Is this function well defined and 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
$$\int_0^1 \tau ^{\frac{1}{2}} k(\tau)d\tau<\infty\label{1} \tag{1}.$$ For example, $k(t)= t^{-\frac{3}{4}}$.
I was wondering if the following assertions are true:
$(1)$ $T$ is well defined on $(0,1)$ and $$\lim_{t \to 0^+}T(t)=\lim_{t \to 1^-}T(t)=0.$$
$(2)$ $T$ is continuous on $[0,1].$
For $$T\left(\frac{1}{2}\right)=T_1\left(\frac{1}{2}\right)+T_2\left(\frac{1}{2}\right),$$ it is easily see that $T_2\left(\frac{1}{2}\right)$ is well defined, but it is not obvious that $T_1\left(\frac{1}{2}\right)$ is well defined. I don't know how to use the assumption \eqref{1}. Also, I think, by using Lebesgue dominated convergence theorem, $$\lim_{t \to 0^+}T(t)=\lim_{t \to 1^-}T(t)=0$$ can be proved, but I couldn't.
I would be grateful if you could give any comments on my questions.
 A: Obviously $T_2$ is well defined and continous in $t$. Also if we replace $T_1$ by $T_1^a$ where we replace the lower integration bound by some $a>0$ then $T_1^a$ is also well defined and continuous in $t$. (This follows from the continuity of all occuring functions on $[a,1]$, as a continuous functions on $[a,b]$ is bounded and thus the integral is finite, and we get continuity quite trivially.)
Thus we need to study the integral close to $0$. We know from $\int_0^1 \tau^{1/2}k(\tau)\,d\tau<\infty$ that $\int_0^a \tau^{1/2}k(\tau)\,d\tau\xrightarrow{a\to 0}0$.
So let’s study $B(t,\tau)=[t(1-\tau)]^{1/2}-(t-\tau)^{1/2} = (t-t\tau)^{1/2} - (t-\tau)^{1/2}$ for $\tau$ close to $0$. At $\tau=0$ this is $0$. Furthermore the derivative to the square of that expression is
$$ -\frac{{\left((t - \tau)^{1/2} t - (t-t\tau)^{1/2}\right)} {\left((t-t\tau)^{1/2} - (t - \tau)^{1/2}\right)}}{((t-t\tau)(t - \tau))^{1/2}} =: D(t,\tau) $$
If $\tau$ is small the denominator is close to $t$, and the numerator is close to $(t^{3/2}-t^{1/2})0=0$. Thus the derivative around $0$ is close to $0$, and at the very least bounded (so $|D(t,\tau)|<C(t)$ for $\tau$ close to $0$).
This means that around $0$ we get $((t-t\tau)^{1/2}-(t-\tau)^{1/2})^2<C(t)\tau$ and thus $(t-t\tau)^{1/2}-(t-\tau)^{1/2}<\sqrt{C(t)}\tau^{1/2}$.
We can then even get this constant independent of $t$: For each $t$ $D(t,\tau)$ is monotonic in $\tau$ for $\tau < t$, as even the numerator is a product of montonic functions and the denominator is falling:
The first factor $t(t-\tau)^{1/2} - (t-t\tau)^{1/2}$ is falling (and thus by incorporating the $-$ sign increasing) as it’s derivative is
$$ -\frac{t {\left((t-t\tau)^{1/2} - (t - \tau)^{1/2}\right)}}{2 \, \left((t-t\tau) (t - \tau)\right)^{1/2}}$$
which is negative, as for $0<t<1$ and $0< \tau\leq t$ we have
$ t-t\tau > t-\tau $ (and in the cases $t=1,\tau=0$ we have equality).
The second factor is simply $B(t,\tau)$. As clearly $D(t,\tau)\geq 0$ we have $B^2(t,\tau)$ is increasing in $\tau$, and as $B(t,\tau)\geq0$ we thus also get $B(t,\tau)$ is increasing in $\tau$.
In the denominator we have $(t-t\tau)$ and $(t-\tau)$, both of which are decreasing in $\tau$. Thus $D$ is a product of nonnegative, increasing factors. Thus $D$ is increasing in $\tau$.
But this then implies that for each $t$ we have $B^2(t,\tau)$ is convex in $\tau$. This means that $B^2(t,\tau)$ lies under the line that is $B^2(t,0)=0$ for $\tau = 0$ and $B^2(t,t)=t(1-t) = t-t^2$.
Then for $0\leq t\leq 1$ we have $0\leq t-t^2\leq t$. So we get: For each $t$ we have $B^2(t,\tau)<\tau$ for $0\leq\tau\leq t$.
Then we get
$$ T_1 \leq \int_0^t \tau^{1/2}k(\tau)\,d\tau$$
This implies that $T_1$ is well defined. Also as the integrand of $T_1$ is bounded on $[0,1]$ by $\tau^{1/2}k(\tau)$ it is continuous in $t$ by dominated convergence.
Also this gives us that $T_1(t)\xrightarrow{t\to0}0$. So take a look at $T_2(t)$. Note that for $t<b$ we get
$$ \int_0^b \tau^{1/2}k(\tau)\,d\tau\geq \int_t^b \tau^{1/2}k(\tau)\,d\tau \geq t^{1/2}\int_t^b k(\tau)\,d\tau $$
Take $b$ small enough so that the left side is smaller than some $\epsilon$. Then for $t<b$ we have
$$ T_2(t) = t^{1/2}\int_b^1 \ldots + t^{1/2}\int_t^b \ldots \leq t^{1/2}\int_b^1 \ldots + \epsilon$$
As the first part goes to $0$ we get
$$ 0\leq \liminf_{t\to0}T_2(t)\leq \limsup_{t\to0} T_2(t)\leq \epsilon $$
With $epsilon\to 0$ you get $\lim_{t\to0}T_2(t)=0$.
Finally
$$\lim_{t\to1} T(t) = T(1) = \int_0^1 ((1-\tau)^{1/2}-(1-\tau)^{1/2})k(\tau)\,d\tau = 0$$
A: For the specific case where $k(t)= t^{-\frac{3}{4}}$
The two integrals do not present  much problems
$$T_1=-\frac{8 \, _2F_1\left(-\frac{3}{4},\frac{1}{2};\frac{1}{4};t\right)}{3 t^{1/4}}+\frac{4}{3}
  t^{3/4} \sqrt{1-t} -\frac{2 \sqrt{\pi } \, \Gamma \left(\frac{5}{4}\right)}{\Gamma
   \left(\frac{7}{4}\right)}t^{3/4}+\frac{8 \sqrt{1-t}}{3 t^{1/4}}$$
$$T_2=\sqrt{t} \left(\frac{2 \sqrt{\pi } \,\Gamma \left(\frac{1}{4}\right)}{3 \Gamma
   \left(\frac{3}{4}\right)}-\frac{2}{3} \left(2 t^{1/4}\sqrt{1-t}
   +B_t\left(\frac{1}{4},\frac{1}{2}\right)\right)\right)$$ and the surprise
$$T=T_1+T_2=\frac{2 \sqrt{\pi } \, \Gamma \left(\frac{1}{4}\right)}{3 \Gamma
   \left(\frac{3}{4}\right)}\left(1-t^{1/4}\right) \sqrt{t}$$  So, assertions $(1)$ and $(2)$ are true.
No problem anywhere for $0 \leq t \leq 1$.
