Let $0 \leq \xi \leq T$, $Y_0 \in \mathcal C([0, 2T])$, $a_1 \in L^2([0,T])$ and consider the following function: $$ Y_{1}(\xi)=\int_{0}^{\xi} a_{1}(\alpha) Y_{0}(\alpha+\xi) d \alpha $$ Now if inside the integral $Y_0$ did not depend on $\xi$ then $Y_1$ would be absolutely continuous. Instead in the case I describe is $Y_1$ continuous?
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Fix $\varepsilon >0$.
Since $Y_0$ is continuous on a compact set, $Y_0$ is uniformly continuous. Let $\delta_1>0$ such that, for all $x_1,x_2\in[0,2T]$, if $|x_1-x_2|\le\delta_1$, then $|Y_0(x_1) - Y_0(x_2)| \le \varepsilon$.
Since $a_1 \in L^2([0,T])$, then $a_1 \in L^1([0,T])$. It follows that there exists $\delta_2 >0$ such that, for every measurable $A \subset [0,T]$ whose Lebesgue measure is less or equal than $\delta_2$, we have $\int_A |a_1(x)| \mathrm{d}x \le \varepsilon$.
Let $\delta := \min(\delta_1, \delta _2)$.
Fix $\zeta,\xi \in [0,T]$ such that $|\xi -\zeta| \le \delta$. WLOG, assume that $\zeta \le \xi$.
We have \begin{align*} \big|Y_1(\zeta) - Y_1(\xi)\big|&= \bigg| \int_0^\zeta a_1(\alpha)Y_0(\alpha+\zeta) \mathrm{d}\alpha - \int_0^\xi a_1(\alpha)Y_0(\alpha+\xi) \mathrm{d}\alpha \bigg| \\ &\le \int_0^\zeta \big|a_1(\alpha)\big| \big|Y_0(\alpha+\zeta) - Y_0(\alpha+\xi)\big| \mathrm{d}\alpha + \int_\zeta ^ \xi |a_1(\alpha)||Y_0(\alpha+\xi)| \mathrm{d}\alpha \\ &\le \int_{0}^{T} |a_1(\alpha)| \mathrm{d}\alpha\cdot \sup_{\alpha \in [0,T]} \big|Y_0(\alpha+\zeta) - Y_0(\alpha+\xi)\big| + \int_{\zeta}^{\xi} |a_1(\alpha)| \mathrm{d}\alpha\cdot \sup_{\alpha \in [0,2T]}|Y_0(\alpha)| \\ &\le \varepsilon \cdot \bigg( \int_{0}^{T} |a_1(\alpha)| \mathrm{d}\alpha + \sup_{\alpha \in [0,2T]}|Y_0(\alpha)| \bigg). \end{align*} Since $\zeta, \xi$ and $\varepsilon$ were arbitrarily chosen, it follows that $Y_1$ is uniformly continuous, hence continuous.
Bob
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