My question appeared from one statement about non-uniqueness for ODE's. Consider a continuous function $\theta:(0,1]\to[0,\infty)$ such that $$ \int^1_0 \frac{\theta(s)}{s}ds=\infty. $$ Of course, there are infinitely many solutions for the Cauchy problem $$ \begin{cases} u'(t)=\frac{1+\theta(t)}{t}u(t) \\ u(0)=0 \end{cases} $$ In fact, $u(t)=ce^{-\int^1_t \frac{1+\theta(s)}{s}ds}$ will be a solution for every $c.$ The problem is to rewrite the equation in the form $u'(t)=f(t,u(t))$ where $f$ is continuous on $[0,1]\times\mathbb{R}.$ This is possible when I know that $\theta(t)e^{-\int^1_t\frac{\theta(s)}{s}ds}\to 0,$ $t\to 0.$
So, the question is whether the condition $\int^1_0 \frac{\theta(s)}{s}ds=\infty$ implies $\theta(t)e^{-\int^1_t\frac{\theta(s)}{s}ds}\to 0,$ $t\to 0?$
$\int_0^1\frac{\theta(s)}{s}\,ds=\infty$ implies $$ \lim_{t\to0^+}\exp\bigl(-\int^1_t\frac{\theta(s)}{s}ds\bigr)=0. $$ Since $\theta$ is continuous, $$ \lim_{t\to0^+}\theta(t)\exp\bigl(-\int^1_t\frac{\theta(s)}{s}ds\bigr)=0. $$ In fact, continuity is not needed. It is enough that $\theta$ is bounded on $[0,a)$ for some $a>0$.