Self-adjoint extensions of Laplacian over half line Let be $A=-\Delta$ the Laplace operator in $L^2(\mathbb{R}^+)$, with domain $D=\mathscr{C}^\infty_0(\mathbb{R}^+)$ the compact supported smooth functions.
This operator $A$ isn't bounded, neither closed; but is symmetric in its domain and so is closable. I know that the operator is not essentially self-adjoint, but what is the domain of $A^*$?
Moreover, analyzing deficiency spaces
$$Z_+ = \ker(A^* -i),\,\,Z_- = \ker(A^*+i)$$ 
I think they should be isometric by complex coniugate, but if I pick a basis $\{u_n\}$ of $Z_+$, how do I prove that $\{\bar{u}_n\}$ is a basis of $Z_-$? Is this sufficient to show that deficiency indices are the same and so conclude that $A$ has self-adjoint extensions?
Thank you very much!
EDIT. 
If my argument above is correct, how does look like a such extension of $A$? Is praticable to classify all the self-adjoint extensions using unitary operators in the well-known manner?
 A: The deficiency indices are the same because of how $\Delta$ commutes with complex conjugation and its domain is closed under complex conjugate. $g \in \mathcal{N}(A^{\star}-iI)$ iff
$$
                 ((A+iI)f,g) = 0,\;\;\; f \in \mathcal{D}(A).
$$
Conjugating the above gives
$$
       ((A-iI)\overline{f},\overline{g})=0,\;\;\; f \in \mathcal{D}(A).
$$
Because the domain of $A$ is invariant under complex conjugate, then the above is equivalent to
$$
               ((A-iI)f,\overline{g})=0,\;\;\; f \in \mathcal{D}(A),
$$
which is equivalent to $(A^{\star}+iI)\overline{g}=0$. If $\{ g_j \}$ is an orthonormal basis of $\mathcal{N}(A^{\star}-iI)$ then $f\in\mathcal{N}(A^{\star}+iI)$ can be written
$$
                \overline{f} = \sum_j (\overline{f},g_j)g_j \\
                  f = \sum_j(f,\overline{g_j})\overline{g_j}.
$$
And $\{\overline{g_j}\}$ is orthonormal.
A closed symmetric operator such as $A^c$ has a selfadjoint extension iff $\mathcal{N}(A^{\star}-iI)$ is unitarily equivalent to $\mathcal{N}(A^{\star}+iI)$. The unitary equivalences are in one-to-one correspondence with the selfadjoint extensions of $A^c$.
Not all selfadjoint extensions are appropriate for $-\Delta$ because one typically wants the domain of such an extension to be closed under conjugation.
As for what the domain looks like, suppose $B$ is a selfadjoint extension of $A$. Then $(B-iI)(B+iI)^{-1}=U$ is unitary. And $A^{\star}$ extends $B$. So $A\preceq B \preceq A^{\star}$. If $f \in \mathcal{N}(A^{\star}-iI)$, then
$$
          Uf = (B-iI)(B+iI)^{-1}=f-2i(B+iI)^{-1}f,
$$
and, because $A^{\star}$ extends $B$,
$$
       (A^{\star}+iI)Uf = (A^{\star}+iI)f-2i(A^{\star}+iI)(B+iI)^{-1}f \\
            = 2if - 2if = 0.
$$
So $U : \mathcal{N}(A^{\star}-iI)\rightarrow\mathcal{N}(A^{\star}+iI)$. Notice also that
$$
         Uf - f = -2i(B+iI)^{-1}f \in \mathcal{D}(B).
$$
In fact, the extended domain is
$$
                \mathcal{D}(B)=\mathcal{D}(A^c)\oplus\{ Uf-f : f\in\mathcal{N}(A^{\star}-iI)\}.
$$
The selfadjoint extensions $B$ are in one-to-one correspondence with unitary maps $U : \mathcal{N}(A^{\star}-iI)\rightarrow\mathcal{N}(A^{\star}+iI)$ through this correspondence. I'll let you work out that details.
If you want the domain of an extension $B$ to be closed under conjugation, then that adds conditions related to the involution of complex conjugation. For your original case,
$$
                \mathcal{N}(-\Delta^{\star}+iI)=\{ \alpha e^{-\sqrt{i}t} : \alpha\in\mathbb{C}\} \\
              \mathcal{N}(-\Delta^{\star}-iI) = \{ \alpha e^{-\overline{\sqrt{i}}t} : \alpha\in\mathbb{C} \}.
$$
Use $\sqrt{i}=\frac{1}{\sqrt{2}}(1+i)$ so that the spaces are in $L^2$. In order to have a selfadjoint extension with a domain closed under conjugation, there is a one parameter choice of domain. The common ones are where you add $\Re e^{-\sqrt{i}t}$ to the domain of $-\Delta^c$ and the other where you add $\Im e^{-\sqrt{i}t}$. These are equivalent to imposing a single endpoint condition at $0$ on $\mathcal{D}(-\Delta^{\star})$: one where you set $f(0)=0$ and the other where you set t $f'(0)=0$. The general endpoint condition is $\cos\alpha f(0)+\sin\alpha f'(0)=0$.
A: Here is a direct approach to finding all selfadjoint extensions of $\Delta$ (sorry if it doesn't answer your questions).
Let $V=\mathcal S(\mathbb R^+)$ be the space of the smooth functions $f$ on $[0,\infty)$ that converge to 0 as $x\to\infty$, together with all its derivatives, faster than any power. For $f,g\in V$ we have 
$$\int_0^\infty(f''g-fg'')dx=f(0)g'(0)-f'(0)g(0)$$
 (as $f''g-fg''=(f'g-fg')'$). Hence for $f,g\in V$ we don't have in general $(f'',g)=(f,g'')$.
For $\lambda\in\mathbb R\cup\{\infty\}$ let $V_\lambda\subset V$ be the space of functions satisfying $f'(0)=\lambda f(0)$ (for $\lambda=\infty$ it's supposed to mean $f(0)=0$), and now we do have $(f'',g)=(f,g'')$ for $f,g\in V_\lambda$. This gives a hope that for every $\lambda$ there is a selfadjoint extension of $\Delta$ - and in fact, this is the classification of selfadjoint extensions.
To see it, let us choose $c>0$, $c\neq-\lambda$, and construct the inverse $T_{\lambda,c}$ of $(\Delta-c^2):V_\lambda\to V$ (showing at the same time that this operator is bijective). A calculation (provided I didn't make a mistake - but the exact result is inessential) gives that
$$(T_{\lambda,c}f)(x)=\int_0^\infty G(x,y)f(y)dy$$
$$G(x,y)=
\begin{cases}\frac{1}{W}u(x)v(y)&(x<y)\\
\frac{1}{W}v(x)u(y)&(x\geq y)
\end{cases}
$$
$$u(x)=(c+\lambda)e^{cx}+(c-\lambda)e^{-cx},\quad v(x)=e^{-cx}$$
$$W=-2c(c+\lambda)$$
(for $\lambda=\infty$ one should take $u(x)=e^{cx}-e^{-cx}$ and $W=-2c$).
Now, as you see, $T_{\lambda,c}:L^2(\mathbb R^+)\to L^2(\mathbb R^+)$ (given by the formula above) is self-adjoint, and so also its inverse is selfadjoint (the domain of the inverse contains $V_\lambda$, which is a dense subspace). Moreover $T_{\lambda,c}\neq T_{\lambda',c}$ if $\lambda\neq\lambda'$, so the extensions of $\Delta$ corresponding to different $\lambda$'s are different.
