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This should be a trivial question for mathematicians but not for typical physicists.

I know that the spectrum of a linear operator on a Banach space splits into the so-called "point," "continuous" and "residual" parts [I gather that no boundedness assumption is needed but I could be wrong]. I further know that the point spectrum coincides with the set of eigenvalues of the operator. It seems from the terminology that the point spectrum is a discrete set of isolated point and that the eigenvalues cannot form a continuum. But I haven't been able to find a clear statement in a math reference about this.

Actually, I'm mostly interested in self-adjoint operators on a Hilbert space; so a simpler version of my question would be: Can a self-adjoint operator have a continuous set of eigenvalues? And if yes, under what conditions do the eigenvalues have to be discrete?

I appreciate any help.

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The answer is yes for certain (easy) self-adjoint operators on a suitable non-seperable Hilbert space. I assume you want to restrict your attention to separable ones? –  Yemon Choi Sep 25 '10 at 1:50
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While this doesn't answer your question about self-adjoint operators, the unilateral backwards shift $S:\ell^2(\mathbb N)\to\ell^2(\mathbb N)$, given by $$ S(x_1,x_2,\dots) = (x_2,x_3,\dots) $$ has lots of eigenvalues -- namely the whole open unit disc. –  Yemon Choi Sep 25 '10 at 1:54
    
Thanks for both of your good comments. Yes I do assume separable Hilbert spaces. The example is also pretty illuminating. –  Mahdiyar Sep 26 '10 at 6:16

8 Answers 8

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Eigenvectors for different eigenvalues of a self-adjoint operator are orthogonal. In a separable Hilbert space, any orthogonal set is countable. So a self-adjoint operator on separable Hilbert space has only countably many eigenvalues. (As noted, this does not mean the spectrum is countable.)

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Such a simple answer! Thanks. This is exactly what I expected for the "simpler version" of my question. Actually Piero D'Ancona also provided the same answer to my question but this one was more to the point and simpler to understand (for someone like me!). –  Mahdiyar Sep 26 '10 at 7:03

You are confusing two notions. First of all, the point spectrum just means eigenvalues; there is no assumption that these form a discrete set. The shift operator is a simple example where the spectrum is "continuous".

The condition for the eigenvalues to be discrete is precicsely that the operator $A:H \to H$ is compact. It is however possible for non-compact self adjoint operators to have a discrete spectrum. The simplest example of this is the orthogonal projection operator $P:H \to Y$ where $Y$ is a closed subspace of the Hilbert space $H$. Here the spectrum is $0$ and $1$.

Bounded self adjoint operators have no residual spectrum but they do indeed have a continuous spectrum. Take any compact operator $A:H \to H$ where dim$H=+\infty$. Then $0$ belongs to the continuous spectrum because otherwise $A:H \to H$ would be invertible, implying that dim$H <\infty$. Continuous = "exists a set of approximate eigenvectors".

If you want a continuous range of spectrum take $Af = x f(x)$ on $L^2([0,1])$. Then the range of the spectrum is just $[0,1]$. There are no eigenvalues for this operator and moreoever since the residual spectrum is empty for self adjoint operator, $[0,1]$ is the spectrum and it is equivalent to the continuous spectrum.

So a final point, continuous just means that $R(A-\lambda I)$ is not dense but that $\lambda$ is not an eigenvalue. It has nothing to do with the actual spectrum being discrete or continuous.

So I think you were mixing up two notions but hopefully I've provided examples for both.

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"take Af=xf(x) on L2([0,1]). Then the range of the spectrum is just [0,1]. These are all eigenvalues." -- Are you sure? –  Yemon Choi Sep 25 '10 at 2:58
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Also, the eigenvalues of a compact operator do not form a discrete set, since they accumulate at zero. –  Yemon Choi Sep 25 '10 at 2:59
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If you take multiplication by the function $f(n) = n$ on the integers with counting measure, then the spectrum coincides with the point spectrum and is the integers, so discrete, but of course the operator is not bounded and so not compact. –  Dick Palais Sep 25 '10 at 3:50
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I thought "discrete" meant "discrete as a subset of the complex plane" but perhaps I've misremembered the terminology –  Yemon Choi Sep 25 '10 at 4:19
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I wouldn't call a set with an accumulation point, "discrete". How would it be different from a set with many accumulation points? –  Martin Argerami Sep 25 '10 at 10:09

One version of the spectral theorem states that if $A$ is a selfadjoint operator on a separable Hilbert space $H$, then we can find a set $X$, a $\sigma$-finite measure on $X$, a unitary operator $U:H\to L^2(X)$, and a measurable function $a:X\to \mathbb{R}$, such that $A=U^*aU$. In other words, any selfadjoint operator, in essence, is nothing but 'multiplication by a real valued function', in suitable 'coordinates'. The spectrum of $A$ coincides with the essential range of the function $a$.

Now, an eigenvalue must be a real number $\lambda$ such that the set $a^{-1}(\lambda)$ has a positive measure. Since $X$ is $\sigma$-finite, the eigenvalues can be at most a countable set. But you can have a continuous spectrum of course, only the points will not be eigenvalues. You can play with real valued functions (and with measures on $X$; any $\sigma$-finite measure is ok) to train intuition.

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Recall that the spectrum of an operator $A$ on a Hilbert space is the set of vales $\lambda$ such that $A-\lambda I$ does not have a bounded inverse . So if $A$ is multiplication by a function in $L^2$ of a measure space, then any point of the essential range of the function is in the spectrum. So, for example (and this is the classic example) the spectrum of multiplication by $x$ on the real line is the whole line. However, a point of the spectrum is not necessarily an eigenvalue. In fact $\lambda$ is an eigenvalue (or in the point spectrum) iff $A- \lambda I$ has a non-trivial null space. (And as others have pointed out, on a separable Hilbert space, the point spectrum is countable.)

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I add this remark because it may be part of what the OP wants. Note that, as to the spectrum of a bounded linear self-adjoint operator on $\ell^2$, of course, it can be any compact set $K$ of $\mathbb{R}$. Just take a diagonal operator where the set of the diagonal elements (eigenvalues) is dense in $K$.

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Just so we have the silly example here. Consider $\ell^2([0,1])$ meaning the set of SEQUENCES $u_{x}$ indexed by a number $x\in [0,1]$. So the scalar product is $$ \langle u, v \rangle = \sum_{x \in [0,1]} \overline{u_x} v_x. $$ In order for $u \in \ell^2([0,1])$, we have that $u_x \neq 0$ for at most countably many $x$. So this is very different from $L^2([0,1])$.

Now consider the operator $$ (Au)_x = u_x. $$ This operator is diagonal and it's eigenvalues are $[0,1]$. The eigenvector corresponding to $x \in [0,1]$ is $$ u_y = \begin{cases} 1, & x=y\\\ 0, & \text{otherwise}.\end{cases} $$ Of course the key to this example and the question is that $\ell^2([0,1])$ is a NON-separable Hilbert space.

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The hyperbolic Laplacian is self adjoint and its spectrum has a continuous component.

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What is a hyperbolic laplacian? Isn't the laplacian elliptic? –  Dorian Sep 25 '10 at 3:45
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The spectrum of the standard Laplace operator on $R^n$ is $\sigma(-\Delta)=[0,+\infty)$. –  Piero D'Ancona Sep 25 '10 at 11:31
    
I would guess that hyperbolic Laplacian means Laplacian on hyperbolic space. –  Emerton Sep 26 '10 at 1:19

The resolvent set is the set of all $\zeta \in \mathbb{C}$ for which $T-\zeta$ is invertible (which means especially that the Range is all of $H$). The spectrum $\Sigma$ is the complement of the resolvent.

Normally, a $\lambda \in \Sigma$ is only called eigenvalue, if a $x \in H$ exists such that $Tx = \lambda x$. The set of those eigenvalues is countable as already argued but not necessarily discrete. For example: A compact operator on a Hilbert space has a countable spectrum, but not necessarily a discrete one; the spectrum has an accumulation point at zero, which may be an eigenvalue.

A noncompact bounded operator may have a countable spectrum consisting only of "true" eigenvalues, but then the eigenvalues -- counted with multiplicity -- have a nonzero accumulation point.

An unbounded operator whose resolvent is compact (for some element of the resolvent set) has a discrete spectrum; this criterion works on Laplacians on compact manifolds, for example. The Laplacian on $L^2(\mathbb{R}^n)$ with usual Lebesgue measure -- which is self-adjoint if $\mathrm{dom}(\nabla) = H^2$ -- however has as spectrum the whole positive real line, but no eigenvalue.

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