# convergence rate in Wiener's approximation theorem

Wiener has the following fantastic results about approximations using translation families:

Given a function $h: \mathbb{R} \to \mathbb{R}$, the set $\{\sum a_i h(\cdot - x_i): a_i, x_i \in \mathbb{R}\}$ is

i) dense in $L^1(\mathbb{R})$ if and only if the Fourier transform of $h$ has no zeros.
ii) dense in $L^2(\mathbb{R})$ if and only if zeros of the Fourier transform of $h$ has zero Lebesgue measure.

After this, a further step is natually about the speed of convergence, i.e., how fast does the error vanishes with respect to the the number of translates. Now let us focus on the $L^1$ case and take $h = \varphi$ to be the standard normal density whose Fourier transform does not vanish on the real line. Given a function $f \in L^1(\mathbb{R})$, the error of the optimal $m$-term approximation is

$\mathop{\inf}\limits_{a_i, x_i \in \mathbb{R}} \left\|f - \sum_{i=1}^m a_i h(\cdot - x_i)\right\|_1$.

My question is whether there is any way to lower bound this quantity. Of course, there won't be any meaningful conclusion without assumptions on $f$ (e.g., $f$ is a finite-mixture of translates of $\varphi$, then it is trivial). So let us consider $f = g * \varphi$ for some smooth $g$ (e.g., $g = \varphi$), where $*$ denotes convolution, that is, $f$ is an infinite"-mixture of translates of $\varphi$. Any idea would be greatly appreciated. If things could be easier using $L^2, L^{\infty}$ or other distance, it should also be helpful.

For the upper bound, there have been many work, the speed of convergence could be $O(m^{-2})$ or even exponential in $m$. For the lower bound, most work consider a min-max setup: for $f$ belonging to a given class of functions, the worst-case convergence rate can never by faster than $O(m^{-2})$. But for a given $f$, there seems to be no known result.

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typo: i) and ii) cannot be true at the same time – Alekk Feb 3 '10 at 3:39
So, you are asking how well g can be approximated by a sum of m delta functions, where the measure is the L^1, L^2, or L^infinty norm of the convolution with phi? – Douglas Zare Feb 3 '10 at 4:26
@ Alekk: I corrected the typo. @ Leonid: Yes. I am aware of that. I'm trying to avoid the trivial case of finite-mixtures. @ Douglas: that is correct. but I'm not sure if that makes things easier to approach. and I'm only concerned about the converse direction (lower bound). – mr.gondolier Feb 3 '10 at 5:04