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Dear community,

I would be happy about any literature or comments on the behaviour of the pointwise product of eigenfunctions of a self-adjoint operator with discrete spectrum, acting on a separable Hilbert space which is closed under pointwise multiplication. The operator I'm actually looking at is a symmetric Markov operator acting on $L^2(\mathcal{A},\mu)$, where $\mathcal{A}$ is some function algebra and $\mu$ the invariant measure.

Some questions I'm especially interested in are:

  1. If you multiply two eigenfunctions, can it happen that the product has an infinite eigenfunction expansion? By "infinite eigenfunction expansion" I mean that it can not be expressed as a finite sum of eigenfunctions.

  2. Somewhat related: If the squares of two eigenfunctions have a finite expansion, respectively, can it happen that the square of the sum of these two eigenfunctions has an infinite expansion?

  3. In the above Markov setting: Is the following "projected Cauchy-Schwarz inequality" always true? $$ \int \operatorname{proj}(fg \mid E)^2 \ \text{d} \mu \leq \sqrt{\int \operatorname{proj}(f^2 \mid E)^2 \ \text{d} \mu} \sqrt{\int \operatorname{proj}(g^2 \mid E)^2 \ \text{d} \mu} $$ Here, $f$ and $g$ are eigenfunctions lying in some common eigenspace, $E$ is another eigenspace and $\operatorname{proj}(f \mid E)$ denotes the projection of $f$ on $E$.

Note that the answer to questions 1 and 2 is no, if the eigenfunctions are orthogonal polynomials.

Thanks for your help,

Simon

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Some care is necessary, Laplacians on many fractals have domains that aren't algebras so even having the product in the domain again is not something to leave unappreciated. –  BSteinhurst Aug 24 '11 at 17:41
    
You are right, thanks for commenting on that. Obviously, I was too quick with my generalizing. I added the condition that the Hilbert space should be closed under multiplication. –  herrsimon Aug 24 '11 at 18:26
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Your question seem to depend strongly on the operator in question. For instance, for "composition operators" (i.e. operators of the form $U_T (f) = f \circ T$, $f \in L^2 (X,\mu)$, $T:X\to X$ "nice"), the pointwise product of two eigenfunctions is also an eigenfunction( as long as it itself lies in $L^2 (X)$) and there are plenty of such operators with discrete spectrum. –  Mark Aug 25 '11 at 17:25
    
Is the question related to expansions of products of representations in terms of irreducible ones? –  Gil Kalai Aug 26 '11 at 10:49
    
@Gil: Maybe it is, but this is not what I had in mind when posing the question –  herrsimon Aug 26 '11 at 11:39
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2 Answers

up vote 7 down vote accepted

It can easily happen that the product of two eigenfunctions has an infinite eigenfunction expansion. Probably this is more typical (in natural problems) than not. For example, the Laplace-Beltrami operator on compact Riemannian manifolds (or suitable values $(\Delta+c)^{-1}$ of its resolvent if one must have a bounded operator) has discrete spectrum, but products of eigenfunctions are rarely eigenfunctions.

For example, on the sphere, (restrictions of) homogeneous harmonic polynomials are the eigenfunctions for the natural Laplacian. Products or squares of harmonic polynomials are rarely harmonic, but do have finite expressions as sums of eigenfunctions. (From my viewpoint, this finiteness is predictable because the irreducibles of the rotation/orthogonal group are all finite-dimensional, because it is compact.)

Either from the a viewpoint of geometric analysis, or from a viewpoint of repn theory, in generality such re-expressions of products would rarely be finite, but should exist under mild hypotheses on the situation. Explicit examples where the decomposition/Fourier coefficients $\langle fg,h\rangle$ are explicitly expressible for triples of eigenfunctions are not so easy to manufacture, and the ones I know (having to do with automorphic forms, zonal spherical harmonics, etc.) I suspect are not in the direction of your interest.

It is also true that there seems to be fairly skimpy literature on such issues.

About your question 2: I don't have an easy example. About question 3: no, there are easy examples of the failure: on $[0,2\pi]$, with $f(x)=e^{ix}$ and $g(x)=e^{-ix}$ (or cosines and sines), and $E$ the space spanned by constants, $fg$ projects to $1$, while both products $f^2$ and $g^2$ project to $0$.

Edit: in response to the questioner's further comment/questions... First, the latter counter-example, with exponential functions, refers to $d^2/dx^2$.

Second, as @Matt Y. noted, someone like me is thinking of automorphic things, which may not be what everyone wants to see as examples... But, further, I think it is not so easy to make any such examples explicit. Even without the conjectured (and probable) typical-non-vanishing and so on, we can find definitive examples of infinite, discrete decompositions in the pseudo-Laplacians Y. Colin de Verdiere used c. 1981 to give another proof of meromorphic continuation of Eisenstein series: these are self-adjoint operators with compact resolvents (so discrete spectrum) on Hilbert spaces of automorphic functions which have, among their eigenfunctions certain truncated Eisenstein series. The integral of three truncated Eisenstein series is not completely trivial to compute, but turns out (e.g., Zagier an others, circa 1978) to be a product/quotient of values of zeta at points that can be controlled directly... so can be made not to vanish. Thus, provably, an infinite-not-finite re-expression of product of two eigenfunctions (for a pseudo-Laplacian...) as sum of eigenfunctions. The "pseudo-Laplacian" is not wildly different from a/the genuine Laplacian...

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Thanks a lot for your answer but unfortunately I can't really follow all your arguments. Indeed, the prod. of two eigenfcts is typically not an eigenfct again, but, as in youLaplacian example, it can be expressed as a finite sum of eigenfcts. In fact, this (trivially) always happens if the eigenfcts are orthonormal polynomials. I'm looking for two eigenfcts whose product is an infinite sum of eigenfcts. Regarding your answer to question 3: For which operator lie $e^{ix}$ and $e^{-ix}$ in the same eigenspace? I think that I didn't formulate my question clear enough and will edit it. –  herrsimon Aug 24 '11 at 16:48
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I think in this answer Paul is alluding to the example of the family of Maass forms invariant on $SL_2(\mathbb{Z}) \backslash \mathbb{H}$. These are eigenfunctions of the Laplacian. There are explicit formulas for $|\langle f^2, g \rangle|^2$ where $f$ and $g$ are such Maass forms, in terms of the central value of certain $L$-functions which are presumably typically nonzero excepting certain trivial vanishings coming from symmetry. This would imply that $f^2$ has to be decomposed as an infinite sum and not a finite sum. –  Matt Young Aug 24 '11 at 17:20
    
Now I get it, thanks for clearing this up for me! –  herrsimon Aug 26 '11 at 9:35
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For question 1, one example of interest comes from the energy eigenfunctions of the one dimensional quantum harmonic oscillator. The Hilbert space is separable, and the Hamiltonian satisfies your conditions. Under suitable normalization, the eigenfunctions have the form $p(x)e^{-\pi x^2}$ for $p$ a Hermite polynomial, and the product of any two then has the form $q(x)e^{-2\pi x^2}$ for $q$ a nonzero polynomial. These products are not given by finite linear combinations of eigenfunctions.

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You're quite right. I was being sloppy. –  S. Carnahan Aug 26 '11 at 7:15
    
Still it's a nice example, thanks! –  herrsimon Aug 26 '11 at 9:36
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