Let $T= -\Delta$ and $D=\{u\in L^2 , \Delta u \in L^2\}$ i want to show that $(T,D)$ is a self-adjoint operator.
The first idea i got is to proof that it is closed and $Ker(T^*\pm i) =\{0\}$ unfortunatly i have been blocked against his closedness , such that i can't calculate that kernel because i haven't the domain of the adjoint.
so the second way was more simple which is to show that $R(T\pm i ) = H$( whole space ) i can proceed but Lax-Milgram theorem and it is ok.
The third one which is the most interresting was to make the Fourier transform of the operator and it will be $Tu=Fu$ such that $F$ is a real measurable function and $D=\{u \in L^2 , Fu \in L^2\}$ but i still confused about the equivalence between the self-adjonction of this two operators.
My other quastion is what if $D= \mathcal{C}^{\infty}_0$. Any help please ?
Let $\partial=-i\frac{d}{dx}$ on the domain $\mathcal{D}(\partial)$ consisting of all locally absolutely continuous functions $f \in L^2(\mathbb{R})$ such that $f'\in L^2$. You can show that $f\in\mathcal{D}(\partial)$ iff $$ f\in L^2, \xi\hat{f}(\xi) \in L^2. $$ Furthermore, $\partial f = (\xi\hat{f})^{\vee}$. So $\partial$ is unitarily equivalent to a multiplication operator on $L^2$. That's enough to show that $\partial$ is selfadjoint on $\mathcal{D}(\partial)$. The Fourier transform is a unitary "diagonalization" of $\partial$. The operator $\partial$ is essentially selfadjoint on $\mathcal{C}_0^{\infty}$.
Your operator $T = -\frac{d^2}{dx^2}$ is selfadjoint on $\mathcal{D}(\partial^2)$, and $T=\partial^2$. The restriction of $T$ to $\mathcal{C}_0^{\infty}$ is essentially selfadjoint.