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?
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.
