Prove Convolution is Commutative

[cassiew]

Homework Statement

Let f,g be two continuous, periodic functions bounded by
$$[-\pi,\pi]$$

Define the convolution of f and g by

$$(f*g)(u)=(\frac{-1}{2\pi})\int_{-\pi}^{\pi}f(t)g(t-u)dt.$$

Show that
$$(f*g)(u)=(g*f)(u)$$

The Attempt at a Solution

I think the way I'm supposed to do this is by interchanging variables, but I'm stuck. If I let k=t-u and try to switch the variables around, I end up with (-1/2pi) times the integral of g(k)f(k+u)dk. Am I doing this wrong? Is there a better way to solve this?

 

[invisible_man]

can you check the integral of f(t)g(t-u) or f(t)g(u-t)

 

[LCKurtz]

Are you sure you stated the problem correctly? Shouldn't the integrand in the convolution be f(t)g(u-t)? That might help.