I know very well that Laplacian in bounded domain has a discrete spectrum. How about Laplacian in $\mathbb{R}^n$?(not in some fancy-shaped unbounded domain, but the whole domain)
Where can I find such results?
Moreover, is there a counterpart of Hilbert-Schmidt theorem for Laplacian in $\mathbb{R}^n$? Hilbert-Schmidt asserts there is a countable set of eigenfunctions $\phi_n$ so that $x=\sum \langle x,\phi_n\rangle \phi_n,\forall x\in H$.
Is there a similar theorem saying $x=\int_0^\infty \langle x,\phi_\lambda\rangle \phi_\lambda\,\mathrm{d}\lambda,\forall x\in H$ where $\phi_\lambda$ is the eigenfunction of Laplacian to spectral value $\lambda$?
Best Answer
Here is another approach. My guiding principle (learned from Reed and Simon's book) is that to understand the spectral theory of self-adjoint operators, you must first understand multiplication operators. So consider the following outline:
Let $(X,\mu)$ be a $\sigma$-finite measure space. (You can take $\mathbb{R}^n$ with Lebesgue measure if you like, but the following arguments look just the same in general.) Let $h : X \to \mathbb{C}$ be measurable, and consider the unbounded multiplication operator $M_h$ on $L^2(X,\mu)$ defined by $M_h f = f h$, whose domain is $D(M_h) := \{f \in L^2(X, \mu) : fh \in L^2(X,\mu)\}$. Show that $M_h$ is densely defined and closed.
Show that $M_h$ is bounded (and everywhere defined) iff $h \in L^\infty(X,\mu)$. (In this case, the operator norm of $M_h$ is $\|h\|_{L^\infty}$.)
Show that if $h$ is a.e. nonzero, then $M_h^{-1} = M_{1/h}$.
Using the previous two facts, show that the spectrum of $M_h$ is the essential range of $h$.
Show that the eigenvalues of $M_h$ (its pure point spectrum) are $\{ \lambda : \mu(h = \lambda) > 0\}$, and that the rest of $\sigma(M_h)$ is continuous spectrum.
There are many other properties of $M_h$ you could prove, but this will do for now.
Suppose $H, K$ are Hilbert spaces, $U : H \to K$ is unitary (i.e. a surjective linear isometry), and $A$ is an unbounded operator on $H$. Then $UAU^{-1}$, with domain $\{ x \in K : U^{-1} x \in D(A)\}$, is an unbounded operator on $K$. Show that $UAU^{-1}$ is respectively closed , densely defined, etc, iff $A$ is.
Show that $\sigma(UAU^{-1}) = \sigma(A)$.
That's enough abstraction for now.
Recall the Plancherel theorem that the Fourier transform $\mathcal{F} : L^2(\mathbb{R}^n,m) \to L^2(\mathbb{R}^n,m)$ is unitary (if appropriately normalized).
Let $\Delta$ be the Laplacian operator $\Delta = -\sum_{i=1}^n \frac{\partial^2}{\partial x_i^2}$, and define $h : \mathbb{R}^n \to \mathbb{R}$ by $h(x) = |x|^2$. If we take the domain of $\Delta$ to be all $L^2$ functions with two weak derivatives in $L^2$ (which gives us a closed densely defined operator) show that $\mathcal{F}^{-1} \Delta \mathcal{F} = M_h$. (Or if you prefer, define the domain of $\Delta$ to be $\mathcal{F}(D(M_h))$. Or first define $\Delta$ on $C^\infty_c(\mathbb{R}^n)$ and then take its closure. Either way you get the same operator.)
Since the essential range of $h$ is clearly $[0,\infty)$, that is the spectrum of $\Delta$. Moreover, since for each $\lambda$ we have $m(h = \lambda) = 0$, it is all continuous spectrum.