[proofplan] The strategy is repeated integration by parts using a first-order differential operator $L$ that reproduces the oscillatory exponential $e^{i\lambda\varphi}$ up to a factor of $\lambda^{-1}$. Concretely, $L f := \frac{1}{i\lambda |\nabla\varphi|^2} \nabla\varphi \cdot \nabla f$ satisfies $L(e^{i\lambda\varphi}) = e^{i\lambda\varphi}$, so each application of the formal adjoint $L^*$ (transferred to the amplitude $\psi$ by integration by parts) gains a factor of $\lambda^{-1}$. The non-vanishing of $\nabla\varphi$ on the compact set $\operatorname{supp}\psi$ secures a uniform lower bound $|\nabla\varphi| \ge c > 0$ there, which keeps the coefficients of $L^*$ smooth and the support unchanged. Iterating $N$ times yields the bound $\lambda^{-N}$. [/proofplan] [step:Establish a uniform lower bound $|\nabla\varphi| \ge c > 0$ on $\operatorname{supp}\psi$] The set $K := \operatorname{supp}\psi \subset \mathbb{R}^n$ is compact because $\psi \in C_c^\infty(\mathbb{R}^n)$. The map \begin{align*} |\nabla\varphi| : \mathbb{R}^n &\to [0, \infty) \\ x &\mapsto \left( \sum_{j=1}^n |\partial_{x_j}\varphi(x)|^2 \right)^{1/2} \end{align*} is continuous because $\varphi \in C^\infty(\mathbb{R}^n)$. By hypothesis $|\nabla\varphi(x)| > 0$ for all $x \in K$. A continuous strictly positive function on a non-empty compact set attains its infimum and the infimum is positive, so there exists $c > 0$ such that \begin{align*} |\nabla\varphi(x)| \ge c \qquad \text{for all } x \in K. \end{align*} [/step] [step:Define the differential operator $L$ and verify $L(e^{i\lambda\varphi}) = e^{i\lambda\varphi}$ on a neighbourhood of $K$] Choose an open set $U \subset \mathbb{R}^n$ with $K \subset U$ and $|\nabla\varphi| \ge c/2$ on $\overline{U}$ (such a $U$ exists because $|\nabla\varphi|$ is continuous and $|\nabla\varphi| \ge c$ on $K$). On $U$, define the smooth vector field \begin{align*} a : U &\to \mathbb{R}^n \\ x &\mapsto \frac{\nabla\varphi(x)}{|\nabla\varphi(x)|^2}, \end{align*} and the first-order differential operator \begin{align*} L : C^\infty(U) &\to C^\infty(U) \\ f &\mapsto \frac{1}{i\lambda} \, a \cdot \nabla f = \frac{1}{i\lambda |\nabla\varphi|^2} \sum_{j=1}^n (\partial_{x_j}\varphi) \, \partial_{x_j} f. \end{align*} Applying $L$ to $e^{i\lambda\varphi}$ gives, by the chain rule, \begin{align*} L(e^{i\lambda\varphi})(x) = \frac{1}{i\lambda |\nabla\varphi(x)|^2} \, \nabla\varphi(x) \cdot \bigl( i\lambda\, \nabla\varphi(x)\, e^{i\lambda\varphi(x)} \bigr) = e^{i\lambda\varphi(x)}. \end{align*} Hence $L(e^{i\lambda\varphi}) = e^{i\lambda\varphi}$ on $U$, and consequently $L^N(e^{i\lambda\varphi}) = e^{i\lambda\varphi}$ on $U$ for every $N \ge 1$. [/step] [step:Compute the formal adjoint $L^*$ and show it preserves $C_c^\infty(U)$] The formal adjoint of $L$ on $C_c^\infty(U)$ with respect to the Lebesgue inner product is \begin{align*} L^* : C_c^\infty(U) &\to C_c^\infty(U) \\ g &\mapsto -\frac{1}{i\lambda} \sum_{j=1}^n \partial_{x_j}\!\left( \frac{\partial_{x_j}\varphi}{|\nabla\varphi|^2} \, g \right). \end{align*} Indeed, for $f, g \in C_c^\infty(U)$ integration by parts (justified because both $f$ and $g$ have compact support inside $U$, so the boundary terms vanish) gives \begin{align*} \int_{\mathbb{R}^n} (Lf)(x)\, g(x)\, d\mathcal{L}^n(x) &= \frac{1}{i\lambda} \sum_{j=1}^n \int_{\mathbb{R}^n} \frac{\partial_{x_j}\varphi}{|\nabla\varphi|^2}\, (\partial_{x_j} f)(x)\, g(x)\, d\mathcal{L}^n(x) \\ &= -\frac{1}{i\lambda} \sum_{j=1}^n \int_{\mathbb{R}^n} f(x)\, \partial_{x_j}\!\left( \frac{\partial_{x_j}\varphi}{|\nabla\varphi|^2}\, g \right)(x)\, d\mathcal{L}^n(x) \\ &= \int_{\mathbb{R}^n} f(x)\, (L^* g)(x)\, d\mathcal{L}^n(x). \end{align*} The operator $L^*$ has smooth coefficients on $U$ (because $\nabla\varphi$ is smooth and $|\nabla\varphi| \ge c/2$ on $\overline{U}$, so $|\nabla\varphi|^{-2}$ is smooth there), and Leibniz expansion gives \begin{align*} L^* g = -\frac{1}{i\lambda}\!\left( a \cdot \nabla g + (\nabla \cdot a)\, g \right), \end{align*} where $a = \nabla\varphi / |\nabla\varphi|^2$ and $\nabla \cdot a$ is its smooth divergence on $U$. Hence $L^* g$ is again smooth and $\operatorname{supp}(L^* g) \subseteq \operatorname{supp} g \subset U$, so $L^*$ maps $C_c^\infty(U)$ into itself. By induction, the same holds for any iterate $(L^*)^N$. [/step] [step:Integrate by parts $N$ times to extract $N$ powers of $\lambda^{-1}$] Fix $N \ge 0$ (the case $N = 0$ follows from the bound $|\int e^{i\lambda\varphi}\psi\, d\mathcal{L}^n| \le \|\psi\|_{L^1}$, which gives the claim with $C_0 = \|\psi\|_{L^1}$, since $|e^{i\lambda\varphi(x)}| = 1$ pointwise and $\psi \in L^1(\mathbb{R}^n)$). For $N \ge 1$, since $\operatorname{supp}\psi \subset K \subset U$, we may regard $\psi \in C_c^\infty(U)$. Using $L^N(e^{i\lambda\varphi}) = e^{i\lambda\varphi}$ on $U$ from the second step, write \begin{align*} \int_{\mathbb{R}^n} e^{i\lambda\varphi(x)}\, \psi(x)\, d\mathcal{L}^n(x) = \int_U \bigl[L^N(e^{i\lambda\varphi})\bigr](x)\, \psi(x)\, d\mathcal{L}^n(x). \end{align*} Iterating the duality $\int (Lf) g\, d\mathcal{L}^n = \int f\, (L^* g)\, d\mathcal{L}^n$ established above $N$ times (each application is legitimate because $(L^*)^k \psi \in C_c^\infty(U)$ for every $0 \le k \le N$, by the previous step), we obtain \begin{align*} \int_{\mathbb{R}^n} e^{i\lambda\varphi(x)}\, \psi(x)\, d\mathcal{L}^n(x) = \int_U e^{i\lambda\varphi(x)}\, \bigl[(L^*)^N \psi\bigr](x)\, d\mathcal{L}^n(x). \end{align*} [/step] [step:Bound the resulting integral and extract the $\lambda^{-N}$ factor] The factor $\frac{1}{i\lambda}$ in front of $L^*$ contributes $(i\lambda)^{-N}$ in absolute value, so writing $L^* = \lambda^{-1} M$ with \begin{align*} M : C_c^\infty(U) &\to C_c^\infty(U) \\ g &\mapsto -\frac{1}{i}\bigl( a \cdot \nabla g + (\nabla \cdot a) g \bigr), \end{align*} yields $(L^*)^N \psi = \lambda^{-N} M^N \psi$, and $M^N \psi \in C_c^\infty(U)$ is independent of $\lambda$. Define \begin{align*} C_N := \int_U |M^N \psi(x)|\, d\mathcal{L}^n(x) = \|M^N \psi\|_{L^1(\mathbb{R}^n)} < \infty, \end{align*} which is finite because $M^N \psi$ is smooth and compactly supported. Taking absolute values in the displayed identity from the previous step and using $|e^{i\lambda\varphi(x)}| = 1$, \begin{align*} \left| \int_{\mathbb{R}^n} e^{i\lambda\varphi(x)}\, \psi(x)\, d\mathcal{L}^n(x) \right| = \lambda^{-N} \left| \int_U e^{i\lambda\varphi(x)}\, M^N \psi(x)\, d\mathcal{L}^n(x) \right| \le \lambda^{-N}\, C_N. \end{align*} This holds for every $\lambda \ge 1$, which is the asserted bound. [/step] [step:Conclude] The constant $C_N = \|M^N \psi\|_{L^1}$ depends on $\varphi$, $\psi$, $n$, and $N$ but not on $\lambda$. The estimate $\left| \int e^{i\lambda\varphi}\, \psi\, d\mathcal{L}^n \right| \le C_N\, \lambda^{-N}$ holds for every $N \ge 0$ and every $\lambda \ge 1$, completing the proof. [/step]