MHB Correcting Constants in Convolution of Fourier Transforms

[Dustinsfl]

I am trying to prove the convolution of the Fourier Transform
$$
(\widehat{f\star g})(\xi) = 2\pi\hat{f}(\xi)\hat{g}(\xi)
$$

 

[Ackbach]

With which definition of the Fourier Transform are you working?

 

[Dustinsfl]

$$
\hat{f}(\xi) = \frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-i\xi x}dx
$$

 

[Ackbach]

So, what have you got so far?

 

[Dustinsfl]

So, what have you got so far?

$$
2\pi\hat{f}(\xi)\hat{g}(\xi) = \frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-i\xi x}g(x)e^{-i\xi x}dx
$$

 

[Ackbach]

And what do you have for the LHS?

 

[Dustinsfl]

And what do you have for the LHS?

Not really sure on how to handle the LHS

 

[Ackbach]

Well, what is the definition of convolution?

 

[Dustinsfl]

Well, what is the definition of convolution?

It isn't the definition of convolution that is throwing me for a loop but the transform of it.
$$
(f\star g)(x) = \int_{-\infty}^{\infty}f(s)g(x-s)ds
$$

 

[Ackbach]

It isn't the definition of convolution that is throwing me for a loop but the transform of it.
$$
(f\star g)(x) = \int_{-\infty}^{\infty}f(s)g(x-s)ds
$$

Great. So the result of this convolution is a function of $x$, correct? Let's say that

$$h(x):=(f\star g)(x) = \int_{-\infty}^{\infty}f(s)g(x-s)\,ds.$$

Now, can you write down the Fourier Transform of $h$?

 

[Dustinsfl]

Great. So the result of this convolution is a function of $x$, correct? Let's say that

$$h(x):=(f\star g)(x) = \int_{-\infty}^{\infty}f(s)g(x-s)\,ds.$$

Now, can you write down the Fourier Transform of $h$?

$$
\hat{h}(\xi)=\frac{1}{2\pi}\int_{-\infty}^{\infty}h(x)e^{-i\xi x}dx = \frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(s)g(x-s)ds\right)e^{-i\xi x}dx
$$
but how do I get to the RHS now?

 

[Ackbach]

I would try putzing around with a change of variable, say, $z=x-s$, and then maybe changing the order of integration. See what that gets you.

 

[Dustinsfl]

I would try putzing around with a change of variable, say, $z=x-s$, and then maybe changing the order of integration. See what that gets you.

$$
\frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(s)g(x-s)ds\right)e^{-i\xi x}dx=
\frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(s)g(z)ds\right)e^{-i\xi (z+s)}dx
$$
Even with a change of integration order, I don't see how it will get to the right result.

 

[Ackbach]

If $z=x-s$, then $x=z+s$. Don't forget to change the differential as well.

 

[Dustinsfl]

$$
\frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(s)g(x-s)ds\right)e^{-i\xi x}dx=
\frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(s)g(z)ds\right)e^{-i\xi (z+s)}dx
$$
Even with a change of integration order, I don't see how it will get to the right result.

\begin{alignat}{3}
\frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(s)g(z)ds\right)e^{-i\xi (z+s)} d(z + s) & = &
\left(\int_{-\infty}^{\infty}f(s)e^{-i\xi s} ds\right)\left(\frac{1}{2\pi}\int_{-\infty}^{\infty}g(z)e^{-i\xi z}d(z + s)\right)\\
& = & \hat{f}(\xi)\left(\frac{1}{2\pi}\int_{-\infty}^{\infty}g(z)e^{-i\xi z}d(z + s)\right)
\end{alignat}

 

[Ackbach]

Try this instead:

\begin{align*}\frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(s)g(x-s)\,ds\right)e^{-i\xi x}\,dx&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(s)g(x-s)e^{-i\xi x}\,ds\,dx\\
&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(s)g(x-s)e^{-i\xi x}\,dx\,ds\\
&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(s)g(z)e^{-i\xi (z+s)}\,dz\,ds.
\end{align*}

Where can you go from here?

 

[Dustinsfl]

Try this instead:

\begin{align*}\frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(s)g(x-s)\,ds\right)e^{-i\xi x}\,dx&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(s)g(x-s)e^{-i\xi x}\,ds\,dx\\
&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(s)g(x-s)e^{-i\xi x}\,dx\,ds\\
&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(s)g(z)e^{-i\xi (z+s)}\,dz\,ds.
\end{align*}

Where can you go from here?

But now we have
$$
\frac{1}{2\pi}\hat{f}(\xi)\hat{g}(\xi) = (\widehat{f\star g})(\xi)
$$

 

[Ackbach]

But now we have
$$
\frac{1}{2\pi}\hat{f}(\xi)\hat{g}(\xi) = (\widehat{f\star g})(\xi)
$$

Check your constants again. To get two Fourier Transforms on the RHS, you need two factors of $2\pi$ in the denominator. How many do you actually have?