Help with 2d convolution formula

[mnb96]

Hello,

I consider two functions f:R^2 \rightarrow R and g:R^2 \rightarrow R, and the two dimensional convolution (f \ast g)(\mathbf{x}) = \int_{\mathbb{R}^2}f(\mathbf{t})g(\mathbf{x-t})d^2\mathbf{t}
I proved using the Fourier transform and the convolution theorem that the convolution of two "rotated" versions of f and g is equivalent to simply taking the convolution (f*g) and rotating it.

However I have troubles proving these statement using only the definition of convolution. I will show my attempt. There must be a mistake somewhere.

I consider an isometry (rotation) \phi:R^2\rightarrow R^2 and the two rotated versions of the functions: f(\phi(\mathbf{x})) and g(\phi(\mathbf{x})).
The convolution would be: \int_{\mathbb{R}^2}f(\phi(\mathbf{t}))g(\mathbf{x}-\phi(\mathbf{t}))d^2\mathbf{t} By setting \mathbf{y}=\phi(\mathbf{t}) and observing that d^2y = d^2t we get: \int_{\mathbb{R}^2}f(\mathbf{y})g(\mathbf{x}-\mathbf{y})d^2\mathbf{y} which is not the expected result. Where is the mistake?

 

[haruspex]

The convolution would be: \int_{\mathbb{R}^2}f(\phi(\mathbf{t}))g(\mathbf{x}-\phi(\mathbf{t}))d^2\mathbf{t}

You mean, the convolution of the rotated functions, f\phi(), g\phi()?
Wouldn't that be \int_{\mathbb{R}^2}f(\phi(\mathbf{t})) g(\phi(\mathbf{x}-\mathbf{t}))d\mathbf{t}
(And \phi(\mathbf{x}-\mathbf{t})=\phi(\mathbf{x})-\phi(\mathbf{t}), right?)

 

[mnb96]

Thanks!
yes. Maybe I was tired and didn't notice that mistake!
The statement to be proven then follows from the linearity of \phi.