Is this function absolutely continuous? I'm studying of a function $T:=T_1+T_2$. Here
$$
T_1(t)=\int_0^t \left([t(1-\tau)]^{\frac{1}{2}}-(t-\tau)^{\frac{1}{2}}\right)k(\tau)d\tau,
$$
$$
T_2(t)=t^{\frac{1}{2}} \int_t^1 (1-\tau)^{\frac{1}{2}} k(\tau)d\tau,$$
and $k$ is a non-negative continuous function on $(0,1]$ satisfying
$$\int_0^1 \tau ^{\frac{1}{2}} k(\tau)d\tau<\infty\label{1} \tag{1}.$$
I was wondering if $T$ is absolutely continuous on $[0,1].$
By Libniz integral rule, $T’(t)$ exists for all $t\in (0,1)$. Consequently, if we prove $T’ \in L^1(0,1)$ and $T(t)=\int_0^t T’(s)ds$ for all $t \in (0,1]$, then $T$ is absolutely continuous on $[0,1].$ But I don’t know how to do it.
Since integrands of $T_1$ and $T_2$ are all bounded by $C \tau^{\frac{1}{2}}$ for some a constant $C$, by \eqref{1}, it can be shown that $T :[0,1] \to [0,\infty)$ is continuous and $T(0)=T(1)=0$ (see 2).
I would be grateful if you could give any comments on my questions.
 A: Write
$$ T(t) = \int_{0}^{1} \Bigl( \sqrt{t(1-\tau)} - \sqrt{(t-\tau)\mathbf{1}_{\{t>\tau\}}} \Bigr) k(\tau) \, \mathrm{d}\tau. $$
Then we expect that $T'(t)$ can be computed by interchanging the order of integration and differentiation. So we introduce another function $S(t)$ that serves as the candidate for $T'(t)$:
$$ S(t) = \int_{0}^{1} \frac{1}{2} \biggl( \sqrt{\frac{1-\tau}{t}} - \sqrt{\frac{1}{t-\tau}\mathbf{1}_{\{t>\tau\}}} \biggr) k(\tau) \, \mathrm{d}\tau $$
Then by noting that, for $0 < t < 1$,
\begin{align*}
&\int_{0}^{t} \int_{0}^{1} \frac{1}{2} \Biggl| \sqrt{\frac{1-\tau}{s}} - \sqrt{\frac{1}{s-\tau}\mathbf{1}_{\{t>\tau\}}} \Biggr| k(\tau) \, \mathrm{d}\tau \mathrm{d}s \\
&= \int_{0}^{t} \int_{0}^{1} \frac{1}{2} \biggl[ \biggl( \sqrt{\frac{1}{s-\tau}} - \sqrt{\frac{1-\tau}{s}} \biggr)\mathbf{1}_{\{s>\tau\}} + \sqrt{\frac{1-\tau}{s}} \mathbf{1}_{\{s<\tau\}} \biggr] k(\tau) \, \mathrm{d}\tau \mathrm{d}s \\
&= \int_{0}^{1} \int_{0}^{1} \frac{1}{2} \biggl[ \biggl( \sqrt{\frac{1}{s-\tau}} - \sqrt{\frac{1-\tau}{s}} \biggr)\mathbf{1}_{\{\tau<s<t\}} + \sqrt{\frac{1-\tau}{s}} \mathbf{1}_{\{s<\tau \wedge t\}} \biggr] k(\tau) \,  \mathrm{d}s\mathrm{d}\tau \\
&= \int_{0}^{1} \Bigl[ \Bigl( \sqrt{t-\tau} - \sqrt{t(1-\tau)} + \sqrt{\tau(1-\tau)} \Bigr)\mathbf{1}_{\{\tau<t\}} + \sqrt{(\tau \wedge t)(1-\tau)} \Bigr] k(\tau) \,  \mathrm{d}\tau
\end{align*}
We can check that
$$ \Bigl( \sqrt{t-\tau} - \sqrt{t(1-\tau)} + \sqrt{\tau(1-\tau)} \Bigr)\mathbf{1}_{\{\tau<t\}} + \sqrt{(\tau \wedge t)(1-\tau)} \leq 2\sqrt{\tau} $$
whenever $0 < t < 1$ and $0 < \tau < 1$, and so, the above integral is finite. This allows us to invoke the Fubini's theorem to write
\begin{align*}
\int_{0}^{t} S(s) \, \mathrm{d}s
&= \int_{0}^{1} \int_{0}^{t} \frac{1}{2} \biggl( \sqrt{\frac{1-\tau}{t}} - \sqrt{\frac{1}{t-\tau}\mathbf{1}_{\{t>\tau\}}} \biggr) k(\tau) \, \mathrm{d}s\mathrm{d}\tau \\
&= \int_{0}^{1} \Bigl( \sqrt{t(1-\tau)} - \sqrt{(t-\tau)\mathbf{1}_{\{t>\tau\}}} \Bigr) k(\tau) \, \mathrm{d}\tau \\
&= T(t).
\end{align*}
Therefore $T$ is absolutely continuous and satisfies $T'(t) = S(t)$.
