Let $X$ be a normed state space, $F:X\to X$ an evolution operator, $f:X\to Y$ an output map, and $x_t=(P_t,C_t,A_t)$. Define the recursive residual $K_F(x)=F(x)-x$. Then $K_F(x^)=0$ iff $F(x^)=x^$, and the corresponding output $I^=f(x^*)$ is invariant under recursion. If $F$ is contractive, the relaxed Picard recursion converges geometrically to the unique fixed point. Independently, a self-adjoint spectral selector $K_C$ determines an invariant subspace $E=\ker K_C$, its orthogonal projector $P$, and the complementary contraction generator $H=I-P$. The variational functional $\mathcal F[x]=\frac12\|Hx\|^2$ generates both discrete and continuous flows whose exact limit is $Px_0$. In the E47 specialization, $\dim\mathcal H=125$, $K_C=(C-6I)(C-30I)$, $\dim\ker K_C=47$, and $\Omega_c=47/125=0.376$.
Define
$$ K_F(x)=F(x)-x. $$
Then
$$ K_F(x^)=0\iff F(x^)=x^*. $$
For $I(t)=f(x_t)$, once $x_t=x^*$,
$$ I(t+1)=f(F(x^))=f(x^)=I(t)=I^*. $$
Thus the output associated with a fixed state is recursion-invariant.
For the declared sign convention $K_F=F-I$, the relaxed Picard step is
$$ x_{n+1}=x_n+\epsilon K_F(x_n)=(1-\epsilon)x_n+\epsilon F(x_n). $$
If
$$ \|F(x)-F(y)\|\le\lambda\|x-y\|,\qquad0\le\lambda<1, $$
then for $0<\epsilon\le1$,
$$ \|x_{n+1}-x^\|\le[1-\epsilon(1-\lambda)]\|x_n-x^\|, $$
with $\rho=1-\epsilon(1-\lambda)<1$, hence $x_n\to x^*$ geometrically.
<aside> ∴
Fixed-point closure: $K_F(x^)=0\iff F(x^)=x^$ and $x_n\to x^\Rightarrow I_n\to f(x^*)$.
</aside>