# Is there a general way to prove this Fourier transform property?

We know that one of the important Fourier transform properties is that, the Fourier transform of a narrow function has a broad spectrum, and vice versa,

We can easily see this in this example, the Fourier transform $F(k)$ of the Delta function $\delta (x-x_0)$ which is a narrow function, has a broad spectrum (sinusoidal function), i.e. $$F(k) = \int_{\infty}^{\infty}\delta (x-x_0)e^{-2\pi i kx}dx = e^{-2\pi i kx_0}$$

Is there a more general way to prove that a Fourier transform of a narrow (or Broad) function has a Broad (or Narrow) spectrum?

I'ma physics student and I think the Uncertainty principle (position-momentum) in some ways proves the above by taking standard deviations, where position and momentum eigen function turn out to be Fourier pairs

• – Batman Apr 8 '16 at 11:02
• Thanks @Batman that seems to be what I'm looking for, but I'll still keep the quick open if anyone else has a different approach – Oswald Apr 8 '16 at 11:28
• I would like to know the specific definitions of "broad" and "narrow" here. – user247327 May 16 '17 at 14:06

You could argue that any narrow (finite) function is just the product of a periodic version of this function with a $\mathrm{rect}$ function. In the Fourier domain, this becomes a convolution of the transform of the original function (which is narrow) with an $\mathrm{si}$ function, thus broadening the original transform. The narrower the original function, the broader the $\mathrm{si}$ function becomes, the broader the transform of the orignal function.