Morlet wavelet time and frequency resolution
Consider shifted and scaled versions of a mother wavelet $\psi(t)$:

$$\psi_{a,b}(t)=\frac{1}{\sqrt{a}}\psi\left(\frac{t-b}{a}\right),\quad a&gt;0,\;b\in\mathbb{R}\tag{1}$$

By the definition of the Fourier transform

$$\Psi(\omega)=\int_{-\infty}^{\infty}\psi(t)e^{-i\omega t}dt$$

it can be shown that the Fourier transform of $\psi_{a,b}(t)$ is

$$\Psi_{a,b}(\omega)=\sqrt{a}e^{-ib\omega}\Psi(a\omega)\tag{2}$$

Note that the scaling factor $a$ appears in the denominator of the argument of $\psi_{a,b}(t)$ and in the "numerator" of the argument of $\Psi_{a,b}(\omega)$. This implies that stretching in the time domain ($a&gt;1$) implies compression in the frequency domain and vice versa. Consequently, if $\Delta t$ and $\Delta f$ denote the "spread" of $\psi(t)$ in time and frequency, it follows from (1) and (2) that the corresponding spreads of $\psi_{a,b}(t)$ - call them $\Delta t_{a}$ and $\Delta f_{a}$ - are given by

$$\Delta t_a=a\Delta t\quad\textrm{and}\quad\Delta f_{a}=\frac{\Delta f}{a}$$

As for the definition of $\Delta t$ and $\Delta f$, there is not way to "prove" the results that you stated. It is just a matter of defining the spread of a function. Since $|\psi(t)|$ is a Gaussian with standard deviation $\sigma=\sqrt{f_b/2}$ it seems natural to define $\Delta t$ as being proportional to the standard deviation, but there is no "correct" way to choose the proportionality constant. Once $\Delta t$ has been defined, the corresponding definition of $\Delta f$ can be obtained from the uncertainty principle.