# Wednesday, May 12

:::{.remark}
Take any $f(\theta)$ which is real-valued on $[0, 2\pi]$ and $2\pi\dash$periodic, we can then write 
\[
f(\theta) = \sum_{k\in \ZZ} c_n e^{in \theta}
.\]
If $f$ is piecewise $C^1$, then 
\[
c_n = \frac 1 {2\pi} \int_0^\pi f(\theta) e^{-i n \theta} ~d\theta
,\]
and this gives the Fourier series of $f$.
Suppose $f$ is holomorphic in $\Omega \ni z_0$, then 
$$
f$(z) = f(z_0 + re^{i\theta}) = \sum a_n (z-z_0)^n = \sum a_n r^n e^{in\theta}
$$
where
$$
a_n 
&= \frac {f^{(n)}(z_0)} {n!} \\
&= \frac 1 {2\pi i} \int_{\bd D_r(z_0)} \frac{f(z)}{(z-z_0)^{n+1}} ~dz \\
&= \frac{1}{2\pi r^n} \int_0^{2\pi} f(z_0 + re^{i\theta}) e^{-i n \theta} ~d\theta \\
&= \frac{c^n}{r^n} \quad n \geq 0
$$

Then $c_n = a_n r^n$ when $n\geq 0$ and 0 otherwise.
Similarly, $c_{-n} = 0$ by just writing 
$$
c_{-n} 
&= \frac{1}{2\pi} \int_0^{2\pi} f(\theta) e^{in\theta} \\
&= \frac{r^n}{2\pi i} \int_{\bd D_r(z_0)} f(z) (z-z_0)^{n-1} ~dz \\
&= 0 \quad n\geq 1
$$

:::

:::{.remark}
Recall that 
\[
L^2 = \theset{f \suchthat \int_0^{2\pi} \abs{f}^2 < \infty}
,\]
and we define $H^2$ is the $F$ that are holomorphic in $D_1$ where $f(\theta) = F(e^{i\theta})$.

Since $H^2 \subset L^2$, there is a projection operator $P: L^2 \to H^2$ called the **Hilbert transform**.

:::

:::{.example title="?"}
$f(\theta) = e^{-\theta}$, then this is in $L^2$ but not $H^2$.
:::

:::{.remark}
For $f$ analytic, the projection is given by $f(\theta) = \sum_{n\in \ZZ} a_n z^n \mapsvia{P} \sum_{n\geq 0} a_n z^n$.
:::

:::{.theorem title="Important, Carleson, 60s"}
\[
f(\theta) \in L^2 \implies \sum_{\abs n < N} c_n e^{in\theta} \converges{N\to\infty}\too f(\theta) \quad \ae
\]
See also Lusin's conjecture, 1920s.
:::

:::{.remark}
This is readable with graduate-level background!
:::



 
Question:
If $f$ is continuous, does the Fourier series converge? Early 1900s, a continuous function with one point where this doesn't hold.

Note that this recovers the MVT:
$$
f(z_0) = a_0 = \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{i\theta}) ~d\theta
$$

## Fourier Transform

Recall that for any $f\in L^1(\RR)$ we define
$$
\hat f(\xi) = \int_\RR f(x) e^{-2\pi i \cdot \xi} ~dx
$$

Note that 
\[
\abs{\hat f(\xi)} \leq \int_\RR \abs{f(x)}~dx \implies \norm{Ff} \leq \norm{f}_1 \text{ uniformly}
,\]
and $\hat f$ is also uniformly continuous.


:::{.remark}
Recall the Riemann-Lebesgue lemma:
\[
\lim_{\abs \xi \to \infty} \abs{\hat f(\xi)} = 0
.\]
:::

:::{.definition title="Schwarz space"}
Define the Schwarz space 
\[
S = \theset{f\in C^\infty \suchthat \sup \abs{x^a \qty{\dd{}{x}}^b f } < \infty}
.\]
:::

:::{.definition title="Tempered distribution"}
Adding a norm makes this a topological vector space, and we can define **tempered distributions** as the elements of $S\dual$.
:::


:::{.remark}
Note that polynomials are in $S\dual$, since we can send $p(x) \to \ell_p: f\mapsto \int f p$.
Similarly, $L^1 \subset S\dual$ by the same argument.
Note that 
\[
\int \hat{g} f = \int g \hat{f}
,\]
and since polynomials aren't in $L^1$ (so the usual formula doesn't converge), we define the transform by this property.

Note that $e^{x^2} \not\in S$, but $e^{-x^2/2}$.
This definition of the Fourier transform yields an isometry on $S$ and extends to a linear operator on $S\dual$ naturally.
:::


:::{.remark}
Next time: proving Fourier inversion formula using the Cauchy Integral formula, namely
$$
f(x) = \int_\RR \hat f(\xi) e^{2\pi i x \cdot \xi} = \frac 1 {2\pi i} \int \frac{f(z)}{z-x}
$$
:::