Radial filters preserved under convolution?
I spent most of my time today trying to answer the following question. Let 
be radial filters in the sense that 
when 
(and likewise for the 
’s). Then is the filter 
also radial?
The answer, it turns out, is no in general (but of course it is true in 
, because then the filters are just even). In the 2-d case, the filters
serve as a counterexample: 
, but 
.
I haven’t yet found a satisfactory reason for this failure — my vague idea is that rotations do not preserve the lattice, but that doesn’t do more than imply the result. To construct this counterexample, I followed Papadakis’ advice of looking at the simplest radial filters of all: filters which are unity on the intersections of the lattice and a sphere and zero elsewhere, as in the examples given above. In the Fourier domain, the convolution product would look like
and if the resulting filter was also radial, it would look like a sum
in the frequency domain. I wrote a Mathematica program that takes 
and 
, calculates the coefficients of the resulting filter, and says if it breaks into such a sum. Now that I think about it, that would have been much easier to do in the spatial domain.
As a corollary of this result, you can see that the following proposition is not true in general: if 
with , there is an 
so that 
and 
.
Possibly relevant posts:
- Distribution function (1/14/2006)
- 2d convolution via frequency domain multiplication (7/11/2006)
- Inner products (6/7/2006)