 Solution

The normalization fixes the critical point at zero and its image at one. It removes arbitrary height and spatial scales, so successive renormalized return maps can be compared in one function space rather than drifting through rescaled copies of the same dynamics. The second iterate near the critical point has a minimum; a negative rescaling reverses it back to a maximum.

Because $f(0)=1$, the period-doubling renormalization operator satisfiesPreservation of normalization therefore forcesThe scale is a functional of $f$, not an independently adjustable constant. In the renormalizable regime it lies in $(-1,0)$. If $h(x)=ax$, then $\mathcal{T}(f)=h^{-1}\circ f^2\circ h$ on the central restrictive interval. Thus topological conjugacy relates one iteration of the renormalized map to two iterations of the original map, and the new critical value is again one. For $f_{\mu }$, $a=1-\mu$, so this regime starts beyond the superstable two-cycle at $\mu =1$.

Normalization alone does not make every normalized unimodal map renormalizable: the central interval and its image must form the appropriate period-two restrictive pair, so that the rescaled second iterate is again an even unimodal map into $[-1,1]$. The operator is considered on the neighbourhood where these additional conditions hold.