---
tag: QDXR02419M
kind: statement
role: lemma
name: Van der Corput lemma
msc: [42B20, 42A38]
uses: [D5HZKE8E1A, 0V33BNKZ9S]
about: [D5HZKE8E1A]
sources:
  - url: https://en.wikipedia.org/wiki/Van_der_Corput_lemma_(harmonic_analysis)
same_as:
  wikidata: Q3229353
  wikipedia: Van der Corput lemma (harmonic analysis)
---

## Statement

Let $k \ge 1$ be an integer.
There is a constant $c_k$, depending only on $k$, with the following property.
Let $a < b$ be real numbers, let $\phi \colon [a, b] \to \R$ be a [[smooth function|0V33BNKZ9S]], and assume that $\abs{\phi^{(k)}(x)} \ge 1$ for all $x \in [a, b]$.
If $k = 1$, assume also that $\phi'$ is monotonic on $[a, b]$.
Then, for every $\lambda > 0$,
$$\Big| \int_a^b e^{i\lambda\phi(x)} \, dx \Big| \le c_k \lambda^{-1/k}.$$

## Remarks

The point is that the bound does not depend on $a$, $b$ or $\phi$.
For $k = 1$, write $e^{i\lambda\phi} = (i\lambda\phi')^{-1} \frac{d}{dx} e^{i\lambda\phi}$ and integrate by parts; monotonicity of $\phi'$ makes the integral of $\abs{(1/\phi')'}$ telescope, and one gets $c_1 = 3$.
For $k \ge 2$, use induction on $k$.
On the set where $\abs{\phi^{(k-1)}} \ge \lambda^{-1/k}$, apply the case $k - 1$ after rescaling.
This set misses at most an interval of length about $\lambda^{-1/k}$, where the integrand is bounded by $1$.

Example: for $\phi(x) = x^2/2$ and $k = 2$, the lemma gives $\abs{\int_a^b e^{i\lambda x^2/2} \, dx} \le c_2 \lambda^{-1/2}$ uniformly in $a, b$; the Fresnel integral $\int_{-\infty}^{\infty} e^{i\lambda x^2/2} \, dx = \sqrt{2\pi/\lambda} \, e^{i\pi/4}$ shows that the power $\lambda^{-1/2}$ cannot be improved.

For $k = 1$ the monotonicity of $\phi'$ is needed.
Without it, no bound $c_1\lambda^{-1}$ holds uniformly in $\phi$ and in the length of the interval.

A corollary adds an amplitude: if $\psi$ is smooth on $[a, b]$, then $\abs{\int_a^b e^{i\lambda\phi}\psi} \le c_k\lambda^{-1/k} \big(\abs{\psi(b)} + \int_a^b \abs{\psi'}\big)$.

## Explanations

### L1 human

An integral of $e^{i\lambda\phi}$ is small when the phase $\phi$ changes fast, because the integrand spins around the circle and cancels itself.
The lemma measures this: if some derivative of $\phi$ of order $k$ is at least $1$, the integral is at most a constant times $\lambda^{-1/k}$.
Higher $k$ means a flatter possible phase and a weaker decay.

### L2 human

This is the basic one-dimensional tool for [[oscillatory integrals|D5HZKE8E1A]].
It gives decay rates for Fourier transforms of measures on curves, such as $\abs{\hat\sigma(\xi)} \lesssim \abs{\xi}^{-1/2}$ for arc length on a curve with nonzero curvature.
In higher dimensions there is no equally clean analogue, and one uses stationary phase or sublevel set estimates instead.