= 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 $u'=R_bu$ and the identity coefficient to $C'=b^DC+c(u)$. Since the total <partition function> is unchanged and $N'=N/b^D$,
$$
C+\mathfrak f(u)=b^{-D}[C'+\mathfrak f(u')],\qquad
\boxed{\mathfrak f(u)=b^{-D}\mathfrak f(u')+g(u),\quad g(u)=b^{-D}c(u).}
$$
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> gives
$$
\mathfrak f(u_0)=b^{-nD}\mathfrak f(u_n)+\sum_{j=0}^{n-1}b^{-jD}g(u_j).
$$
When a regular background $A(u)$ solves $A(u)=b^{-D}A(u')+g(u)$, the singular remainder $f_s=\mathfrak f-A$ satisfies the homogeneous scaling equation $f_s(u)=b^{-D}f_s(u')$. Thus \b[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.
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