= Solution
Fix an integer radius $m\geq0$ and put $r=\lfloor m/2\rfloor$. Starting with $A_x$, scan the $2r$ layers of $K^r$ from right to left. In each layer retain precisely those bond <orthogonal projections> whose supports meet the current support; multiply them onto the current operator and enlarge that support to include their bonds. Discard every other projection: it commutes through the current operator and acts as the identity on $|\psi_0\rangle$, by <frustration freeness>.
If $L_r$ denotes the ordered product of the retained projections, define
$$
\boxed{A_{x(m)}=L_r A_x.}
$$
Each layer expands the support by at most one lattice spacing. Therefore the <local projector cone in a frustration-free chain> lies within radius $2r\leq m$ of $x$, and
$$
K^rA_x|\psi_0\rangle=A_{x(m)}|\psi_0\rangle,\qquad
\|A_{x(m)}\|\leq\|A_x\|,\qquad
\langle\psi_0|A_{x(m)}=\langle\psi_0|A_x.
$$
The last equality holds because every retained projection also fixes the ground-state bra.
For $r\geq1$, the stronger estimate from (i) gives
$$
\|(P_0A_x-A_{x(m)})|\psi_0\rangle\|
\leq\|A_x\|q^{2r-1}
\leq q^{-2}\|A_x\|q^m.
$$
For $r=0$, choose $A_{x(m)}=A_x$; the error is at most $\|A_x\|$, so the same bound holds. Hence
$$
\boxed{\|(P_0A_x-A_{x(m)})|\psi_0\rangle\|
\leq q^{-2}\|A_x\|e^{-\alpha m}.}
$$
Here $\alpha$ is exactly the value in (i), and the prefactor is independent of $m$ and chain length. Counting $K^r$ as $r$ rather than $2r$ layers would give an incorrect support claim; the <powers of a product of two orthogonal projections> are what avoid a lost factor of two in the decay rate.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-67-projector-cone.png]
{title=Retained bond projectors in the six-layer cone of a single-site operator under three applications of K}
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