Question about convolutions

[AxiomOfChoice]

I know that the convolution of two functions $f(x)$ and $g(x)$ is given by

$$(f * g)(y) = \int_{\mathbb R} f(x)g(y-x) dx.$$

But what if I'm trying to convolve a function $f(x)$ with a function $g(x + az)$, where $a$ is some constant? Is it just

$$(f*g)(y) = \int_{\mathbb R} f(x)g(y - x + az) dx.$$

If so, why? I can't seem to find a definition of the convolution that makes this obvious.

 

[Mute]

Convolutions are symmetric in f and g:

$$(f \star g)(y) = (g \star f)(y);$$

you could thus write the convolution of f(x) and g(x + az) as

$$\int_{\mathbb{R}} dx~f(y-x)g(x+az),$$

which is perhaps easier to see.

 

[mathman]

Your question is a little puzzling. How does z relate to x in g(x+az)?

 

[AxiomOfChoice]

Okay, let me see if I can be more specific. What I'm really trying to do is to take the Fourier transform...in the $x$ variable...of the product $f(x) g(ax + y)$, where $x,y\in \mathbb R$ are independent (hence unrelated) variables. So I'm interested in

$$\int_{-\infty}^{\infty} e^{-i p x} f(x) \cdot g(ax + y) \ dx.$$

I know that Fourier transforms turn products into convolutions...but is it that straightforward in this case?

 

[mathman]

Okay, let me see if I can be more specific. What I'm really trying to do is to take the Fourier transform...in the $x$ variable...of the product $f(x) g(ax + y)$, where $x,y\in \mathbb R$ are independent (hence unrelated) variables. So I'm interested in

$$\int_{-\infty}^{\infty} e^{-i p x} f(x) \cdot g(ax + y) \ dx.$$

I know that Fourier transforms turn products into convolutions...but is it that straightforward in this case?

What you need is the opposite. Fourier transform of a convolution is product of Fourier transforms of the individual items. Once you've done that, take the back transform to get what you are looking for.