# Help with 2d convolution formula

1. Nov 23, 2012

### 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?

2. Nov 23, 2012

### haruspex

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?)

3. Nov 24, 2012

### mnb96

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