Show self-adjointness with eigenvalue expansion.

Question

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.

Answer 1

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 $A_{\mbox{new}}$.