 Solution

Since $f(0)=0$, the fundamental theorem of calculus gives $f=T{f}^{\prime }$, with the same oriented integral on the negative half-interval. Set $h(t)=t^2$. The adjoint calculation gives the real-valued functionChoose any value at zero. This function is bounded and hence belongs to $L^2(-\pi ,\pi ]$. Since it is real, the adjoint operator identity yieldsOne can also check this by integration by parts separately on each half-interval: $g^{\prime }=-t^2$, $g(\pm \pi )=0$, and the two possible jump terms at zero vanish because $f(0)=0$.