Show self-adjointness with eigenvalue expansion. I was wondering if anybody here knows how to show that the negative Laplacian is self-adjoint on the 2 nd order Sobolev space of the two-sphere? I read that it is a rather cumbersome calculation, but I also don't see how one should start doing this calculation, so if you have a good reference for this or if you like to give a proof here, this would totally answer my question. 
 A: In your remarks, you asked about two specific issues.
Question 1: Why is the closure of $\mathcal{C}^{\infty}(S)$ under the graph norm of the Laplacian the same as $H^{2}(S)$, even though the first derivative terms are not present in this norm $\|f\|+\|\Delta f\|$?
The Laplacian in spherical coordinates is
$$
\frac{1}{r^{2}}\frac{\partial}{\partial r}r^{2}\frac{\partial}{\partial r}
   + \frac{1}{r^{2}\sin\theta}\frac{\partial}{\partial\theta}\sin\theta\frac{\partial}{\partial\theta}
   + \frac{1}{r^{2}\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}},
$$
and the restriction of this operator to functions at $r=1$ which do not depend on $r$ is the correct Laplacian $\Delta_{S}$ on the spherical manifold for the unit sphere $S$:
$$
         \Delta_{S} = \frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\sin\theta\frac{\partial}{\partial\theta}
   + \frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}.
$$
The Sobolev norm that I gave you on the sphere does not require the first derivatives because the manifold has no boundary. Assuming $f$ is in $\mathcal{C}^{\infty}(S)$,
and noting that $dS=\sin\theta\,d\theta\,d\phi$, integration by parts gives
$$
     (-\Delta_{S}f,f) = \|\nabla f\|^{2} = \int_{S}\left(\left|\frac{\partial f}{\partial\theta}\right|^{2}+\left|\frac{1}{\sin\theta}\frac{\partial f}{\partial\phi}\right|^{2}\right)dS
$$
This happens because of working on a manifold without boundary, where there are no boundary evaluation terms after integrating by parts. This requires an argument, but can easily be achieved by splitting along a closed curve such as a great circle.
Therefore, when you consider the closure of the graph of $-\Delta_{S}$ from the $\mathcal{C}^{\infty}(S)$ functions, you see that no explicit mention of first derivatives is needed because the following is continuous with respect to the graph norm of $-\Delta_{S}$:
$$   (f,f)+(-\Delta_{S}f,f)=\|f\|_{H^{1}}^{2}. $$
This is the expected case for the typical compact manifold without boundary.
Question 2: I'll modify this question a bit, and get back to yours below. Let $\{ e_{n} \}$ be an orthonormal basis of a Hilbert space $X$ and let $\{ \lambda_{n} \}$ be a sequence of real numbers. Then why is $Ax = \sum_{n}\lambda_{n}(x,e_{n})e_{n}$ selfadjoint on the domain $\mathcal{D}(A)=\{ x : \sum_{n}\lambda_{n}^{2}|(x,e_{n})|^{2} < \infty \}$?
It is easy to verify that $(Ax,y)=(x,Ay)$ for $x,y\in\mathcal{D}(A)$. Next it is shown that $A\pm iI$ are surjective. To do this, let $y\in X$ be given, and define
$$
              x_{\pm} = \sum_{n}\frac{1}{\lambda_{n}\pm i}(y,e_{n})e_{n}
$$
This is well-defined because the scalar coefficients are uniformly bounded by $1$. And $(x_{\pm},e_{n})=(y,e_{n})/(\lambda_{n}\pm i)$ for all $n$. Hence,
$$
\begin{align}
        \sum_{n}|\lambda_{n}(x_{\pm},e_{n})|^{2} & =\sum_{n} \left|\frac{\lambda_{n}}{\lambda_{n}\pm i}\right|^{2}|(y,e_{n})|^{2} \\
      & = \sum_{n}\frac{\lambda_{n}^{2}}{\lambda_{n}^{2}+1}|(y,e_{n})|^{2} \le \|y\|^{2}
\end{align}
$$
Therefore $x_{\pm} \in \mathcal{D}(A)$ and, and one can check that
$$
          (A\pm iI)x_{\pm} = y.
$$
This is enough to imply that the symmetric operator $A$ is selfadjoint. First, the domain is dense because it includes all finite linear combinations of $\{ e_{n}\}$. So the adjoint $A^{\star}$ is well-defined with $\mathcal{D}(A)\subseteq\mathcal{D}(A^{\star})$ because of the symmetry of $A$.
To show that $A$ is selfadjoint, we suppose $z \in \mathcal{D}(A^{\star})$ and show that $z \in \mathcal{D}(A)$. For such $z$,
$$
         ((A-iI)x,z) = (x,(A^{\star}+iI)z),\;\;\; x\in\mathcal{D}(A).
$$
Because $A+iI$ is surjective, there exists $x'\in\mathcal{D}(A)$ such that
$$
           (A^{\star}+iI)z = (A+iI)x'.
$$
Now the above gives
$$
         ((A-iI)x,z)=(x,(A+iI)x')=((A-iI)x,x').
$$
(The second equality follows from symmetry of $A$ on its domain.) Therefore,
$z-x'$ is orthogonal to $(A-iI)\mathcal{D}(A)=X$, which gives $z=x'\in\mathcal{D}(A)$. So $A^{\star}=A$.
Original Question 2: Suppose $A$ is symmetric on its domain with a complete orthonormal basis of eigenvectors $\{ e_{n}\}$ with eigenvalues $\{\lambda_{n}\}$. Why is $A$ selfadjoint on the domain as described in Question 1?
I should have said that the closure $A^{c}$ of $A$ is selfadjoint on this domain. If $A$ is already closed, then $A$ itself is selfadjoint. This is a technicality that does not change anything because a densely defined symmetric $A$ is always closable to a symmetric $A^{c}$, a fact which is easily checked from the symmetry relation. And this $A$ is densely defined because of the complete orthonormal basis of eigenfunctions.
To see that $x$ in the proposed domain is in the domain of $A^{c}$, first note that $x_{k}=\sum_{n=1}^{k}(x,e_{n})e_{n}$ is in the domain of $A$ and $Ax_{k}=\sum_{n=1}^{k}(x,e_{n})\lambda_{n}e_{n}$. By definition of this domain, $x_{k}$ converges to $x$ and $Ax_{k}$ converges to $y=\sum_{n}\lambda_{n}(x,e_{n})e_{n}$ which puts $\langle x,y\rangle$ in the graph of the closure $A^{c}$ with $A^{c}x=y$ as proposed. By what was shown before, this definition of $A$ is selfadjoint.
Let $A_{\mbox{new}}$ be the definition given using the orthonormal basis. In terms of graph inclusions,
$$
                      A_{\mbox{new}} \preceq A^{c} \preceq (A^{c})^{\star}.
$$
The first inclusion was just shown, and the second follows from symmetry of $A^{c}$. Therefore, taking adjoints, and recalling that $(A^{c})^{\star\star}=A^{c}$ because $A^{c}$ is closed, you get
$$
              A^{c} \preceq (A^{c})^{\star} \preceq A_{\mbox{new}}^{\star}=A_{\mbox{new}}
$$
Putting those two chains together forces $A^{c} = A_{\mbox{new}}$ to be selfadjoint and equal to the nice new operator $A_{\mbox{new}}$.
