What is spectrum for Laplacian in $\mathbb{R}^n$?
Solution 1:
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.
Solution 2:
Preliminaries: I will use the following sign convention for the Laplacian: $ \Delta u := - \sum_{j=1}^n \frac{\partial^2 u }{\partial x_j^2}.$ $\Delta$ is an unbounded operator on $L^2(\mathbb{R}^n)$. To define the domain of $\Delta$, recall that for $u \in C^\infty_0$, $$\mathcal{F}(\Delta u)(\xi) = 4\pi^2 |\xi|^2 \hat{u}(\xi).$$ (I use $\mathcal{F}(\phi)$ and $\hat{\phi}$ interchangeably to denote the Fourier transform of a function $\phi$.) Let's take the domain of $\Delta$ to be $$D(\Delta) := \{ u \in L^2 ~:~ 4\pi^2 |\xi|^2 \hat{u}(\xi) \in L^2 \}.$$ This makes $\Delta$ a closed unbounded operator.
Recall that the resolvent set of $\Delta$, denoted by $\rho(\Delta)$, is defined as the set of complex numbers $\lambda$ such that $\lambda I - \Delta$ is a bijection $D(\Delta) \to L^2$ (note: in this case, the closed graph theorem gives that the resolvent $(\lambda I - \Delta)^{-1}$ is necessarily bounded). The spectrum of $\Delta$, denoted by $\sigma(\Delta)$, is defined as the complement of $\rho(\Delta)$.
My goal is to sketch a proof of the following claim: \begin{equation} \sigma(\Delta) = [0, \infty) \end{equation}
First note that for $\lambda \in \mathbb{C} \backslash [0, \infty)$, $\lambda$ is in the resolvent set since we can easily invert $\lambda I - \Delta$ using the Fourier transform: $$(\lambda I - \Delta) u = f \iff (\lambda - 4\pi^2|\xi|^2)\hat{u} = \hat{f} \iff \hat{u} = (\lambda - 4\pi^2|\xi|^2)^{-1} \hat{f} .$$ Thus for $f \in L^2$, $(\lambda I - \Delta)^{-1} f = \mathcal{F}^{-1}\left((\lambda - 4\pi^2|\xi|^2)^{-1} \hat{f}\right)$. (Of course, this only works when $\lambda$ is not a nonnegative real number.) This proves that $\sigma(\Delta) \subset [0, \infty)$.
Now we'll show that $[0, \infty) \subset \sigma(\Delta)$. Let $\lambda \in [0, \infty)$. To prove that $\lambda \in \sigma(\Delta)$, it suffices to exhibit a sequence of functions $u_k$ in $D(\Delta)$ such that $$\frac{||u_k||_{L^2}}{||(\lambda I - \Delta)u_k||_{L^2}} \to \infty \text{ as } k \to \infty $$ as this shows that the resolvent $(\lambda I - \Delta)^{-1}$ cannot possibly be bounded. To that end, pick a point $x_0 \in \mathbb{R}^n$ such that $|x_0|^2 = \lambda$, and define u(x) by $$u(x) = e^{ix_0 \cdot x}.$$ (Note that $u$ is an eigenfunction for the Laplacian with eigenvalue $\lambda = |x_0|^2$, but $u$ is not in $L^2$.) Choose a sequence of cutoff functions $\phi_k \in C_0^\infty$ such that $0 \leq \phi_k \leq 1$, $\text{supp}(\phi_k) \subset B_{k+1}(0)$, and $\phi_k(x) \equiv 1$ for $x \in B_k(0)$. We can choose the $\phi_k$'s so that all their first two partial derivatives are uniformly bounded in $k$. Define the sequence $u_k \in C_0^\infty \subset L^2$ by $$u_k := \phi_k u.$$ Then $(\lambda I - \Delta)u_k$ is supported in the annulus $B_{k+1}(0) \backslash B_k(0)$, which has volume that is $O(k^{n-1})$ as $k \to \infty$. The set $\{ (\lambda I - \Delta)u_k \}$ is uniformly bounded in the sup norm, so there exists $C$ such that $$||(\lambda I - \Delta)u_k ||_{L^2}^2 \leq C k^{n-1}.$$ On the other hand, $|u_k (x)| \equiv 1$ for $x \in B_k(0)$, so we have the upper bound $$||u_k||_{L^2}^2 \geq \text{vol}(B_k(0)) = \omega_n k^n.$$ The last two inequalities together give the desired result.
Solution 3:
Hilbert-Schmidt operators are compact. The resolvent $(\Delta-\lambda I)^{-1}$ is not compact and $\Delta$ is not compact. Otherwise you would end up discrete spectrum, which you do not have.
The spectral resolution of the identity for $-\Delta$ is not as simple as $f=\int_{0}^{\infty}(f,\phi_{\lambda})\phi_{\lambda}d\lambda$ because there are so many approximate eigenfunctions with eigenvalue $r > 0$ that they are indexed by a vector $\vec{\xi}$ on a sphere of radius $\sqrt{r}$: $$ -\Delta(e^{i\vec{\xi}\cdot\vec{x}}) =|\vec{\xi}|^{2}e^{i\vec{\xi}\cdot \vec{x}} $$ Every $\phi_{\vec{\xi}}(\vec{x})=e^{i\vec{\xi}\cdot\vec{x}}$ for $|\vec{\xi}|=\sqrt{r}$ is a classical solution of $-\Delta\phi_{\vec{\xi}} = r\phi_{\vec{\xi}}$. The expansion that you want is obtained from the Fourier transform by writing the inversion integral as an outer integral over a radius, and an inner integral over the surface of a sphere of radius $r$: $$ f(\vec{x})=\frac{1}{(2\pi)^{n/2}}\int_{0}^{\infty}\left(\int_{|\vec{\xi}|=r}e^{i\vec{\xi}\cdot\vec{x}}\hat{f}(\vec{\xi})dS(\vec{\xi})\right)dr, $$ where $dS$ is the surface measure on the spherical shell. Classically speaking (not necessarily in $L^{2}(\mathbb{R}^{n})$): $$ -\Delta \left(\int_{|\vec{\xi}|=r}e^{i\vec{\xi}\cdot\vec{x}}\hat{f}(\vec{\xi})dS(\vec{\xi})\right) = r^{2}\left(\int_{|\vec{\xi}|=r}e^{i\vec{\xi}\cdot\vec{x}}\hat{f}(\vec{\xi})dS(\vec{\xi})\right) $$ The spectrum of $-\Delta$ is $[0,\infty)$ and it is all continuous spectrum. The spectral measure $E$ for $-\Delta$ is given on any closed interval $[a,b]\subset [0,\infty)$ by the expression $$ E[a,b]f = (\chi_{\sqrt{a} \le|\vec{\xi}|\le \sqrt{b}}\hat{f}(\vec{\xi}))^{\wedge}. $$ The spectral theorem for $-\Delta$ is $$ -\Delta f = \int_{0}^{\infty} \lambda dE(\lambda)f,\;\;\; f\in\mathcal{D}(-\Delta). $$
Solution 4:
This post
Reason for Continuous Spectrum of Laplacian
might be helpful, which tackled the case of $\mathbb{R}^{1}$.