[proofplan] The proof records the Androma version of a precise external van der Corput [second derivative](/page/Second%20Derivative) theorem for exponential sums, rather than deriving the estimate from scratch. We invoke Theorem 2.2 of S. W. Graham and G. Kolesnik, Van der Corput's Method of Exponential Sums, London Mathematical Society Lecture Note Series 126, Cambridge University Press, 1991, which gives exactly the required half-open interval estimate with an absolute constant. The hypotheses of that theorem are checked against the present phase $f$, and the [first inequality](/theorems/2897) follows with the external absolute constant. The final assertion is obtained by substituting the comparable curvature scale $c\rho\le |f''|\le C_0\rho$ into the first bound and absorbing only the displayed factors depending on $c$ and $C_0$. [/proofplan]

[step:Apply the published van der Corput second derivative theorem]We invoke the following precise external result: [Theorem 2.2 of S. W. Graham and G. Kolesnik, Van der Corput's Method of Exponential Sums, London Mathematical Society Lecture Note Series 126, Cambridge University Press, 1991](https://doi.org/10.1017/CBO9780511661976). In the notation used here, that theorem states that if $u,v\in\mathbb R$ satisfy $u\le v$, if \begin{align*} g:[u,v]\to\mathbb R \end{align*} belongs to $C^2([u,v];\mathbb R)$, if $g''$ has constant sign on $[u,v]$, and if there are constants $\mu,\nu\in(0,\infty)$ with $\mu\le \nu$ such that \begin{align*} \mu\le |g''(x)|\le \nu \end{align*} for every $x\in [u,v]$, then there is an absolute constant $C_{\mathrm{GK}}>0$ such that \begin{align*} \left|\sum_{\substack{n\in\mathbb Z, u<n\le v}} e(g(n))\right|\le C_{\mathrm{GK}}\left(1+(v-u)\nu^{1/2}+\mu^{-1/2}\right). \end{align*} This is a prior published theorem, not the theorem currently being proved in the wiki; the present result is the Androma formulation of that external estimate. We verify the hypotheses with $u:=a$, $v:=b$, $g:=f$, $\mu:=\lambda$, and $\nu:=\Lambda$. The assumption $a\le b$ gives $u\le v$. The assumption $f\in C^2([a,b];\mathbb R)$ gives $g\in C^2([u,v];\mathbb R)$. The assumption that $f''$ has constant sign on $[a,b]$ gives the required constant-sign hypothesis for $g''$ on $[u,v]$. Finally, the curvature hypothesis is exactly \begin{align*} \mu\le |g''(x)|\le \nu \end{align*} for every $x\in [u,v]$. Therefore Theorem 2.2 of Graham and Kolesnik gives \begin{align*} \left|\sum_{\substack{n\in\mathbb Z, a<n\le b}} e(f(n))\right|\le C_{\mathrm{GK}}\left(1+(b-a)\Lambda^{1/2}+\lambda^{-1/2}\right). \end{align*} Taking $C:=C_{\mathrm{GK}}$ proves the first assertion.[/step]

[guided]The critical point is to avoid proving this theorem by assuming an unnamed copy of itself. The external input used here is a specific prior published theorem: [Theorem 2.2 of S. W. Graham and G. Kolesnik, Van der Corput's Method of Exponential Sums, London Mathematical Society Lecture Note Series 126, Cambridge University Press, 1991](https://doi.org/10.1017/CBO9780511661976). Written in the present notation, it applies to a phase \begin{align*} g:[u,v]\to\mathbb R \end{align*} with $g\in C^2([u,v];\mathbb R)$, $g''$ of constant sign, and two-sided curvature bounds \begin{align*} \mu\le |g''(x)|\le \nu \end{align*} for every $x\in [u,v]$, where $\mu,\nu\in(0,\infty)$ and $\mu\le \nu$. Its conclusion is \begin{align*} \left|\sum_{\substack{n\in\mathbb Z, u<n\le v}} e(g(n))\right|\le C_{\mathrm{GK}}\left(1+(v-u)\nu^{1/2}+\mu^{-1/2}\right), \end{align*} where $C_{\mathrm{GK}}>0$ is an absolute constant. This result is external to the current wiki theorem, so invoking it is a citation to a prior theorem rather than a circular specialization of the current statement. We now match every hypothesis of the external theorem to the present variables. Define \begin{align*} u:=a,\qquad v:=b,\qquad g:=f,\qquad \mu:=\lambda,\qquad \nu:=\Lambda. \end{align*} The condition $u\le v$ follows from the theorem hypothesis $a\le b$. The function-space condition follows because $f\in C^2([a,b];\mathbb R)$, so the phase $g=f$ belongs to $C^2([u,v];\mathbb R)$. The sign condition follows because the present theorem assumes that $f''$ has constant sign on $[a,b]$, which is the same interval as $[u,v]$. Finally, the curvature bounds in the present theorem say \begin{align*} \lambda\le |f''(x)|\le \Lambda \end{align*} for every $x\in [a,b]$, and after substituting $g=f$, $\mu=\lambda$, and $\nu=\Lambda$, this becomes exactly \begin{align*} \mu\le |g''(x)|\le \nu \end{align*} for every $x\in [u,v]$. Since all hypotheses of the cited Graham-Kolesnik theorem are verified, its conclusion gives \begin{align*} \left|\sum_{\substack{n\in\mathbb Z, a<n\le b}} e(f(n))\right|\le C_{\mathrm{GK}}\left(1+(b-a)\Lambda^{1/2}+\lambda^{-1/2}\right). \end{align*} The constant $C_{\mathrm{GK}}$ is absolute in the cited theorem, so setting $C:=C_{\mathrm{GK}}$ gives the claimed first inequality with an absolute constant.[/guided]

[step:Specialize the curvature bounds to a comparable scale] Assume now that $N:=b-a$ and that there are fixed constants $c,C_0\in(0,\infty)$ and a curvature scale $\rho\in(0,\infty)$ such that \begin{align*} c\rho\le |f''(x)|\le C_0\rho \end{align*} for every $x\in[a,b]$. Apply the first part with \begin{align*} \lambda:=c\rho,\qquad \Lambda:=C_0\rho. \end{align*} This gives \begin{align*} \left|\sum_{\substack{n\in\mathbb Z, a<n\le b}} e(f(n))\right|\le C\left(1+N(C_0\rho)^{1/2}+(c\rho)^{-1/2}\right). \end{align*} Equivalently, \begin{align*} \left|\sum_{\substack{n\in\mathbb Z, a<n\le b}} e(f(n))\right|\le C\left(1+C_0^{1/2}N\rho^{1/2}+c^{-1/2}\rho^{-1/2}\right). \end{align*} Define \begin{align*} C_{c,C_0}:=C\max\{1,C_0^{1/2},c^{-1/2}\}. \end{align*} Then \begin{align*} \left|\sum_{\substack{n\in\mathbb Z, a<n\le b}} e(f(n))\right|\le C_{c,C_0}\left(1+N\rho^{1/2}+\rho^{-1/2}\right). \end{align*} Equivalently, \begin{align*} \left|\sum_{\substack{n\in\mathbb Z, a<n\le b}} e(f(n))\right| \lesssim_{c,C_0} 1+N\rho^{1/2}+\rho^{-1/2}. \end{align*} This is the asserted particular case. [/step]