Energy normalization across wavelet subbands According to the following reference: A Really Friendly Guide to Wavelets, © C. Valens, 1999
Equation 3, the wavelets are generated from the mother wavelet by scaling and translation. S is the scaling factor. The equation is divided by the square root of S for energy normalization across the different bands. I can not understand why this division achieves energy normalization.

According to the same reference, equation 14,  From Fourier theory we know that compression in time is equivalent to stretching the spectrum and shifting it upwards. I can understand spectrum stretching in this equation, but where is the spectrum shifting in the equation?1
The report under reference is A Really Friendly Guide to Wavelets by Clemens Valens, 1999.
The title of the question might be misleading. The energy normalization concerns the wavelets themselves, required to have the same energy, whatever the scaling and the shift. This is useful for theoretical and practical reasons (algorithmic/computational stability,  and several statistical properties).
If function $\psi(t)$ is of finite energy $\|\psi\|_2^2$ (or square integrable), then by a change of variable $t\mapsto st$, you will find that  $\psi\left(\frac{t-\tau}{s}\right)$ has finite energy as well, and that $\|\psi\left(\frac{t-\tau}{s}\right)\|_2^2 = |s|\|\psi\|_2^2$. For the wavelets to have the same energy, you should divide by $1/\sqrt{|s|}$.
An  interpretation is the following. The energy of function is homogeneous to a rectangle with a time span horizontally, and a square in amplitude vertically.
You want to keep the area of the rectangles constant for each wavelet. So if you dilate the time by $s&gt;0$, you should contract the square in amplitude by $1/s$, then the amplitude by $1/\sqrt{s}$.
For question 2, the spectrum will be stretched/shifted accordingly, if you compute the Fourier transform of $\psi\left(\frac{t-\tau}{s}\right)$. The scaling property of Fourier is mentioned in Clemens Valens tutorial at Equation 14:
$$\mathcal{F}\left(f(at)\right) = \frac{1}{|a|}F\left(\frac{\omega}{a}\right)\,.$$