Let $f : [0,1] \to \mathbb{R}$ be continous and define the operator $T$ in the following way
$$Tf(x)=\frac{1}{\sqrt{\pi}} \int_0^x \frac{f(t)}{\sqrt{x-t}} \, \text{d}t \, . $$
(i) Prove $Tf$ is continuous on $[0,1]$.
(ii) Prove that $$T^2f(x)= \int_0^xf(t)\,\text{d}t\,, \ \ \ \forall x \in [0,1]\,. $$
I think I got the first part, using the limit under the integral sign theorem. However, I am stuck on part (ii).
On
Another kind of answer.
Put first $f_m(t)=t^m$, where $m$ is a real $\geq 0$. Then using the change of variable $t=ux$ we get easily that $T(f_m)(x)=c_m x^{m+1/2}$, with $$\sqrt{\pi}c_m= \int_0^1 u^m(1-u)^{-1/2}du=B(m+1,1/2)=\Gamma(m+1)\Gamma(1/2)/\Gamma(m+3/2)$$
Hence $$T^2(f_m)=(m+1)c_mc_{m+1/2}\int_0^x f_m(t)dt$$ It is easy to see (using $\Gamma(1/2)=\sqrt{\pi}$) that $(m+1)c_mc_{m+1/2}=1$. Hence we get the formula for all $f_m$, and by linearity for all $f\in \mathbb{R}[x]$. But as $T^2$ and $G$ defined by $G(f)(x)=\int_0^x f(t)dt$ are continous function of $f$, and $\mathbb{R}[x]$ is dense in $C([0,1], \mathbb{R})$, we have the equality for all $f$.
We have
\begin{align} T^2f(x) &= T(Tf)(x) = \frac{1}{\sqrt{\pi}}\int_0^x \frac{Tf(t)}{\sqrt{x-t}}\, dt\\ &=\frac{1}{\sqrt{\pi}}\int_0^x \frac{1}{\sqrt{x-t}} \cdot \frac{1}{\sqrt{\pi}}\int_0^t \frac{f(u)}{\sqrt{t-u}}\, du\, dt\\ &= \frac{1}{\pi}\int_0^x f(u)\, du \int_u^x (x - t)^{-1/2}(t -u)^{-1/2}\, dt\tag{*}\\ \end{align}
By the change of variable $v = x - t$,
$$\int_u^x (x - t)^{-1/2}(t - u)^{-1/2}\, dt = \int_0^{x - u} v^{-1/2}(x - u - v)^{-1/2}\, dv$$
By the change of variable $v = (x - u)w$,
$$\int_0^{x-u} v^{-1/2}(x - u - v)^{-1/2}\, dv = \int_0^1 w^{-1/2}(1 - w)^{-1/2}\, dw$$
Show that $\int_0^1 w^{-1/2}(1 - w)^{-1/2}\, dw = \pi$. Then (*) becomes $\int_0^x f(u)\, du$.