 Solution

Write $\mathfrak{f}(u)$ for the physical free energy per site excluding the identity term, so $F/N=C+\mathfrak{f}(u)$. Blocking sends $u$ to $u^{\prime }=R_{b}u$ and the identity coefficient to $C^{\prime }=b^{D}C+c(u)$. Since the total partition function is unchanged and $N^{\prime }=N/b^D$,The function $g$ records the field-independent energy generated by eliminated modes, including their entropy and normalization. It is the identity-operator contribution to renormalization-group free energy; it must be tracked to recover the complete thermodynamics, even though it does not change normalized spin probabilities.

Iterating the additive free-energy recursion under blocking givesWhen a regular background $A(u)$ solves $A(u)=b^{-D}A(u^{\prime })+g(u)$, the singular remainder $f_s=\mathfrak{f}-A$ satisfies the homogeneous scaling equation $f_s(u)=b^{-D}{f}_s(u^{\prime })$. Thus the full free energy contains an inhomogeneous contribution, while the background-subtracted singular part is homogeneous. A finite initial identity energy $C_0$ adds a regular normalization, not a new magnetization singularity.