Van der Corput's Lemma and its application
Van der Corput's Lemma:
For $\lambda > 0, \phi \in C^2([a, b])$ such that $\left| \phi''(x) \right| \geq 1,$
$$\left| \int_{a}^{b} e^{i\lambda\phi(x)} \textit{dx} \ \right| \leq \frac{2\sqrt{2}}{\lambda^{1/2}}.$$Proof of Van der Corput's Lemma:
Note that it is enough to show for the case $\phi''(x) \geq 1.$
To solve this problem, we want to use Problem 1.1 on this blog. We know that
For $\lambda > 0, \phi \in C^1([a, b])$ with $\left| \phi'(x) \right| \geq 1$ and $\phi'(x)$ is monotonic,
$$\left| \int_{a}^{b} e^{i\lambda\phi(x)} \textit{dx} \ \right| \leq \frac{2}{\lambda}.$$With a slight change, we get for $\left| \phi'(x) \right| \geq \delta>0,$
$$\left| \int_{a}^{b} e^{i\lambda\phi(x)} \textit{dx} \ \right| = \left| \int_{a}^{b} e^{i\lambda\delta\frac{\phi(x)}{\delta}} \textit{dx} \ \right| \leq \frac{2}{\lambda\delta}.$$Now, by the final note of Problem 1.1, the problem occurs when $\phi'(x)$ vanishes. If $\phi'(x)$ doesn't vanish on $(a, b)$, we get the following graph since $\phi''(x) \geq 1.$
For $\lambda > 0, \phi \in C^2([a, b])$ such that $\left| \phi''(x) \right| \geq 1,$
$$\left| \int_{a}^{b} e^{i\lambda\phi(x)} \textit{dx} \ \right| \leq \frac{2\sqrt{2}}{\lambda^{1/2}}.$$Proof of Van der Corput's Lemma:
Note that it is enough to show for the case $\phi''(x) \geq 1.$
To solve this problem, we want to use Problem 1.1 on this blog. We know that
For $\lambda > 0, \phi \in C^1([a, b])$ with $\left| \phi'(x) \right| \geq 1$ and $\phi'(x)$ is monotonic,
$$\left| \int_{a}^{b} e^{i\lambda\phi(x)} \textit{dx} \ \right| \leq \frac{2}{\lambda}.$$With a slight change, we get for $\left| \phi'(x) \right| \geq \delta>0,$
$$\left| \int_{a}^{b} e^{i\lambda\phi(x)} \textit{dx} \ \right| = \left| \int_{a}^{b} e^{i\lambda\delta\frac{\phi(x)}{\delta}} \textit{dx} \ \right| \leq \frac{2}{\lambda\delta}.$$Now, by the final note of Problem 1.1, the problem occurs when $\phi'(x)$ vanishes. If $\phi'(x)$ doesn't vanish on $(a, b)$, we get the following graph since $\phi''(x) \geq 1.$
Since $\left| \phi'(x) \right| \geq \delta>0$ and $\phi'(x)$ is monotonic for $x>a+\delta,$ we get the following inequality.
$$\left| \int_{a}^{b} e^{i\lambda\phi(x)} \textit{dx} \ \right| \leq \left| \int_{|x-a|>\delta} e^{i\lambda\phi(x)} \textit{dx} \ \right| + \left| \int_{|x-a|<\delta} e^{i\lambda\phi(x)} \textit{dx}\ \right| \leq \frac{2}{\lambda\delta} + 2\delta.$$On the other hand, if $\phi'(x)$ vanishes at $x_0 \in (a, b),$ we get the following graph since $\phi''(x) \geq 1.$
Since $\left| \phi'(x) \right| \geq \delta>0$ and $\phi'(x)$ is monotonic for $|x-a| > \delta,$ we get the following inequality.
$$\left| \int_{a}^{b} e^{i\lambda\phi(x)} \textit{dx} \ \right| \leq \left| \int_{|x-a|>\delta} e^{i\lambda\phi(x)} \textit{dx} \ \right| + \left| \int_{|x-a|<\delta} e^{i\lambda\phi(x)} \textit{dx}\ \right| \leq \frac{2}{\lambda\delta} + 2\delta.$$We know that $2/\lambda\delta + 2\delta$ has the minimal value $2\sqrt{2}/\lambda^{1/2}$ when $2/\lambda\delta = 2\delta.$ Thus, take $\delta = \lambda^{-1/2}$ for above cases then we get
$$\left| \int_{a}^{b} e^{i\lambda\phi(x)} \textit{dx} \ \right| \leq \frac{2\sqrt{2}}{\lambda^{1/2}}.$$
Problem (Application of Van der Corput's Lemma):
Show that,
$$\left| \int_{-1}^{1} e^{i[\xi_1t + \xi_2t^2]} \textit{dt} \ \right| \leq \frac{C}{|\xi_1\xi_2|^{1/2}}.$$
댓글
댓글 쓰기