1
$\begingroup$

Consider the Hilbert space $L^2(\Omega)$ over some Euclidean domain $\Omega$. Let $F=\{f_i;i\in\mathbb N\}$ be an orthonormal basis of this space consisting of functions in $L^2(\Omega)\cap L^4(\Omega)$. What is the space spanned by the pointwise squares, $s_i(x)=|f_i(x)|^2$? In particular, can $S_F=\{s_i;i\in\mathbb N\}$ span the whole space?

The functions $s_i$ are not necessarily linearly independent, but it seems that for a "generic basis" they should be. Linear independence is not important to me, however, but only the space spanned. If $\Omega$ is replaced with a set of two elements and the counting measure, the pointwise squares of basis vectors fail to span the space only if the basis vectors are essentially $\frac1{\sqrt2}(1,\pm1)$.

On the other hand, $\{\sin(kx);k>0\}\cup\{\cos(kx),k\geq0\}$ is an orthogonal (normalization is irrelevant for the problem) basis for $L^2((-\pi,\pi))$, but the squares span the same space as $\{\cos(2kx),k\geq0\}$. We are only left with "one fourth" of the original dimension.

Are there some conditions on the basis $F$ that guarantees that the closed span of $S_F$ is all of $L^2(\Omega)$? I have not seen any results in this direction, so any insight would be welcome. The spaces can be real or complex.

$\endgroup$
  • 1
    $\begingroup$ Some more examples: The Haar basis on $[0,1]$ or $\mathbb R$ works, complex exponentials never work, Hadamard type bases don't work and Gabor-type bases also don't work. $\endgroup$ – Dirk Sep 23 '16 at 12:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.