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](http://scirate.com/who.php?id=1003.2133&what=comments) regarding a recently translated [paper by von Neumann](http://arXiv.org/abs/1003.2133). There the commenter, Matt Hastings, points out some related results.


In particular, the [Balian-Low theorem](http://en.wikipedia.org/wiki/Balian%E2%80%93Low_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. 



Simultaneous time-frequency concentration of orthonormal sequences?
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](http://www.ams.org/mathscinet-getitem?mr=950087)).


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](http://www.springerlink.com/content/j365534350378065/) 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.


