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

Is there a constant $C$ (depending on $\alpha, \beta$) such that $\sup_{t \in [0, 1]} f(t) \le C$?

Thank you so much for your elaboration!

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$\newcommand\al\alpha\newcommand\be\beta$The answer is yes.

Indeed, using the inequalities $u^\be\le\max(1,u)\le1+u$ for $u=f(s)\ge0$, we see that for $t\in[0,1]$ the inequality $$f(t) \le \al+ \int_0^t (t-s)^{-1/2} [f(s) + f(s)^\be] \,ds$$ implies $$f(t) \le \al+2t^{1/2}+2 \int_0^t (t-s)^{-1/2} f(s) \,ds.$$

It remains to use Theorem 2.2, which yields $$f(t) \le (\al+2t^{1/2})E_{1/2}(2\sqrt\pi\,t^{1/2}) \le E_{1/2}(2\sqrt\pi)(\al+2)<573503\,(\al+2)=:C$$ for $t\in[0,1]$, where $E_b$ is the Mittag--Leffler function, which is an entire function given by series $E_b(z):=\sum_{k=0}^\infty\frac{z^k}{\Gamma(bk+1)}$ for complex $z$.

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  • $\begingroup$ Thank you for your answer! I wonder if we can obtain a tight upper bound in the sense that if $\alpha=0$ then $f=0$. $\endgroup$
    – Akira
    Commented Mar 26 at 20:47
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    $\begingroup$ @Akira : I think this can be done, say by an iteration argument and using Jensen's inequality. However, as your present question has been answered, I suggest you post the additional question separately. $\endgroup$
    – Iosif Pinelis
    Commented Mar 26 at 20:58

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