81. Why exact boundary preservation can have zero noise radius
On this page · Permalink · Source line 3241
An exact certificate can be correct while having no tolerance to unrestricted perturbations. This is a geometrical fact, not a failure of the certificate checker.
Proposition 72: Boundary fragility under mass preserving perturbations
Assume \(d\geq1\). Let \(F\) preserve \(\mathcal L_d\). Suppose a normalized input \(z=(1,v)\) has a boundary image \(Fz=(m,w)\) with \(m=\|w\|_2\). For every \(\epsilon>0\), there is a linear perturbation \(\Delta\) with \(\lambda\Delta=0\) and
such that \((F+\Delta)z\notin\mathcal L_d\). Here the base norm is \(\|(m,w)\|_{\mathcal L_d}=\max(|m|,\|w\|_2)\).
Read proof · Recorded Lean scope · Visual guide: Uncertainty on a curved domain · Permalink · Source line 3245
Proof
If \(w\ne0\), put \(e=w/\|w\|_2\). If \(w=0\), choose any spatial unit vector \(e\). Define \(\Delta=\epsilon(0,e)\lambda\). Its mass row vanishes and its value on every normalized input is \((0,\epsilon e)\), of base norm \(\epsilon\). If \(w\ne0\), the new spatial norm is \(m+\epsilon\). If \(w=0\), the original output mass is zero and the new spatial norm is \(\epsilon\). Either image lies outside the cone. \(\square\)
The positive spatial dimension is essential. When \(d=0\), the state is mass alone and \(\lambda\) is the identity on \(\mathbb R\). The condition \(\lambda\Delta=0\) forces \(\Delta=0\), so a perturbation of prescribed norm \(\epsilon>0\) cannot exist. This explicit boundary was missing from the earlier wording. All previously worked examples have two spatial coordinates and are unaffected.
For a continuous operator family, a constant spatial perturbation of this form preserves every integrated mass row and produces the same obstruction at an observation with a boundary image. It need not preserve the algebraic restrictions of a narrower uncertainty class. Perturbing a positive congruence factor and rebuilding its congruence, for example, is different from perturbing an arbitrary predictive matrix.
In the preceding exact continuous example, the direct operator at \(u=0\) is \(I/4\). Every normalized boundary state has a boundary image. There is therefore no positive unrestricted matrix-error radius on that domain, even though the original family is exactly valid. A robust theorem that silently treats its zero slack as positive would be false.