Let $I=[0,b)$, $b< \infty$. Suppose $u$ is a positive bounded measurable function on $I$. $v(s)$ is a positive, smooth function on $I$. Note that $u(b),v(b)$ may be $0$. Suppose that $$ u(t) \leqslant u(t_0) +\int_{t_0}^t u(s) \frac{v'(s)}{v(s)}ds $$ for any $0<t_0<t <b$. Then does the inequality below holds? $$ \frac{u(s_2)}{u(s_1)} \leqslant \frac{v(s_2)}{v(s_1)} $$ for any $0 \leqslant s_1 <s_2<b$.

If not, what other conditions should we add to $v(s)$, is $v'(s)\leqslant 0$ enough?

  • $\begingroup$ Hint: The usual Gronwall fails if u is multiplied by a negative number in the integral. $\endgroup$ – Fan Zheng May 2 '15 at 14:49
  • $\begingroup$ @FanZheng:If $v'(s) \geqslant 0$, I know it's right, then you mean if $v'(s) \leqslant 0$, it's wrong? $\endgroup$ – mafan May 3 '15 at 2:00
  • $\begingroup$ Sorry, I didn't see that t_0 is also arbitrary, is it? $\endgroup$ – Fan Zheng May 3 '15 at 4:53
  • $\begingroup$ @FanZheng:Yes, it's arbitrary. $\endgroup$ – mafan May 4 '15 at 8:55

If $u$ is e.g. Lipschitz the desired inequality is true. Rewrite the original inequality as $$\frac{u(t)-u(t_0)}{t-t_0} \leq \frac{1}{t-t_0}\int_{t_0}^t u(\log v)'\,ds.$$ Taking $t$ to $t_0$ we get $$(\log u)' \leq (\log v)'.$$ Integrating gives the desired inequality.

If $u$ is not continuous, note that $\limsup_{t \rightarrow t_0^+} u(t) \leq u(t_0)$ by the given inequality, so the same arguments as above lead to $$\frac{\overline{D}^+ u}{u} \leq (\log v)'.$$ (Here $\overline{D}^+w(x_0) = \limsup_{x \rightarrow x_0^+} (w(x)-w(x_0))/(x-x_0)$ and similarly for $\underline{D}^+$ with $\liminf$ instead). It is easy to verify that $$\overline{D}^+(\log u) \leq \frac{\overline{D}^+ u}{u},$$ so we have $$0 \leq \underline{D}^+(\log v) - \overline{D}^+(\log u) \leq \underline{D}^+(\log v - \log u),$$ and $\log(v) - \log(u)$ is thus increasing, giving the inequality.

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    $\begingroup$ :We should not assume $u$ is continuous. $\endgroup$ – mafan May 4 '15 at 8:57
  • $\begingroup$ I updated my response. Intuitively, the only discontinuities can be downward jumps, which only help the inequality. $\endgroup$ – Connor Mooney May 4 '15 at 17:47
  • $\begingroup$ :If $log(u)$ is continuous, combining with the right upper (Dini) derivative non-negative, we can get $log(v)-log(u)$ is increasing, now you get the right lower (Dini) derivative non-negative, and $u$ is almost continuous, I wonder whether we can get it's increasing? $\endgroup$ – mafan May 5 '15 at 9:01

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