Komorebi Space

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1 \partnote{From approximation to exact realization}{A vanishing defect is a statement about measurements. A limiting realization also requires control of the candidate itself. This part follows the movement of approximate answers, the closure of admissible correction images, and the meaning of the ambient completion.}
2 \chap{Exactification without a fixed contraction rate}\label{sec:variable-exactification}
3 The constructed linear repair in Section~\ref{sec:split} is a useful special case of a more general principle. One can prove convergence without one fixed contraction factor, without a linear update rule, and without a globally Lipschitz iteration. What must not be lost is control of the total movement.
4 
5 \subsection{The two limits that must be established}
6 Let \((M,\dist)\) be a complete metric space. Let \(\omega:M\to[0,\infty)\) be a nonnegative residual function, and call a point exact when \(\omega(x)=0\). We regard \(\omega\) as a real-valued function when discussing continuity. Assume it is lower semicontinuous, so that every sublevel set \(\{x:\omega(x)\le a\}\) is closed.
7 
8 For an approximation sequence \(f_n\), the statements
9 \[
10  \omega(f_n)\longrightarrow0
11  \quad\text{and}\quad
12  f_n\longrightarrow x
13 \]
14 are different. The first concerns the observed discrepancy. The second concerns the mathematical candidate in the chosen metric. Neither implies the other without additional hypotheses.
15 
16 \begin{theorem}[Summable movement gives exactification]\label{KS71}
17 Suppose \(b_n\ge0\), \(\sum_{n=0}^{\infty}b_n<\infty\), and
18 \[
19  \dist(f_n,f_{n+1})\le b_n,\qquad \omega(f_n)\longrightarrow0.
20 \]
21 Then there is a limit \(x\in M\) satisfying
22 \[
23  f_n\longrightarrow x,\qquad\omega(x)=0,\qquad
24  \dist(f_n,x)\le\sum_{k=n}^{\infty}b_k.
25 \]
26 \end{theorem}
27 \begin{proof}
28 For integers \(m>n\), apply the triangle inequality successively along the finite chain \(f_n,f_{n+1},\ldots,f_m\). It gives
29 \[
30  \dist(f_n,f_m)\le\sum_{k=n}^{m-1}\dist(f_k,f_{k+1})
31  \le\sum_{k=n}^{m-1}b_k\le\sum_{k=n}^{\infty}b_k.
32 \]
33 Given \(\varepsilon>0\), convergence of the nonnegative series supplies \(N\) for which the final tail is below \(\varepsilon\) whenever \(n\ge N\). Interchanging \(m,n\) when necessary proves the Cauchy condition for all pairs of indices at least \(N\). Completeness, not residual decay, is the hypothesis that now supplies a limit \(x\in M\).
34 
35 Keep \(n\) fixed. The reverse triangle inequality says
36 \[
37  \bigl|\dist(f_n,f_m)-\dist(f_n,x)\bigr|\le\dist(f_m,x)\longrightarrow0.
38 \]
39 Passing to the limit in the uniform tail bound proves the claimed distance certificate at that \(n\).
40 
41 It remains to prove exactness of the limit. Fix \(\varepsilon>0\). Since \(\omega(f_m)\to0\), an entire tail of the sequence belongs to the sublevel set \(C_\varepsilon=\{z:\omega(z)\le\varepsilon\}\). Lower semicontinuity makes this set closed, so the limit of that tail belongs to it. Hence \(\omega(x)\le\varepsilon\) for every positive \(\varepsilon\). If \(\omega(x)>0\), choosing \(\varepsilon=\omega(x)/2\) contradicts this inequality. Since \(\omega\) is nonnegative, \(\omega(x)=0\). We have not used, or asserted, continuity of \(\omega\) or the stronger equality \(\omega(f_m)\to\omega(x)\) for arbitrary convergent inputs.
42 \end{proof}
43 \source{KS71}
44 The formal theorem is slightly more general: it is stated in a complete pseudometric space. In a metric space the limit of this fixed sequence is unique. That does not mean different admissible approximation procedures select the same exact point.
45 
46 \subsection{Variable rates and a movement budget}
47 Let \(q_n,C_n\ge0\), and define the prefix product
48 \[
49  Q_0=1,\qquad Q_{n+1}=q_nQ_n.
50 \]
51 Assume
52 \begin{equation}\label{eq:variable-rate}
53  \omega(f_{n+1})\le q_n\omega(f_n),\qquad
54  \dist(f_n,f_{n+1})\le C_n\omega(f_n).
55 \end{equation}
56 
57 \begin{theorem}[Variable-rate exactification]\label{KS72}
58 If \(Q_n\to0\) and \(\sum_n C_nQ_n<\infty\), then \((f_n)\) has an exact limit and
59 \begin{equation}\label{eq:variable-tail}
60  \dist(f_n,x)\le\omega(f_0)\sum_{k=n}^{\infty}C_kQ_k.
61 \end{equation}
62 For constant \(q_n=q\in[0,1)\) and constant \(C_n=C\), the sharper restarted estimate is
63 \[
64  \dist(f_n,x)\le\frac{C}{1-q}\,\omega(f_n).
65 \]
66 \end{theorem}
67 \begin{proof}
68 The estimate \(\omega(f_n)\le Q_n\omega(f_0)\) follows by induction. At \(n=0\) it is equality. If it holds at \(n\), nonnegativity of \(q_n\) permits multiplication and gives
69 \[
70  \omega(f_{n+1})\le q_n\omega(f_n)
71  \le q_nQ_n\omega(f_0)=Q_{n+1}\omega(f_0).
72 \]
73 Because \(Q_n\to0\) and \(\omega(f_0)\) is a fixed finite number, the residual tends to zero. The movement assumption then gives
74 \[
75  \dist(f_n,f_{n+1})\le C_nQ_n\omega(f_0).
76 \]
77 This is a summable nonnegative budget by the hypothesis on \(\sum C_nQ_n\). All assumptions of Theorem~\ref{KS71} now hold, and its tail certificate is exactly \eqref{eq:variable-tail}. If \(\omega(f_0)=0\), the same estimates make every step zero and every residual zero. No cancellation of that value was used.
78 
79 For constant rates, the global estimate can be sharpened by restarting at the current index. Fix \(n\). A new induction on \(j\) gives
80 \[
81  \omega(f_{n+j})\le q^j\omega(f_n),\qquad
82  \dist(f_{n+j},f_{n+j+1})\le Cq^j\omega(f_n).
83 \]
84 Sum the finite chain from \(n\) to \(n+m\), and then let \(m\to\infty\) using the already established limit. The geometric series equals \(1/(1-q)\), including \(q=0\), where its first term is one and all later terms vanish. This proves the restarted bound. The variable-rate proof itself requires only nonnegative \(q_n\), decay of their cumulative products, and summability of the movement weights. It does not insert an unstated upper bound \(q_n<1\) at each step.
85 \end{proof}
86 \source{KS72}
87 The individual factors \(q_n\) need not all be below one. Temporary residual increases are permitted if the prefix products decay and the weighted movements remain summable. Conversely, a decaying product is not enough when its movement budget diverges.
88 
89 \subsection{A convergent path with no uniform geometric factor}
90 In \(\R^2\) with the maximum norm, fix \(s\in\R\) and take
91 \[
92  f_n=(s,(n+1)^{-1}),\qquad \omega(u,r)=|r|.
93 \]
94 The protected coordinate is the first one. Direct calculation gives
95 \[
96  q_n=\frac{n+1}{n+2},\quad Q_n=\frac1{n+1},\quad
97  C_n=\frac1{n+2},\quad
98  \dist(f_n,f_{n+1})=\frac1{(n+1)(n+2)}.
99 \]
100 No constant \(q<1\) bounds all the ratios \(q_n\), since they tend to one. Nevertheless
101 \[
102  \sum_{k=n}^{\infty}C_kQ_k
103  =\sum_{k=n}^{\infty}\left(\frac1{k+1}-\frac1{k+2}\right)
104  =\frac1{n+1}.
105 \]
106 The limit is \((s,0)\), and the error bound is an equality. The model shows why the variable-rate theorem is not merely a rephrasing of a fixed geometric contraction.
107 
108 \plate{25_summable_path}{Polynomial residual decay can still give an exact limit with a sharp error certificate. The residual $1/(n+1)$ is not summable, but the actual movement $1/((n+1)(n+2))$ is. The theorem uses the movement series.}{fig:summable-path}
109 
110 \subsection{Residual bursts can be compatible with convergence}
111 Set \(q_{2k}=2\), \(q_{2k+1}=1/8\). Then
112 \[
113  Q_{2k}=4^{-k},\qquad Q_{2k+1}=2\cdot4^{-k}.
114 \]
115 For \(f_n=(s,Q_n)\), the residual doubles at the start of every pair of steps. Choose \(C_{2k}=1\) and \(C_{2k+1}=7/8\). These give equality in the movement bound, and
116 \[
117  \sum_{n=0}^{\infty}C_nQ_n
118  =\sum_{k=0}^{\infty}\left(1+\frac74\right)4^{-k}
119  =\frac{11}{3}.
120 \]
121 Thus \(f_n\to(s,0)\) despite infinitely many increases. Starting at an even index \(2k\), the remaining movement budget is \(\frac{11}{3}4^{-k}\). At an odd index \(2k+1\), it is \(\frac83 4^{-k}\). These bounds are larger than the actual distance because the path repeatedly moves away and returns.
122 
123 \plate{26_residual_bursts}{The residual increases at every even-to-odd step, but each two-step product is $1/4$. A convergent product and a summable movement budget control the whole path. Monotonicity of individual residual values is not assumed by the variable-rate theorem.}{fig:residual-bursts}
124 
125 \subsection{Protected meaning survives a controlled limit}
126 Return to Banach and normed linear spaces, with continuous linear maps \(d:E\to R\) and \(p:E\to S\). Put \(\omega(f)=\norm{df}\). The residual is continuous, hence lower semicontinuous.
127 
128 \begin{proposition}[Controlled paths kill the actual obstruction]\label{KS73}
129 Suppose \(E\) is complete, \(\norm{df_n}\to0\), \(p(f_n)=p(f_0)\), and \(\dist(f_n,f_{n+1})\le b_n\) for a summable nonnegative budget. Then there is \(x\) with
130 \[
131  f_n\to x,\qquad dx=0,\qquad p(x)=p(f_0),
132 \]
133 and the constrained obstruction \([df_0]\in R/d(\ker p)\) vanishes.
134 \end{proposition}
135 \begin{proof}
136 The function \(f\mapsto\norm{df}\) is continuous because \(d\) is bounded linear and the norm is continuous. It is therefore an admissible lower semicontinuous residual for Theorem~\ref{KS71}. That theorem supplies a limit \(x\in E\), together with \(\norm{dx}=0\), which means \(dx=0\).
137 
138 Continuity of \(p\) gives \(p(f_n)\to p(x)\). The left side is the constant sequence with value \(p(f_0)\). Uniqueness of limits in the normed target therefore gives \(p(x)=p(f_0)\). Define the actual vector \(a=f_0-x\). Linearity, rather than a closure argument, now yields
139 \[
140  p(a)=p(f_0)-p(x)=0,\qquad da=df_0-dx=df_0.
141 \]
142 Thus \(a\in\ker p\) and \(df_0\in d(\ker p)\). Membership in this algebraic image is precisely what makes the class \([df_0]\) zero. Merely knowing membership in the closure of that image would not justify the same conclusion.
143 \end{proof}
144 \source{KS73}
145 This proposition converts a controlled approximation argument into a statement about the algebraic quotient defined earlier. Merely showing that \(df_n\to0\) would not produce the admissible vector \(f_0-x\), because there might be no candidate limit \(x\).
146 
147 \subsection{When two repair paths select the same point}
148 \begin{proposition}[Path independence requires a comparison]\label{KS74}
149 Let \(f_n\to x\) and \(g_n\to y\) in a metric space. If
150 \[
151  \dist(f_n,g_n)\le\beta_n,\qquad\beta_n\to0,
152 \]
153 then \(x=y\). In contrast, exactness and preservation of a nonzero observable alone do not force a unique answer.
154 \end{proposition}
155 \begin{proof}
156 For each \(n\), the triangle inequality gives
157 \[
158  \dist(x,y)\le\dist(x,f_n)+\dist(f_n,g_n)+\dist(g_n,y)
159  \le\dist(x,f_n)+\beta_n+\dist(g_n,y).
160 \]
161 Given any \(\varepsilon>0\), choose \(n\) large enough that the three terms on the right are each below \(\varepsilon/3\). Thus \(\dist(x,y)\le\varepsilon\) for every positive \(\varepsilon\), so the distance is zero. The metric separation axiom gives \(x=y\). In a pseudometric space only zero distance would follow without an additional separation hypothesis.
162 
163 For the negative assertion let \(E=\R\times(\R\times\R)\), \(d(s,u,r)=r\), and \(p(s,u,r)=s\). Both \(x=(1,0,0)\) and \(y=(1,1,0)\) have zero defect and protected value one, but their middle coordinates differ. They are distinct points of the same feasible fiber. Constant sequences at those points meet exactness and preservation but not a vanishing comparison bound, so these conditions alone cannot imply path independence.
164 \end{proof}
165 \source{KS74}
166 A stronger worked example uses actual convergent paths from the \emph{same} initial input:
167 \[
168  f_n^{(a)}=(s,a(1-2^{-n}),2^{-n})\qquad(a\in\R).
169 \]
170 All paths begin at \((s,0,1)\), keep \(p=s\), and have residual \(2^{-n}\). Their maximum-norm movements are \(\frac12\max\{|a|,1\}2^{-n}\), so every path satisfies geometric exactification. Their limits are \((s,a,0)\). For different \(a,b\),
171 \[
172  \dist(f_n^{(a)},f_n^{(b)})=|a-b|(1-2^{-n})\longrightarrow|a-b|,
173 \]
174 not zero. The missing uniqueness hypothesis is visible in the exact direction along which the paths separate.
175 
176 \plate{27_path_limits}{All paths start at the same candidate, have the same residual at each step, and preserve the same coordinate $s$. They converge to different exact points because the coordinate $u$ is unprotected. A vanishing residual is not a path-comparison budget.}{fig:path-limits}
177 
178 \subsection{Changing the sampling schedule does not preserve a rate}
179 A residual profile \(\eta_n\) indexed by resolution is not an iteration trajectory. Nevertheless a useful convergence property is invariant under a cofinal schedule when the profile is monotone.
180 
181 \begin{proposition}[Cofinal sampling of a decreasing profile]\label{KS75}
182 Suppose \(\eta_n\ge0\) is nonincreasing and \(\phi:\N\to\N\) is monotone and unbounded. Then
183 \[
184  \eta_n\to0\quad\Longleftrightarrow\quad\eta_{\phi(n)}\to0.
185 \]
186 Without the monotonicity of \(\eta\), the reverse implication can fail even for a cofinal schedule.
187 \end{proposition}
188 \begin{proof}
189 The schedule tends to infinity, so convergence of the full profile implies convergence of the sampled one. Conversely, choose a sampled value below any prescribed \(\varepsilon>0\). Every later index of the original profile has no larger value, by monotonicity. This proves convergence of the full profile.
190 
191 For failure without monotonicity, take \(\eta_{2n}=0\), \(\eta_{2n+1}=1\), and \(\phi(n)=2n\). All sampled values vanish, but infinitely many original values equal one.
192 \end{proof}
193 \source{KS75}
194 Even in the decreasing case, the numerical rate changes. The profile \(\eta_n=1/(n+1)\), sampled at \(\phi(n)=2^n-1\), becomes \(2^{-n}\). This is exponential decay in the \emph{number of observations}, not a proof of a faster correction process at the original resolution scale.
195 
196 \subsection{Three hypotheses that cannot simply be dropped}
197 \textbf{Summable movement.} In maximum-norm \(\R^3\), let
198 \[
199  f_n=(s,H_n,(n+1)^{-1}),\qquad
200  H_n=\sum_{j=1}^{n}\frac1j.
201 \]
202 The first coordinate is protected and the residual is the last coordinate. It tends to zero with the same ratios as the polynomial example. The movement equals \(1/(n+1)\), however, and the unprotected coordinate diverges. Here \(C_n=1\), so \(\sum C_nQ_n\) is the divergent harmonic series. Residual decay and semantic preservation hold, but exactification of this path fails.
203 
204 \textbf{Completeness.} In the metric space \((0,\infty)\), take \(f_n=1/(n+1)\) and \(\omega(x)=x\). The steps are summable and the residual tends to zero, but there is no limit in this space. Completing the space adds the missing zero. It is essential to say whether that completion belongs to the intended model.
205 
206 \textbf{Lower semicontinuity.} In \(\R\), define \(\omega(0)=1\) and \(\omega(x)=0\) for \(x\ne0\). The same positive sequence converges with summable movement and residual zero at every step, yet its limit is not exact. The zero sublevel set is not closed. This example isolates why convergence of measurements along a sequence must interact correctly with the ambient topology.

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