**Proof plan.** The idea is to construct a first-order differential operator $L$ satisfying $L(e^{i\lambda\varphi}) = e^{i\lambda\varphi}$ (Claim 1), then repeatedly integrate by parts: each application of $L$ transfers a derivative from the exponential onto the amplitude and introduces a factor of $\lambda^{-1}$. The non-stationary hypothesis $\nabla\varphi \neq 0$ on $\operatorname{supp}(a)$ ensures the operator $L$ has smooth, bounded coefficients. The formal transpose $L^t$ is computed explicitly (Claim 2) and shown by induction to satisfy $\|(L^t)^m a\|_{L^1} \leq C_m\lambda^{-m}$ (Claim 3). After $m$ integrations by parts, the oscillatory integral equals $\int e^{i\lambda\varphi}(L^t)^m a\,d\mathcal{L}^n$, from which the bound $|I(\lambda)| \leq C_m\lambda^{-m}$ follows by the triangle inequality.
**Step 1 (The [integration](/page/Integral)-by-parts operator).**
Define the vector field $V: \mathbb{R}^n \to \mathbb{R}^n$ by
\begin{align*}
V(y) := \frac{\nabla\varphi(y)}{|\nabla\varphi(y)|^2}, \qquad y \in \operatorname{supp}(a).
\end{align*}
Since $\nabla\varphi(y) \neq 0$ on $\operatorname{supp}(a)$ and $\varphi \in C^\infty$, the vector field $V$ is smooth on an open neighbourhood of the compact [set](/page/Set) $\operatorname{supp}(a)$, with $\|V\|_{L^\infty(\operatorname{supp}(a))} \leq (\inf_{\operatorname{supp}(a)}|\nabla\varphi|)^{-1} < \infty$.
Define the first-order differential operator $L$ acting on smooth [functions](/page/Function) $u: \mathbb{R}^n \to \mathbb{C}$ by
\begin{align*}
Lu := \frac{1}{i\lambda}\,V \cdot \nabla u = \frac{1}{i\lambda}\sum_{j=1}^n V_j(y)\,\partial_j u(y) = \frac{1}{i\lambda}\,\frac{\nabla\varphi(y) \cdot \nabla u(y)}{|\nabla\varphi(y)|^2}.
\end{align*}
[claim:Key Identity]
$L(e^{i\lambda\varphi}) = e^{i\lambda\varphi}$ on $\operatorname{supp}(a)$.
[/claim]
[proof]
By the chain rule, $\nabla(e^{i\lambda\varphi(y)}) = i\lambda\,(\nabla\varphi(y))\,e^{i\lambda\varphi(y)}$. Substituting:
\begin{align*}
L(e^{i\lambda\varphi}) = \frac{1}{i\lambda}\,\frac{\nabla\varphi \cdot (i\lambda\,\nabla\varphi\,e^{i\lambda\varphi})}{|\nabla\varphi|^2} = \frac{i\lambda\,|\nabla\varphi|^2}{i\lambda\,|\nabla\varphi|^2}\,e^{i\lambda\varphi} = e^{i\lambda\varphi}.
\end{align*}
[/proof]
**Step 2 (The formal transpose).**
[claim:Formula For The Formal Transpose]
The formal transpose $L^t$, defined by the identity $\int_{\mathbb{R}^n}(Lu)\,b\,d\mathcal{L}^n = \int_{\mathbb{R}^n}u\,(L^t b)\,d\mathcal{L}^n$ for all $u \in C^\infty(\mathbb{R}^n)$ and $b \in C_c^\infty(\mathbb{R}^n)$ with $\operatorname{supp}(b) \subseteq \operatorname{supp}(a)$, is given by
\begin{align*}
L^t b = -\frac{1}{i\lambda}\,\nabla \cdot (b\,V) = -\frac{1}{i\lambda}\left(V \cdot \nabla b + b\,\nabla \cdot V\right).
\end{align*}
[/claim]
[proof]
Expanding the left-hand side:
\begin{align*}
\int_{\mathbb{R}^n}(Lu)\,b\,d\mathcal{L}^n = \frac{1}{i\lambda}\sum_{j=1}^n\int_{\mathbb{R}^n} V_j\,(\partial_j u)\,b\,d\mathcal{L}^n.
\end{align*}
Integrating by parts in $y_j$ (the [boundary](/page/Boundary) terms vanish because $b$ has compact support):
\begin{align*}
= -\frac{1}{i\lambda}\sum_{j=1}^n\int_{\mathbb{R}^n} u\,\partial_j(V_j\,b)\,d\mathcal{L}^n = \int_{\mathbb{R}^n} u\left(-\frac{1}{i\lambda}\sum_{j=1}^n \partial_j(V_j\,b)\right)d\mathcal{L}^n.
\end{align*}
Identifying $L^t b = -\frac{1}{i\lambda}\sum_j \partial_j(V_j b) = -\frac{1}{i\lambda}\nabla \cdot (bV)$. The product rule gives $\nabla \cdot (bV) = V \cdot \nabla b + b\,\nabla \cdot V$.
[/proof]
**Step 3 (Iterated transpose and $L^1$ bound).**
[claim:Iterated Transpose Bound]
For every $m \in \mathbb{N}$, there exists a constant $C_m > 0$ such that
\begin{align*}
\|(L^t)^m a\|_{L^1(\mathbb{R}^n)} \leq C_m\,\lambda^{-m},
\end{align*}
where $C_m$ depends on $m$, on [derivatives](/page/Derivative) of $a$ up to order $m$, on derivatives of $\varphi$ up to order $m+1$, and on $c := \inf_{\operatorname{supp}(a)}|\nabla\varphi| > 0$.
[/claim]
[proof]
We prove by induction on $m$ that $(L^t)^m a$ has the form
\begin{align*}
(L^t)^m a(y) = \frac{1}{(i\lambda)^m}\sum_{|\alpha| \leq m} c_\alpha^{(m)}(y)\,\partial^\alpha a(y),
\end{align*}
where each $c_\alpha^{(m)}: \mathbb{R}^n \to \mathbb{C}$ is smooth on $\operatorname{supp}(a)$ and depends only on $\varphi$ and its derivatives (not on $a$), with $\|c_\alpha^{(m)}\|_{L^\infty(\operatorname{supp}(a))} < \infty$.
**Base case ($m = 1$).** From Claim 2:
\begin{align*}
L^t a = -\frac{1}{i\lambda}(V \cdot \nabla a + a\,\nabla \cdot V) = \frac{1}{i\lambda}\left(-\sum_{j=1}^n V_j\,\partial_j a - (\nabla \cdot V)\,a\right).
\end{align*}
This has the required form with $c_{e_j}^{(1)} = -V_j$ for each unit multi-index $e_j$ (i.e. $|\alpha| = 1$) and $c_0^{(1)} = -\nabla \cdot V$ for $|\alpha| = 0$. Since $V = \nabla\varphi/|\nabla\varphi|^2$ is smooth on $\operatorname{supp}(a)$ (as $|\nabla\varphi| \geq c > 0$ there), each $c_\alpha^{(1)}$ is smooth and bounded on $\operatorname{supp}(a)$.
**Inductive step.** Assume $(L^t)^m a = (i\lambda)^{-m}\sum_{|\alpha| \leq m}c_\alpha^{(m)}\partial^\alpha a$. Apply $L^t$:
\begin{align*}
(L^t)^{m+1}a = L^t\!\left((L^t)^m a\right) = -\frac{1}{i\lambda}\nabla \cdot \left((L^t)^m a \cdot V\right).
\end{align*}
Substituting the inductive hypothesis and using the product rule: $\nabla \cdot \bigl((i\lambda)^{-m}\sum_\alpha c_\alpha^{(m)}(\partial^\alpha a)\,V\bigr) = (i\lambda)^{-m}\sum_\alpha \nabla \cdot (c_\alpha^{(m)}(\partial^\alpha a)\,V)$. For each term, the product rule gives
\begin{align*}
\nabla \cdot (c_\alpha^{(m)}(\partial^\alpha a)\,V) = V \cdot \nabla(c_\alpha^{(m)}\partial^\alpha a) + c_\alpha^{(m)}(\partial^\alpha a)\,\nabla \cdot V.
\end{align*}
Expanding $V \cdot \nabla(c_\alpha^{(m)}\partial^\alpha a) = (V \cdot \nabla c_\alpha^{(m)})\partial^\alpha a + c_\alpha^{(m)}\sum_j V_j \partial_j(\partial^\alpha a)$. The second term produces derivatives $\partial^{\alpha + e_j} a$ with $|\alpha + e_j| = |\alpha| + 1 \leq m + 1$. All other terms involve $\partial^\alpha a$ with $|\alpha| \leq m$, multiplied by smooth functions of $\varphi$ and its derivatives. The prefactor is $(i\lambda)^{-(m+1)}$.
Therefore $(L^t)^{m+1}a = (i\lambda)^{-(m+1)}\sum_{|\beta| \leq m+1}c_\beta^{(m+1)}\partial^\beta a$, with each $c_\beta^{(m+1)}$ smooth and bounded on $\operatorname{supp}(a)$ (smoothness follows from $V, \nabla \cdot V, \nabla c_\alpha^{(m)}$ all being smooth, and boundedness from $|\nabla\varphi| \geq c > 0$).
**Conclusion.** Since $a \in C_c^\infty(\mathbb{R}^n)$, every derivative $\partial^\alpha a$ is bounded and supported in $\operatorname{supp}(a)$. Therefore:
\begin{align*}
\|(L^t)^m a\|_{L^1} \leq \frac{1}{\lambda^m}\sum_{|\alpha| \leq m}\|c_\alpha^{(m)}\|_{L^\infty(\operatorname{supp}(a))}\|\partial^\alpha a\|_{L^1} \leq C_m\,\lambda^{-m}.
\end{align*}
The constant $C_m$ is finite because the sum is finite (finitely many multi-indices with $|\alpha| \leq m$), each $\|c_\alpha^{(m)}\|_{L^\infty}$ is finite (smooth on a compact set), and each $\|\partial^\alpha a\|_{L^1}$ is finite ($a \in C_c^\infty$).
[/proof]
**Step 4 (Conclusion).**
Set $I(\lambda) := \int_{\mathbb{R}^n} e^{i\lambda\varphi(y)}\,a(y)\,d\mathcal{L}^n(y)$. Using Claim 1, write $e^{i\lambda\varphi} = L(e^{i\lambda\varphi})$, then iterate: $e^{i\lambda\varphi} = L^m(e^{i\lambda\varphi})$ for any $m \in \mathbb{N}$ (applying Claim 1 $m$ times). Substituting into the integral and applying the transpose identity from Claim 2 $m$ times (each [integration by parts](/theorems/210) is justified because $(L^t)^j a \in C_c^\infty(\mathbb{R}^n)$ for each $j$, since $L^t$ maps compactly supported smooth functions to compactly supported smooth functions, so all boundary terms vanish):
\begin{align*}
I(\lambda) = \int_{\mathbb{R}^n} L^m(e^{i\lambda\varphi})\,a\,d\mathcal{L}^n = \int_{\mathbb{R}^n} e^{i\lambda\varphi}\,(L^t)^m a\,d\mathcal{L}^n.
\end{align*}
Taking absolute values and using $|e^{i\lambda\varphi}| = 1$:
\begin{align*}
|I(\lambda)| \leq \int_{\mathbb{R}^n}|(L^t)^m a|\,d\mathcal{L}^n = \|(L^t)^m a\|_{L^1} \leq C_m\,\lambda^{-m},
\end{align*}
where the last inequality is Claim 3.
