# Proving Fourier transform-related identities

I'm trying to find proofs for the following identities:

1. $$\quad F({f \cdot g}) = F({f}) \ast F({g})$$
2. $$\quad F({f(\alpha x)})=\frac{1}{|\alpha|} \cdot F(f(\frac{k}{\alpha}))$$
where $$F$$ denotes the Fourier transform.

I'm aware that 1) is a form of the convolution theorem, but I struggle to find a proof of it and instead I always just find the proof for the form $$F(f \ast g)=F(f) \cdot F(g)$$. I can't really find a way to proof this form since I don't know how to express $$F({f}) \ast F({g})$$ in integrals.

For 2) I think I know how to start, but I can't go on from here: $$F(f(\alpha x))= \int _{-\infty}^\infty f(\alpha x) \exp(-2\pi ikx) \, dx = \int_{-\infty}^\infty f(u) \exp\left(-2\pi i\frac{k}{\alpha}u\right) \, du$$

Any help or just a link would be greatly appreciated.

• Feb 25, 2017 at 14:36
• @KennyWong Thanks! do you have any hint on the second one, as well?
– Max
Feb 25, 2017 at 14:40
• Yes. In fact your method for the second one is fine. Your $\int_{-\infty}^\infty f(u) \exp(-2\pi i \frac k \alpha u) du$ is precisely $F(f)(\frac k \alpha)$. But I think you accidentally missed the factor of $1 / | \alpha |$, coming from $dx = du/|\alpha |$. Feb 25, 2017 at 14:54
• @KennyWong You're totally right. Thanks a lot.
– Max
Feb 25, 2017 at 14:56

\begin{align} \mathcal{F}(f)(\alpha t) & := \int f(x) e^{-2\pi i x \cdot \alpha t} \, dx \\ & = \int f(x) e^{-2\pi i (\alpha x) \cdot t} \, dx \end{align}

Taking the change of variables $$y = \alpha x$$, this is

$$\int f\left(\frac{y}{\alpha}\right) e^{-2 \pi i y t} \frac{dy}{\alpha}, \text{ if } \alpha > 0.$$

If $$\alpha < 0$$ then the order of the integral gets reversed and you'll end up with an extra $$-$$ out front, giving you

$$\frac{1}{|\alpha|} \int f\left(\frac{y}{\alpha}\right) e^{-2\pi i y t} \, dy,$$ as claimed.

\begin{align} \int _{-\infty}^\infty f(\alpha x) \exp(-2\pi ikx) \, dx \ne \int_{-\infty}^\infty f(u) \exp\left(-2\pi i\frac{k}{\alpha}u\right) \, & du \\[10pt] & \updownarrow \\[10pt] \int _{-\infty}^\infty f(\alpha x) \exp(-2\pi ikx) \, dx = \int_{-\infty}^\infty f(u) \exp\left(-2\pi i\frac{k}{\alpha}u\right) \, & \frac{du} {|\alpha|} \end{align} Two points:

• That denominator needs to be there.
• Why the absolute value sign? If $$\alpha>0$$ then that's not needed. If $$\alpha<0,$$ then as $$x$$ goes from $$-\infty$$ to $$+\infty,$$ $$u$$ goes from $$+\infty$$ to $$-\infty,$$ so omitting the absolute value and interchanging $$\pm\infty$$ would be correct. Then interchanging those two multiplies the whole thing by $$-1,$$ and putting the absolute value sign there also multiplies the whole thing by $$-1.$$