A singular Gronwall inequality Let $f : [0,T] \to \Bbb{R}^+$ be a continuous function such that $f(0)=0 $ and :
$$
f(t)\le C\int_0^t s^{-1}f(s) ds,\; \forall t\in [0,T]
$$
for some constant $C>0.$ Is it true that $f(t)=0,\; \forall t\in [0,T]?$
 A: No, for
$$
f(t)=t,
$$
we have
$f(0)=0$ and
$$
\int_0^t s^{-1}f(s)\,ds=t=f(t),\quad\forall t>0,
$$
so (your constant is $1$, and you actually have equality)
$$
f(t)\leq 1\int_0^t s^{-1}f(s)\,ds,\quad\forall t>0.
$$
A: If we set $g(t)=\frac{f(t)}{t}$ and $G(t)=\int_{0}^{t}g(u)\,du$, we have the inequality:
$$ t\cdot g(t)\leq C\cdot G(t)\tag{1} $$
or:
$$ \frac{g(t)}{G(t)}\leq \frac{C}{t}\tag{2} $$
hence integrating both sides over $[\varepsilon,x]$ we have:
$$ \log(G(t))-\log(G(\varepsilon))\leq C\left(\log(t)-\log(\varepsilon)\right)\tag{3}$$
or:
$$ G(t)\leq G(\varepsilon)\left(\frac{t}{\varepsilon}\right)^C\tag{4} $$
but as shown by mickep, by choosing $g\equiv 1\neq 0$ and $C=1$ we have that $(1)$ is fulfilled.
A: Here's a related family of counter examples to mickep's example:
(I'm writing your equation in differential form)
Let $T < 1$, and $C \geq 1$.
Then if $f(t) = t$, then $f'(t) \leq C f(t)/t$ on $[0,T]$. However, on $(0,T]$, $f(t) > t^C$.
This is helpful because on might guess (as I did) that $f'(t) \leq C f(t) /t$ on $[0,T]$ implies that $f(t) \leq t^C$ for short times, but this example shows that this is not true.
