I am going through Chapter 4 (Fourier Transforms) of Stein's Complex Analysis text and cannot follow a certain part of Theorem 2.1 which appears as if it should be trivial.
First, for $a>0$, Stein defines a class of functions $\mathcal{F}_a$ such that $f \in \mathcal{F}_a$ if:
- The function $f$ is holomorphic in the horizontal strip $$S_a = \{z \in \mathbb{C} : |\text{Im}(z)| < a \}$$
- There exists a constant $A > 0$ such that $$|f(x + iy)| \leq \frac{A}{1+x^2} \quad \text{for all } x \in \mathbb{R} \text{ and } |y| < a.$$
Now here is the theorem itself:
If $f$ belongs to the class $\mathcal{F}_a$ for some $a > 0$, then $|\hat{f}(\xi)| \leq B e^{-2\pi b |\xi|}$ for any $0 \leq b < a$.
To prove this for $0 < b <a$ and the case $\xi > 0$ Stein shifts the contour of integration (the real line) down by $b$, and defines a function $g(z) = f(z)e^{-2\pi i z \xi}$. He shows that as $R$ tends to infinity, the integrals of $g$ over the two vertical sides converge to zero: $$\Biggl |\int_{-R-ib}^{-R} g(z) dz \Biggr | \leq \int_0^b |f(-R-it)e^{-2\pi i (-R-it)\xi} | dt \\\leq \int_0^b\frac{A}{R^2} e^{-2\pi t \xi}dt$$
Here is the illustration in the text illustrating the contour in question:
Now I understand the first inequality holds from your basic integral inequalities. The second I am less sure of, though it clearly uses the property of moderate decrease, I am just now sure how, and also how the argument of the exponential changes as well to remove $R$.
Also, how come it suffices to prove the integral over the vertical segments go to zero and that the integral of the interior is 0? Wouldn't we also have to show that the lower of the two horizontal segments also go to 0? Thanks!
The moderate decrease is used to bound $$ |f(-R-it)|\leq \dfrac{A}{1+R^2} \leq \dfrac{A}{R^2}. $$ On the other hand, for the exponential we have $$ |e^{-2\pi i(-R-it)\xi}|= |e^{2\pi iR\xi}||e^{-2\pi t\xi}|= e^{-2\pi t\xi}, $$ where we distributed the $-2\pi i\xi$ term in the exponent, and used that $|e^{i\theta}|=1$ for any $\theta \in \mathbb{R}$.