# Thread: Operation of convoulution is commutative

1. ## Operation of convoulution is commutative

Hi,

If we define the convolution of two functions $f,g:\mathbb{R}^n \longrightarrow \mathbb{R}$ to be $f* g(x) = \int_{\mathbb{R}^n} f(x-y)g(y)dy$ then this operation is supposed to commute, i.e. $f* g = g* f$. When you change variables though, surely you get a minus sign???

Any help would be greatly appreciated.

2. Originally Posted by markwolfson16900
Hi,

If we define the convolution of two functions $f,g:\mathbb{R}^n \longrightarrow \mathbb{R}$ to be $f* g(x) = \int_{\mathbb{R}^n} f(x-y)g(y)dy$ then this operation is supposed to commute, i.e. $f* g = g* f$. When you change variables though, surely you get a minus sign???

Any help would be greatly appreciated.
Define $t=x-y$. Then $\,dy=-\,dt$. But then the "limits of integration" are reversed in the $\displaystyle\int_{\mathbb{R}^n}$ part. So to get them in the right order, we "flip" the limits and then make that result negative.

Like for instance, consider the case we're integrating over $\mathbb{R}$. Then its clear that if we consider the convolution $f*g(t) = \displaystyle\int_0^t f(t-\tau)g(\tau)\,d\tau$ and make the same change of variables, say $s=t-\tau$, then we get $\,d\tau=-\,ds$. But then our limits of integration change positions: $\displaystyle\int_{t}^{0}f(s)g(t-s)(-\,ds)$. So now, we flip the limits of integration to get $\displaystyle\int_0^t f(s)g(t-s)\,ds=g*f(t)$

Thus, I believe a similar idea holds in the n dimensional case; so, in other words: $\displaystyle\int_{\mathbb{R}^n}f(x-y)g(y)\,dy \xrightarrow{t=x-y}{} -\int_{\mathbb{R}^n}f(t)g(x-t)(-\,dt)=\int_{\mathbb{R}^n}f(t)g(x-t)\,dt=g*f$.

I hope this helps!