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The mother wavelet is defined as
$$\Psi_{a,\tau}(t) = \lvert a\rvert^{-1/2}\Psi\left(\frac{t-\tau}{a}\right)\tag{1}$$
in which the factor
$$ \lvert a\rvert^{-1/2}\tag{2}$$
is supposed to serve that
$$\lVert\Psi_{a,\tau}\rVert = \lVert\Psi\rVert\tag{3}$$
is valid.
Does someone have an idea how $(3)$ is to be understood?
My answer is for real scale $a$ and the fact that wavelet transform is usually defined in $L_2$ with norm
$$||\Psi(\tau)|| = \int_\mathbb{R} \Psi(\tau)\Psi^*(\tau)\mathrm{d}\tau $$
So
$$||\Psi_{a,t}(\tau)|| = \int_\mathbb{R} \frac{1}{|a|}\Psi(\frac{\tau-t}{a})\Psi^*(\frac{\tau-t}{a})\mathrm{d}\tau$$
Set $\tau' = \frac{\tau-t}{a} \implies d\tau' = d\tau / a \implies d\tau = a \times d\tau'$
$$||\Psi_{a,t}(\tau)|| = \int_\mathbb{R} \frac{1}{|a|}\Psi(\tau')\Psi^*(\tau') \times a \times \mathrm{d}\tau' = \int_\mathbb{R} \Psi(\tau')\Psi^*(\tau') \mathrm{d}\tau' = ||\Psi(\tau)||$$
Update for the question "How does it not change the norm considering that it could introduce a negative sign with the norm being sign-less?"
We develop stuff for the case that scaling factor $a < 0$.
with $A > 0$ $$||\Psi_{a,t}(\tau)|| = \lim_{A \to \infty}\int_{(-A-t)/a}^{(A-t)/a} \frac{1}{|a|}\Psi(\tau')\Psi^*(\tau') \times (-|a|) \times \mathrm{d}\tau' \\ = \lim_{A \to \infty}\int_{(A-t)/a}^{(-A-t)/a} \Psi(\tau')\Psi^*(\tau') \times \mathrm{d}\tau' \\ = \lim_{A \to \infty}\int_{(-A+t)/|a|}^{(A+t)/|a|} \Psi(\tau')\Psi^*(\tau') \times \mathrm{d}\tau' = \int_{-\infty}^{+\infty} \Psi(\tau')\Psi^*(\tau') \times \mathrm{d}\tau'$$
Wavelets play differents role in functional spaces, especially as unconditional bases (see What are unconditional bases and which wavelets have this property?). In $L_p$ spaces, if $|\psi|^p$ is integrable, the translation operator $t\to t-\tau$ preserves the norm, while the dilation $t\to t/a$ induces a scale factor of $|a|^{1/p}$, which can be corrected to have the same $\|\|_p$ norm for all functions $\psi\left(\frac{t-\tau}{a}\right)$.
The case $p=2$ is especially important (orthonormality, duality), hence the $1/\sqrt{a}$ factor.
