# Bound of an oscillatory integral from Stein's Harmonic Analysis book

On Stein's Harmonic Analysis Real-variable methods, orthogonality, and oscillatory integrals'' (5.13, page 363) there is the following statement. Let $\phi$ be a real homogeneous polynomial on $\mathbb{R}^n$ of degree $k \geq 2$ that is non-degenerate, in the sense that $\det \left( \frac{\partial^2 \phi}{\partial x_i \partial x_j} \right) \not = 0$ whenever $\mathbf{x} \not = 0$. Then if $\psi \in C_0^{\infty}$, $$\int_{\mathbb{R}^n} e^{i (\lambda \phi(\mathbf{x}) + \boldsymbol{\xi} \cdot \mathbf{x} )} \psi(\mathbf{x}) d \mathbf{x} = O((|\lambda| + |\boldsymbol{\xi}|)^{-n/k}).$$

The statement is without proof and I was wondering about how I can prove this. As I am not too familiar with this area, I was wondering if someone could provide a proof or any assistance would be appreciated. Thank you very much.

• Looks roughly like en.wikipedia.org/wiki/Van_der_Corput_lemma_(harmonic_analysis) But isn't this topic (oscillatory integrals) discussed at great length in Stein's book, so just try and see what you can do with these methods. – Christian Remling Dec 3 '17 at 17:34
• @ChristianRemling Right, I was guessing it should be possible using some of the methods in the book. So I was hoping maybe for more specific tip on exactly which result to use (or the proof). – Johnny T. Dec 3 '17 at 18:01
• Have you tried reading the preceding results to see if those help? Do you have a copy of the book in its entirety? – Yemon Choi Dec 4 '17 at 3:46

Something which may help to observe is that you can immediately restrict to the subset of $\mathbb{R}^n$ given by $\text{supp}(\psi)\cap \{\lambda\nabla \phi(x)+|\xi|^2=0\}$, by appealing to the non-stationary phase lemma. A version of this result is as follows: Given an oscillatory integral of the form $I_{\psi,\varphi}(\omega)=\int_{\mathbb{R}^d} \psi(x)e^{i\omega\varphi(x)}dx$, satisfying that $\nabla\varphi(x)\neq 0$ on $\text{supp}(\psi)$, then we have $$I_{\psi,\phi}(\omega)=\int_{\mathbb{R}^d} \psi(x)e^{i\omega\varphi(x)}dx=\mathcal{O}(\omega^{-N})\hspace{5mm}\forall N>0.$$ To apply this lemma to your problem, first set $\varphi(x)=\lambda\phi(x)+\xi\cdot x$. Then $\nabla \varphi(x)=\lambda\nabla \phi(x)+\xi$. So away from this set $\text{supp}(\psi)\cap \{\lambda\nabla \phi(x)+\xi=0\}$ you already have much better decay than you are trying to prove above.
The hypothesis that $\text{det}\left(\frac{\partial \phi}{\partial x_i\partial x_j}\right)\neq 0$ implies that the equation $\{\lambda\nabla \phi(x)+\xi=0\}$ has isolated solutions, and intersecting it with the compact support of $\psi$ will be finite. So, it's these points which should contribute to the (less than rapid) decay you have in this statement. It looks like Stein gives a brief argument for why this should be the case at the top of page 364, although these are modeled on his proofs for decay of the fourier transform for surface measures of finite type.
• btw what is meant by $|\xi|$ here? – Johnny T. Dec 11 '17 at 10:59
• This is the norm of the vector $\xi$, $|\xi|^2=\xi_1^2+\ldots \xi_n^2$. Answering this makes me realize there was a typo in my formula for $\nabla \varphi(x)$. I have updated my answer to reflect this. – Hadrian Quan Dec 11 '17 at 18:32