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 $\lambda$?

  • $\begingroup$ The spectrum is all nonnegative real numbers: $[0,\infty)$. I don't have a reference handy for this, though. $\endgroup$ – Phillip Andreae Apr 23 '14 at 21:33
  • $\begingroup$ By the way, my comment above is for the positive semidefinite Laplacian $-\sum_j \frac{\partial^2}{\partial x_j^2}$ (which is the negative of some people's usual convention). I found some notes with a proof that the spectrum is $[0, \infty)$, and I will try to post an answer explaining it later. $\endgroup$ – Phillip Andreae Apr 24 '14 at 13:01
  • $\begingroup$ Thanks for your attention. By the way, is $[0,\infty)$ all continuous spectrum? $\endgroup$ – user33869 Apr 24 '14 at 14:00
  • $\begingroup$ Sorry for the delay. I have posted an answer below. I didn't address your last two questions in my answer, but I think the answer to both is no. My guess is the entire spectrum is continuous spectrum, but I'm not sure on that. $\endgroup$ – Phillip Andreae May 2 '14 at 23:14

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.

  • $\begingroup$ Nice! This is a more abstract and general than my answer above. It reminded me that $M_h$, where $h = |x|^2$ (in Fourier transform land), is precisely the multiplication operator guaranteed by the spectral theorem. $\endgroup$ – Phillip Andreae May 3 '14 at 16:48
  • $\begingroup$ @user49048: Actually, yes and no. Some statements of the spectral theorem guarantee that a self-adjoint operator is unitarily equivalent to a multiplication operator on a finite measure space $(X,\mu)$. In that case, if you start with the self-adjoint operator $\Delta$, the measure space $(X,\mu)$ you get is much harder to visualize; the construction is roughly analogous to that of the Stone-Cech compactification. $\endgroup$ – Nate Eldredge May 3 '14 at 23:03

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}

  1. 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 $\sigma(\Delta) \subset [0, \infty)$.

  2. 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.


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 $$ -\Delta f = \int_{0}^{\infty} \lambda dE(\lambda)f,\;\;\; f\in\mathcal{D}(-\Delta). $$

  • $\begingroup$ Hi @TrialAndError. I didn't find a way to contact you otherwise than by letting you a comment (I used an old answer of you to not bother other users). I just wanted to ask you some references in Spectral Theory: I saw many of your answers about this topic, and I remember one of your comments saying that (and I agree) you were surprised that there was not really a reference for ST and functional calculus. I've learned them from many references (Reed&Simon, Rudin, Yosida, Kato, Davies ) but none of them is really exhaustive (especially for the functional calculus). Any suggestion ? Thanks! $\endgroup$ – Nicolas May 10 '16 at 16:21
  • $\begingroup$ @Nicolas : I can't imagine much that wouldn't be covered in one or more of the references you listed. What aspect of the Spectral Theorem are you hoping to better understand? $\endgroup$ – Disintegrating By Parts May 10 '16 at 17:07
  • $\begingroup$ I was afraid you say this... It is not really a question of learning something new but rather gathering some important results (mostly coming from the functional calculus). For example, I don't remember having read results such those you've mentionned in the above answer of you (spectrum and spectral measure for $-\Delta$) in standard references (I've found them in some exercises of different books or in my old lectures). It is obvious that there does not exist any general formula giving you the spectral measures of a normal operator (I do not consider Stone's formula), but... $\endgroup$ – Nicolas May 10 '16 at 18:08
  • $\begingroup$ ... but I expected finding some of important examples (such as the laplacian operator defined on some standard domains, like $\mathbb{R}^n$ and the unit sphere) dealt in at least one book. I am really surprised of this fact. But you are not the first person that gives me this answer, so I believe that such a book does not exist. Thanks anyway for your attention. $\endgroup$ – Nicolas May 10 '16 at 18:13
  • $\begingroup$ @Nicolas : I was also surprised when I started looking--almost no Functional Analysis text deals with the specifics of actual ODEs and PDEs, not even the classical ODEs and PDEs. Or, if they do, they just state a few results and wave their hands. I posted here how to get the Fourier transform from Stone's formula because I had not seen even that much in a text, which is really surprising. The complete $L^2$ theory (including Parseval's identity) for the Fourier transform can be derived from first principles using Stone's formula. That's why I posted it. Let me know if you find anything. $\endgroup$ – Disintegrating By Parts May 10 '16 at 18:36

This post

Reason for Continuous Spectrum of Laplacian

might be helpful, which tackled the case of $\mathbb{R}^{1}$.


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