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Let $\beta \in (0, 1)$. We assume $f : [0, 1] \to [0, \infty)$ is a measurable and bounded function such that $$ f(t) \le \int_0^t (t-s)^{-\frac{1}{2}} [f(s) + |f(s)|^{\beta}] \, \mathrm d s, \quad \forall t \in [0, 1]. $$

I would like to ask if $f=0$?

Thank you so much for your elaboration!

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$$ f(t) = \varepsilon t^A$$ will be a counterexample if $\varepsilon>0$ is small enough and $A > 1$ is large enough (one basically needs $A \beta + 1/2 \leq A$ and $\varepsilon \leq \varepsilon^\beta \int_0^1 (1-s)^{-1/2} s^A\ ds$).

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  • $\begingroup$ I am sad that it is not true but thank you so much for your answer! $\endgroup$
    – Akira
    Commented Mar 27 at 8:18
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    $\begingroup$ One can compare with the initial value problem $f'(t) = f(t)^\beta, f(0)=0$ (or equivalently $f(t) = \int_0^t f(s)^\beta\ ds$), which famously has a non-trivial solution $f(t) = C t^{\frac{1}{1-\beta}}$ where $C = (1-\beta)^{1/(1-\beta)}$ when $0 < \beta < 1$. $\endgroup$
    – Terry Tao
    Commented Mar 27 at 18:07
  • $\begingroup$ @TerryTao , Can we utilize the Banach fixed-point theorem, to approach that problem ? $\endgroup$
    – zeraoulia rafik
    Commented Mar 27 at 22:54

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