Fourier Analysis 1.1 - Solution

Suppose that $\lambda > 0, \phi \in C^1([a, b])$ with $\left| \phi'(x) \right| \geq 1$ and $\phi'(x)$ is monotonic. Then

(1) prove that $$\left| \int_{a}^{b} e^{i\lambda\phi(x)} \textit{dx} \ \right| \leq \frac{2}{\lambda}.
$$(2) Can we extend the above result to the case of multiple integral?

Proof of (1):
Note that it is enough to show for the case $\phi'(x) \geq 1,$ and we know that if $\left| \phi'(x) \right| \geq 1$ and $\phi'(x)$ is monotonic, $\phi''(x)$ exists a.e. and $\phi''(x)$ is always positive or always negative.

Now, by integration by parts
$$\begin{align}
\left| \int_{a}^{b} e^{i\lambda\phi(x)} dx \right|
&= \left| \int_{a}^{b} \frac{e^{i\lambda\phi(x)}}{i\lambda\phi'(x)}i\lambda\phi'(x) \textit{dx} \ \right|\\
&= \left| \left[ \frac{e^{i\lambda\phi(x)}}{i\lambda\phi'(x)} \right]^b_a + \frac{1}{i\lambda} \int_{a}^{b} e^{i\lambda\phi(x)} \frac{\textit{d}}{\textit{dx}} \left( \frac{1}{\phi'(x)} \right) \textit{dx} \ \right|\\
&\leq \frac{1}{\lambda} \left( \left| \frac{e^{i\lambda\phi(b)}}{\phi'(b)} - \frac{e^{i\lambda\phi(a)}}{\phi'(a)} \right| + \left| \frac{1}{\phi'(b)} - \frac{1}{\phi'(a)} \right| \right)\\
&\leq \frac{1}{\lambda} \left( \frac{1}{\phi'(b)} + \frac{1}{\phi'(a)}+ \left| \frac{1}{\phi'(b)} - \frac{1}{\phi'(a)} \right| \right)\\
&\leq \frac{2}{\lambda}.
\end{align}$$ Note that by condition $\left| \phi'(x) \right| \geq 1$, we could escape the problem of $\phi'(x) = 0.$