# Thursday November 14

Let $e_n(x) \definedas e^{2\pi i n x}$ for all $n\in \ZZ$ and $x\in [0,1]$.

**Theorem:**
$\theset{e_n}_{n\in \ZZ}$ is an orthonormal basis for $L^2([0, 1])$.

> Note that $L_2([0, 1]) = L^2(\Pi) = \theset{f\in L^2 \suchthat f(0) = f(1)}$, since this only modifies a function at one point and we are identifying functions that agree almost everywhere.

## Fourier Series

**Definition:**
For any $f\in L^1(\Pi)$, we define its *Fourier coefficients* 
$$
\hat{f}(n) \definedas \int_0^1 f(x) e^{-2\pi i n x} ~dx \quad \forall n \in \ZZ
$$

> Note that this resembles $\inner{f}{e_n}$, although this is not an inner product space.

**Definition:**
The *Fourier series* of $f$ is defined as 
$$
\hat f(x) \definedas \sum_{n\in \ZZ} \hat{f}(n) e^{-2\pi i n x}
$$

> Note that this isn't necessarily *equal* to $f$, this only makes sense if the partial sums are converging to $f$ in some sense.

Define the $N$th partial sum 
$$
S_Nf(x) \definedas \sum_{\abs{n} \leq N} \hat{f}(n) e^{2\pi i n x}
.$$

> Remark: We have $L_2(\Pi) \subseteq L^1(\Pi)$, so the Fourier coefficients *do* make sense here as an inner product for all $f\in L_2(\Pi)$.

**Some consequences:**

- By Riesz-Fischer, given any $\theset{a_n} \in \ell^2(\ZZ)$, there is a function $f\in L^2(\Pi)$ such that $\hat{f}(n) = a_n$ for all $n$.

- By Parseval, 
$$
\sum_{n\in \ZZ} \abs{\hat{f}(n)}^2 = \int_0^1 \abs{f(x)}^2~dx
$$

- $\lim_{N\to\infty} \norm{S_Nf - f}_2 = 0$, i.e. the Fourier series of $f$ converges to $f$.

An answer: the Fourier series equals $f$ in an $L^2$ sense, but recall that pointwise equality was a hard theorem proved only 50 or so years ago!

> Remark: note that for the Fourier transform, when $f\in L^1$ and $\hat{f} \in L^1$, we have 
$$
f(x) = \int \hat{f} e^{-2\pi i x}~dx
,$$ 
and we can get analogous statements here.

## Uniform Convergence of Fourier Series

**Theorem:**

If $f\in L^1(\Pi)$ and $\theset{\hat{f}(n)} \in \ell^1$, then $S_Nf \mapsvia{u} f$ *uniformly* on $\Pi$.

**Corollaries:**

1. If $f\in C^1(\Pi)$, then $S_N f \to f$ uniformly on $\Pi$.

2. 
$$
f\in C(\Pi) \text{ and } f' \in L^2(\Pi) \implies S_Nf \mapsvia{u} f \text{ on } \Pi.
$$ 

    > Note that the first condition alone is not sufficient: there exists a continuous function whose Fourier series diverges at a point.

*Proof:*
Exercise

> So if $f\in C^1$, it is *equal* to its Fourier series. Everyone should know this fact!

The following is a beautiful and fundamentally amazing fact:

**Fact (The Riemann-Localization Principle):**

If $f\in L^1(\Pi)$ and $f$ is constant on some neighborhood of $x \in \Pi$, then $S_Nf(x) \to f(x)$ pointwise at this particular $x$.

> Note that computing the Fourier coefficients requires integrating over the entire circle, but somehow the behavior of the function elsewhere doesn't matter at $x$!

*Proof of theorem:*

Since $\theset{\hat{f}(n)} \in \ell^1(\ZZ)$, this gives us what we need to apply the $M$ test.
So $S_Nf \mapsvia{u} g$ uniformly for some continuous $g$.

How can we argue $g = f$? 
Consider

\begin{align*}
\hat{g}(n) 
&= \int_0^1 \left( \sum_{m\in \ZZ} \hat{f}(m) e^{2\pi i m x} \right) e^{-2\pi i n x} ~dx \\
&= \sum_{m\in \ZZ} \hat{f}(m) \int_0^1 e^{-2\pi i x (m-n)} \\
&= \sum_{m\in \ZZ} \hat{f}(m) \indic{m = n} \\
&= \hat{f}(m)
,\end{align*}

so the question is now whether 
$$
\hat{g}(n) = \hat{f}(n) ~\forall n \implies g = f \text{ almost everywhere }.
$$

> Note: if we accept this fact at face value, this proof only requires undergraduate analysis and uses facts about uniform convergence allowing sums to commute with integrals.

This will be true if $f \in \ell^2$. 
Why is this the case? 

It's not strict inclusion, since 
$$
L^2([0, 1]) \subseteq L^1([0, 1]) \text{ but } \ell^1(\ZZ) \not\subseteq \ell^2(\ZZ)
.$$

We can use the fact that if $\sum \abs{f_n} < \infty$, then $f_n \to 0$, and in particular $f_n$ is bounded.
So we have

\begin{align*}
\left( \sum \abs{f_n}^2 \right)^{1/2} \leq \sum \abs{f_n} < \infty
.\end{align*}


But then since $\ell^1 \subseteq \ell^2 \implies \theset{\hat{f}(n)} \in \ell^2 \implies f\in L^2$ by Riesz-Fischer and completeness.

$\qed$

*An alternative proof:
*
Again suppose $\theset{\hat{f}(n)} \in \ell^1(\ZZ)$, $S_Nf \to g$.

Since $f\in L^2(\Pi)$ (because of the second assumption), we have $S_Nf \to f$ in $L^2$.

> Use the fact that $f_n \to f \in L^2$ (or $L^1$) implies that a **subsequence converges almost everywhere.**

So $f=g$ almost everywhere.
$\qed$

## Dual of a Vector Space

See notes on web page.

- Definition: Linear functional.

- Examples.

- Definition: Dual space. 
  - Don't need $X$ to be complete, just need a normed vector space.

- Theorem: A linear functional is continuous iff it is *bounded*.

- Definition: Bounded functional.
  - Can't define as $\abs{L(x)} \leq C$ (since linearity allows pulling out scalars), so we can just restrict attention to $\theset{x\suchthat \norm{x} \leq 1}$.

- $X\dual$ is always a vector space

- $X\dual$ is always a Banach space with norm $L \mapsto \sup_{x\in X} \abs{L(x)}$.

- $y \mapsto L_y \definedas \inner{\wait}{y}$ is a conjugate linear isometry $H \to H\dual$ which is surjective.

- Riesz Representation Theorem (for Hilbert spaces)
  - Use the fact that $L\in H\dual \implies \ker L$ is a closed subspace.
  - $z\in M^\perp$, look at $\inner{ (Lx)z - (Lz)x }{z}$.


Upcoming:

- A bit about $L^p$ spaces

- Dual of $L^1$, dual of $L^\infty$.

- Abstract measure theory

- Hahn-Banach, Radon-Nikodym, and Lebesgue Density from the perspective of differentiation theorems.