# 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 $$f(t)=0,\; \forall t\in [0,T]?$$

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) $$f(t)\leq 1\int_0^t s^{-1}f(s)\,ds,\quad\forall t>0.$$

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.

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.