Working on my thesis I need to prove or disprove that there exists $1/2 < \alpha <1$ and $C$ independent of $\epsilon \in (0,1]$
$$\int_0^s \int_r^s \frac{1}{\epsilon} (s-r)^{-\alpha}(s-\tau)^{-\alpha}e^{-\lambda (s-r)}e^{-\lambda (s-\tau)}e^{-\delta (\tau-r)/\epsilon}\,\mathrm{d}\tau\,\mathrm{d}r < C$$ where $s,r \in [0,T]$, $T, \lambda>0, \delta>0$ are fixed and $1>\epsilon >0$ is a parameter.
Then by Holder with $3/2, 3$ coniugates
$$\leq \left [\int_0^{s} \int_r^{s} (s-\tau)^{-3/2\alpha}(s-r)^{-3/2\alpha}e^{-3/2\lambda (s-\tau)}e^{-3/2\lambda (s-r)} dz d\nu\right ]^{2/3} \\ \times \left [\int_0^{s} \int_r^{s} e^{-3\delta (\tau-r)/\epsilon} \frac{1}{\epsilon^3}\,\mathrm{d}z\,\mathrm{d}\nu \right]^{1/3}. $$ Now the first integral is finite and independent of $\epsilon$ if $1/2 < \alpha < 2/3$. But the second integral seems to depend on $\epsilon$. (I can't use Holder wit $p=2$ since I have $\alpha> 1/2$). Do you see other ways or the claim is wrong?
The claim is not true. We have
\begin{align*} &\frac{1}{\varepsilon}\int_0^s\int_r^s(s-r)^{-a}(s-\tau)^{-a}e^{-\lambda(s-r)}e^{-\lambda(s-\tau)} e^{-\delta(\tau-r)/\varepsilon}d\tau dr\\ \ge &\frac{e^{-2\lambda s}}{\varepsilon} \int_{0}^s (s-r)^{-2a} \int_{r}^s e^{-\delta(\tau-r)/\varepsilon}= \frac{1}{\delta}\int_0^s (s-r)^{-2a}(1-e^{-\delta(s-r)/\varepsilon})dr\\ \ge &\frac{1}{\delta} \int_{s-\varepsilon}^s(s-r)^{-2a}(1-e^{-\delta(s-r)/\varepsilon})dr\ge \frac{\varepsilon^{-2a}}{\delta}\int_{s-\varepsilon}^s(1-e^{-\delta(s-r)/\varepsilon})dr\\ =&\frac{\varepsilon^{-2a}}{\delta}(\varepsilon-\frac{\varepsilon}{\delta}+\frac{\varepsilon}{\delta}e^{-\delta})=\frac{\varepsilon^{1-2a}}{\delta}(1-1/\delta+e^{-\delta}/\delta)\to\infty \end{align*}
as $\delta\to0$, since $a>1/2$ and $1-1/\delta+e^{-\delta}>0$.