Sequential repetition amplifies completeness and soundness gaps

Fix $p$ with $\max \{s,1-c,0\}<p<1\mathrm{/}2$. On a yes-input, use one honest strategy afresh in each repetition, ignoring previous copies. Let $Y_i$ indicate rejection in copy $i$. On a no-input, fix any adaptive cheating strategy and instead let $Y_i$ indicate acceptance. In both cases, conditional on the full history of earlier copies, $\mathrm{Pr}(Y_i=1\mid \text{history})\le p$: on yes-inputs this is completeness, and on no-inputs the previous history fixes a legitimate strategy for the next fresh-coin copy, so soundness applies. The verifier is reset between copies.

Put $t=1\mathrm{/}(2p)>1$. If the conditional success probability of $Y_i$ is $u\le p$, then the conditional average of $t^{2Y_i}$ is $1-u+u{t}^2\le 1-p+p{t}^2$. Grouping finite sums by complete previous-copy histories and inducting on $j$ therefore gives $\mathbb{E}[t^{2{\sum }_{i=1}^j{Y}_i}]\le (1-p+p{t}^2{)}^j$, starting with expectation one at $j=0$. This argument allows dependence between copies.

On ${\sum }_{i=1}^k{Y}_i\ge k\mathrm{/}2$ one has $t^{2{\sum }_i{Y}_i}\ge t^k$. Summing this pointwise bound over the finite coin space gives $\mathrm{Pr}({\sum }_i{Y}_i\ge k\mathrm{/}2)\le q^k$, where $q=(1-p+p{t}^2)\mathrm{/}t=1\mathrm{/}2+2p(1-p)=1-2(1\mathrm{/}2-p{)}^2$ satisfies $0<q<1$. For odd $k$, either a majority rejection on a yes-input or a majority acceptance on a no-input is contained in this event.

Write $d=q^{-1}-1>0$. Induction using $(1+d{)}^{k+1}\ge (1+kd)(1+d)\ge 1+(k+1)d$ gives $q^k\le 1\mathrm{/}(1+kd)$. Choose an odd integer $k$ making this at most $\epsilon$. For fixed $c,s,\epsilon$ it is a constant independent of input length, so the repeated verifier retains polynomial time, communication and randomness bounds. The yes-input honest strategy and the uniform bound against every no-input cheating strategy give the two asserted errors.