Stationary Phase with controlled phase $|\int_0^Te^{\frac{i}{\epsilon}\phi_j(t)}dt| = O(\epsilon)$

Question

My question concerns the stationary phase theorem when we only know control information on the phase. Suppose that $\phi_a(t)$ and $\phi_b(t)$ are phases such that $$\phi_a(t) = a_1 + a_2t$$ $$\phi_b(t) = b_1 + b_2t$$ with $$ a_1 > b_1 > 0$$ $$ b_2 > a_2> 0$$ And suppose that $\phi(t,\epsilon) = \phi_1(t) + \epsilon \phi_2(t/\epsilon)$ is bounded by $$\phi_a(t) \leq \phi(t,\epsilon) \leq \phi_b(t),\ \forall (t,\epsilon)\in [0,T]\times (0,1] $$ where $$\phi_a(T) = \phi_b(T) $$ and suppose that $0<C_1 \leq \phi'(t,\epsilon)$ on $[0,T]$. The non-stationary phase theorem implies that $$A_\epsilon(\phi_j(t)) := |\int_0^Te^{\frac{i}{\epsilon}\phi_j(t)}dt| = O(\epsilon) , \text{ for } j\in\{a,b\}$$ Is it necessarily true that $A_\epsilon(\phi(t,\epsilon)) = O(\epsilon)$. For simplicity suppose that $\phi_2 \in C^\infty \cap L^\infty$. The complication arises as $\phi'$ is not necessarily monotone and $\phi'' = O(\epsilon^{-1})$

Answer 1

Answer: no, that is not necessarily true. Take as a counterexample $$ \phi = 1 + t + \epsilon\,\sin \frac{t}{\epsilon}, $$ which can be bounded without any problems by appropriately chosen lines $\phi_a$ and $\phi_b$. However, using the stationary phase approximation (see e.g. here), you can infer that the integral is actually $\mathcal{O}(1)$ instead of $\mathcal{O}(\epsilon)$. This is basically because the number of local maxima and minima of $\phi$ scales with $\frac{1}{\epsilon}$, which precisely counters the prefactor $\frac{2 \pi \epsilon}{f''(x_0)}$.

For the example above, my leading order calculations give me $$ \int_0^T e^{\frac{i}{\epsilon} \phi(s)}\text{d}s = \frac{1}{2}K^2 \left(1-\sin \frac{2}{\epsilon^2}\right) \sqrt{2 \pi} \epsilon^2 + \text{higher order terms}, $$ where $K$ is the number of zeroes of $\cos\frac{t}{\epsilon}$ for $0\leq t \leq T$. Since $K$ scales with $\frac{1}{\epsilon}$ (to be precise, $K = \lfloor \frac{T}{\pi \epsilon} \rfloor$ ), one concludes that $$ \int_0^T e^{\frac{i}{\epsilon} \phi(s)}\text{d}s = \mathcal{O}(1). $$

Answer 2

The counter example by Frits Veerman needs to be modified to be $$\phi'(t) = 1 + 2t + \epsilon\sin(t/\epsilon)$$ to satisfy $\phi' \geq C_1 >0$. However let $T=1$ and $\epsilon = \frac{1}{2n\pi}$

$$ \int_0^1e^{\frac{i}{\epsilon}(1+2t + \epsilon\sin(t/\epsilon))}dt = \epsilon e^{\frac{i}{\epsilon}}\int_0^{\frac{1}{\epsilon}}e^{i(2t + \sin(t))}dt\\ = \frac{1}{2\pi n}\sum_{k=0}^{n-1}\int_{2\pi k}^{2\pi (k+1)}e^{i(2t + \sin(t))}dt$$

and $$2\pi J_2(1) = \int_{2\pi k}^{2\pi(k+1)}e^{i(2t + \sin(t))}dt \ \text{ for } k\in \mathbb{Z} $$ Where $J_2$ is the Bessel function of the 2nd kind. Thus $$ \int_0^1e^{\frac{i}{\epsilon}(1+2t + \epsilon\sin(t/\epsilon))}dt = J_2(1) = O(1)$$ for $\epsilon = \frac{1}{2\pi n}$