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Let $f(x)$ be differentiable over $[0,+\infty)$ and satisfy $f(0)=0$. If there exists a constant $c>0$ such that $|f'(x)|\leq c|f(x)|$ holds for every $x \in [0,+\infty)$, prove $f(x)\equiv 0.$

This is a version of Bellman-Gronwall inequality.

mengdie1982
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  • See also https://math.stackexchange.com/q/509876/42969. – Martin R Mar 04 '20 at 15:46
  • @MartinR Sir, the problem in your link is not the same as mine . The proof there depend on the existence of the maximum, which is not trivial for the present one. – mengdie1982 Mar 04 '20 at 15:48
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    It suffices to show that $f = 0$ on every interval $[0, b]$, because then $f = 0$ on $[0, \infty)$. – Martin R Mar 04 '20 at 15:48
  • You can try the following, which seem different from the solutions provided by @Martin R:

    You have $$ f(x)^2=2\int_0^x f^{\prime}(t)f(t)dt\leq 2c\int_0^x (f(t))²dt$$ Now put $F(x)=\int_0^x f(t)^2dt$, $G(x)=F(x)\exp(-2cx)$, note that $G(0)=0$, $G(x)\geq 0$, and compute $G^{\prime}(x)$.

     – Kelenner Mar 04 '20 at 15:49
  • I consider using the auxilliary function $g(x):=e^{-cx}f(x)$, but how to deal with the absolute value sign? – mengdie1982 Mar 04 '20 at 15:51
  • @Kelenner Why do you not consider the absolute value sign？ – mengdie1982 Mar 04 '20 at 16:03
  • You have

    You have $$ f(x)^2=|2\int_0^x f^{\prime}(t)f(t)dt|\leq2 \int_0^x |f^{\prime}(t)||f(t)|dt \leq ...$$

     – Kelenner Mar 04 '20 at 16:15
  • @Kelenner Thanks, but there exists a question left. $f'(x)$ is integrable ( in the meaning of Riemanne)? This seems to be not given from the assumption condition. – mengdie1982 Mar 04 '20 at 16:53
  • @mengdie1982 Yes, you have to suppose that $f^{\prime}(x)$ is for example continuous to have the integrability. – Kelenner Mar 04 '20 at 16:59
  • @Kelenner You say the statement does not hold without the continuity of $f'(x)$? – mengdie1982 Mar 04 '20 at 17:04
  • @mengdie1982 No; I say that my idea works only with the hypothesis of integrability (for example if $f^{\prime}$ is continuous ) – Kelenner Mar 04 '20 at 17:13
  • @Kelenner: It seems that your idea works with the only assumption that $f$ is differentiable: https://math.stackexchange.com/a/3569551/42969. – Martin R Mar 04 '20 at 20:31
  • @mengdie1982: I have added an answer to the other question which covers infinite intervals directly. – Martin R Mar 04 '20 at 21:02

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