%%%%		-*- Mode: LaTeX -*-
%%%% Author:    J. Robert Buchanan
%%%% Address:   Department of Mathematics
%%%%		Millersville University
%%%%            P.O. Box 1002
%%%%            Millersville, PA 17551-0302
%%%% Phone:     717-871-7305
%%%% FAX:	717-871-7948
%%%% Email:     Robert.Buchanan@millersville.edu
%%%%
%%%% Date:      06/30/2022
%%%%
%%%% Purpose:   Dirichlet Convergence Theorem
%%%%

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\title{Dirichlet Convergence Theorem}
\subtitle{MATH 467 \textit{Partial Differential Equations}} % (optional)


\author % (optional, use only with lots of authors)
{J~Robert~Buchanan}
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\institute[Millersville University of Pennsylvania]
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{
  Department of Mathematics
}
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  \titlepage
\end{frame}

%\begin{frame}
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% Since this a solution template for a generic talk, very little can
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%   15 and 30 frames, all told.

%\section{Introduction}

\begin{frame}\frametitle{Objectives}

In this lesson we will sketch proofs of
\begin{itemize}
\item the Dirichlet Convergence Theorem for Fourier series, and
\item the theorem regarding the uniform convergence of Fourier series.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Dirichlet Convergence Theorem}

\begin{theorem}
Assume that $f(x)$ is a
piecewise smooth function on the interval $[-L,L]$ extended to
$(-\infty,\infty)$ periodically with period $2L$.
Then the  Fourier series of $f(x)$ converges for all $x$ to the value
\[
\frac{1}{2} \left( f(x+) + f(x-) \right)	.
\]
\end{theorem}

\end{frame}
%%

\begin{frame}\frametitle{First Necessary Lemma}

In order to prove Dirichlet's convergence theorem for Fourier series we will
need the following lemma.
\begin{lemma}
For any fixed $L>0$ and piecewise continuous function $g(x)$
on interval $[0,L]$,
\[
\lim_{N \to\infty}
\int_{0}^{L} g(t) \sin \frac{(2N+1) \pi t}{2L} \, dt
	= 0	.
\]
\end{lemma}

\end{frame}
%%

\begin{frame}\frametitle{Proof}

Let $g(x)$ be piecewise continuous on $[0,L]$.
\begin{align*}
\lefteqn{\int_{0}^{L} g(t) \sin \frac{(2N+1) \pi t}{L} \, dt}	\\
	&=
\int_{0}^{L} \left[ g(t) \sin \frac{\pi t}{L} \right]
\cos \frac{2 N \pi t}{L} \, dt +
\int_{0}^{L} \left[ g(t) \cos \frac{\pi t}{L} \right]
\sin \frac{2 N \pi t}{L} \, dt
\end{align*}
\uncover<2->{%
Consider the even $2L$-periodic extension of $g(t) \sin (\pi t/L)$ and
the $2N$th coefficient in the Fourier cosine series for this
extension.
\begin{align*}
a_{2N} &= \frac{2}{L} \int_{0}^{L} g(t) \sin \frac{\pi t}{L} \cos
\frac{2 N \pi t}{L} \, dt	\\
\frac{L}{2} a_{2N} &= \int_{0}^{L} g(t) \sin \frac{\pi t}{L} \cos
\frac{2 N \pi t}{L} \, dt	\\
\lim_{N \to \infty} \frac{L}{2} a_{2N} &=
\lim_{N \to \infty} \int_{0}^{L} g(t) \sin \frac{\pi t}{L} \cos
\frac{2 N \pi t}{L} \, dt = 0
\end{align*}
In a similar manner the other integral converges to 0 as $N \to \infty$.
}

\end{frame}
%%

\begin{frame}\frametitle{$N$th Partial Sum Revisited}

\begin{align*}
S_{N}(x)
	= {} & \frac{a_{0}}{2} + \sum_{n=1}^{N} \left( a_{n} \cos \frac{n \pi x}{L} +
              b_{n} \sin \frac{n \pi x}{L} \right)	\\
	= {} & \frac{1}{2L} \int_{-L}^{L} f(t) \, dt +
\frac{1}{L} \sum_{n=1}^{N} \left[ \int_{-L}^{L} f(t) \cos \frac{n \pi
              t}{L} \, dt \right] \cos \frac{n \pi x}{L}	\\
	& +
\frac{1}{L} \sum_{n=1}^{N} \left[ \int_{-L}^{L} f(t) \sin \frac{n \pi
              t}{L} \, dt \right] \sin \frac{n \pi x}{L}	\\
	= {} & \frac{1}{2L} \int_{-L}^{L} f(t) \, dt	\\
	& +
\frac{1}{L} \sum_{n=1}^{N} \int_{-L}^{L} f(t) \left[ \cos \frac{n \pi
            t}{L} \cos \frac{n \pi x}{L} + \sin \frac{n \pi t}{L} \sin
            \frac{n \pi x}{L} \right] \, dt	\\
	= {} & \frac{1}{2L} \int_{-L}^{L} f(t) \, dt +
\frac{1}{L} \sum_{n=1}^{N} \int_{-L}^{L} f(t) \cos \frac{n \pi (t-x)}{L} \, dt
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Dirichlet Kernel}

\begin{align*}
S_{N}(x)
	&=
\frac{1}{2L} \int_{-L}^{L} f(t) \, dt +
\frac{1}{L} \sum_{n=1}^{N} \int_{-L}^{L} f(t) \cos \frac{n \pi (t-x)}{L} \, dt	\\
	&=
\frac{1}{L} \int_{-L}^{L} f(t) \left[ \frac{1}{2} + \sum_{n=1}^{N}
              \cos \frac{n \pi (t-x)}{L} \right] \, dt	\\
	&=
\frac{1}{L} \int_{-L}^{L} f(t) D_{N}(t-x) \, dt
\end{align*}
where
\[
D_{N}(z) = \frac{1}{2} + \sum_{n=1}^{N} \cos \frac{n \pi z}{L}
\]
is called the \textbf{Dirichlet Kernel}.

\end{frame}
%%

\begin{frame}\frametitle{Closed Form of Dirichlet Kernel (1 of 3)}

\begin{align*}
D_{N}(z) &= \frac{1}{2} + \sum_{n=1}^{N} \cos \frac{n \pi z}{L}	\\
D_{N}(z) \sin \frac{\pi z}{L}
	&=
\frac{1}{2} \sin \frac{\pi z}{L} +
\sum_{n=1}^{N} \cos \frac{n \pi z}{L} \sin \frac{\pi z}{L}	\\
	&=
\frac{1}{2} \sin \frac{\pi z}{L} +
\frac{1}{2} \sum_{n=1}^{N} \left[ \sin \frac{(n+1) \pi z}{L}  - \sin
              \frac{(n-1) \pi z}{L} \right]
\end{align*}
according to a product-to-sum formula.
The summation telescopes to produce
\[
D_{N}(z) \sin \frac{\pi z}{L} = \frac{1}{2} \left[ \sin \frac{(N+1)
    \pi z}{L} + \sin \frac{N \pi z}{L} \right]	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Closed Form of Dirichlet Kernel (2 of 3)}

\begin{align*}
\lefteqn{D_{N}(z) \sin \frac{\pi z}{L}}	\\
	&=
\frac{1}{2} \left[ \sin \frac{(N+1)
    \pi z}{L} + \sin \frac{N \pi z}{L} \right]	\\
	&=
\frac{1}{2} \left[ \sin \frac{2 (N+1)
    \pi z}{2 L} + \sin \frac{2 N \pi z}{2 L} \right]	\\
	&=
\frac{1}{2} \left[ \sin \left( \frac{(2 N+1) \pi z}{2 L} + \frac{\pi
              z}{2 L} \right) +
\sin \left( \frac{(2 N+1) \pi z}{2 L}
              - \frac{\pi z}{2 L} \right) \right]	\\
	&=
\sin \frac{(2N+1) \pi z}{2L} \cos \frac{\pi z}{2L}
\end{align*}
using a product-to-sum formula.

\end{frame}
%%

\begin{frame}\frametitle{Closed Form of Dirichlet Kernel (3 of 3)}

\begin{align*}
D_{N}(z) \sin \frac{\pi z}{L}
	&=
\sin \frac{(2N+1) \pi z}{2L} \cos \frac{\pi z}{2L}	\\
D_{N}(z)
	&=
\frac{\sin \frac{(2N+1) \pi z}{2L} \cos \frac{\pi z}{2L}}{\sin
              \frac{\pi z}{L}}
\quad\mbox{(if $z \ne k \, L$ with $k \in \mathbb{Z}$)}	\\
	&=
\frac{\sin \frac{(2N+1) \pi z}{2L} \cos \frac{\pi z}{2L}}{\sin
              \frac{2 \pi z}{2 L}}	\\
	&=
\frac{\sin \frac{(2N+1) \pi z}{2L} \cos \frac{\pi z}{2L}}{2 \sin
              \frac{\pi z}{2 L} \cos \frac{\pi z}{2 L}}	\\
D_{N}(z)	&=
\frac{\sin \frac{(2N+1) \pi z}{2L}}{2 \sin
              \frac{\pi z}{2 L}}
\end{align*}
using the double-angle formula for the sine and assuming $z \ne 2 k L$.

\end{frame}
%%

\begin{frame}\frametitle{Remarks (1 of 2)}

Consider the following limit:
\begin{align*}
\lim_{z \to 2 k L} D_{N}(z)
	&=
\lim_{z \to 2 k L} \frac{\sin \frac{(2N+1) \pi z}{2L}}{2 \sin
              \frac{\pi z}{2 L}}	\\
	&=
\lim_{z \to 2 k L} \frac{\frac{(2N+1) \pi}{2L} \cos \frac{(2N+1) \pi
              z}{2L}}{\frac{\pi}{L} \cos \frac{\pi z}{2L}}	\\
	&=
\lim_{z \to 2 k L} \frac{(2N+1) \cos \frac{(2N+1) \pi
              z}{2L}}{2 \cos \frac{\pi z}{2L}}	\\
	&=
\frac{2N+1}{2}
\end{align*}
Thus the limit of the Dirichlet Kernel exists as $z \to 2 k L$.

\end{frame}
%%

\begin{frame}\frametitle{Remarks (2 of 2)}

\begin{itemize}
\item It can also be shown that
\[
\lim_{z \to (2k+1) L} D_{N}(z) = \frac{(-1)^{N}}{2}
\]
and thus the closed form of the Dirichlet Kernel is defined for all
real numbers.
\item Integrating the original definition of the Dirichlet Kernel over
  $[0,L]$ yields
\[
\int_{0}^{L} D_{N}(z) \, dz = \frac{L}{2}	.
\]
\item The Dirichlet Kernel is a $2L$-periodic and even function.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Second Necessary Lemma}

\begin{lemma}
If $g(x)$ is a piecewise smooth
function on $[0,L]$, then
\[
\lim_{N \to\infty} \int_{0}^{L} g(t) D_{N}(t) \, dt =
\frac{L}{2} g(0+).
\]
\end{lemma}

\end{frame}
%%

\begin{frame}\frametitle{Return to Proof of Dirichlet Convergence Theorem}

Recall,
\[
S_{N}(x) = \frac{1}{L} \int_{-L}^{L} f(t) D_{N}(t-x) \, dt	.
\]
\begin{itemize}
\item Since $f(t) D_{N}(t-x)$ is $2L$-periodic, then
\[
S_{N}(x) = \frac{1}{L} \int_{x-L}^{x+L} f(t) D_{N}(t-x) \, dt	.
\]
\item Make the substitution $s = t-x$,
\begin{align*}
S_{N}(x)
	&=
\frac{1}{L} \int_{-L}^{L} f(x+s) D_{N}(s) \, ds	\\
	&=
\frac{1}{L} \int_{-L}^{0} f(x+s) D_{N}(s) \, ds +
\frac{1}{L} \int_{0}^{L} f(x+s) D_{N}(s) \, ds	\\
	&=
\frac{1}{L} \int_{0}^{L} f(x-s) D_{N}(s) \, ds +
\frac{1}{L} \int_{0}^{L} f(x+s) D_{N}(s) \, ds	.
\end{align*}
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Proof}

Define,
\begin{align*}
I_{N}(x) &= \frac{1}{L} \int_{0}^{L} f(x-s) D_{N}(s) \, ds	\\
J_{N}(x) &= \frac{1}{L} \int_{0}^{L} f(x+s) D_{N}(s) \, ds	.
\end{align*}
By the second necessary lemma,
\begin{align*}
\lim_{N \to\infty} I_{N}(x)
	&=
\lim_{N \to \infty} \frac{1}{L} \int_{0}^{L} f(x-s) D_{N}(s) \, ds =
\frac{1}{L} \cdot \frac{L}{2} f(x-) =
\frac{1}{2} f(x-)	\\
\lim_{N \to\infty} J_{N}(x)
	&=
\lim_{N \to \infty} \frac{1}{L} \int_{0}^{L} f(x+s) D_{N}(s) \, ds =
\frac{1}{L} \cdot \frac{L}{2} f(x+) =
\frac{1}{2} f(x+)	,
\end{align*}
and thus
\[
\lim_{N \to \infty} S_{N}(x) =
\frac{1}{2} \left[ f(x-) + f(x+) \right]	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Fourier Series Uniform Convergence}

Now we can turn to the issue of the uniform convergence of the Fourier
series for piecewise smooth, $2L$-periodic functions.

\begin{theorem}
Assume that $f(x)$ is
continuous on $[-L,L]$ with $f(-L) = f(L)$ and $f'(x)$
is piecewise continuous on
$[-L,L]$, then the Fourier series of $f(x)$
converges uniformly to $f(x)$ on $[-L,L]$.
\end{theorem}

\end{frame}
%%

\begin{frame}\frametitle{Differentiation of Fourier Series}

Recall the theorem:
\begin{theorem}
Assume that $f(x)$ is continuous on $(-L,L)$
with $f(-L+) = f(L-)$, and $f'(x)$ is piecewise continuous on
$(-L,L)$.
Then the Fourier series of $f(x)$ can be differentiated term-by-term.
\end{theorem}

\vspace*{0.10in}

\uncover<2->{%
\textbf{Remark}: this theorem does not depend on the uniform
convergence of the Fourier series.
}

\end{frame}
%%

\begin{frame}\frametitle{Fourier Coefficients of $f'(x)$}

Suppose
\[
f'(x) \sim \sum_{n=1}^{\infty} \left( \alpha_{n} \cos \frac{n \pi
    x}{L} + \beta_{n} \sin \frac{n \pi x}{L} \right)	,
\]
then
\begin{align*}
\alpha_{n} &= \frac{n \pi b_{n}}{L}	\\
\beta_{n} &= -\frac{n \pi a_{n}}{L}
\end{align*}
where $a_{n}$ and $b_{n}$ are the Fourier coefficients for $f(x)$ on
$[-L,L]$.

\end{frame}
%%

\begin{frame}\frametitle{Trigonometric Relationship}

Suppose we write
\begin{align*}
A \cos \theta + B \sin \theta
	&=
R \cos \left( \theta + \delta \right)	\\
	&=
R \cos \delta \cos \theta - R \sin \delta
              \sin \theta	\\
\uncover<2->{%
A &= R \cos \delta	\\
B &= -R \sin \delta
}
\end{align*}
\uncover<3->{%
which implies
\begin{align*}
R^{2} &= A^{2} + B^{2}	\\
\delta &= -\tan^{-1} (B/A)	.
\end{align*}
\uncover<4->{%
Furthermore
\[
\left| A \cos \theta + B \sin \theta \right| \le R
\]
for all $\theta$.
}
}

\end{frame}
%%

\begin{frame}\frametitle{Bound}

Consider
\begin{align*}
\left| f(x) - S_{N}(x) \right|
	&= \left| \sum_{k=N+1}^{\infty} \left( a_{k} \cos \frac{k \pi x}{L} +
              b_{k} \sin \frac{k \pi x}{L} \right) \right|	\\
\uncover<2->{%
	&\le \sum_{k=N+1}^{\infty} \left| a_{k} \cos \frac{k \pi x}{L} +
              b_{k} \sin \frac{k \pi x}{L} \right|	\\
\uncover<3->{%
	&\le \sum_{k=N+1}^{\infty} \sqrt{a_{k}^{2} + b_{k}^{2}}	\\
	&= \sum_{k=N+1}^{\infty} \sqrt{\frac{L^{2}}{k^{2} \pi^{2}}}
   \sqrt{\alpha_{k}^{2} + \beta_{k}^{2}}
}
}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Cauchy-Schwarz Inequality}

\[
\left( \sum_{k=1}^{n} x_{k} y_{k} \right)^{2}
	\le
\left( \sum_{k=1}^{n} x_{k}^{2} \right) \left( \sum_{k=1}^{n} y_{k}^{2} \right)
\]
Application of the Cauchy-Schwarz inequality to
\begin{align*}
\left| f(x) - S_{N}(x) \right|^{2}
	&\le \left( \sum_{k=N+1}^{\infty} \sqrt{\frac{L^{2}}{k^{2} \pi^{2}}}
              \sqrt{\alpha_{k}^{2} + \beta_{k}^{2}} \right)^{2}	\\
	&\le \left( \sum_{k=N+1}^{\infty} \frac{L^{2}}{k^{2} \pi^{2}} \right)
\left( \sum_{k=N+1}^{\infty} (\alpha_{k}^{2} + \beta_{k}^{2}) \right)
  \\
\uncover<2->{%
\left| f(x) - S_{N}(x) \right|
	&\le \left( \sum_{k=N+1}^{\infty} \frac{L^{2}}{k^{2} \pi^{2}} \right)^{1/2}
\left( \sum_{k=N+1}^{\infty} (\alpha_{k}^{2} + \beta_{k}^{2}) \right)^{1/2}	\\
	&= \frac{L}{\pi} \left( \sum_{k=N+1}^{\infty} \frac{1}{k^{2}} \right)^{1/2}
\left( \sum_{k=N+1}^{\infty} (\alpha_{n}^{2} + \beta_{n}^{2}) \right)^{1/2}	.
}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Bound for $p$-Series}

Using an improper integral,
\[
\sum_{k=N+1}^{\infty} \frac{1}{k^{2}} \le
\int_{N+1}^{\infty} \frac{1}{x^{2}} \, dx =
\frac{1}{N}	.
\]

\vspace*{0.10in}

\uncover<2->{%
Thus
\begin{align*}
\left| f(x) - S_{N}(x) \right|
	&\le \frac{L}{\pi} \left( \sum_{k=N+1}^{\infty} \frac{1}{k^{2}} \right)^{1/2}
\left( \sum_{k=N+1}^{\infty} (\alpha_{n}^{2} + \beta_{n}^{2}) \right)^{1/2}	\\
	&\le \frac{L}{\pi \sqrt{N}} \left( \sum_{k=N+1}^{\infty} (\alpha_{n}^{2}
                + \beta_{n}^{2}) \right)^{1/2}	\\
	&\le \frac{\sqrt{L}}{\pi \sqrt{N}} \left( \int_{-L}^{L} (f'(x))^{2} \, dx \right)^{1/2}
\end{align*}
by Bessel's inequality.
This inequality holds for all $x \in [-L,L]$ and thus the Fourier
series for $f$ converges uniformly to $f$.
}

\end{frame}
%%

\begin{frame}\frametitle{Homework}

\begin{itemize}
\item Read Section $3.11$--$3.12$
\item Exercises: 21--25
\end{itemize}

\end{frame}
%%

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%%%%
