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
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.