$\, f \in L^2([0,1],\mathbb{C})$
show that $T : f \to Tf, \, f \in L^2([0,1],\mathbb{C})$ is continuous, $[T(f)](x) = ie^{i\pi x}(\int_0^x e^{-i\pi t}f(t)dt - \int_x^1 e^{-i\pi t}f(t)dt)$
the operator is linear so it suffices to show the boundedness only
$[T(f)](x) \overline{[T(f)](x) } = (\int_0^x \cos{(\pi t)}f(t)dt - \int_x^1 \cos{(\pi t)}f(t)dt)^2 + (\int_0^x \sin{(\pi t)}f(t)dt - \int_x^1 \sin{(\pi t)}f(t)dt)^2 $
then by Holder Inequality : $$|[T(f)](x) \overline{[T(f)](x) }| \leq 2x\|f\|^2_{L^2} +2(1-x)\|f\|^2_{L^2} + 4 \sqrt{x(1-x)} \|f\|^2_{L^2} \leq C \|f\|^2_{L^2} $$ and that implies $\|[T(f)](x)||_{L^2}^2 \leq C \|f\|^2_{L^2} $
now suppose $(f_n)_{n \geq 0} \subset L^2([0,1],\mathbb{C}) $ such that $\|f_n\|_{L^2} \leq 1$
also suppose that that sequence of functions converges $\textbf{weakly}$ to some function in $L^2$ called $f$
from that we can say that $[T(f_n)](x) \to [T(f)](x)$
now all the previous points are supposed to help me prove that $(T(f_n))_{n \geq 0}$ converges $\textbf{strongly}$ to $T(f)$
but I fail to see how, any help will be greatly appreciated.
Use Theorem 8.2.3 in https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&ved=2ahUKEwjWt7j83t3fAhXKso8KHSR3BecQFjACegQIBRAC&url=http%3A%2F%2Fpeople.math.gatech.edu%2F~heil%2Fmetricnote%2Fchap8.pdf&usg=AOvVaw3hD60cdJ-gOIAiuX4Jr5hh
to conclude that $T$ is a compact operator. Hence any subsequence of $\{Tf_n\}$ has a further subsequence converging in the norm. The limit has to be $Tf$ because $f_n \to f$ weakly implies $Tf_n \to Tf$ weakly. Hence the entire sequence $\{Tf_n\}$ converges in the norm to $Tf$.