Time-Bandwidth Product The following text is cited from a textbook, "Spotlight Mode Synthetic Aperture Radar: A Signal Processing Approach", I would like to ask if anyone knows the proof to the following statements, as the proof is not outlined in the book, and I am unable to prove it myself.

The function $f(t)$ is represents a pulse envelope waveform.



The effective duration of a pulse envelope waveform is given as:

$$T_{e}=\frac{\int_{-\infty}^{\infty}f(t)dt}{f(0)}$$

A corresponding measure of effective bandwidth of the pulse envelope waveform is given by:

$$B_{e}=\frac{1}{2\pi}\cdot \frac{\int_{-\infty}^{\infty}F(\omega )d\omega }{F(0)}$$

From the defining equation for the Fourier transform, the above measures can be shown to have a product which is constant.

$$B_{e}T_{e}=1$$

I have read elsewhere that the above is the time-bandwidth product. I have no idea what that means or what the physical significance of it is. Can anyone shed light on how the last statement can actually be proven? Thank you!
Note that these definitions of effective duration and effective bandwidth are only useful for very specific signals, namely for (real-valued) low-pass signals that are even and centered around $t=0$. This implies that their Fourier transform is also real-valued, even and centered around $\omega=0$, and, consequently, the same definition of width can be used. 

Now for the proof, which is straightforward. Take the definition of the Fourier transform and its inverse:

$$F(\omega)=\int_{-\infty}^{\infty}f(t)e^{-j\omega t}dt\tag{1}$$

$$f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{j\omega t}d\omega\tag{2}$$

From $(1)$ and $(2)$ we get

$$F(0)=\int_{-\infty}^{\infty}f(t)dt\tag{3}$$

and

$$f(0)=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)d\omega\tag{4}$$

from which we get

$$T_e=\frac{F(0)}{f(0)}\tag{5}$$

$$B_e=\frac{f(0)}{F(0)}\tag{6}$$

The result follows immediately.

For information on more general definitions of duration/bandwidth and on the uncertainty principle take a look at this answer.