[proofplan] We prove the two transformations separately. For the A-process, we apply van der Corput differencing, estimate each differenced correlation sum by the assumed exponent-pair bound with derivative parameter comparable to $hZ/N$, and choose the differencing length $H$ to balance the diagonal term against the correlation term. For the B-process, we use the stationary phase duality built into the admissible class, apply the assumed exponent-pair estimate to the dual datum whose length and derivative parameters are interchanged, and then collect the square-root factor. [/proofplan]

[step:Apply van der Corput differencing to reduce the A-process to correlation sums]Fix $\varepsilon>0$. Choose a number $\eta>0$ small enough that every exponent loss denoted $O_{\kappa,\lambda}(\eta)$ below is at most $\varepsilon$. Let $(I,U,N,Z,f)\in\mathcal A$, and define \begin{align*} S:=\sum_{n\in I}e(f(n)). \end{align*} For an integer $H$ with $1\le H\le N$, the A-process stability hypothesis, equivalently the van der Corput differencing inequality stated in the hypothesis, gives \begin{align*} |S|^2\lesssim \frac{N^2}{H}+\frac{N}{H}\sum_{1\le h<H}\left|\sum_{n\in I_h}e(\Delta_h f(n))\right|, \end{align*} where the implicit constant depends only on the admissibility constants. For each $h$, the datum associated to $\Delta_h f$ has length parameter at most a constant multiple of $N$ and derivative parameter comparable to $hZ/N$. Applying the assumed exponent-pair estimate to this admissible datum gives \begin{align*} \left|\sum_{n\in I_h}e(\Delta_h f(n))\right|\lesssim_{\eta} \left(\frac{hZ}{N}\right)^{\kappa+\eta}N^{\lambda+\eta}. \end{align*} Since $\kappa+\eta\ge 0$, the elementary power-sum bound \begin{align*} \sum_{1\le h<H}h^{\kappa+\eta}\lesssim_{\kappa,\eta} H^{\kappa+\eta+1} \end{align*} yields \begin{align*} |S|^2\lesssim_{\eta} \frac{N^2}{H}+Z^{\kappa+\eta}N^{\lambda-\kappa+1+O(\eta)}H^{\kappa+O(\eta)}. \end{align*}[/step]

[guided]The purpose of the A-process is to replace the original sum by sums with differenced phases. We begin by fixing an admissible datum $(I,U,N,Z,f)\in\mathcal A$ and writing \begin{align*} S:=\sum_{n\in I}e(f(n)). \end{align*} For an integer $H$ satisfying $1\le H\le N$, the A-process hypothesis says precisely that van der Corput differencing is available inside the class. Thus, using the differencing inequality from [citetheorem:9058], we have \begin{align*} |S|^2\lesssim \frac{N^2}{H}+\frac{N}{H}\sum_{1\le h<H}\left|\sum_{n\in I_h}e(\Delta_h f(n))\right|. \end{align*} Here \begin{align*} I_h:=\{n\in I:n+h\in I\} \end{align*} and \begin{align*} \Delta_h f:U_h\to\mathbb R,\qquad x\mapsto f(x+h)-f(x). \end{align*} The first term $N^2/H$ is the diagonal cost of differencing; the second term contains the genuine oscillation of the shifted correlations. Now we estimate each correlation sum using the hypothesis that $(\kappa,\lambda)$ is an exponent pair. The A-process stability assumption says that the datum associated to $\Delta_h f$ has length parameter at most a constant multiple of $N$ and derivative parameter comparable to $hZ/N$. Therefore, for the small auxiliary loss $\eta>0$, \begin{align*} \left|\sum_{n\in I_h}e(\Delta_h f(n))\right|\lesssim_{\eta} \left(\frac{hZ}{N}\right)^{\kappa+\eta}N^{\lambda+\eta}. \end{align*} Substituting this into the differencing inequality gives \begin{align*} |S|^2\lesssim_{\eta}\frac{N^2}{H}+\frac{N}{H}Z^{\kappa+\eta}N^{\lambda-\kappa+O(\eta)}\sum_{1\le h<H}h^{\kappa+\eta}. \end{align*} Because $\kappa+\eta\ge 0$, the sum of powers is bounded by a constant depending only on $\kappa$ and $\eta$ times $H^{\kappa+\eta+1}$. Hence \begin{align*} |S|^2\lesssim_{\eta} \frac{N^2}{H}+Z^{\kappa+\eta}N^{\lambda-\kappa+1+O(\eta)}H^{\kappa+O(\eta)}. \end{align*} This is the key inequality: it has one term decreasing in $H$ and one term increasing in $H$, so the next step is to choose $H$ near the balancing point.[/guided]

[step:Choose the differencing length to obtain the A-transform exponents]Define the balancing parameter \begin{align*} H_0:=\left(N^{\kappa+1-\lambda}Z^{-\kappa}\right)^{1/(\kappa+1)}. \end{align*} If $H_0<2$, then either $Z^\kappa\gtrsim N^{\kappa+1-\lambda}$ or, in the degenerate case $\kappa=0$ and $\lambda<1$, the condition $H_0<2$ forces $N$ to be bounded because then $H_0=N^{1-\lambda}$. In either situation the elementary bound $|S|\le |I|\lesssim N$ gives \begin{align*} |S|\lesssim Z^{\kappa/(2\kappa+2)}N^{(\kappa+\lambda+1)/(2\kappa+2)}. \end{align*} Since $N,Z\ge 1$, this is stronger than the required estimate with an $\varepsilon$-loss. Assume now that $H_0\ge 2$. Since $Z^{-\kappa}\le 1$ and $1\le N^\lambda$, we have \begin{align*} H_0^{\kappa+1}=N^{\kappa+1-\lambda}Z^{-\kappa}\le N^{\kappa+1}. \end{align*} Thus $H_0\le N$. Choose $H\in\mathbb N$ with $1\le H\le N$ and $H\asymp H_0$. Substituting this choice into the estimate from the previous step balances the two main terms and gives \begin{align*} |S|^2\lesssim_{\eta} Z^{\kappa/(\kappa+1)+O(\eta)}N^{(\kappa+\lambda+1)/(\kappa+1)+O(\eta)}. \end{align*} Taking square roots gives \begin{align*} |S|\lesssim_{\eta} Z^{\kappa/(2\kappa+2)+O(\eta)}N^{(\kappa+\lambda+1)/(2\kappa+2)+O(\eta)}. \end{align*} By the choice of $\eta$ and by $N,Z\ge 1$, \begin{align*} |S|\le C_\varepsilon Z^{\kappa/(2\kappa+2)+\varepsilon}N^{(\kappa+\lambda+1)/(2\kappa+2)+\varepsilon} \end{align*} for a constant $C_\varepsilon>0$ depending on $\varepsilon$, $\kappa$, $\lambda$, and the admissibility constants, but not on the datum. Hence $A(\kappa,\lambda)$ is an exponent pair for $\mathcal A$.[/step]

[guided]The parameter $H$ controls the tradeoff created by differencing. The estimate from the previous step has a decreasing term $N^2/H$ and an increasing term of size $Z^{\kappa+\eta}N^{\lambda-\kappa+1+O(\eta)}H^{\kappa+O(\eta)}$, so the natural balancing scale is obtained by ignoring the harmless $\eta$-losses and solving \begin{align*} \frac{N^2}{H}=Z^\kappa N^{\lambda-\kappa+1}H^\kappa. \end{align*} This gives \begin{align*} H_0:=\left(N^{\kappa+1-\lambda}Z^{-\kappa}\right)^{1/(\kappa+1)}. \end{align*} If $H_0<2$, then $Z^\kappa\gtrsim N^{\kappa+1-\lambda}$. The elementary estimate $|S|\le |I|\lesssim N$ then implies \begin{align*} |S|\lesssim Z^{\kappa/(2\kappa+2)}N^{(\kappa+\lambda+1)/(2\kappa+2)}. \end{align*} Since $N,Z\ge 1$, this is stronger than the required estimate after inserting the allowed $\varepsilon$-loss. Now suppose $H_0\ge 2$. We must check that the balancing scale is allowed in the differencing inequality, which requires $H\le N$. Since $Z^{-\kappa}\le 1$ and $1\le N^\lambda$, we have \begin{align*} H_0^{\kappa+1}=N^{\kappa+1-\lambda}Z^{-\kappa}\le N^{\kappa+1}. \end{align*} Therefore $H_0\le N$, so we may choose an integer $H$ with $1\le H\le N$ and $H\asymp H_0$. Substituting this choice into the A-process estimate balances the two main terms and yields \begin{align*} |S|^2\lesssim_{\eta} Z^{\kappa/(\kappa+1)+O(\eta)}N^{(\kappa+\lambda+1)/(\kappa+1)+O(\eta)}. \end{align*} Taking square roots gives \begin{align*} |S|\lesssim_{\eta} Z^{\kappa/(2\kappa+2)+O(\eta)}N^{(\kappa+\lambda+1)/(2\kappa+2)+O(\eta)}. \end{align*} The auxiliary number $\eta$ was chosen so that these $O_{\kappa,\lambda}(\eta)$ exponent losses are at most the prescribed $\varepsilon$. Hence \begin{align*} |S|\le C_\varepsilon Z^{\kappa/(2\kappa+2)+\varepsilon}N^{(\kappa+\lambda+1)/(2\kappa+2)+\varepsilon}. \end{align*} This is exactly the exponent-pair estimate for $A(\kappa,\lambda)$.[/guided]

[step:Apply the B-process duality and estimate the dual sum]Let $(I,U,N,Z,f)\in\mathcal A$ and define \begin{align*} S:=\sum_{n\in I}e(f(n)). \end{align*} By the B-process stability hypothesis, equivalently the [one-dimensional stationary phase](/theorems/6985) transformation stated in the hypothesis in this admissible framework, there exists a dual admissible datum $(J,V,Z,N,g)\in\mathcal A$ such that \begin{align*} |S|\lesssim \left(\frac{N}{Z}\right)^{1/2}\left|\sum_{m\in J}e(g(m))\right|+E, \end{align*} where the endpoint error $E$ is dominated by the final B-process bound. Applying the assumed exponent-pair estimate to the dual datum, whose length parameter is $Z$ and derivative parameter is $N$, gives \begin{align*} \left|\sum_{m\in J}e(g(m))\right|\le C_\varepsilon N^{\kappa+\varepsilon}Z^{\lambda+\varepsilon}. \end{align*} Therefore \begin{align*} |S|\lesssim_{\varepsilon} \left(\frac{N}{Z}\right)^{1/2}N^{\kappa+\varepsilon}Z^{\lambda+\varepsilon}+E. \end{align*} By the endpoint-error domination assumption, \begin{align*} E\lesssim_{\varepsilon} Z^{\lambda-1/2+\varepsilon}N^{\kappa+1/2+\varepsilon}. \end{align*} Combining the two estimates yields, for some constant $C'_\varepsilon>0$ depending on $\varepsilon$, $\kappa$, $\lambda$, and the admissibility constants, \begin{align*} |S|\le C'_\varepsilon Z^{\lambda-1/2+\varepsilon}N^{\kappa+1/2+\varepsilon} \end{align*} for every admissible datum $(I,U,N,Z,f)\in\mathcal A$. Hence \begin{align*} B(\kappa,\lambda)=\left(\lambda-\frac{1}{2},\kappa+\frac{1}{2}\right) \end{align*} is an exponent pair for $\mathcal A$.[/step]

[guided]For the B-process, the admissibility hypothesis supplies the whole stationary phase reduction. Starting from an admissible datum $(I,U,N,Z,f)\in\mathcal A$, define \begin{align*} S:=\sum_{n\in I}e(f(n)). \end{align*} The B-process stability assumption, equivalently the one-dimensional stationary phase transformation in [citetheorem:9059] within this admissible framework, produces a dual admissible datum $(J,V,Z,N,g)\in\mathcal A$. The important point is that the length and derivative parameters are interchanged: the dual datum has length parameter $Z$ and derivative parameter $N$. The same assumption gives \begin{align*} |S|\lesssim \left(\frac{N}{Z}\right)^{1/2}\left|\sum_{m\in J}e(g(m))\right|+E, \end{align*} where $E$ denotes the endpoint contribution. We now apply the assumed exponent-pair estimate to the dual datum. Since $(J,V,Z,N,g)$ belongs to $\mathcal A$, the hypothesis for $(\kappa,\lambda)$ gives \begin{align*} \left|\sum_{m\in J}e(g(m))\right|\le C_\varepsilon N^{\kappa+\varepsilon}Z^{\lambda+\varepsilon}. \end{align*} Multiplying by the square-root stationary phase factor gives \begin{align*} \left(\frac{N}{Z}\right)^{1/2}\left|\sum_{m\in J}e(g(m))\right|\lesssim_\varepsilon Z^{\lambda-1/2+\varepsilon}N^{\kappa+1/2+\varepsilon}. \end{align*} The B-process hypothesis also states that the endpoint term is dominated by this same final bound: \begin{align*} E\lesssim_\varepsilon Z^{\lambda-1/2+\varepsilon}N^{\kappa+1/2+\varepsilon}. \end{align*} Combining the dual-sum estimate with the endpoint estimate yields \begin{align*} |S|\le C'_\varepsilon Z^{\lambda-1/2+\varepsilon}N^{\kappa+1/2+\varepsilon}. \end{align*} This is precisely the exponent-pair estimate with transformed pair $B(\kappa,\lambda)=(\lambda-1/2,\kappa+1/2)$.[/guided]

[step:Combine the two transformations] The preceding steps prove that, for every $\varepsilon>0$, every admissible datum in $\mathcal A$ satisfies the exponent-pair estimate with exponents \begin{align*} \left(\frac{\kappa}{2\kappa+2},\frac{\kappa+\lambda+1}{2\kappa+2}\right) \end{align*} and also with exponents \begin{align*} \left(\lambda-\frac{1}{2},\kappa+\frac{1}{2}\right). \end{align*} These are exactly $A(\kappa,\lambda)$ and $B(\kappa,\lambda)$, respectively. Thus both transformed pairs are exponent pairs for $\mathcal A$. [/step]