Van der Corput oscillatory integral estimates in one dimension

First derivative on an interval. Write $F=e^{2\pi i\lambda \phi }$ and $g=1\mathrm{/}{\phi }^{\mathrm{\prime }}$ on $[u,v]\subset J$. The continuous derivative has constant sign; $g$ is monotone of that sign and $\mathrm{\mid }g\mathrm{\mid }\le {\lambda }_1^{-1}$. Stieltjes integration by parts gives ${\int }_u^{v}F=(2\pi i\lambda {)}^{-1}(g(v)F(v)-g(u)F(u)-{\int }_u^{v}Fdg)$. Its modulus is at most $(2\pi \lambda {)}^{-1}(\mathrm{\mid }g(v)\mathrm{\mid }+\mathrm{\mid }g(u)\mathrm{\mid }+\mathrm{\mid }g(v)-g(u)\mathrm{\mid })\le (\pi \lambda {\lambda }_1{)}^{-1}$. This stronger bound implies the displayed pure-phase estimate in (a), and also bounds the primitive on every subinterval. Endpoint inclusion does not change the integral.

First-derivative amplitude bound. On each component $(u,v)$ of $\{a\mathrm{\ne }0\}$, $a$ vanishes at the endpoints and the hypotheses on $\mathrm{supp}a$ give the estimate of step 1.1 on every subinterval. Thus $P(x)={\int }_u^{x}F$ has modulus at most $(\pi \lambda {\lambda }_1{)}^{-1}$, and ordinary integration by parts gives ${\int }_u^{v}Fa=-{\int }_u^{v}P{a}^{\mathrm{\prime }}$. Summing gives $\mathrm{\mid }\int Fa\mathrm{\mid }\le (\pi \lambda {\lambda }_1{)}^{-1}\mathrm{\parallel }{a}^{\mathrm{\prime }}{\mathrm{\parallel }}_1$. The components are canonically at most countable by assigning to each its first rational in a fixed enumeration; the sum is justified by $\int \mathrm{\mid }a\mathrm{\mid }<\mathrm{\infty }$ and the sum of ${\int }_u^v\mathrm{\mid }{a}^{\mathrm{\prime }}\mathrm{\mid }$ being at most $\mathrm{\parallel }{a}^{\mathrm{\prime }}{\mathrm{\parallel }}_1$. No reciprocal of ${\phi }^{\mathrm{\prime }}$ is used in gaps outside the support.

Higher-derivative interval estimate. Suppose $\mathrm{\mid }{\phi }^{(k)}\mathrm{\mid }\ge \gamma >0$ throughout a bounded interval $J$, $k\ge 2$. We prove that every subinterval has pure-phase integral bounded by $C_k(\lambda \gamma {)}^{-1\mathrm{/}k}$, independently of its length. The derivative ${\phi }^{(k)}$ has constant sign, so ${\phi }^{(k-1)}$ is monotone. For $\tau >0$, the set $\mathrm{\mid }{\phi }^{(k-1)}\mathrm{\mid }\le \tau$ is an interval of length at most $2\tau \mathrm{/}\gamma$, by the fundamental theorem. Its complement has at most two intervals on which $\mathrm{\mid }{\phi }^{(k-1)}\mathrm{\mid }\ge \tau$. For $k=2$, step 1.1 applies there because ${\phi }^{\mathrm{\prime }}$ is monotone. For $k>2$, use induction with lower bound $\tau$. The resulting bound is $2\tau \mathrm{/}\gamma +2C_{k-1}(\lambda \tau {)}^{-1\mathrm{/}(k-1)}$. Set $\tau ={\gamma }^{(k-1)\mathrm{/}k}{\lambda }^{-1\mathrm{/}k}$ to obtain the asserted bound. The same reasoning on any subinterval proves the primitive bound required below.

Higher-derivative amplitude bound. On every component $(u,v)$ of $\{a\mathrm{\ne }0\}$, the hypothesis $\mathrm{\mid }{\phi }^{(k)}\mathrm{\mid }\ge 1$ holds throughout that interval. Step 2.2 with $\gamma =1$ gives a primitive $P(x)={\int }_u^{x}F$ bounded by $C_k{\lambda }^{-1\mathrm{/}k}$. Since $a(u)=a(v)=0$, integration by parts yields $\mathrm{\mid }{\int }_u^{v}Fa\mathrm{\mid }\le C_k{\lambda }^{-1\mathrm{/}k}{\int }_u^v\mathrm{\mid }{a}^{\mathrm{\prime }}\mathrm{\mid }$. Sum over the canonically countable components as in step 2.1 to obtain $\mathrm{\mid }\int Fa\mathrm{\mid }\le C_k{\lambda }^{-1\mathrm{/}k}\mathrm{\parallel }{a}^{\mathrm{\prime }}{\mathrm{\parallel }}_1$. This stronger estimate implies (b) with the stated constant dependence, even for disconnected support; no lower derivative bound in its gaps is assumed.

Quadratic phase. For $\phi =\pm x^2\mathrm{/}2$, the second derivative has modulus one on every interval. Apply step 2.2 directly on one interval containing $\mathrm{supp}a$ and integrate against its bounded primitive. This gives $\mathrm{\mid }\int e^{\pm \pi i\lambda x^2}a\mathrm{\mid }\le C_0{\lambda }^{-1\mathrm{/}2}\mathrm{\parallel }{a}^{\mathrm{\prime }}{\mathrm{\parallel }}_1$, which implies the stated Fresnel estimate with $\mathrm{\parallel }a{\mathrm{\parallel }}_{\mathrm{\infty }}+\mathrm{\parallel }{a}^{\mathrm{\prime }}{\mathrm{\parallel }}_1$. No scale-dependent cutoff derivative enters.