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Standard convolutions from $-\infty$ to $\infty$ are commutative. But, let's say we have

$$\int_{-R}^{+R}f(\tau)g(t-\tau)d\tau.$$

Is this equivalent to

$$\int_{-R}^{+R}f(t-\tau)g(\tau)d\tau?$$

I make a substitution $u = t-\tau$ giving us

$$\int_{t+R}^{t-R}f(t-u)g(u)du$$

and then I get stuck.

Testing says NO https://www.wolframalpha.com/input?i=integral+%28x%29%28y-x%29%5E2dx+from+-R+to+R%2C+integral+%28y-x%29x%5E2dx+from+-R+to+R

But wikipedia's misleading image to the upper right suggests yes https://en.wikipedia.org/wiki/Convolution.

This is further confusing for the following reason:

Let $f$ and $g$ be continuous on $[-R,R]$ and let $f$ and $g$ be $0$ outside of $[-R,R].$ Then $f*g$ and $g*f$ both clearly exist.

Wikipedia then goes on to say "continuous functions of compact support, are closed under the convolution, and so also form commutative associative algebras." further adding to the confusion.

I've put a lot of time into trying to google this and go through the math.

askquestions2
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1 Answers1

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As mentioned in my comments above, if you fix an interval and do not impose further conditions on $f,g$, then the convolution will not be commutative. As an example, we can consider $f=1_{[-1;1]}, g=1_{[99;101]}$. We have (here $t=100, R=1$) $$ \int_{-1}^1 f(\tau) g(100-\tau) d\tau = \int_{-1}^1 1 d\tau =1 $$ whereas $$ \int_{-1}^1 f(100-\tau) g(\tau) d\tau = \int_{-1}^1 0 d\tau =0. $$

As mentioned in Proving commutativity of convolution $(f \ast g)(x) = (g \ast f)(x)$ commutativity is restored if we consider periodic functions (note that we need the right period here to make this work).

Finally if we assume that $f,g$ vanish outside of $[-R;R]$, then we can also show that the convolution commutes. We have

$$ (f*g)(x) = \int_{-R}^R f(y) g(x-y) dy = \int_{-R}^{R+x} f(y) g(x-y) dy $$

as $f,g$ vanish outside of $[-R;R]$. After the substitution $\tau = x-y$ we obtain

$$ - \int_{x+R}^{-R} f(x-\tau) g(\tau) d\tau = \int_{-R}^{x+R} f(x-\tau) g(\tau) d\tau. $$

As $f,g$ vanish outside of $[-R;R]$ we get that this is equal to

$$ \int_{-R}^R f(x-\tau) g(\tau) d\tau = (g*f)(x). $$

Severin Schraven
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