# Friday March 6th

## The Fourier Transform

Recall $\hat f(\xi) =  \int_\RR f(x) e^{2\pi i x\ cdot \xi} ~dx$.
Define $\mcf_a = ??$.

Definition (Decay)
:  $f\in \mcf_a$ iff
  1. $f$ is holomorphic in the strip $S_a = \theset{z = x + iy \suchthat \abs{y} < a}$.
  2. There exists an $A>0$ such that $\abs{f(x + iy)} < \frac{A}{1+x^2}$.

Examples:

- $e^{-z^2} \in \mcf_a$ for all $a$
- $\frac{1}{c^2 + z^2} \in \mcf_a$ for all $a > c$
- $\frac {1}{\cosh(\pi z)} \in \mcf_a$ for $a< \frac 1 2$.

Lemma
: If $f\in \mcf_a$, then $f^{(n)}(z) \in \mcf_b$ for all $b < a$.

Theorem (Boundedness of ?? Functions)
: If $f\in \mcf_a$, then $\abs{\hat f(\xi)} \leq B e^{-2\pi b \abs \xi}$ for some constants $b, B$.

Proof
:   If $\xi = 0$,
    \begin{align*}
    \abs{ \hat f(\xi) }
    &= \abs{\int_\RR f(x) e^{2\pi i x \cdot \xi}} \\
    &\leq \int_\RR \abs{f(x)} ~dx \\
    &\leq A \int_\RR \frac 1 {1+x^2} ~dx \\
    &= A\pi
    .\end{align*}

    For $\xi > 0$, integrate over the box $[-R, R]\cross i[-b, 0]$:

    \begin{center}
    \input{figures/RectangleIntegrate.tikz}
    \end{center}

    Define $g(z) = f(z) e^{-2\pi i z \cdot \xi}$.
    The integral over the rectangle is zero, since $g$ is holomorphic, so we can equate
    $$
    \int_R^{R-ib} f(z) e^{-2\pi i z \cdot \xi} ~dz = \int_0^b f(R - it) e^{-2\pi i (R-it)\cdot \xi} (-i)~dt
    $$
    We can use the estimate in $\mcf_a$ to obtain
    \begin{align*}
    \int_0^b \cdots 
    &\leq \int_0^b \frac{A}{1+R^2} e^{-2\pi s \xi} ~ds \\
    &\leq O(R^{-2})
    .\end{align*}


    Then
    
    \begin{align*}
    \int_\RR f(x) e^{-2\pi i x \cdot \xi} ~d\xi 
    &= \int_{-\infty - ib}^{\infty - ib} \cdots ~dz\\
    &= \int_\RR f(x-ib) e^{2\pi i (x - ib)\cdot \xi} ~dx \\ 
    &\leq \int_\RR \frac{A}{1+x^2} e^{-2\pi b \xi} ~dx \\
    &= A\pi e^{-2\pi b \xi}
    ,\end{align*}

    so we can take $B = A\pi$.

    For $\xi > 0$, the same argument works with the rectangle above the axis.

## Fourier Inversion

Theorem (Fourier Inversion)
: If $f\in \mcf_a$, then $f(x) = \int \hat f(\xi) e^{2\pi i x\cdot \xi} ~d\xi$.

Proof
:  Letting $L_1 = \theset{x-ib}$ and $L_2 = \theset{x+ib}$
    \begin{align*}
    I 
    &= \int_0^\infty \hat f \cdots + \int_{-\infty}^0 \hat f \cdots \\
    &= \int_0^\infty e^{2\pi i x \cdot \xi} \qty{ \int_{L_1} f(z) + e^{-2\pi i z \cdot \xi} ~dz } ~d\xi
    + \int_{\infty}^0 e^{2\pi i x \cdot \xi} \qty{ \int_{L_1} f(z) + e^{-2\pi i z \cdot \xi} ~dz } ~d\xi \\
    &= \int_{L_1} \int_0^\infty e^{2\pi i x \xi - 2\pi i (s-ib)\xi} ~d\xi ~ds 
    + \int_{L_2} f(z) \int_{-\infty}^0 e^{2\pi i x \cdot \xi - 2\pi i (s + ib)\xi} ~d\xi ds\\
    & \quad \quad\text{ by absolute convergence, where } z = s-ib \\
    &= \int_{L_1} f(z) \int_0^\infty e^{2\pi i (x - s + ib)\xi} ~d\xi ~ds 
    + \int_{L_2} f(z) \int_{-\infty}^0 e^{2\pi i (x - s + ib)\xi} ~d\xi ~ds \\
    &= \int_{L_1} f(z) \frac{1}{2\pi i (x - i + ib)} ~ds
    + \int_{L_2} f(z) \frac{1}{2\pi i (x - s - ib)} \\
    &= \frac{1}{2\pi i} \int \frac{f(z)} {z - x} ~dz \\
    &= f(x)
    ,\end{align*}

    noting that 

    \begin{align*}
    \int_0^\infty e^{as} ~ds = \frac 1 a \quad\text{for }\Re(a) > 0
    .\end{align*}

Note the similar trick: for $\xi < 0$, move up, and $\xi > 0$ move down to form a rectangle.
Use the fact that integration along the vertical edges is zero.