# Can the system of shifts of an $L^2(\mathbb{R})$ function be an ONB?

In Wavelet theory, one constructs wavelet bases via translations a dialations of an $L^2$ function... Is it possible for some set of translations alone to form an Orthonormal Basis? That is:

1. Does there exist $w \in L^2 (\mathbb{R})$ and $S$ a subset of real numbers so that the family of translates $\{w(t-s)\}_{s\in S}$ is an ONB for $L^2$?

2. If so, can we pick $S$ to be only integers? If not, can we pick $S$ to be only integers if we relax the requirement to only be a complete system rather than an ONB?

UPDATE: For the case $S =\mathbb{R}$, we can see it is not complete by noticing the fourier transforms are all $\hat{w}$ times $1-$periodic functions... and the result is easy from there. Part 1 is still open!

ONB? Surely not, not that I have a proof. Otoh if $S$ is dense then there does exist $f$ such that the closed span of the translates of $f$ by elements of $S$ is all of $L^2$. I suspect that if $S$ is not dense there cannot exist such an $f$; see below for why I suspect this.
Say $\tau_x f(t)=f(t-x)$.
Say $f\in L^2$ satisfies $\hat f\ne0$ almost everywhere. If $g$ is orthogonal to $\tau_x f$ for every $x\in S$ then the convolution $f*h$ vanishes on $S$, where $h(t)=g(-t)$. Suppose $S$ is dense. Since $f*h$ is continuous it follows that $f*h=0$. Hence $\hat f\hat h=0$, hence $\hat h=0$, so $g=0$.
Now if $S$ is not dense there does exist $\phi\in L^1$ with $\phi\ne0$ such that $\hat\phi$ vanishes on $S$. This doesn't quite show that $\tau_x f$ for $x\in S$ cannot span $L^2$, but it seems like evidence to that effect; saying those translates of $f$ do not span $L^2$ is the same as saying that we can take $\phi=\hat f\hat g$ for some $g\in L^2$.
• Thanks for your input! I can prove that a (almost everywhere) nonzero Fourier transform is necessary and sufficient for completeness in the case that $S=\mathbb{R}$. I also suspect the answer to be no, mostly because this is a fairly strict condition, and changing $S=\mathbb{R}$ to $S=\mathbb{Z}$ is a huge leap. To proceed by contradiction, we know the nonzero Fourier tranform is still necessary... but I'm not sure where to go from there. – Spanky Dec 9 '15 at 19:06