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Does there exist an orthonormal basis of square-integrable functions (either $L^2(\mathbb{R})$ or $L^2(\mathbb{C})$) such that the sequence of functions has bounded variance, and also the sequence consisting of the Fourier transform of each function also has bounded variance?

Some background:

This question came up in a comment on SciRate regarding a recently translated paper by von Neumann. There the commenter, Matt Hastings, points out some related results.

In particular, the Balian-Low theorem states that this can't exist for any Gabor basis, i.e. one which is composed of time and frequency translates of a given fiducial $L^2$ function. If there were a generalization of this theorem to arbitrary bases, it would prove that such a sequence can't exist.

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Is von Neumann's footnote 10 on page 4 relevant? –  Jonas Meyer Mar 12 '10 at 4:36
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After reading Matt Hastings's comment more closely, there is a high likelihood that I have misunderstood your question. Do you mean to say that there is a uniform bound on the variances? I.e., do you mean that the sequence of variances is bounded, as opposed to "every function has bounded variance"? Matt Hastings's second point seems to refer to uniform bounds, whereas the Balian-Low theorem is just about (lack of) finiteness of the individual terms. For just the latter, either my or Yemon's answer should cover it, but will you please clarify? –  Jonas Meyer Mar 12 '10 at 6:09
    
I updated the question to reflect that I mean a uniform bound on the variance. –  Steve Flammia Mar 12 '10 at 17:10
    
Thank you. Even though the question has an affirmative answer, I'm still curious about what von Neumann had in mind exactly. E.g., does that sketch he gave in a footnote lead to another example? (But I'm not curious enough to investigate it myself.) –  Jonas Meyer Mar 13 '10 at 1:19
    
I am also curious about what von Neumann had in mind. Thanks for the great answer Jonas! –  Steve Flammia Mar 13 '10 at 1:35

2 Answers 2

up vote 5 down vote accepted

Such orthonormal bases do exist, as proved in:

Bourgain, J. A remark on the uncertainty principle for Hilbertian basis. J. Funct. Anal. 79 (1988), no. 1, 136--143 (MathSciNet link).

The theorem says that for each $\rho>1/2$ there is an orthonormal basis for $L^2(\mathbb{R})$ such that all of the variances of the basis elements and their Fourier transforms are less than $\rho$. After the statement Bourgain remarks:

Thus Balian’s strong uncertainty principle does not hold for a nonperiodic basis.

It is remarkable that this appears to have been discovered (rediscovered?) well after von Neumann's time. Powell proved more recently the result that Matt Hastings mentioned, namely that in such a case the sequence of means of the orthonormal basis is unbounded.


My old answer, posted before reading Matt Hastings's comment led me to the correct question, was to the question of whether all of the variances can be finite. It was this:

Yes, because you can take an orthonormal basis in the Schwartz space by applying Gram-Schmidt to a countable $L^2$ dense subset of the Schwartz space.

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I should add that von Neumann seems to state the existence and gives a hint as to how to construct an example, but I haven't gone through those details. –  Jonas Meyer Mar 12 '10 at 8:40

The Fourier transform on $L^2({\mathbb R})$ has an complete set of eigenvectors (that is, there is an o.n. basis of $L^2({\mathbb R})$ consisting of eigenfunctions for the FT, and they are all in the Schwartz space). Does this do what you want?

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In a way this is a special case of my answer, in that it works for the same reason if I understand the question properly. But it is nice to have a concrete basis with extra properties. –  Jonas Meyer Mar 12 '10 at 4:36
    
I agree Jonas. In fact, I rather suspect that taking $f_n(t) =t^n e^{-t^2/2}$ and applying Gram-Schmidt ought to give something like the Hermite functions, possibly the Hermite functions themselves. Your answer has the benefit of giving a conceptual reason (i.e. the fact that Schwartz space gets mapped to itself by the FT). –  Yemon Choi Mar 12 '10 at 5:30
    
This answers the question based on how it now reads, but I now strongly suspect that boundedness of the sequences of variances is desired, based both on Matt Hastings's comment and on the actual content of the von Neumann paper to which he refers. The sequence of variances of the Hermite functions is unbounded. –  Jonas Meyer Mar 12 '10 at 8:47

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