We have the operator $T$ on $L_2[0,1]$ which is defined as $Tf = y$ where $y$ is the solution to the ODE $y^{\prime \prime} + y^\prime = f$ with boundary conditions $y(0)=0, y(1) = 1$
Show that $T$ is compact and if $T = T^\ast$? (self-adjoint)
My take on this was to solve the ODE by substituting $w = y^\prime$, which gave me the solution: $$ y(x) = c_2 + \int^x_a \bigg( c_1 e^{-t} + e^{-t}\int^t_a e^\xi f(\xi) d\xi \bigg) dt .$$
So I think this integral expression would be my $T$-operator where I put $f$ to obtain $y$ $(Tf = y)$. A linear operator on normed spaces is compact if the image is relatively compact, i.e. the closure of the image is compact. One way is to look at if the operator maps bounded sequences onto convergent sub-sequences, but I'm not sure with this particular operator.