I am stuck with the following exercise:
Show that the operator $A= -d^2x$ with $D(A) =\{f \in L_2[0,1]:f,f' \in C[0,1] \,with\, f'' \in L_2,\, f(0)=f(1)=0 , f'(0)=f'(1)=0 \}$
Is symmetric but not self-adjoint.
I am aware of all the definitions, but I just can't figure it out.
Thanks a lot for your help.
You need to assume $f'$ is absolutely continuous as well in order for integration by parts to be applied. Define $\mathcal{D}'$ to be the subspace on $L^2[0,1]$ consisting of all $g\in L^2[0,1]$ be such that $g\in C^1[0,1]$ and $g'$ is absolutely continuous with $g''\in L^2$. Then you can use integration by parts twice and use the fact that $f(0)=f'(0)=f(1)=f'(1)=0$ to show that $$ \langle Af,g\rangle = \langle f,-g''\rangle,\;\; f\in\mathcal{D}(A), g\in\mathcal{D}'. $$ In particular, the above holds for all $g\in\mathcal{D}(A)$ because $\mathcal{D}(A)\subset \mathcal{D}'$. Hence $A$ is symmetric. Because this holds for all $f\in\mathcal{D}(A)$ (which is a dense subspace of $L^2[0,1]$,) it follows that $g\in\mathcal{D}(A^*)$ and $A^*g=-g''$. Therefore $$ \mathcal{D}(A) \subsetneq\mathcal{D}'\subseteq\mathcal{D}(A^*), $$
which shows that $A$ is symmetric, but not self-adjoint.