BQP is unchanged by polynomially small bounded error

If the promise problem is in BQP, take $\alpha =2\mathrm{/}3$, $\beta =1\mathrm{/}3$, and $s=3$. Use the all-length rational target $\widehat{\delta }(n)=2^{-q(n)}$ when $q(n)\ge 2$ and $\widehat{\delta }(n)=1\mathrm{/}4$ otherwise, and apply [F2]. Retain the amplified circuit and predicate where $q(n)\ge 2$, and retain the original circuit and predicate at the remaining lengths; this polynomial-time length test preserves full-list uniformity. The rational values have polynomial bit length because $q$ is polynomially bounded, and the gap is $1\mathrm{/}3$, so this gives the second family at those lengths. At any length with $q(n)=0$, the target bounds are $p\ge 0$ on YES and $p\le 1$ on NO; with $q(n)=1$, they are $p\ge 1\mathrm{/}2$ and $p\le 1\mathrm{/}2$. The original $2\mathrm{/}3,1\mathrm{/}3$ bounds imply each of these finite exceptional cases.

Conversely, choose $N$ such that $q(n)\ge 2$ for all $n\ge N$. At those lengths, $1-2^{-q(n)}\ge 3\mathrm{/}4\ge 2\mathrm{/}3$ and $2^{-q(n)}\le 1\mathrm{/}4\le 1\mathrm{/}3$, so the high-confidence family meets the BQP bounds. There are finitely many binary strings of length less than $N$. Change the total classical predicate on this finite set to accept each member of $Y$ and reject each member of $N$, while leaving it unchanged off the promise and on longer strings. The finite table is a fixed finite description and adds only constant time and space. The circuit generator is unchanged, so the modified family remains full-list uniform and satisfies [F1] on every promised input.

For the more general thresholds, choose a fixed $N$ beyond both the finitely many gap exceptions and the lengths with $q(n)<2$. To satisfy the all-length input conditions of [F2], define polynomial-time rational functions $\widehat{\alpha },\widehat{\beta }$ to equal $2\mathrm{/}3,1\mathrm{/}3$ below $N$ and $\alpha ,\beta$ at and above $N$; set $\widehat{s}(n)=s(n)+3$; and define $\widehat{\delta }(n)=1\mathrm{/}4$ below $N$ and $2^{-q(n)}$ at and above $N$. These functions meet the endpoint, positive-gap and $0<\widehat{\delta }\le 1\mathrm{/}2$ conditions at every length. Before amplification, change the base predicate below $N$ to accept exactly the YES members of the finite set of short inputs and reject its NO members, independently of the measured string; leave it unchanged at longer lengths. This is legal because the predicate API is $A_n(x,z)$ and explicitly receives $x$. The fixed finite table gives base probabilities $1$ on short YES inputs and $0$ on short NO inputs, so the corrected base family meets $\widehat{\alpha },\widehat{\beta }$ at every length. Apply [F2] to this corrected base family with these functions. For $n\ge N$ the resulting circuits have the desired error and $\mathrm{\ell }(2^{q(n)})=q(n)+1$, so $r(n)=2(s(n)+3{)}^2(q(n)+2)$ is polynomial. At the finitely many lengths below $N$, replace the amplified predicate by a finite table that answers each promised input exactly; this table is a fixed constant-size addition to a polynomial-time predicate. The resulting family is full-list uniform and meets the target on every promised input. The finite table is an existence argument for this fixed promise problem; no algorithm for recovering its short YES membership from an arbitrary family description is asserted.