Fourier transform 4 times = original function (from Bracewell book)

I was glancing through "The Fourier Transform & Its Applications" by Ronald Bracewell, which is a good intro book on Fourier Transforms. In it, he says that if you take the FT of a function 4 times, you get back the original function, i.e. $$F\left( F\left( F\left( F\left( g(x) \right) \right) \right) \right) = g(x)\,. $$

Could someone kindly show me how this is possible? I'm assuming the above statement is for complex x, and this has something to do with $i^0=1$, $i^1=i$, $i^2=-1$, $i^3 = -i$, $i^4=1$?

Thank you for your enlightenment.

Answer

I'll use the non-unitary Fourier transform (but this is not important, it's just a preference):

$$X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-i\omega t}dt\tag{1}$$

$$x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{i\omega t}d\omega\tag{2}$$

where (1) is the Fourier transform, and (2) is the inverse Fourier transform.

Now if you formally take the Fourier transform of $X(\omega)$ you get

$$\mathcal{F}\{X(\omega)\}=\mathcal{F}^2\{x(t)\}=\int_{-\infty}^{\infty}X(\omega)e^{-i\omega t}d\omega\tag{3}$$

Comparing (3) with (2) we have

$$\mathcal{F}^2\{x(t)\}=2\pi x(-t)\tag{4}$$

So the Fourier transform equals an inverse Fourier transform with a sign change of the independent variable (apart from a scale factor due to the use of the non-unitary Fourier transform).

Since the Fourier transform of $x(-t)$ equals $X(-\omega)$, the Fourier transform of (4) is

$$\mathcal{F}^3\{x(t)\}=2\pi X(-\omega)\tag{5}$$

And, by an argument similar to the one used in (3) and (4), the Fourier transform of $X(-\omega)$ equals $2\pi x(t)$. So we obtain for the Fourier transform of (5)

$$\mathcal{F}^4\{x(t)\}=2\pi\mathcal{F}\{X(-\omega)\}=(2\pi)^2x(t)\tag{6}$$

which is the desired result. Note that the factor $(2\pi)^2$ in (6) is a consequence of using the non-unitary Fourier transform. If you use the unitary Fourier transform (where both the transform and its inverse get a factor $1/\sqrt{2\pi}$) this factor would disappear.

In sum, apart from irrelevant constant factors, you get

$$x(t)\overset{\mathcal{F}}{\Longrightarrow} X(\omega)\overset{\mathcal{F}}{\Longrightarrow} x(-t)\overset{\mathcal{F}}{\Longrightarrow} X(-\omega)\overset{\mathcal{F}}{\Longrightarrow} x(t)$$