The proof proceeds by induction on $k$. The main reduction shows it suffices to bound the pure exponential [integral](/page/Integral) $\int_a^b e^{i\lambda\varphi(x)}\, d\mathcal{L}^1(x)$ by $c_k \lambda^{-1/k}$; the full result with amplitude $f$ follows by a single [integration by parts](/theorems/210). **Step 1 (Reduction to pure exponential).** Let $\operatorname{supp}(f) \subset [a,b]$ and define $G(x) = \int_a^x e^{i\lambda\varphi(y)}\, d\mathcal{L}^1(y)$. Integration by parts gives \begin{align*} \int_a^b e^{i\lambda\varphi(x)} f(x)\, d\mathcal{L}^1(x) = [G(x)f(x)]_a^b - \int_a^b G(x) f'(x)\, d\mathcal{L}^1(x) = -\int_a^b G(x) f'(x)\, d\mathcal{L}^1(x), \end{align*} since $G(a) = 0$ and $f(b) = 0$ (as $f \in C^1_c$). Therefore $|\int e^{i\lambda\varphi} f\, d\mathcal{L}^1| \le \sup_{x \in [a,b]} |G(x)| \cdot \|f'\|_{L^1}$. It suffices to prove $|G(x)| \le c_k \lambda^{-1/k}$ for all $x$. **Step 2 (Base case $k = 1$).** Assume $|\varphi'| \ge 1$ on $[a,b]$ with $\varphi'$ monotone. Writing $e^{i\lambda\varphi(x)} = \frac{1}{i\lambda\varphi'(x)} \frac{d}{dx}(e^{i\lambda\varphi(x)})$ and integrating by parts: \begin{align*} \left|\int_a^b e^{i\lambda\varphi(x)}\, d\mathcal{L}^1(x)\right| &= \frac{1}{\lambda}\left|\left[\frac{e^{i\lambda\varphi(x)}}{i\varphi'(x)}\right]_a^b - \int_a^b e^{i\lambda\varphi(x)} \frac{d}{dx}\!\left(\frac{1}{\varphi'(x)}\right) d\mathcal{L}^1(x)\right| \\ &\le \frac{1}{\lambda}\left(\frac{1}{|\varphi'(b)|} + \frac{1}{|\varphi'(a)|} + \int_a^b \left|\frac{d}{dx}\!\left(\frac{1}{\varphi'}\right)\right| d\mathcal{L}^1\right). \end{align*} Since $\varphi'$ is monotone, $1/\varphi'$ is also monotone, so $\int_a^b |d(1/\varphi')/dx|\, d\mathcal{L}^1 = |1/\varphi'(b) - 1/\varphi'(a)|$. The total is at most $2/|\varphi'(b)| + 2/|\varphi'(a)| \le 4$ (using $|\varphi'| \ge 1$). Thus $c_1 = 4$. **Step 3 (Inductive step).** Assume the result holds for some $k \ge 1$ with constant $c_k$. Suppose $|\varphi^{(k+1)}| \ge 1$ on $[a,b]$. Since $\varphi^{(k+1)}$ does not change sign (being at least $1$ or at most $-1$ in absolute value), the [function](/page/Function) $\varphi^{(k)}$ is monotone. Let $x_0 \in [a,b]$ be the point where $|\varphi^{(k)}|$ is minimised. For any $\delta > 0$, split $[a,b]$ into two regions: (i) **Near $x_0$:** On the interval $[x_0 - \delta, x_0 + \delta] \cap [a,b]$, the trivial bound $|\int e^{i\lambda\varphi}\, d\mathcal{L}^1| \le 2\delta$ applies. (ii) **Away from $x_0$:** On $[a, x_0 - \delta]$ and $[x_0 + \delta, b]$, the [mean value theorem](/theorems/186) gives $|\varphi^{(k)}(x)| \ge |x - x_0| \cdot 1 \ge \delta$ (since $|\varphi^{(k+1)}| \ge 1$). Writing $\lambda\varphi = (\delta\lambda) \cdot (\delta^{-1}\varphi)$ and noting $|(\delta^{-1}\varphi)^{(k)}| \ge 1$, the inductive hypothesis gives $|\int e^{i\lambda\varphi}\, d\mathcal{L}^1| \le c_k (\delta\lambda)^{-1/k}$ on each sub-interval. Combining: $|\int_a^b e^{i\lambda\varphi}| \le 2\delta + 2c_k (\delta\lambda)^{-1/k}$. Choosing $\delta = \lambda^{-1/(k+1)}$ to balance the two terms: $2\lambda^{-1/(k+1)} + 2c_k \lambda^{-1/(k+1)} = (2 + 2c_k)\lambda^{-1/(k+1)}$. Thus $c_{k+1} = 2c_k + 2$, completing the induction.