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

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

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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.

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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.

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