I have been struggling with this problem..

Q. Let $f(x)$, $x\geq 0$, be a non-negative continuous function, and let $F(x)=\int_0^x f(t) dt$, $x\geq0$. If for some $c>0$, $f(x)\leq cF(x)$ for all $x\geq 0$, then show that $f(x)=0$ for all $x\geq0$ .

I have tried everything in my ability, but in vain. I get a feeling that this can be solved using Mean Value theorem. Any ideas? Please help!!


Let $$\phi(t) = e^{-ct} F(t)$$ Then $\phi(0) = 0$, and $\phi(t) \ge 0$ for all $ t \ge 0$.

Furthermore, $$\phi'(t) = e^{-ct}(F'(t) - c F(t)) \le 0$$hence $\phi(t) = \int_0^t \phi'(\tau) d \tau \le 0$, and so $\phi(t) =0 $ for all $t \ge 0$.

If $ϕ(t)=0$ for all $t≥0$ , then $F(t)=0$ for all $t≥0$ . Since $f$ is continuous, $F$ is differentiable and $F ′ =f$ , hence $f=0$.

  • $\begingroup$ As simple, short and elegant as possible. +1 $\endgroup$ – DonAntonio Jan 16 '14 at 19:02
  • $\begingroup$ @copper.hat:Very nice. Just a simple request. Can you also write the last step from $\phi(x)=0$ to $f(x)=0$? I understand it intuitively but would like to see it formally $\endgroup$ – Abhimanyu Arora Feb 13 '14 at 11:45
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    $\begingroup$ @AbhimanyuArora: If $\phi(t) = 0$ for all $t \ge 0$, then $F(t) = 0$ for all $t \ge 0$. Since $f$ is continuous, $F$ is differentiable and $F'=f$, hence $f=0$. $\endgroup$ – copper.hat Feb 13 '14 at 16:10

Write your assumption as $F'(x) \leq c F(x)$. You know that $F(0)=0$. Fix $x_0>0$ and define $$ M_0 = \sup_{x \in [0,x_0]} F(x), \quad M_1 =\sup_{x \in [0,x_0]} F'(x). $$ Then, by the Mean Value Theorem, for every $x \in [0,x_0]$, $F(x) \leq M_1 x_0 \leq c x_0 M_0$. If $c x_0 < 1$, then $F(x)=0$ for every $x \in [0,x_0]$. Now you start from $x_0$ and go on.

Bibliography. W. Rudin. Principles of mathematical analysis. Chapter 5, exercise 26.

  • $\begingroup$ I might be making a mistake, but it seems that by taking $f(x)=e^x -1$ and C=2, I get that $f(x) \leq 2F(x)$. What am I doing wrong? $\endgroup$ – voldemort Jan 16 '14 at 18:24
  • $\begingroup$ $F(x)=e^x-x-1$. Try drawing $y=f(x)$ and $y=2F(x)$: the graphs intersect. $\endgroup$ – Siminore Jan 16 '14 at 18:38
  • $\begingroup$ Oops- I had made a mistake in integrating! After all this time you would think I should have got it correct. Thanks for pointing it out, and really apologize for wasting your time. $\endgroup$ – voldemort Jan 16 '14 at 18:39
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    $\begingroup$ Don't worry :-) $\endgroup$ – Siminore Jan 16 '14 at 18:39

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