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\begin{document}
\title{Notes from Fourier Analysis (MATH ) Summer 2005}
\author{http://www.tangentspace.net/mathnotes/fourieranalysis.pdf}
\maketitle

\paragraph{Warning} These notes are nowhere near fleshed out. I'm still typing in material from the start of class, and am proceeding by putting in the statement of the most important stuff. When that's all done, I'm going to go back and put proofs for these, and then add the less important details and comments. I'll remove this warning when the process is complete.

\section{Fourier Series and their convergence}

\begin{thm}[Weierstrass Approximation Theorem]
If $f(\theta)$ is a $2\pi$-periodic function on $[-\pi,\pi]$, then $f$ is the uniform limit of a sequence of trigonometric polynomials of the form $\sum_{-N}^N a_n e^{in\theta}$
\end{thm}

\begin{proof}
\end{proof}

The \emph{Poisson kernel}, defined as 
$$ P_r(\theta) = \frac{1}{2\pi} \sum_{-\infty}^\infty r^{|n|} e^{in\theta} = \frac{1}{2\pi} \frac{1-r^2}{1-2 r \cos \theta + r^2}, \quad 0 \leq r < 1,$$
has the following properties:
\begin{enumerate}[i)]
\item $P_r(\theta) \geq 0$
\item $P_r(\theta) = P_r(-\theta)$
\item $P_r(\theta)$ is monotonically decreasing on $[0,\pi]$
\item $\displaystyle \max_{-\pi \leq \theta \leq \pi} P_r(\theta) = \frac{1}{2\pi} \frac{1+r}{1-r} $
\item  $\displaystyle \min_{-\pi \leq \theta \leq \pi} P_r(\theta) = \frac{1}{2\pi} \frac{1-r}{1+r} $
\item $\int_{-\pi}^\pi P_r(\theta)\,d\theta = 1$
\item For all $\epsilon, \delta > 0$, there is a $r_0$ satisfying $ 0 < r_0 < 1$ such that for all $r$ satisfying $r_0 < r < 1$,
$$ \int_{-\pi}^{-\delta} P_r(\theta)\,d\theta + \int_\delta^\pi P_r(\theta)\,d\theta < \epsilon. $$
\end{enumerate}

\begin{proof}
\end{proof}

\begin{thm}
Let $f$ be a $2\pi$-periodic continuous function on $[-\pi, \pi]$. Let 
$$ u(r, \theta) = \frac{1}{2\pi} \int_{-\pi}^\pi \frac{1-r^2}{1-2r\cos(\theta - t) + r^2} f(t) \, dt = \frac{1}{2\pi} \int_{-\pi}^\pi P_r(\theta-t) f(t)\,dt. $$
Then $\lim_{r \rightarrow 1} u(r, \theta) = f(\theta)$ uniformly.
\end{thm}

\begin{rem}
Note $u(r, \theta) = \sum_{-\infty}^\infty \alpha_n r^{|n|} e^{in\theta} $.
\end{rem}

\begin{proof}
\end{proof}

Let $f$ be real-valued, $f \sim \sum_{-\infty}^\infty \alpha_n e^{in\theta}$, then
$$ \alpha_n = \frac{1}{2\pi}\int_{-\pi}^\pi f(\theta) e^{-in\theta} \,d\theta = \frac{1}{2\pi}\overline{\int_{-\pi}^\pi f(\theta)e^{in\theta} \,d\theta } = \frac{1}{2\pi} \overline{ \int_{-\pi}^\pi f(\theta) e^{-i(-n)\theta} \,d\theta } = \overline{\alpha_{-n}}. $$

A Hilbert space $H$ is a vector space over a field $F$, equipped with an inner product $\langle \cdot, \cdot \rangle : H \times H \rightarrow F$ which has the following properties:
\begin{enumerate}[i)]
 \item nonnegativity and nondegeneracy: $\langle x, x \rangle \geq 0$ for all $x \in H$, and $ \langle x, x \rangle = 0 \Leftrightarrow x = 0 $ 
 \item linearity: $\langle ax + y, z \rangle = a \langle x, z \rangle + \langle y, z \rangle $ for all $x,y,z \in H$ and $a \in F$
 \item conjugate symmetry: $\langle x, y \rangle = \overline{\langle y, x \rangle}$ for all $x,y \in H$
\end{enumerate}
and equipped with a norm $\|x\| = \sqrt{\langle x, x \rangle}$ derived from this inner-product. The norm has the properties:
\begin{enumerate}[i)]
 \item $\|x \| \geq 0$ and $\|x \| = 0 \Leftrightarrow x = 0$ for all $x \in H$
 \item $\|ax\| = |a| \|x\|$ for all $x \in H$ and $a \in F$
 \item $\|x + y\| \leq \|x\| + \|y\|$ for all $x,y \in H$
\end{enumerate}
Finally, $H$ is complete with respect to the norm $\|\cdot\|$: every Cauchy sequence converges to an element in the space. Succintly, a Hilbert space is a complete normed inner-product space.

The concept of a Hilbert space is a generalization of a finite dimensional vector space. The inner-product and norm together give Hilbert spaces a geometry: the inner-product provides a concept of an angle between two vectors, in the same way dot products do in finite dimensional Euclidean spaces, while the norm gives a measure of length. In these notes, the field $F$ is taken to be $\R$ or $\C$.

It is ocassionally useful to know $\|\cdot\|$ has the following properties:
\begin{enumerate}[i)]
 \item $ \left| \|x\| - \|y\| \right| \leq \|x - y\| $
 \item $ x_n \rightarrow x \Leftrightarrow \|x_n - x\| \rightarrow 0 $
\end{enumerate}

\begin{thm}
Let $H$ be a Hilbert space and $M \subset H$ be a closed subspace of $H$, then $H = M \oplus M^\perp$, where $M^\perp = \{ y \in H : \langle x, y \rangle = 0,\; \forall x \in M \} $ is called the orthogonal complement of $M$. Furthermore $M^\perp$ is closed. 
\end{thm}

A complete orthonormal system $\{e_i\}_{i \in I}$ in $H$ is a collection of vectors satisfying
\begin{enumerate}[i)]
 \item $\{e_i\}$ is an orthonormal set
 \item $\Span\{e_i\}$ is dense in $H$: for every $x \in H$, there exists a sequence $\{x_n\} \subset \Span\{e_i\}$ satisfying $\| x_n - x \| \rightarrow 0$ as $n \rightarrow \infty$.
\end{enumerate}

To develop the theory of Fourier series, we will consider the Hilbert space 
$$L^2[-\pi, \pi] = \left\{ f: [-\pi, \pi] \rightarrow \C\, : \, \int_{-\pi}^\pi |f|^2 < \infty \right\} $$
(abbreviated herein as $L^2$) equipped with the inner-product 
$$ \langle f, g \rangle = \frac{1}{2\pi} \int_{-\pi}^\pi f(\theta) \overline{g(\theta) } \,d\theta.$$

\begin{thm}
$ \{ e^{in\theta} \}_{n \in \Z} $ is a complete orthonormal set in $L^2$.
\end{thm}

\begin{proof}
\end{proof}

\begin{thm} Let $\{e_n\}_{n \in \Z}$ be a complete orthonormal sequence in a Hilbert space $H$. Then each element $x \in H$ can be uniquely written as 
$$ x = \sum_{n=-\infty}^\infty \langle x, e_n \rangle e_n, $$
known as the Fourier series of $x$ in $H$, where the infinite sum should be interpreted as saying
$$ \| x - \sum_{j=-N}^N \langle x, e_j \rangle \| \rightarrow 0 \text{ as } N \rightarrow \infty$$
\end{thm}

Note that the equality in the above theorem is only the most commonly used concept of equality under special circumstances. For instance, in the case of $L^2$, the above theorem \emph{does not} guarantee pointwise convergence; there are functions whose Fourier series diverge at each point.

\begin{proof}
\end{proof}

\begin{cor}
For all $x \in H$, $\|x\|^2 = \sum_{n=\infty}^\infty |\langle x, e_n \rangle|^2 $
\end{cor}

\begin{proof}
\end{proof}

\begin{thm}[Best approximations]
Take $x \in H$, $x = \sum_{-\infty}^\infty \langle x, e_n \rangle e_n$, and let $M_n = \bigvee_{k = -n}^n e_n $ be the subspace spanned by a finite subsequence of the orthonormal complete sequence $\{e_n\} \subset H$. Then $ p_n = \sum_{-n}^n \langle x, e_n \rangle e_n $ is the best approximation to $f$ in $M_n$:
$$ \|f - p_n \| \leq \|f - h \| \quad \text{ for all } h \in M_n. $$
\end{thm}

\begin{proof}
\end{proof}

The uniqueness of Fourier coefficients in $L^2$ follows from Hilbert space theory, but $L^1$ is not a Hilbert space, so proving the uniqueness of Fourier coefficients there requires some effort. Dr. Singh's proof of the Unicity Theorem uses invariant subspaces, and is more geometrical than the standard development.

Let $S: L^2 \rightarrow L^2$ be the continuous linear mapping defined by $S: f \mapsto e^{i\theta} f(\theta)$. The commutant of $S$, $\{S\}^\prime$, is the set of continuous linear transformations from $L^2$ to $L^2$ which commute with $S$:
$$ \{S\}^\prime = \{ T:L^2 \rightarrow L^2 \,:\, TS = ST \text{ and } T \text{ is continuous and linear } \}. $$


$L^\infty$ is the set of essentially bounded functions: those which are bounded except on sets of measure 0; given $\varphi \in L^\infty$, let $\|\varphi\|_\infty$ be the essential supremum of $\varphi$. An alternate characterization of $L^\infty$ is
$$ L^\infty = \{ \varphi \in L^2 : \varphi f \in L^2 \text{ for all } f \in L^2 \}. $$
Define for each $\varphi \in L^\infty$ the continuous linear transformation $M_\varphi : L^2 \rightarrow L^2$ defined by $M\varphi(f) = \varphi f$. Since $M_\varphi$ and $S$ are defined by pointwise multiplication, $SM_\varphi = M_\varphi S$, so $L^\infty \subset \{S\}^\prime$. In particular, $M_p$ for any trigonometric polynomial $p(\theta) = \alpha_0 + \alpha_1 e^{i\theta} + \cdots + \alpha_n e^{in \theta}$ commutes with $S$; notice that $M_p$ can also be expressed as a sum of powers of $S$: $M_p = \alpha_0 I + \alpha_1 S + \cdots + \alpha_n S^n$. As a special case, $S^{-1} = M_{e^{-i\theta}} \in \{S\}^\prime$. 

Each $T \in \{S\}^{-1}$ can be characterized by a function $\varphi_T \in L^2$ as follows: $T : 1 \in L^2 \mapsto \varphi_T \in L^2 $, so 
$$ T(e^{i\theta}) = T(S(1)) = S(T(1)) = e^{i\theta} \varphi_T, $$
and likewise $T(e^{i\theta}) = e^{i\theta} \varphi_T $. Let $p(\theta) = \alpha_{-n}e^{-in\theta} + \alpha_{-n+1}e^{i (-n + 1) \theta} + \cdots + \alpha_0 + \alpha_1 e^{i\theta} + \cdots + \alpha_n e^{in\theta} \in L^2$, then 
$$ T(p(\theta)) = T(M_p(1)) = T( (\alpha_{-n}S^{-n} + \alpha_{-n+1}S^{-n+1} + \cdots + \alpha_0 I + \cdots + \alpha_n S^n )(1)) = M_p(T(1)) = p(\theta)\varphi_T. $$
Let $f$ be an arbitrary element in $L^2$, then there is a sequence of trigonometric polynomials $\{f_N = \sum_{n=-N}^N \alpha_n e^{in\theta} \}_{N \in \N}$ converging to $f$. Since $T$ is continuous, $T(f_N) = \varphi_T f_N \rightarrow T(f)$; $\varphi_T f_N \rightarrow \varphi_T f$, so $T(f) = \varphi_T f$. So for all $T \in \{S\}^\prime$, there is a $\varphi_T \in L^\infty$ such that $T = M_{\varphi_T}$. Therefore $\{S\}^\prime = \{M_\varphi : \varphi \in L^\infty\}$.

A closed subspace $M$ of $L^2$ such that $S(M)=M$ is said to be doubly invariant. A simple example of a doubly invariant subspace is the space $ M = \chi_I L^2$ where $I$ is an interval in $[-\pi, \pi]$:
$$ S(M) = e^{i\theta} M = e^{i\theta} \chi_I L^2 = \chi_I e^{i\theta} L^2 = \chi_I L^2 = M. $$
In this case, $M$ is the set of functions in $L^2$ which vanish outside of $I$.

In fact, every doubly invariant subspace can be described as the subset of $L^2$ which vanishes outside some fixed set!

\begin{thm}
Let $M$ be a doubly invariant subspace of $L^2$. Then there exists a set $I$ of positive measure such that $M = \chi_I L^2 $.
\end{thm}

\begin{proof}
\end{proof}

\begin{thm}[Unicity theorem for $L^1$]
Let $f \in L^1[-\pi, \pi]$ and suppose $\int_{-\pi}^\pi f(\theta) e^{-in\theta} \, d\theta = 0$ for all $n \in \Z$, then $f = 0$.
\end{thm}

\begin{proof}
\end{proof}

\end{document}