Analysis Problem of the Day 88

Today’s problem appeared as Problem 9 on Stanford’s Spring 2017 Analysis Qual:

Problem 9. Suppose $f \in C^\infty(\mathbb{R}^n;\mathbb{C})$ with non-negative imaginary part is such that $\text{Im }f(x) =0, x \in K \implies \nabla f(x) \not = 0,$ where $K \subseteq \mathbb{R}^n$ is some compact set. Show that for all $u \in C^\infty(\mathbb{R}^n)$ with support in $K,$ for all $N \geq 0$ there exists a $C=C(N,K,u)>0$ such that

$\left|\int e^{i \omega f(x)} u(x)dx\right| \leq C \omega^{-N}, \quad \omega >1.$

Solution: Split the integral over the part where the imaginary part of $f$ is at least $\epsilon$ and less than $\epsilon$ for some small $\epsilon>0.$ If $\text{Im }f >0,$ then $e^{i \omega f(x)} = e^{-\omega \text{Im}f(x)} e^{i \omega \text{Re}f(x)} \to 0$ as $\omega \to \infty.$ In particular, the decay is exponential, i.e. since $u$ is smooth and compactly supported, one has

$|e^{i \omega f(x)} u(x)| \leq e^{-\epsilon \omega} \|u\|_\infty = C(K,u) e^{-\epsilon \omega} \leq C(K,u,N,\epsilon) \omega^{-N}$

for all $N \geq 0$ (since exponential decay is faster than algebraic decay). Now, when $\text{Im }f < \epsilon,$ we note that near a point $x \in \{x: \text{Im }f(x) = 0\},$ we have the Taylor series approximation $f(x+h) = f(x) + \nabla f(x) \cdot h +o (h).$ It follows that $e^{i \omega f(x+h)} u(x+h)= e^{i \omega f(x)} e^{i \nabla f(x) \cdot \omega h + o(h)} u(x+h) .$ By performing a change of variables to integrate locally in $h,$ using a compactness argument to select finitely many center points, and shrinking $\epsilon$ enough to control the error term in Taylor’s theorem at each center point, we thus get that

$\int_{\text{Im}f(x+h)<\epsilon} e^{i\omega f(x+h)}u(x+h) dh \sim \widehat{v}(-\omega (\nabla f(x)+f(x)))+o(h),$

where $v(x)=u(x+h)$ and $\widehat{v}$ is the Fourier transform of $v.$ Now, since $v$ is smooth and compactly supported, by the properties of the Fourier transform, one has that $|\widehat{v}(-\omega (\nabla f(x)+f(x)))| \leq C(K,u,N)\omega^{-N}$ for all $N \geq 0.$ It follows that by sending $h \to 0,$ one gets

$\int e^{i \omega f(x)} u(x)dx = \int_{\text{Im }f > \epsilon} e^{i \omega f(x)} u(x)dx +\int_{\text{Im }f \leq \epsilon} e^{i \omega f(x)} u(x)dx \lesssim C(u,N,K) \omega^{-N}$

as $\omega \to \infty,$ which demonstrates the desired claim.

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