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Let $T>0$, $\alpha,\beta>0$ and consider a non-negative continuous function $x$ on $[0,T]$ such that for all $t \in [0,T]$ one has $$x(t) \leq \alpha+\beta\left(\int_0^t x(s)\,\mathrm ds \right)^{1/2}.$$ Does anyone knows what kind of Grönwall inequality I can get from this ? It would be fantastic if I can get something like $x(t) \leq Ct$.

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  • $\begingroup$ It looks to me like it invites very much to try pulling the fundamental theorem of calculus, but I would not bet my freedom on it. $\endgroup$ – mathreadler Mar 21 '17 at 12:42
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The inequality implies $$ x(t) \le \alpha + \epsilon \int_0^tx(s) ds + \frac{\beta}{4 \epsilon} $$ for any $\epsilon > 0$. By the usual Gronwall inequality, this implies $$ x(t) \le \left(\alpha + \frac{\beta}{4 \epsilon} \right) e^{\epsilon t} $$ for any $t > 0,\epsilon > 0$. Now set $\epsilon = t^{-1}$ to obtain $$ x(t) \le \left(\alpha + \frac{\beta}{4} t \right)e $$ You may be able to get better constants by choose $\epsilon = \delta t^{-1}$ with the right $\delta > 0$.

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By an appropriate change of variables we may assume without loss of generality that $\alpha=\beta=1$. I will also asume that $x(t)\ge0$. Let $u(t)=\int_0^t x(s)\,ds$. Then $$ u'(t)\le1+\sqrt u\implies\frac{u'}{1+\sqrt u}\le1. $$ Integrating and using $u(0)=0$ we get $$ 2\sqrt{u}-2\log(1+\sqrt u)\le t. $$ Let $f(u)$ be the function on the right hand side of the above inequality. It is a concave function, and satisfies the following inequalities: $$ f(u)\ge\begin{cases} 2(1-\log2)u & \text{if }0\le u\le1,\\ 1-2\log2+\sqrt u &\text{if }u>1. \end{cases} $$ From here we deduce that $u(t)\le C_1\,t$ where $0\le u\le1$ and $u(t)\le C_2\,t^2$ where $u>1$. This means that for $x$ we have bounds of the form $$ x(t)\le\begin{cases} 1+A\,\sqrt t & \text{if $t$ small,}\\ B\,t &\text{if $t$ large,} \end{cases} $$ for some constants $A$ and $B$.

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