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Perhaps the most fundamental differential operator on Euclidean space is the Laplacian
The Laplacian is a linear translation-invariant operator, and as such is necessarily diagonalised by the Fourier transform
Indeed, we have
for any suitably nice function (e.g. in the Schwartz class; alternatively, one can work in very rough classes, such as the space of tempered distributions, provided of course that one is willing to interpret all operators in a distributional or weak sense).
Because of this explicit diagonalisation, it is a straightforward manner to define spectral multipliers of the Laplacian for any (measurable, polynomial growth) function , by the formula
(The presence of the minus sign in front of the Laplacian has some minor technical advantages, as it makes positive semi-definite. One can also define spectral multipliers more abstractly from general functional calculus, after establishing that the Laplacian is essentially self-adjoint.) Many of these multipliers are of importance in PDE and analysis, such as the fractional derivative operators , the heat propagators , the (free) Schrödinger propagators , the wave propagators (or and , depending on one’s conventions), the spectral projections , the Bochner-Riesz summation operators , or the resolvents .
Each of these families of multipliers are related to the others, by means of various integral transforms (and also, in some cases, by analytic continuation). For instance:
- Using the Laplace transform, one can express (sufficiently smooth) multipliers in terms of heat operators. For instance, using the identity
(using analytic continuation if necessary to make the right-hand side well-defined), with being the Gamma function, we can write the fractional derivative operators in terms of heat kernels:
- Using analytic continuation, one can connect heat operators to Schrödinger operators , a process also known as Wick rotation. Analytic continuation is a notoriously unstable process, and so it is difficult to use analytic continuation to obtain any quantitative estimates on (say) Schrödinger operators from their heat counterparts; however, this procedure can be useful for propagating identities from one family to another. For instance, one can derive the fundamental solution for the Schrödinger equation from the fundamental solution for the heat equation by this method.
- Using the Fourier inversion formula, one can write general multipliers as integral combinations of Schrödinger or wave propagators; for instance, if lies in the upper half plane , one has
In a similar vein, if , then
- Using the Cauchy integral formula, one can express (sufficiently holomorphic) multipliers in terms of resolvents (or limits of resolvents). For instance, if , then from the Cauchy integral formula (and Jordan’s lemma) one has
- The imaginary part of is the Poisson kernel , which is an approximation to the identity. As a consequence, for any reasonable function , one has (formally, at least)
Among other things, this type of formula (with replaced by a more general self-adjoint operator) is used in the resolvent-based approach to the spectral theorem (by using the limiting imaginary part of resolvents to build spectral measure). Note that one can also express as .
Remark 1 The ability of heat operators, Schrödinger propagators, wave propagators, or resolvents to generate other spectral multipliers can be viewed as a sort of manifestation of the Stone-Weierstrass theorem (though with the caveat that the spectrum of the Laplacian is non-compact and so the Stone-Weierstrass theorem does not directly apply). Indeed, observe the *-algebra type properties
Because of these relationships, it is possible (in principle, at least), to leverage one’s understanding one family of spectral multipliers to gain control on another family of multipliers. For instance, the fact that the heat operators have non-negative kernel (a fact which can be seen from the maximum principle, or from the Brownian motion interpretation of the heat kernels) implies (by (1)) that the fractional integral operators for also have non-negative kernel. Or, the fact that the wave equation enjoys finite speed of propagation (and hence that the wave propagators have distributional convolution kernel localised to the ball of radius centred at the origin), can be used (by (3)) to show that the resolvents have a convolution kernel that is essentially localised to the ball of radius around the origin.
In this post, I would like to continue this theme by using the resolvents to control other spectral multipliers. These resolvents are well-defined whenever lies outside of the spectrum of the operator . In the model three-dimensional case , they can be defined explicitly by the formula
whenever lives in the upper half-plane , ensuring the absolute convergence of the integral for test functions . (In general dimension, explicit formulas are still available, but involve Bessel functions. But asymptotically at least, and ignoring higher order terms, one simply replaces by for some explicit constant .) It is an instructive exercise to verify that this resolvent indeed inverts the operator , either by using Fourier analysis or by Green’s theorem.
Henceforth we restrict attention to three dimensions for simplicity. One consequence of the above explicit formula is that for positive real , the resolvents and tend to different limits as , reflecting the jump discontinuity in the resolvent function at the spectrum; as one can guess from formulae such as (4) or (5), such limits are of interest for understanding many other spectral multipliers. Indeed, for any test function , we see that
Both of these functions
solve the Helmholtz equation
as , leading also to the Sommerfeld radiation condition
where is the outgoing radial derivative. Indeed, one can show using an integration by parts argument that is the unique solution of the Helmholtz equation (6) obeying (8) (see below). is known as the outward radiating solution of the Helmholtz equation (6), and is known as the inward radiating solution. Indeed, if one views the function as a solution to the inhomogeneous Schrödinger equation
and using the de Broglie law that a solution to such an equation with wave number (i.e. resembling for some amplitide ) should propagate at (group) velocity , we see (heuristically, at least) that the outward radiating solution will indeed propagate radially away from the origin at speed , while inward radiating solution propagates inward at the same speed.
known as the limiting absorption principle:
for all , where depends only on , and is the weighted norm
This principle allows one to extend the convergence (9) from test functions to all functions in the weighted space by a density argument (though the radiation condition (8) has to be adapted suitably for this scale of spaces when doing so). The weighted space on the left-hand side is optimal, as can be seen from the asymptotic (7); a duality argument similarly shows that the weighted space on the right-hand side is also optimal.
We prove this theorem below the fold. As observed long ago by Kato (and also reproduced below), this estimate is equivalent (via a Fourier transform in the spectral variable ) to a useful estimate for the free Schrödinger equation known as the local smoothing estimate, which in particular implies the well-known RAGE theorem for that equation; it also has similar consequences for the free wave equation. As we shall see, it also encodes some spectral information about the Laplacian; for instance, it can be used to show that the Laplacian has no eigenvalues, resonances, or singular continuous spectrum. These spectral facts are already obvious from the Fourier transform representation of the Laplacian, but the point is that the limiting absorption principle also applies to more general operators for which the explicit diagonalisation afforded by the Fourier transform is not available. (Igor Rodnianski and I are working on a paper regarding this topic, of which I hope to say more about soon.)
In order to illustrate the main ideas and suppress technical details, I will be a little loose with some of the rigorous details of the arguments, and in particular will be manipulating limits and integrals at a somewhat formal level.
A few days ago, I found myself needing to use the Fredholm alternative in functional analysis:
Theorem 1 (Fredholm alternative) Let be a Banach space, let be a compact operator, and let be non-zero. Then exactly one of the following statements hold:
- (Eigenvalue) There is a non-trivial solution to the equation .
- (Bounded resolvent) The operator has a bounded inverse on .
Among other things, the Fredholm alternative can be used to establish the spectral theorem for compact operators. A hypothesis such as compactness is necessary; the shift operator on , for instance, has no eigenfunctions, but is not invertible for any unit complex number . The claim is also false when ; consider for instance the multiplication operator on , which is compact and has no eigenvalue at zero, but is not invertible.
It had been a while since I had studied the spectral theory of compact operators, and I found that I could not immediately reconstruct a proof of the Fredholm alternative from first principles. So I set myself the exercise of doing so. I thought that I had managed to establish the alternative in all cases, but as pointed out in comments, my argument is restricted to the case where the compact operator is approximable, which means that it is the limit of finite rank operators in the uniform topology. Many Banach spaces (and in particular, all Hilbert spaces) have the approximation property that implies (by a result of Grothendieck) that all compact operators on that space are almost finite rank. For instance, if is a Hilbert space, then any compact operator is approximable, because any compact set can be approximated by a finite-dimensional subspace, and in a Hilbert space, the orthogonal projection operator to a subspace is always a contraction. (In more general Banach spaces, finite-dimensional subspaces are still complemented, but the operator norm of the projection can be large.) Unfortunately, there are examples of Banach spaces for which the approximation property fails; the first such examples were discovered by Enflo, and a subsequent paper of by Alexander demonstrated the existence of compact operators in certain Banach spaces that are not approximable.
I also found out that this argument was essentially also discovered independently by by MacCluer-Hull and by Uuye. Nevertheless, I am recording this argument here, together with two more traditional proofs of the Fredholm alternative (based on the Riesz lemma and a continuity argument respectively).
[This is a (lightly edited) repost of an old blog post of mine, which had attracted over 400 comments, and as such was becoming difficult to load; I request that people wishing to comment on that puzzle use this fresh post instead. -T]
This is one of my favorite logic puzzles, because of the presence of two highly plausible, but contradictory, solutions to the puzzle. Resolving this apparent contradiction requires very clear thinking about the nature of knowledge; but I won’t spoil the resolution here, and will simply describe the logic puzzle and its two putative solutions. (Readers, though, are welcome to discuss solutions in the comments.)
— The logic puzzle —
There is an island upon which a tribe resides. The tribe consists of 1000 people, with various eye colours. Yet, their religion forbids them to know their own eye color, or even to discuss the topic; thus, each resident can (and does) see the eye colors of all other residents, but has no way of discovering his or her own (there are no reflective surfaces). If a tribesperson does discover his or her own eye color, then their religion compels them to commit ritual suicide at noon the following day in the village square for all to witness. All the tribespeople are highly logical and devout, and they all know that each other is also highly logical and devout (and they all know that they all know that each other is highly logical and devout, and so forth).
Of the 1000 islanders, it turns out that 100 of them have blue eyes and 900 of them have brown eyes, although the islanders are not initially aware of these statistics (each of them can of course only see 999 of the 1000 tribespeople).
One day, a blue-eyed foreigner visits to the island and wins the complete trust of the tribe.
One evening, he addresses the entire tribe to thank them for their hospitality.
However, not knowing the customs, the foreigner makes the mistake of mentioning eye color in his address, remarking “how unusual it is to see another blue-eyed person like myself in this region of the world”.
What effect, if anything, does this faux pas have on the tribe?
Note 1: For the purposes of this logic puzzle, “highly logical” means that any conclusion that can logically deduced from the information and observations available to an islander, will automatically be known to that islander.
Note 2: Bear in mind that this is a logic puzzle, rather than a description of a real-world scenario. The puzzle is not to determine whether the scenario is plausible (indeed, it is extremely implausible) or whether one can find a legalistic loophole in the wording of the scenario that allows for some sort of degenerate solution; instead, the puzzle is to determine (holding to the spirit of the puzzle, and not just to the letter) which of the solutions given below (if any) are correct, and if one solution is valid, to correctly explain why the other solution is invalid. (One could also resolve the logic puzzle by showing that the assumptions of the puzzle are logically inconsistent or not well-defined. However, merely demonstrating that the assumptions of the puzzle are highly unlikely, as opposed to logically impossible to satisfy, is not sufficient to resolve the puzzle.)
Note 3: An essentially equivalent version of the logic puzzle is also given at the xkcd web site. Many other versions of this puzzle can be found in many places; I myself heard of the puzzle as a child, though I don’t recall the precise source.
Below the fold are the two putative solutions to the logic puzzle. If you have not seen the puzzle before, I recommend you try to solve it first before reading either solution.