Let $H=L^2(0,\pi)$ and T such that $T:D(T)\subset H \rightarrow H$ defined by: $$D(T)=\{v\in H^2(0, π):v(0)= v'(0) = v(π) = v′(π)=0\}$$ $$Tv=v''\;\text{with}\;v\in D(T)$$ Prove that T is a symmetric operator, find $D(T^*)$ and $T^*$. At last, prove that $T$ is not self-adjoint.
What I have tried:
I have already proved that the operator T is symmetric, which is not difficult, but I'm stuck in obtaining the adjoint. In my class we saw the example with the weak derivative operator and as what I have seen the steps to solve the problems are:
First we need to define an operator like this $\lambda_v(\phi)=\int^{\pi}_{0}\phi''vdx$, then we note that this operator is linear and continuous. Then by the B.L.T Theorem, we extend it ($\overline{\lambda_v}$),such that we have this: $$|\lambda_v|\leq C||w|| $$ Where the norm is in $L^2$ and C is a constant. Then by the Riesz representation Theorem, we note that we get a unique $h_v$ such that it satisfies this: $$\overline{\lambda_v}(w)=(w,h_v)\forall w\in L^2$$ After this, it should be simple to determine who is $D(T^*)$, and then it would be immediate to obtain $T^*$ but I do all of the above and can´t go any further. Any help would be really appreciated.