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\newtheorem{theorem}{Theorem}
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{criterion}[theorem]{Criterion}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}[theorem]{Exercise}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{problem}[theorem]{Problem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{solution}[theorem]{Solution}
\newtheorem{summary}[theorem]{Summary}
\newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\input{tcilatex}

\begin{document}


\begin{center}
Fourier Series in Complex notation
\end{center}

\bigskip 
\begin{align*}
\sin (x) &= \frac{e^{ix}-e^{-ix}}{2i} \qquad \cos (x)= \frac{e^{ix}+e^{-ix}}{%
2} \\
&= -\frac{i}{2}\left( e^{ix}-e^{-ix}\right)
\end{align*}

\bigskip \noindent So

\begin{align*}
\varphi(x) &=\frac{A_0}{2}+\sum_{n=1}^{\infty } A_n \cos \left( \frac{n\pi x%
}{l}\right) + B_n \sin \left( \frac{n\pi x}{l}\right) \qquad -l<x<l \\
&= \frac{A_0}{2}+\sum_{n=1}^{\infty }A_{n}\frac{e^{in\pi x/l}+e^{-in\pi x/l}%
}{2}-\frac{i}{2}B_{n}\left( e^{in\pi x/l}-e^{-in\pi x/l}\right) \\
&= \frac{A_{0}}{2}+\sum_{n=1}^{\infty }\frac{A_{n}-iB_{n}}{2}e^{in\pi x/l}+%
\frac{A_{n}+iB_{n}}{2}e^{-in\pi x/l} \\
&= \sum_{n=-\infty }^{\infty }C_{n}e^{\frac{in\pi x}{l}}\qquad \qquad C_{n}=%
\begin{cases}
\frac{A_{n}-iB_{n}}{2} & n>0 \\ 
\frac{A_{-n}+iB_{-n}}{2} & n<0 \\ 
\frac{A_{0}}{2} & n=0%
\end{cases}
\\
\text{So} \quad C_{-n}& =C_{n}^{\ast }
\end{align*}

\begin{align*}
\int_{-l}^{l}e^{\frac{in\pi x}{l}}e^{-\frac{im\pi x}{l}}dxdx& =%
\begin{cases}
0 & m\neq n \\ 
2l & m=n%
\end{cases}
\\
C_{n}& =\frac{1}{2l}\int_{-l}^{l}\varphi (x)e^{-\frac{in\pi x}{l}}dx
\end{align*}

\newpage

\begin{center}
Sturm-Liouville
\end{center}

\begin{align*}
\text{(*) \ \ }-\frac{d}{dx}\left( p(x)\frac{du}{dx}\right) +q(x)u(x)&
=\lambda m(x)u(x)\qquad a<x<b \\
p(x)>0\quad q(x)& \geq 0\quad m(x)>0
\end{align*}
\noindent and all quantities are real. \bigskip

\begin{definition}
$\lambda $ is the eigenvalue and $u$ is the eigenfunction
\end{definition}

\medskip

\begin{definition}
A homogeneous boundary condition is \textbf{symmetric} if%
\begin{equation*}
p\left[ fg^{\prime}-f^{\prime} g \right]_{a}^{b}=0
\end{equation*}
\end{definition}

\bigskip

\noindent Examples:

\begin{itemize}
\item Dirichlet

\item Neumann

\item periodic

\item Robin (?)
\end{itemize}

\begin{align*}
a_{l}u(a)+b_{l}\frac{du}{dx}(a) &= 0 \\
a_{r}u(b)+b_{r}\frac{du}{dx}(b) &= 0
\end{align*}

\noindent then%
\begin{equation*}
p\left[ fg^{\prime }-f^{\prime }g\right] _{a}^{b}=f(a)\left[ -\frac{a_{l}}{%
b_{l}}g(a)\right] -\left( -\frac{a_{l}}{b_{l}}\right) f(a)g(a)=0
\end{equation*}

\begin{definition}
Inner product%
\begin{equation*}
(u,v)=\int_{a}^{b}m(x)u(x)v(x)dx\qquad m>0
\end{equation*}
\end{definition}

\newpage

\begin{center}
Example
\end{center}

\begin{equation*}
p=1\quad q=0\quad m=1
\end{equation*}

\bigskip

Then with Dirichlet conditions we have%
\begin{align*}
-\frac{d^{2}u}{dx^{2}} &= \lambda u(x)\qquad 0<x<l \\
u(0) &= u(l)=0
\end{align*}

\noindent If $\lambda >0$ then%
\begin{equation*}
u(x)=A\cos (\sqrt{\lambda }x)+B\sin (\sqrt{\lambda }x)
\end{equation*}

\noindent Using $u(0)=0$ we get 
\begin{equation*}
u(x)=B\sin (\sqrt{\lambda }x)
\end{equation*}%
\noindent and using $u(l)=0$ 
\begin{align*}
\lambda &= \left( \frac{n\pi }{l}\right) ^{2} \\
u(x) &= B\sin (\frac{n\pi }{l}x)
\end{align*}%
\noindent So we have an infinite number of eigenvalues/eigenfunctions.

\begin{center}
Green's Identities
\end{center}

\bigskip

\noindent First Identity:%
\begin{equation*}
\int_{a}^{b}\frac{d}{dx}\left( p(x)\frac{du}{dx}\right)
v(x)dx+\int_{a}^{b}p(x)\frac{du}{dx}\frac{dv}{dx}dx=\left[ p(x)\frac{du}{dx}v%
\right] _{a}^{b}
\end{equation*}

\bigskip

\noindent Second Identity:%
\begin{equation*}
\int_{a}^{b}-\frac{d}{dx}\left( p(x)\frac{du}{dx}\right) v(x)dx+\int_{a}^{b}%
\frac{d}{dx}\left( p(x)\frac{dv}{dx}\right) u(x)dx=\left[ p(x)\left( -\frac{%
du}{dx}v+\frac{dv}{dx}u\right) \right] _{a}^{b}
\end{equation*}

\bigskip

\noindent In multi-dimensions this generalizes to \ ($p=1)$

\begin{align*}
\iiint_{D}\left( \nabla u\cdot \nabla v+v\Delta u\right) dV &=
\iint_{\partial D}v\frac{\partial u}{\partial n}dS \\
\iiint_{D}\left( v\Delta u-u\Delta v\right) dV &= \iint_{\partial D}\left( v%
\frac{\partial u}{\partial n} -u\frac{\partial v}{\partial n}\right) dS
\end{align*}

\bigskip

\begin{proof}
\noindent From the divergence theorem%
\begin{align*}
\iiint \text{div}(F)dV &= \iint F\cdot ndS \\
\text{div}(v\text{ grad }u) &= \nabla u\cdot \nabla v+v\Delta u \\
\text{So}\iiint \left( \nabla u\cdot \nabla v+v\Delta u\right) dV &= \iiint 
\text{ div}(v\text{ grad }u)dV \\
&= \iint v\text{ grad }u\cdot ndS=\iint v\frac{\partial u}{\partial n}dS
\end{align*}
\end{proof}

\bigskip

\noindent Interchange $u$ and $v$ and subtract to get the second identity.

\newpage

\noindent From (*)

\bigskip

\begin{align*}
\text{(1) \ }-\frac{d}{dx}\left( p(x)\frac{du}{dx}\right) +q(x)u(x) &=
\lambda _{1}m(x)u(x) \\
\text{(2) \ }-\frac{d}{dx}\left( p(x)\frac{dv}{dx}\right) +q(x)v(x) &=
\lambda _{2}m(x)v(x)
\end{align*}

\noindent Multiply (1) by $v$ and (2) by $u$ and subtract and integrate

\bigskip

\begin{equation*}
\int_{a}^{b}\left[ -v(x)\frac{d}{dx}\left( p(x)\frac{du}{dx}\right) +u(x)%
\frac{d}{dx}\left( p(x)\frac{dv}{dx}\right) \right] dx=\left( \lambda
_{1}-\lambda _{2}\right) \int_{a}^{b}m(x)u(x)v(x)dx
\end{equation*}

\medskip

\noindent Integrate by parts (Green's theorem in multidimensions)

\medskip

\begin{align*}
\int_{a}^{b}\left[ \frac{dv}{dx}\left( p(x)\frac{du}{dx}\right) -\frac{du}{dx%
}\left( p(x)\frac{dv}{dx}\right) \right] dx+\left[ p\left( \frac{du}{dx}v-%
\frac{dv}{dx}u\right) \right] _{a}^{b} &= \left( \lambda _{1}-\lambda
_{2}\right) \int_{a}^{b}m(x)u(x)v(x)dx \\
\left[ p\left( \frac{du}{dx}v-\frac{dv}{dx}u\right) \right] _{a}^{b} &=
\left( \lambda _{1}-\lambda _{2}\right) \int_{a}^{b}m(x)u(x)v(x)dx
\end{align*}

\noindent If the boundary conditions are symmetric then 
\begin{equation*}
\left( \lambda _{1}-\lambda _{2}\right) \int_{a}^{b}m(x)u(x)v(x)dx=0
\end{equation*}

\noindent Hence, if $\lambda _{1}\!\neq \!\lambda _{2}$

\bigskip

\begin{theorem}
For symmetric boundary conditions , if $\lambda _{1}\!\neq \!\lambda _{2}$
then $(u,v)=0$

\noindent If $\lambda_1 \!=\!\lambda_2 $ then we have a subspace and we can
choose an orthogonal basis. \medskip \noindent\ This is done by Gram Schmidt
\end{theorem}

\newpage

\begin{center}
Gram-Schmidt
\end{center}

\bigskip

If $\left\{ \psi _{k}(x)\right\} $ is a linearly independent basis then we
can construct an orthonormal basis that spans the same space.

\bigskip

\begin{itemize}
\item $\varphi _{1}(x)=\psi _{1}(x)$

\item $\varphi _{2}(x)=\psi _{2}(x)-\frac{(\psi _{2},\psi _{1})}{(\varphi
_{1},\varphi _{1})}\varphi _{1}$

Then $(\varphi _{2},\varphi _{1})=(\psi _{2},\varphi _{1})-\frac{(\psi
_{2},\varphi _{1})}{(\varphi _{1},\varphi _{1})}(\varphi _{1},\varphi
_{1})=0 $

\item $\varphi _{k}(x)=\psi _{k}(x)-\dsum\limits_{j=1}^{k-1}\frac{(\psi
_{k},\varphi _{j})}{(\varphi _{j},\varphi _{j})}\varphi _{j}$
\end{itemize}

\bigskip

\noindent Hence, we can consider all the solutions of (*) to be orthogonal
to each other.

\noindent So consider a sequence of orthogonal solutions of (*) $\left\{
\varphi _{k}(x)\right\} .$Then if%
\begin{align*}
f(x)& =\dsum\limits_{n}a_{n}\varphi _{n}(x) \\
(f,\varphi _{m})& =\sum_{n}a_{n}\left( \varphi _{n},\varphi _{m}\right) \\
& =a_{m}\left( \varphi _{m},\varphi _{m}\right)
\end{align*}

\noindent So%
\begin{equation*}
a_{m}=\frac{(f,\varphi _{m})}{\left( \varphi _{m},\varphi _{m}\right) }=%
\frac{\int_{a}^{b}m(x)f(x)\varphi _{m}(x)dx} {\int_{a}^{b}m(x)\varphi
_{m}^{2}(x)dx}
\end{equation*}

\bigskip

\begin{theorem}
\noindent If p,q,m are real and the boundary conditions are symmetric then
there are no complex eigenvalues
\end{theorem}

\medskip

\begin{proof}
\begin{align*}
\text{(1) \ }-\frac{d}{dx}\left( p(x)\frac{du}{dx}\right) +q(x)u(x)&
=\lambda m(x)u(x) \\
\text{(2) \ }-\frac{d}{dx}\left( p(x)\frac{d\overline{u}}{dx}\right) +q(x)%
\overline{u}(x)& =\overline{\lambda }m(x)\overline{u}(x)
\end{align*}

\noindent As before multiply first equation by $\overline{u}$ , the second
by $u$, subtract and integrate. Then

\begin{equation*}
\int_{a}^{b}\left[ -\overline{u}(x)\frac{d}{dx}\left( p(x)\frac{du}{dx}%
\right) +u(x)\frac{d}{dx}\left( p(x)\frac{d\overline{u}}{dx}\right) \right]
dx=\left( \lambda -\overline{\lambda }\right) \int_{a}^{b}m(x)u(x)\overline{u%
}(x)dx
\end{equation*}

\noindent Again integrate by parts and use the symmetry of the boundary
conditions.%
\begin{align*}
\left( \lambda -\overline{\lambda }\right) \int_{a}^{b}m(x)|u(x)|^{2}dx &= 0
\\
\text{So }\lambda &= \overline{\lambda } \qquad \text{i.e. } \lambda \text{
is real}
\end{align*}
\end{proof}

\noindent If $u$ is complex then its real and imaginary components are
solutions.

\bigskip

\begin{center}
Negative Eigenvalues
\end{center}

\begin{equation*}
\text{(*) \ \ }-\frac{d}{dx}\left( p(x)\frac{du}{dx}\right)
+q(x)u(x)=\lambda m(x)u(x)\qquad a<x<b
\end{equation*}

\noindent By Green's first identity we have for all $v$

\begin{equation*}
\int_{a}^{b}\frac{d}{dx}\left( p(x)\frac{du}{dx}\right)
v(x)dx=-\int_{a}^{b}p(x)\frac{du}{dx}\frac{dv}{dx}dx+\left[ p(x)\frac{du}{dx}%
v\right] _{a}^{b}
\end{equation*}

\noindent Choose $v=u$ and assume symmetric boundary conditions. Then

\begin{equation*}
\int_{a}^{b}u(x)\frac{d}{dx}\left( p(x)\frac{du}{dx}\right)
dx=-\int_{a}^{b}p(x)\left( \frac{du}{dx}\right) ^{2}dx\leq 0
\end{equation*}

\noindent Using the ODE we get

\begin{equation*}
\int_{a}^{b}u(x)\left[ q(x)u(x)-\lambda m(x)u(x)\right] dx=-\int_{a}^{b}p(x)%
\left( \frac{du}{dx}\right) ^{2}dx
\end{equation*}

\noindent So%
\begin{align*}
\lambda \int_{a}^{b}m(x)u^{2}(x)dx &= \int_{a}^{b}q(x)u^{2}(x)dx+
\int_{a}^{b}p(x)\left( \frac{du}{dx}\right)^2dx \\
\lambda &= \frac{\int_{a}^{b}q(x)u^{2}(x)dx+ \int_{a}^{b}p(x)\left( \frac{du%
}{dx}\right) ^{2}dx} {\int_{a}^{b}m(x)u^{2}(x)dx}\geq 0
\end{align*}

\noindent We can have equality only if $q(x)=0$ and $\frac{du}{dx}=0$

\noindent \textbf{Note}: All these proofs work equally well in
multidimensions using Green's theorem instead of integration by parts.

\newpage

\begin{center}
Completeness
\end{center}

\bigskip

\begin{theorem}
There are an infinite number of eigenvalues for (*) and $\lambda_{n}%
\rightarrow \infty$. \newline
\noindent Furthermore%
\begin{align*}
f(x) &= \sum c_{n}\varphi _{n}(x) \\
c_{n} &= \frac{(f,\varphi _{n})}{(\varphi _{n},\varphi _{n})}
\end{align*}
\end{theorem}

\begin{center}
\bigskip \textbf{Convergence:}
\end{center}

\begin{equation*}
\sum_{n=1}^{\infty }\varphi _{n}(x)\overset{?}{\rightarrow }f(x)
\end{equation*}

\begin{definition}
Pointwise Convergence:%
\begin{equation*}
\lim_{N\rightarrow \infty }\left\vert f(x)-\sum_{n=1}^{N}\varphi
_{n}(x)\right\vert =0\qquad \text{for every x}
\end{equation*}
\end{definition}

\begin{definition}
Uniform Convergence: 
\begin{equation*}
\lim_{N\rightarrow \infty }\max_{a\leq x\leq b}\left\vert
f(x)-\sum_{n=1}^{N}\varphi _{n}(x)\right\vert =0\qquad \text{for every x}
\end{equation*}
\end{definition}

\begin{definition}
$L_{2}$ (root mean square)%
\begin{equation*}
\lim_{N\rightarrow \infty }\int_{a}^{b}\left\vert f(x)-\sum_{n=1}^{N}\varphi
_{n}(x)\right\vert ^{2}dx=0
\end{equation*}
\end{definition}

\noindent Uniform convergence implies pointwise convergence.

\noindent Uniform convergence implies root mean square convergence.

\bigskip

\begin{center}
Examples
\end{center}

\begin{itemize}
\item 
\begin{equation*}
f(x)=x^{n}\qquad 0 \le x \le 1
\end{equation*}

Then%
\begin{equation*}
x_{n}\rightarrow 
\begin{cases}
0 & 0 < x < 1 \\ 
1 & x=1%
\end{cases}%
\end{equation*}

\item 
\begin{align*}
f_{n}(x)& =(1-x)x^{n-1}=x^{n-1}-x^{n} \\
\sum_{n=1}^{N}f_{n}(x)& =\sum_{n=1}^{N}(x^{n-1}-x^{n})=1-x^{N}\rightarrow
1\qquad \text{as\ }N\rightarrow \infty
\end{align*}%
so we have pointwise convergence.

\noindent However%
\begin{equation*}
\max_{0\leq x\leq 1}\left\vert 1-\left( 1-x^{N}\right) \right\vert
=\max_{0\leq x\leq 1}\left\vert x^{N}\right\vert =1\neq 0
\end{equation*}

\noindent So we don't have uniform convergence.

\noindent For $L_2$ we have%
\begin{equation*}
\int_{0}^{1}\left\vert x^{N}\right\vert ^{2}dx= \frac{1}{2N+1}\rightarrow 0
\end{equation*}

\noindent So we have $L_2$ convergence.
\end{itemize}

\newpage

\begin{theorem}
If

\begin{itemize}
\item $f, f^{\prime}, f^{\prime \prime}$ exist and are continuous in $a \le
x \le b$ i.e. $f\epsilon C^2 [a,b]$

\item f satisfies the boundary conditions
\end{itemize}

then 
\begin{equation*}
f(x)=\sum_{n=1}^{\infty }a_{n}\varphi _{n}(x)\qquad \text{converges uniformly%
}
\end{equation*}
\end{theorem}

\begin{theorem}
If\ $\int_{a}^{b}f^{2}(x)dx<\infty $

then $f(x)=\tsum\limits_{n=1}^{\infty }a_{n}\varphi _{n}(x)$ \qquad
converges in $L_{2}$
\end{theorem}

\begin{theorem}
For sine and cosine series only.

If

\begin{itemize}
\item $f$ is continuous on $a\leq x\leq b$

\item $f^{^{\prime }}$is piecewise continuous

Then the series converges pointwise. If $f$ and $f^{^{\prime }}$are
piecewise continuous then%
\begin{equation*}
\sum a_{n}\varphi _{n}(x)\rightarrow \frac{f(x+)+f(x-)}{2}
\end{equation*}
\end{itemize}
\end{theorem}

\begin{theorem}
Integration: \ \noindent If formally 
\begin{equation*}
f(x)\leftrightarrow \frac{A_{0}}{2}+\dsum\limits_{n=1}^{\infty }A_{n}\cos
\left( \frac{n\pi x}{l}\right) +B_{n}\sin \left( \frac{n\pi x}{l}\right)
\qquad \text{\textbf{not necessarily convergent}}
\end{equation*}

\noindent Then%
\begin{equation*}
\tint\limits_{-l}^{x}f(y)dy=\frac{A_{0}}{2}(x+l)+\frac{l}{\pi }%
\dsum\limits_{n=1}^{\infty }\left[ \frac{A_{n}}{n}\sin \left( \frac{n\pi t}{l%
}\right) -\frac{B_{n}}{n}\cos \left( \frac{n\pi t}{l}\right) \right]
_{-l}^{x}\quad \text{is convergent}
\end{equation*}
\end{theorem}

\noindent We note that for differentiation it is the opposite i.e. the
derivative of a convergent series may not converge.

\noindent Example: expanding $x$ in a sine series we have 
\begin{equation*}
x=2\dsum\limits_{n=1}^{\infty }\frac{l}{n\pi }(-1)^{n+1}\sin \left( \frac{%
n\pi x}{l}\right) \quad 0\leq x\leq l
\end{equation*}

\noindent Differentiating we get%
\begin{equation*}
1=2\dsum\limits_{n=1}^{\infty }(-1)^{n+1}\cos \left( \frac{n\pi x}{l}\right)
\quad 0\leq x\leq l
\end{equation*}

\noindent This is certainly \textbf{NOT} the cosine series of $1$ which is
just $1$ . In fact this series does not converge!

\bigskip

\noindent We begin with the proof of convergence in least squares. Restating
the theorem we have

\begin{theorem}
If $\varphi _{n}$ are the eigenfunctions of a Sturm-Liouville problem with
symmetric boundary conditions and $||f||<\infty $. Then 
\begin{equation*}
||f-\sum_{n=1}^{N}a_{n}\varphi _{n}||\rightarrow 0\qquad a_{n}=\frac{%
(f,\varphi _{n})}{(\varphi _{n},\varphi _{n})}
\end{equation*}
\end{theorem}

\begin{theorem}
Let $\varphi _{n}$ be an orthogonal set and $||f||<\infty $, Then the choice
of constants $c_{n}$ that minimizes $||f-\tsum\limits_{n\leq N}c_{n}\varphi
_{n}||$ is $c_{n}=a_{n}$
\end{theorem}

\begin{proof}
Assume for simplicity that all quantities are real%
\begin{align*}
E_{N}& =||f-\sum_{n\leq N}c_{n}\varphi _{n}||^{2}=\tint |f(x)-\sum_{n\leq
N}c_{n}\varphi _{n}(x)|^{2}dx \\
& =\tint |f(x)|^{2}-2\sum_{n\leq N}c_{n}\tint f(x)\varphi
_{n}(x)dx+\tsum\limits_{n}\tsum\limits_{m}c_{n}c_{m}\int \varphi
_{n}(x)\varphi _{m}(x)dx \\
& =||f||^{2}-2\sum_{n\leq N}c_{n}(f,\varphi _{n})+\sum_{n\leq
N}c_{n}^{2}||\varphi _{n}(x)||^{2} \\
& =||f||^{2}+\sum_{n\leq N}||\varphi _{n}(x)||^{2}\left[ c_{n}-\frac{%
(f,\varphi _{n})}{||\varphi _{n}(x)||^{2}}\right] ^{2}-\sum_{n\leq N}\frac{%
(f,\varphi _{n})^{2}}{||\varphi _{n}(x)||^{2}}
\end{align*}

To minimize we can only "play" with $c_{n}$ . Since the middle term is
positive we minimize $E_{N}$ if 
\begin{equation*}
c_{n}=\frac{(f,\varphi _{n})}{||\varphi _{n}(x)||^{2}}=a_{n}
\end{equation*}

Then%
\begin{equation*}
E_{N}=||f||^{2}-\sum_{n\leq N}\frac{(f,\varphi _{n})^{2}}{||\varphi
_{n}(x)||^{2}}=||f||^{2}-\sum_{n\leq N}A_{n}^{2}||\varphi _{n}(x)||^{2}\geq 0
\end{equation*}

So we have Bessel's inequality. If $||f||^{2}<\infty $ then%
\begin{equation*}
\sum_{n\leq N}\frac{(f,\varphi _{n})^{2}}{||\varphi _{n}(x)||^{2}}\leq
||f||^{2}
\end{equation*}
\end{proof}

\begin{center}
Parseval's Equality
\end{center}

\begin{theorem}
The Fourier Series converges to $f(x)$ in $L_{2}$ if and only if%
\begin{equation*}
\sum_{n\leq N}\frac{(f,\varphi _{n})^{2}}{||\varphi _{n}(x)||^{2}}%
=\sum_{n\leq N}a_{n}^{2}||\varphi _{n}(x)||^{2}=||f||^{2}
\end{equation*}
\end{theorem}

\begin{definition}
A sequence $\left\{ {\varphi _{n}(x)}\right\} $ is complete if Parseval's
equality holds whenever $||f||^{2}<\infty $
\end{definition}

\begin{center}
Riemann-Lebesque Theorem
\end{center}

\begin{theorem}
If (a) $f\in C^{1}$

or (b) $||f||_{L^{2}}<\infty $

\noindent Then%
\begin{equation*}
\lim_{n\rightarrow \infty }\tint\limits_{-l}^{l}f(x) 
\begin{cases}
\sin (\frac{n\pi x}{l}) \\ 
\cos (\frac{n\pi x}{l})%
\end{cases}
dx=0
\end{equation*}
\end{theorem}

\begin{proof}
\begin{enumerate}
\item integration by parts

\item In a Fourier series $B_{n}=\tint\limits_{-l}^{l}f(x)\sin (\frac{n\pi x%
}{l})dx$ by Bessel's inequality $B_{n}\rightarrow 0$
\end{enumerate}
\end{proof}

\begin{center}
Example
\end{center}

\noindent Consider $f(x)=1$ on $(0,\pi )$. We find that%
\begin{equation*}
1=\tsum\limits_{\text{n odd}}\frac{4}{n\pi }\sin (nx)
\end{equation*}

\noindent So by Parseval's equality%
\begin{align*}
\tint\limits_{0}^{\pi }1^{2}dx &= \tsum\limits_{\text{n odd}}\left( \frac{4}{%
n\pi }\right) ^{2}\frac{\pi }{2} \\
\tsum\limits_{\text{n odd}}\frac{1}{n^{2}} &= \frac{\pi ^{2}}{8}
\end{align*}

\bigskip

\begin{center}
Pointwise Convergence
\end{center}

\noindent If 
\begin{align*}
f(x) &= \frac{A_{0}}{2}+\dsum\limits_{n=1}^{\infty }A_{n}\cos \left(
nx\right) +B_{n}\sin \left( nx\right) \\
A_{n} &= \frac{1}{\pi }\tint\limits_{-\pi }^{\pi }f(y)\cos (ny)dy\qquad
n=0,1,2,3... \\
B_{n} &= \frac{1}{\pi }\tint\limits_{-\pi }^{\pi }f(y)\sin (ny)dy\qquad
n=1,2,3...
\end{align*}

\begin{center}
\bigskip

\bigskip

Dirichlet kernel
\end{center}

\noindent Consider the partial sum%
\begin{align*}
S_{N} &= \frac{A_{0}}{2}+\dsum\limits_{n=1}^{N}A_{n}\cos \left( nx\right)
+B_{n}\sin \left( nx\right) \\
&= \frac{1}{2\pi }\tint\limits_{-\pi }^{\pi }\left[ 1+2\dsum%
\limits_{n=1}^{N}\left( \cos (nx)\cos (ny)+\sin (nx)\sin (ny)\right) \right]
f(y)dy \\
&= \frac{1}{2\pi }\tint\limits_{-\pi }^{\pi }\left[ 1+2\dsum%
\limits_{n=1}^{N}\cos (nx-ny)\right] f(y)dy \\
&= \frac{1}{2\pi }\tint\limits_{-\pi }^{\pi }K_{N}(x-y)f(y)dy
\end{align*}

\noindent where%
\begin{equation*}
K_{N}(\theta )=1+2\dsum\limits_{n=1}^{N}\cos (n\theta )= \frac{\sin (N+\frac{%
1}{2})\theta }{\sin (\frac{\theta }{2})}
\end{equation*}

\begin{proof}
Use $\cos (\theta )=\frac{e^{in\theta }+e^{-in\theta }}{2}$and get geometric
series.
\end{proof}

\bigskip

\noindent Now let $\theta =y-x$. Then%
\begin{align*}
S_{N}& =\frac{1}{2\pi }\tint\limits_{-\pi }^{\pi }K_{N}(\theta )f(x+\theta
)d\theta \\
S_{N}(x)-f(x)& =\frac{1}{2\pi }\tint\limits_{-\pi }^{\pi }K_{N}(\theta )%
\left[ f(x+\theta )-f(x)\right] d\theta \\
& =\frac{1}{2\pi }\tint\limits_{-\pi }^{\pi }g(\theta )\sin (N+\frac{1}{2}%
)\theta \,d\theta \qquad \text{where }g(\theta )=\frac{f(x+\theta )-f(x)}{%
\sin (\frac{\theta }{2})}
\end{align*}

\noindent Let $\phi _{n}(\theta )=\sin (N+\frac{1}{2})\theta $. By Bessel's
inequality we have%
\begin{equation*}
\dsum\limits_{n=1}^{\infty }\frac{|(g,\phi _{n})|}{||\phi _{n}||^{2}}=\frac{1%
}{\pi }\dsum\limits_{n=1}^{\infty }|(g,\phi _{n})|\leq
||g||^{2}=\tint\limits_{-\pi }^{\pi }\frac{\left[ f(x+\theta )-f(x)\right]
^{2}}{\sin ^{2}(\frac{\theta }{2})}d\theta
\end{equation*}

\noindent By L'hopital's rule the integrand is finite at $\theta =0$ . Hence
it is bounded everywhere and the integral exists. Since the sum converges
each term much approach zero and so%
\begin{equation*}
|(g,\phi _{n})|=\tint\limits_{-\pi }^{\pi }g(\theta )\sin (N+\frac{1}{2}%
)\theta \,d\theta \rightarrow 0
\end{equation*}

\bigskip

\begin{center}
Gibbs Phenomena
\end{center}

\bigskip

\noindent If instead we are interested in uniform convergence we need to
analyze%
\begin{equation*}
\lim_{N\rightarrow \infty }\max_{x}|S_{N}(x)-f(x)|
\end{equation*}%
\noindent One can show that if the function $f(x)$ has a discontinuity at $%
x=x_{0}$ then in fact this limit is nonzero and is about 9\% of the size of
the jump on either side.

\end{document}