014 at 14:49
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To add to the very nice techniques already used, here's another way of seeing that $f = 0$. Put $M_x = \sup_{t\in[0,x]}{|f(t)|}$. (Actually, as @V.Rosetto notes, $f$ is increasing and so $M_x = f(x)$, which simplifies things!) For each $x>0$, the mean value theorem furnishes a $t\in(0,x)$ such that $f(x) = xf'(t)$. But then \begin{align*} M_x = \sup_{t\in[0,x]}|f(t)| \leq \sup_{t\in[0,x]} t|f'(t)| \leq \sup_{t\in[0,x]} t|f(t)|\leq xM_x \tag{1} \end{align*} for $x>0$. Then for $x\in(0,1)$, we must have $M_x = 0$, lest $(1)$ lead to the contradictory inequality $M_x<M_x$. Thus $f = 0$ on $[0,1)$.

But now $x\mapsto f(x-1)$ satisfies the same conditions as $f$ on $[0,\infty)$, and so it vanishes on $[0,1)$ as well. This means that $f$ vanishes on $[0,2)$. Now consider $x\mapsto f(x-2)$, and so on.

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  • $\begingroup$ As $f$ is increasing, one can also write $f(x)=xf'(t)\leq xf(t)\leq xf(x)$ and draw the same conclusion. Very nice (+1) ! $\endgroup$
    – Tom-Tom
    Jan 23, 2014 at 15:48
  • $\begingroup$ @V.Rossetto Excellent point! I'll add that in there. $\endgroup$ Jan 23, 2014 at 16:10
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Let $x_0\in[0,x]$ be a point in which $f'$ assumes a maximum. Then

$$f(x) = f(x) - f(0) = \int_0^xf'(t)dt \le xf'(x_0) \le xf(x_0) \le xf(x)$$

so $f$ must be $0$ on $[0,1)$ and by continuity $f(1) = 0$. Since $f(x - 1)$ satisfies properties 1), 2), 3), the result follows by induction.

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