12. A bounded infinite-resolution bridge
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Finite reconstruction identities do not alone prove convergence of an unbounded or infinite-dimensional model. A useful bounded extension can nevertheless be established directly.
Proposition 13: Strong fixed-horizon refinement limit
Let \(X\) be a Hilbert space, \(L:X\to X\) bounded, and \(P_n\) orthogonal projections converging strongly to \(I\). Put \(L_n=P_nLP_n\). For every fixed \(x\in X\) and every fixed nonnegative integer \(k\),
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Proof
Since \(\|P_n\|\leq1\),
Also \(\|L_n\|\leq\|L\|\). Induct on \(k\), using
The first term tends to zero by the uniform norm bound and induction, and the second by strong convergence on the fixed vector \(L^kx\). \(\square\)
If estimated operators \(\widehat L_n\) are uniformly bounded and \((\widehat L_n-L_n)x\to0\) for every fixed \(x\), the same argument gives fixed-horizon convergence for \(\widehat L_n\). These are explicit assumptions, not conclusions of the finite learning procedure.
This result is strong convergence, not operator-norm convergence and not a bound uniform in time. On \(\ell^2\), finite coordinate projections converge strongly to \(I\) while \(\|I-P_n\|=1\) for every \(n\). Unbounded operators require domain control. Adaptive data-dependent partitions and fitted coefficients require additional consistency arguments. These remain separate mathematical tasks.