I have 3D volume that is convoluted with a 3D blur function. Both are positive and integrate to a finite value. I can see experimentally (meaning playing with matlab) that this is true:
$\int_{-a}^{a }(g\star f)(x,y,z)dz=(\int_{-a}^{a}g(x,y,z)dz)\star (\int_{-b}^{b }f(x,y,z)dz)$
I see that wikipedia (http://en.wikipedia.org/wiki/Convolution#Integration , http://en.wikipedia.org/wiki/Fubini%27s_theorem#Corollary) has the following property of convolution:
$\int_{R}^{ }(g\star f)(x)dx=(\int_{R}^{ }g(x)dx) (\int_{R}^{ }f(x)dx)$
How could this be proven for my case? Sorry I am just a programmer if this is obvious.
edit: doing a little more math, but still not there.
so I expand out the convolution to give the following integral
$$h(x,y)=\int_{\mathbb{Z}}^{ }\int_{\mathbb{X}'}^{ }\int_{\mathbb{Y}'}^{ }\int_{\mathbb{Z}'}^{ } g(x',y',z')f(x-x',y-y',z-z')dx'dy'dz'dz$$
I then rearrange the integrals to put the integrals over z in the center (is this correct?)
$$h(x,y)=\int_{\mathbb{X}}^{ }\int_{\mathbb{Y}'}^{ }\int_{\mathbb{Z}}^{ }\int_{\mathbb{Z}'}^{ } g(x',y',z')f(x-x',y-y',z-z')dz'dzdx'dy'$$
I then ignore the x and y integrals for a moment, to get
$$\int_{\mathbb{Z}}^{ }\int_{\mathbb{Z}'}^{ } g(x',y',z')f(x-x',y-y',z-z')dz'dz$$
which matches the equation from the wikipedia page. Both functions are always positive, so I can change this to
$$\int_{\mathbb{Z}}^{ }\int_{\mathbb{Z}'}^{ } \left | g(x',y',z') \right |\left |f(x-x',y-y',z-z') \right |dz'dz$$
since g does not have any z in it
$$\int_{\mathbb{Z}}^{ } \left | g(x',y',z') \right |\left [ \int_{\mathbb{Z}'}^{ }\left |f(x-x',y-y',z-z') \right |dz\right ]dz'$$
which gives: $$\int_{\mathbb{Z}}^{ } \left | g(x',y',z') \right | \left \| f \right \|_{1} dz' = \left \| g \right \|_{1}\left \| f \right \|_{1}$$
(I have no idea what $\left \| \right \|_{1}$ means or how this step works, hard to find a symbol on google!) So putting back the integrals of x' and y', this gives
$$H(x,y)=\int_{\mathbb{X}'}^{ } \int_{\mathbb{Y}'}^{ }\left \| g \right \|_{1} \left \| f \right \|_{1} dx'dy' =???$$ which is where I lost it