<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}

\begin{document}

\parindent=0pt

\newcommand{\ds}{\displaystyle}

\begin{center}

FOURIER SERIES

\end{center}

We consider trigonometrical series which have period $2\pi$.
Consider the series

$$\frac{1}{2}a_0+\sum_{n=1}^\infty(a_n\cos nx+b_n\sin nx)$$

and suppose it converges uniformly with respect to $x$ in
$[-\pi,\pi]$ with sum $f(x)$, which will be continuous in
$[-\pi\pi]$, and periodic - $2\pi$. Then

\begin{eqnarray*}
\int_{-\pi}^\pi
f(x)\,dx&amp;=&amp;\frac{1}{2}a_0\int_{-\pi}^\pi\,dx+\sum_{n=1}^\infty\int_{-\pi}^\pi(a_n\cos
nx+b_n\sin nx)\,dx=\pi a_0\\ \int_{-\pi}^\pi f(x)\cos
mx\,dx&amp;=&amp;\frac{1}{2}a_0\int_{-\pi}^\pi\cos mx\,dx\\ &amp;&amp;
+\sum_{n=1}^\infty\int_{-\pi}^\pi(a_n\cos nx+a_n\sin nx)\cos
mx\,dx\\ &amp;=&amp;a_m\int_{-\pi}^\pi\cos^2mx\,dx=\pi a_m
\end{eqnarray*}

Similarly $\int_{-\pi}^\pi f(x)\sin mx\,dx=\pi b_m$

Hence

\begin{eqnarray}
a_m&amp;=&amp;\frac{1}{\pi}\int_{-\pi}^\pi f(f)\cos mx\,dx\ \
m=0,1,\ldots\\ b_m&amp;=&amp;\frac{1}{\pi}\int_{-\pi}^\pi f(x)\sin mx\,dx\
\ m=1,2,\ldots
\end{eqnarray}

We can write the coefficients in the form

$$a_n+ib_n=\frac{1}{\pi}\int_{-\pi}^\pi f(x)e^{inx}\,dx\ \
n=1,2,\ldots$$

Now suppose $f(x)\in {\cal{L}}(-\pi\pi)$ and is periodic - $2\pi$
define constants $a_n,\ b_n$ by the relations (1), (2) above
(Euler, Fourier formulae)

This gives rise to the series

$$\frac{1}{2}a_0+\sum_{n=1}^\infty(a_n\cos nx+b_n\sin nx)\sim
f(x)$$

called the Fourier Series of $f(x)$.

If $f$ is an even function $b_m=0\ a_m=\frac{2}{\pi}\int_0^\pi
f(x)\cos mx\,dx$

If $f$ is an odd function $a_m=0\ b_m=\frac{2}{\pi}\int_0^\pi
f(x)\sin mx\,dx$

\begin{description}

\item[Problem 1]
Suppose $f$ continuous and periodic and periodic -$2\pi$ and
$f(x)\sim\frac{1}{2}a_0+\sum_{n=1}^\infty(a_n\cos nx+b_n\sin nx)$.

Suppose the Fourier Series converges to $\phi(x)$ uniformly in
$[-\pi\ \pi]$. Is it true the $f(x)=\phi(x)$?

By the definition of the coefficients, we get

$$\int_{-\pi}^\pi
f(x)e^{inx}\,dx=\int_{-\pi}^\pi\phi(x)e^{inx}\,dx\ \
n=0,1,2,\ldots$$

where $f$ and $\phi$ are continuous.

Does it follow that $f(x)=\phi(x)$ for all $x$?

\item[Problem 2]
Can 2 different continuous functions, periodic - $2\pi$ have the
same Fourier Series.

\item[Problem 3]
Is the Fourier Series of a continuous periodic function
convergent?

\item[Theorem 1 Riemann Lebesgue]
If $f\in{\cal{L}}(a\ b)$ then $\int_a^bf(x)e^{i\lambda x}\,dx\to0$
as $\lambda\to\infty$.

\item[Proof]
We first prove the result for $f$ continuous. Define, without loss
of generality, $f(x)=f(b)\ x&gt;b,f(x)=f(a)&lt;a$ then $f$ is continuous
everywhere.

\begin{eqnarray*}
I&amp;=&amp;\int_a^b f(x)e^{i\lambda x}\,dx\\
&amp;=&amp;-\int_a^bf(x)e^{i\lambda\left(x+\frac{\pi}{\lambda}\right)}\,dx\\
&amp;=&amp;-\int_{a+\frac{\pi}{\lambda}}^{b+\frac{\pi}{\lambda}}f(t-\frac{\pi}{\lambda})e^{i\lambda
t}\,dt\\
&amp;=&amp;-\int_a^bf\left(t-\frac{\pi}{\lambda}\right)e^{i\lambda
t}\,dt+\int_a^{a+\frac{\pi}{\lambda}}f(t-\frac{\pi}{\lambda})e^{i\lambda
t}\,dt\\
&amp;&amp;-\int_b^{b+\frac{\pi}{\lambda}}f\left(t-\frac{\pi}{\lambda}\right)e^{i\lambda
t}\,dt\\ &amp;=&amp;I_1+I_2-I_3
\end{eqnarray*}

Therefore
$$2I=\int_a^b\left\{f(x)-f\left(x-\frac{\pi}{\lambda}\right)\right\}e^{i\lambda
x}\,dx+I_2-I_3$$

$$\left|\int_a^b\left\{f(x)-f\left(x-\frac{\pi}{\lambda}\right)\right\}e^{i\lambda
x}\,dx\right|\leq\int_a^b\left|f(x)-f\left(x-\frac{\pi}{\lambda}\right)\right|\,dx$$

$\to0$ as $\lambda\to\infty$ by uniform continuity of $f$.

$$|I_2|\leq\int_a^{a+\frac{\pi}{\lambda}}\left|f\left(t-\frac{\pi}{\lambda}\right)\right|\,dx&lt;\frac{\pi}{\lambda}M\
M=\sup_{a\leq x\leq b}f(x)&lt;\infty$$

provided $\frac{\pi}{\lambda}&lt;b-a$ therefore $I_2\to0$ as
$\lambda\to\infty$.

Sum $I_3\to0$ as $\lambda\to\infty$

Therefore $I\to0$ as $\lambda\to\infty$.

Now if $f\in{\cal{L}}(ab)$ and $\varepsilon&gt;0$ is given, there is
a function $\phi(x)=\phi_\varepsilon(x)$, continuous in $[a\ b]$
such that $\int_a^b|f(x)-\phi(x)|\,dx&lt;\varepsilon$. Therefore

\begin{eqnarray*}
\int_a^bf(x)e^{i\lambda x}\,dx&amp;=&amp;\int_a^b\phi(x)e^{i\lambda
x}+\int_a^b\left\{f(x)-\phi(x)\right\}e^{i\lambda x}\,dx\\
&amp;=&amp;I_1+I_2
\end{eqnarray*}

$|I_2|&lt;\varepsilon$ for all $\lambda\ |I_1|&lt;\varepsilon\
\lambda&gt;\lambda_0$ by the first part of the proof. Hence the
result.

\item[Corollary 1]
If $a\leq a'\leq b'\leq b$ then $\int_{a'}^{b'}f(x)e^{i\lambda
x}\,dx\to0$ as $\lambda\to\infty$ uniformly in $a'\ b'$

\item[Corollary 2]
$$a_n+ib_n=\frac{1}{\pi}\int_{-\pi}^\pi f(x)e^{i\lambda
x}\,dx\to0\textrm{ as }n\to \infty.$$

This result may be proved in a more elementary way if $f$ is
continuous as follows.

\begin{eqnarray*}
\pi(a_n+ib_n)&amp;=&amp;\int_{-\pi}^\pi f(x)e^{inx}\,dx\\
&amp;=&amp;-\int_{-\pi}^\pi f(x)e^{in\left(x+\frac{\pi}{n}\right)}\,dx\\
&amp;=&amp;-\int_{-n+\frac{\pi}{n}}^{n+\frac{\pi}{n}}f\left(t-\frac{\pi}{n}\right)e^{int}\,dt\\
&amp;=&amp;-\int_{-\pi}^\pi f\left(x-\frac{\pi}{n}\right)e^{n\lambda x}\\
&amp;&amp; \textrm{ by periodicity of integrand.}\\\textrm{ Therefore }
2\pi(a_n+ib_n)&amp;=&amp;\int_{-\pi}^\pi\left\{f(x)-f\left(x-\frac{\pi}{n}\right)\right\}e^{inx}\,dx\\
\textrm{ therefore }
|2\pi(a_n+ib_n)|&amp;\leq&amp;\int_{-\pi}^\pi\left|f(x)-f\left(x-\frac{\pi}{n}\right)\right|\,dx
\end{eqnarray*}

$\to0$ as $n\to\infty$ by uniform continuity of the integrand.

\item[Convergence of Fourier Series at $t=x$]
Suppose $f(t)\in{\cal{L}}(-\pi\ \pi)$ and is periodic - $2\pi$ and
suppose $f(t)\sim\frac{1}{2}a_0+\sum_{n=1}^\infty(a_n\cos
nt+b_n\sin nt)$

\begin{eqnarray*}
S_n\equiv S_n(x)\equiv S_n(x;f)&amp;=&amp;\frac{1}{2}a_0+\sum_{\nu=1}^n
a_\nu\cos\nu x+b_\nu\sin\nu x\\ &amp;=&amp;\frac{1}{2\pi}\int_{-\pi}^\pi
f(t)\,dt+\sum_{\nu=1}^n\frac{\cos\nu x}{\pi}\int_{-\pi}^\pi
f(t)\cos\nu t\,dt\\ &amp;&amp;+\sum_{\nu=1}^n\frac{\sin \nu
x}{\pi}\int_{-\pi}^\pi f(t)\sin\nu t\,dt\\
&amp;=&amp;\frac{1}{\pi}\int_{-\pi}^\pi
f(t)\left\{\frac{1}{2}\sum_{\nu=1}^n(\cos\nu x\cos\nu t+\sin\nu
x\sin\nu t)\right\}\,dt\\ &amp;=&amp;\frac{1}{\pi}\int_{-\pi}^\pi
f(t)\left\{\frac{1}{2}\sum_{\nu=1}^n\cos\nu(t-x)\right\}\,dt\\
&amp;=&amp;\frac{1}{\pi}\int_{-\pi}^\pi
f(t)\left\{\frac{\sin\left(n+\frac{1}{2}\right)(t-x)}{2\sin\frac{1}{2}(t-x)}\right\}\,dt\\
&amp;&amp;\textrm{(Dirichlet's Integral)}\\
&amp;=&amp;\frac{1}{\pi}\int_{-\pi-x}^{\pi-x}f(x+u)\frac{\sin\left(n+\frac{1}{2}\right)u}{2\sin\frac{1}{2}u}\,du\\
&amp;=&amp;\frac{1}{\pi}\int_{-\pi}^\pi
f(x+u)\frac{\sin\left(n+\frac{1}{2}\right)u}{2\sin\frac{1}{2}u}\,du
\end{eqnarray*}

The sum $\frac{1}{2}+\sum_1^n\cos\nu
u=\frac{\sin\left(n+\frac{1}{2}\right)u}{2\sin\frac{1}{2}u}=D_n(u)$
is called Dirichlet's Kernel.

$$\frac{1}{\pi}\int_{-\pi}^\pi
f(x+u)D_n(u)\,du=\frac{1}{\pi}\int_0^\pi\left\{f(x+u)+f(x-u)\right\}D_n(u)\,du$$

 We have at $t=x$

\begin{eqnarray*}
f(t)=f(x+u)&amp;=&amp;\frac{1}{2}\{f(x+u)+f(x-u)\}+\frac{1}{2}\{f(x+u)-f(x-u)\}\\
&amp;=&amp;\phi_x(u)+\psi_x(u)
\end{eqnarray*}

$\phi_x=$even part of $f(t)$ with respect to $t=x$

$\psi_x=$odd part of $f(t)$ with respect to $y=x$.

Dirichlet's integral becomes
$\frac{2}{\pi}\int_0^\pi\phi_x(t)D_n(t)\,dt$.

\item[Theorem 2]
The convergence of a Fourier series is a \lq local' property of
the function when $f(t)\in{\cal{L}}(-\pi,\pi)$ and is periodic -
$2\pi$.

\item[Proof]
Let $0&lt;\delta&lt;\pi$

\begin{eqnarray*}
\pi S_n(x)&amp;=&amp;\int_{-\pi}^\pi f(x+t)D_n(t)\,dt\\
&amp;=&amp;\int_{-\pi}^{-\delta}+\int_{-\delta}^\delta+|int_\delta^\pi=I_1+I_2+I_3\\
I_3&amp;=&amp;\int_\delta^\pi\frac{f(x+u)}{2\sin\frac{1}{2}u}\sin(n+\frac{1}{2})u\,du
\end{eqnarray*}

$\to 0$ by Riemann Lebesgue theorem as $n\to\infty$ since
$\frac{f(x+u)}{2\sin\frac{1}{2}u}\in{\cal{L}}(\delta \pi)$

Similarly $I_1\to0$ as $n\to\infty$.

Hence the convergence of the Fourier series at $t=x$ depends only
on the behaviour of $f$ in an arbitrarily small interval about
$t=x$.

\item[Theorem 3]
If $f\in{\cal{L}}(-\pi\ \pi)$ and is periodic - $2\pi$, and $a&lt;b$,
then the uniform convergence of the Fourier Series in $[a\ b]$
depends only on the function in any interval $(a-\delta,a+\delta)\
\delta&gt;0$.

\item[Proof]
Suppose $a\leq x\leq b$ and $\delta&gt;0$

\begin{eqnarray*} \pi S_n(x)&amp;=&amp;\int_{-\pi}^\pi f(x+t)D_n(t)\,dt\\
&amp;=&amp;\int_{-\pi+x}^{\pi+x} f(t)D_n(t-x)\,dt\\
&amp;=&amp;\int_{\pi+x}^{x-\delta}+\int_{x-\delta}^{x+\delta}+\int_{x+\delta}^{x+\pi}=I_1+I_2+I_3\\
I_3&amp;=&amp;\int_{x+\delta}^{x+\pi}f(t)\frac{\sin\left(n+\frac{1}{2}\right)(t-x)}{2\sin\frac{1}{2}(t-x)}\,dt\\
&amp;=&amp;\frac{1}{2\sin\frac{1}{2}\delta}\int_{x+\delta}^\zeta
f(t)\sin\left(n+\frac{1}{2}\right)(t-x)\,dt\textrm{ 2nd MVT}\\
&amp;=&amp;\frac{\cos\left(n+\frac{1}{2}\right)x}{2\sin\frac{1}{2}\delta}\int_{x+\delta}^\zeta
f(t)\sin\left(n+\frac{1}{2}\right)t\,dt\\ &amp;&amp;-
\frac{\sin\left(n+\frac{1}{2}\right)x}{2\sin\frac{1}{2}\delta}\int_{x+\delta}^\zeta
f(t)\cos\left(n+\frac{1}{2}\right)t\,dt
\end{eqnarray*}

$a\leq a+\delta\leq x+\delta\leq\zeta\leq x+\pi\leq b+\pi$

$\left|\cos\left(n+\frac{1}{2}\right)x\right|\leq1\ \
\left|\sin\left(n+\frac{1}{2}\right)x\right|\leq1$

Therefore $I_3\to0$ uniformly with respect to $x$ as $n\to\infty$
by Riemann Lebesgue theorem. Similarly $I_1\to0$ uniformly with
respect to $x$ as $n\to\infty$.

hence the uniform convergence depends only on the behaviour of $f$
in $(x-\delta,x+\delta)\ a\leq x\leq b$ i.e. in
$(a-\delta,b+\delta)$.

\item[Note]
$$S_n=\frac{2}{\pi}\int_0^\pi\phi_x(t)D_n(t)\,dt.$$

$s$ - const is even therefore $S=\frac{2}{\pi}\int_o^\pi
sD_n(t)\,dt$ therefore

$$S_n-S=\frac{2}{\pi}\int_0^\pi\{\phi_x(t)-S\}D_n(t)\,dt$$

\item[Theorem 4 Dimi's Test]
If $f\in{\cal{L}}(-\pi\ \pi)$ and is periodic - $2\pi$ and if
$\exists S|\frac{\phi_x(t)-S}{t}\in{\cal{L}}(0\ \delta)\ \delta&gt;0$
then the Fourier series of $f$ converges to $S$ at $t=x$.

\item[Proof]

\begin{eqnarray*}
\frac{\pi}{2}(S_N(x)-S)&amp;=&amp;\int_0^\pi(\phi_x(t)-S)D_n(t)\,dt\\
&amp;=&amp;\int_0^\pi\frac{\phi_x(t)-S}{2\sin\frac{1}{2}t}\sin\left(n+\frac{1}{2}\right)t\,dt\\
\frac{\phi_x(t)-S}{2\sin\frac{1}{2}t}&amp;=&amp;\frac{\phi_x(t)-S}{t}.\frac{t}{2\sin\frac{1}{2}t}.
\end{eqnarray*}

If $g(t)=\frac{t}{2\sin\frac{1}{2}t}\ o&lt;t\leq\pi$, and $g(0)=1$
then $g$ is continuous in $[0\ \pi]$ Therefore
$\frac{\phi_x(t)-S}{2\sin\frac{1}{2}t}\in{\cal{L}}(0,\delta)$ and
also $\frac{\phi_x(t)-S}{2\sin\frac{1}{2}t}\in{\cal{L}}(\delta\
\pi)$ therefore
$\frac{\phi_x(t)-S}{2\sin\frac{1}{2}t}\in{\cal{L}}(0,\ \pi)$.

So by Riemann-Lebesgue Theorem $\frac{\pi}{2}(S_n(x)-S)\to0$ as
$n\to\infty$ i.e. $S_n(x)\to S$ as $n\to\infty$.

\item[Corollary I]
If $f(t)\in{\cal{L}}(-\pi\ \pi)$ and is periodic - $2\pi$ and if
$f(t)$ is differentiable at $t=x$, then the Fourier series
converges to $f(x)$ at $t=x$.

\item[Proof]

$$\frac{\phi_x(t)-f(x)}{t}=\frac{f(x+t)-f(x)}{2t}+\frac{f(x-t)-f(x)}{2t}$$

$\to\frac{1}{2}f'(x)-\frac{1}{2}f'(x)=0$ as $t\to0$ therefore
$\exists\delta&gt;0|\left|\frac{\phi_x(t)-f(x)}{t}\right|&lt;1\ 0\leq
t\leq \delta$ therefore
$\frac{\phi_x(t)-f(x)}{t}\in{\cal{L}}(0,\delta)$.

Therefore by Dimi's test the Fourier series converges to $f(x)$.

\item[Corollary II Lipschitz Condition]
$f(t)\in{\cal{L}}(-\pi\ \pi)$ and periodic - $2\pi$ and
$f(x+t)-f(x)=O|t|^\alpha$ for some $\alpha\neq0$ as $t\to0$ then
the Fourier series converges to $f(x)$ at $t=x$.

\item[Proof]

\begin{eqnarray*}
|\phi_x(u)-f(x)|&amp;\leq&amp;K|U|^\alpha\\ \textrm{therefore
}\frac{|\phi_x(u)-f(x)|}{|U|}&amp;\leq&amp;K|U|^\alpha\\ \textrm{therefore
}\left|\int_0^\delta\frac{\phi_x(u)-f(x)}{u}\,du\right|&amp;\leq&amp;K\int_0^\delta|U|^{\alpha-1}\,du=K\frac{\delta^\alpha}{\alpha}&lt;\infty
\end{eqnarray*}

Hence Dimi's condition is satisfied.

\item[Modified forms of Dirichlet's integral]
if $f(t\in{\cal{L}}(-\pi\ \pi)$ and periodic - $2\pi$, we have

$$S_n(x)-S=\frac{2}{\pi}\int_0^\pi\{\phi_x(t)-S\}\frac{\sin\left(n+\frac{1}{2}\right)t}{2\sin\frac{1}{2}t}\,dt$$

\begin{enumerate}

\item
$\frac{1}{2\sin\frac{1}{2}t}$ may be replaced by $\frac{1}{t}$
with error O(1).

\begin{eqnarray*}
h(t)&amp;=&amp;\frac{1}{2\sin\frac{1}{2}t}-\frac{1}{t}\\
&amp;=&amp;\frac{t-2\sin\frac{1}{2}t}{2t\sin\frac{1}{2}t}\\
&amp;=&amp;\frac{t-2\left(\frac{1}{2}t-\left(\frac{1}{2}\right)^3\frac{t^3}{3!}+\ldots\right)}{2t\left(\frac{1}{2}t-\ldots\right)}\\
&amp;\sim&amp;2\frac{\left(\frac{1}{2}\right)^3}{3!}\frac{t^3}{t^2}=\frac{1}{24}t\textrm{
as }t\to0
\end{eqnarray*}

Therefore

$$\frac{2}{\pi}\int_0^\pi\{\phi_x(t)-S\}h(t)\sin\left(n+\frac{1}{2}\right)t\,dt$$

$\to0$ as $n\to\infty$ by Riemann Lebesgue Theorem.

\item
We may replace $\sin\left(n+\frac{1}{2}\right)t$ by $\sin nt$ with
error $O(1)$.

\begin{eqnarray*}
\sin\left(n+\frac{1}{2}\right)t-\sin
nt&amp;=&amp;2\cos\left(n+\frac{1}{4}\right)t\sin\frac{1}{4}t\\
\textrm{therefore
}&amp;&amp;\frac{2}{\pi}\int_0^\pi\frac{\{\phi_x(t)-S\}}{t}\left\{\sin\left(n+\frac{1}{2}\right)t-\sin
nt\right\}\,dt\\
&amp;=&amp;\frac{2}{\pi}\int_0^\pi\{\phi_x(t)-S\}\frac{2\sin\frac{1}{4}t}{t}.\cos\left(n+\frac{1}{4}\right)t\,dt
\end{eqnarray*}

$\to 0$ as $n\to\infty$ by Riemann-Lebesgue Theorem.

\end{enumerate}

\item[Functions of bounded variables]
Let $f(x)$ be defined on $[a,b]$.

Let $\Delta_N:a=x_0&lt;x_1&lt;\ldots&lt;x_n=b$ and define
$V_\Delta=\sum_{\nu=1}^n|f(x_\nu)-f(x_{\nu-1})|$

Let $\delta f(x_\nu)=f(x_\nu)-f(x_{\nu-1})$.

Define $\delta^+f(x_\nu)=\frac{1}{2}\left[|\delta
f(x_\nu)|+\delta(fx_\nu)\right]=\begin{array}{ll}\delta
f(x_\nu)&amp;\textrm{if}&gt;0\\0&amp;\textrm{otherwise}\end{array}$

Define $\delta^+f(x_\nu)=\frac{1}{2}\left[|\delta
f(x_\nu)|-\delta(fx_\nu)\right]=\begin{array}{ll}-\delta
f(x_\nu)&amp;\textrm{if}&gt;0\\0&amp;\textrm{otherwise}\end{array}$

Define $P_\Delta=\sum_{\nu=1}^n\delta^+f(x_\nu),\
N_\Delta=\sum_{\nu=1}^n\delta^-f(x_\nu)$

Then $P_\Delta+N_\Delta=V_\Delta\ \ P_\Delta-N_\Delta=f(b)-f(a)$

Define $P=P_a^bf(x)=\sup_{\Delta}P_\Delta$ to be the positive
variation and $N=N_a^bf(x)=\sup_{\Delta}N_\Delta$ to be the
negative variation.

$P_a^b+N_a^b=V_a^b\ \ P_a^b-N_a^b=f(b)-f(a)$

Define $V(x)=V_a^x\ \ P(x)=P_a^x\ \ N(x)=N_a^x$

\item[Theorem 5]
If $f$ is B.V. in $[a\ b]$ i.e. $V_a^b&lt;\infty$ then $\exists$
functions $f_1$ and $f_2$ both increasing and bounded
$|f(x)=f_1(x)-f_2(x)\ \ [a\ b]$

\item[Proof]
$P(x)-N(x)=f(x)-f(a)$ therefore $f(x)=f(a)+P(x)-N(x)$.

$P(x)$ and $N(x)$ are increasing and bounded. Let
$f_1(x)=f(a)+P(x)\ f_2(x)=N(x)$.

\item[Corollary]
A function of BV in $[a\ b]$ is continuous p.p. in $[a\ b]$.

\item[Theorem 6 Jordan's Condition]
If $f(t)\in{\cal{L}}(-\pi\ \pi)$ and is periodic $-2\pi$ and $f$
is B.V in $[x-\delta,x+\delta]$ for some $\delta&gt;0$ then the
Fourier series of $f$ converges at $t=x$ to
$\frac{1}{2}[f(x+)+f(x-)]$.

\item[Proof]
$f(t)$ B.V. in $[x-\delta,x\delta]\Rightarrow\phi_x(t)$ B.V. in
$[0\ \delta]$.

\begin{eqnarray*}
S_m(x)-\frac{1}{2}\{f(x+)+f(x-)\}&amp;=&amp;\frac{2}{\pi}\int_0^\pi\{\phi_x(t)-\phi_x(0+)\}D_n(t)\,dt\\
&amp;=&amp;\frac{2}{\pi}\int_0^\pi\{\phi_x(t)-\phi_x(0+)\}\frac{\sin
nt}{t}\,dt+O(1)
\end{eqnarray*}

as $n\to\infty$.

By Theorem 5, $\phi_x(t)=\psi_x(t)=\theta_x(t),\ \psi,\theta$
increasing and bounded so that $\psi_x(t)-\psi_x(0+)$ increasing
and $\geq0$ in $[0\ \delta]$ and $\to0$ as $t\to0+$.

Therefore suppose without loss of generality that $\phi_x(t)$
increasing and bounded in $[0\ \delta]$ and
$\phi_x(t)-\phi_x(0+)&lt;\varepsilon$ for $0\leq t\leq\eta&lt;\delta$.

\begin{eqnarray*}
I&amp;=&amp;\int_0^\pi\{\phi_x(t)-\phi_x(0+)\}\frac{\sin nt}{t}\,dt\\
&amp;=&amp;\int_0^\eta+\int_\eta^\pi=I_1+I_2\\
|I_1|&amp;=&amp;\left|\int_0^\eta\{\phi_x(t)-\phi_x(0+)\}\frac{\sin
nt}{t}\,dt\right|\\
&amp;=&amp;\{\phi_x(\eta)-\phi_x(0+)\}\left|\int_{\zeta}^\eta\frac{\sin
ni}{t}\,dt\right|\textrm{ 2nd M.V.T}\\
&amp;\leq&amp;\varepsilon\left|\int_{n\zeta}^{n\eta}\frac{\sin
u}{u}\,du\right|\\ &amp;\leq    &amp;^2M\varepsilon
\end{eqnarray*}

for all $n\geq 0$ since $\int_0^x\frac{\sin u}{u}$ is continuous
and $\to\frac{\pi}{2}$ as $x\to+\infty$ and is hence bounded by
$M$. $I_2\to0$ as $n\to\infty$ by Riemann-Lebesgue theorem. Hence
the result.

\item[Examples]

\begin{enumerate}

\item

$$f(t)=\frac{1}{\log\left|\frac{1}{t}\right|}\ \ (0\ \delta]\ \
f(0)=0$$

satisfies Jordan's condition for $t\in[-\delta\ \delta]$ but
doesn't satisfy Dimi's condition, since

$$\frac{\phi_0(t)-S}{t}=\frac{\left(\log\frac{1}{t}\right)^{-1}-S}{t}=\frac{1}{t\log\frac{1}{t}}-\frac{s}{t}$$

which cannot be integrated down to the origin.

\item

$$g(t)=|t|^\alpha\sin\left|\frac{1}{t}\right|\ \ o&lt;\alpha\leq 1$$

satisfies Dimi's condition

$$\frac{\phi_0(t)-0}{t}=\left|\frac{t^\alpha\sin\frac{1}{t}}{t}\right|\leq|t|^{\alpha-1}\in{\cal{L}}(0\
\delta]$$

But not Jordan's condition.

\begin{eqnarray*}
\phi_0(t)&amp;=&amp;t^\alpha\sin\frac{1}{t}\\
\phi_0\left(\frac{1}{\left(2n+\frac{1}{2}\right)\pi}\right)&amp;=&amp;\left[\frac{1}{\left(2n+\frac{1}{2}\right)\pi}\right]^\alpha\\
\phi_0\left(\frac{1}{\left(2n+\frac{1}{2}\right)\pi}\right)&amp;=&amp;-\left[\frac{1}{\left(2n+\frac{1}{2}\right)\pi}\right]^\alpha\\
V_{\frac{1}{2N}}^1\phi_0(t)&amp;\geq&amp;N_{\frac{1}{2N}}^1=\sum_{1}^{N-1}\left\{\frac{1}{\left(2n+\frac{1}{2}\right)\pi}\right\}^\alpha+\left\{\frac{1}{\left(2n-\frac{1}{2}\right)\pi}\right\}^\alpha\\
&amp;\geq&amp;2\pi^{-\alpha}\sum_1^{N-1}\frac{1}{(2n-1)\alpha}
\end{eqnarray*}

$\to\infty$ as $N\to\infty$.

\end{enumerate}

\item[Theorem 7]
If $f(t)$ is B.V. $[-\pi\ \pi]$ and periodic $2\pi$ then the
Fourier series converges boundedly in $[-\pi,\pi]$.

\item[Proof]
We prove $\exists M||s_n(x)|&lt;M$ for all $n$ and $x$

$S_n(x)\to\phi_x(0+)$ for all $x$ by Jordan's test. We can suppose
without loss of generality $f(t)&gt;0$ increasing and bounded in
$(-\pi\ \pi)$

\begin{eqnarray*}
|S_n(x)|&amp;=&amp;\left|\frac{1}{\pi}\int_{-\pi}^\pi
f(t)D_n(t-x)\right|=\frac{1}{\pi}f(\pi)\left|\int_\zeta^\pi
D_n(t-x)\,dt\right|\\
&amp;\leq&amp;f(\pi)(4\sqrt{\pi}+\frac{1}{2}(\pi-\zeta))\leq
f(\pi)(4\sqrt{\pi}+\frac{1}{2}\pi)=M
\end{eqnarray*}

\item[Theorem 8]
Of $f(t)$ is B.V in $[a\ b]\ 0&lt;b-a&lt;2\pi,\ f\in{\cal{L}}(-\pi\
\pi)$ periodic - $2\pi$ then the Fourier series converges
boundedly in $[a+\eta,b-\eta]\ 0&lt;\eta&lt;\frac{b-a}{2}$.

\item[Proof]
$S_n(x)\to\phi_x(0+)\ a&lt;x&lt;b$ by Jordan's test.

Suppose without loss of generality that $f(t)$ increasing and $&gt;0$
in $[a\ b]$.

$$S_n(x)=\frac{1}{\pi}\int_{-\pi}^\pi f(t)D_n(t-x)\,dt$$

Choose $\eta$ and suppose $x\in[a+\eta,b-\eta]$

\begin{eqnarray*}
S_n(x)&amp;=&amp;\frac{1}{\pi}\int_a^{a+2\pi}f(t)D_n(t-x)\,dt\textrm{ by
periodicity}\\
&amp;=&amp;\frac{1}{\pi}\int_a^b+\frac{1}{\pi}\int_b^{a+2\pi}=I_1+I_2\\
|I_1|&amp;=&amp;\frac{f(b)}{\pi}\left|\int_\zeta^bD_n(t-x)\,dt\right|\textrm{
2nd M.V.T.}\\
&amp;\leq&amp;\frac{f(b)}{\pi}\left[4\sqrt{\pi}+\frac{1}{2}\pi\right]=L\textrm{
for all }x,\eta\\
|I_2|&amp;=&amp;\left|\frac{1}{\pi}\int_b^{a+2\pi}f(t)\frac{\sin\left(n+\frac{1}{2}\right)(t-x)}{2\sin\frac{1}{2}(t-x)}\,dt\right|\\
&amp;\leq&amp;\frac{1}{\pi}\int_a^{a+2\pi}|f(t)|\frac{1}{|2\sin\frac{1}{2}(t-x)|}\,dt
\end{eqnarray*}

Now $\eta\leq t-x\leq2\pi-\eta$ therefore
$\frac{1}{2}\eta\leq\frac{1}{2}(t-x)\leq\pi-\frac{1}{2}\eta$
therefore

$$|I_2|\leq\frac{1}{2\pi}\frac{1}{2\sin\frac{1}{2}\eta}\int_b^{a+2n}|f(t)|
\,dt\leq\frac{1}{\pi}\frac{1}{2\sin\frac{1}{2}\eta}\int_{-\pi}^\pi|f(t)|\,dt=C(\eta)$$

Hence the result.

\item[Theorem 9]
If $f\in{\cal{L}}(-\pi\ \pi)$ and is periodic - $2\pi$, and if $f$
B.V. in $[a\ b]\ a&lt;b$ then if $f(t)$ is continuous in $[a\ b]$
then the Fourier series converges uniformly to $f(t)$ in
$[a+\eta,b-\eta]$.

\item[Proof]
Suppose without loss of generality that $b-a&lt;\pi$, and $f(t)$
increasing and $&gt;0$ in $[a\ b]$.

Let $0&lt;\eta&lt;\frac{b-a}{2}$

Choose $\delta$ such that

\begin{description}

\item[(i)]
$0&lt;\delta&lt;\eta$

\item[(ii)]
$0\leq f(t_2)-f(t_1)&lt;\varepsilon$, whenever $\varepsilon&gt;0$ is
given and $0&lt;t_2-t_1&lt;\delta,\ \ t_1,t_2\in[a\ b]$

\end{description}

Suppose $x\in[a+\eta,b-\eta]$

\begin{eqnarray*}
S_n(x)-f(x)&amp;=&amp;\frac{1}{\pi}\int_{-\pi}^\pi
D_n(t-x)[f(t)-f(x)]\,dt\\
&amp;=&amp;\frac{1}{\pi}\int_{x-\pi}^{x+\pi}=\frac{1}{\pi}\left\{
\int_{x-\pi}^{x-\delta}+\int_{x-\delta}^x+\int_x^{x+\delta}+\int_{x+\delta}^{x+\pi}\right\}\\
&amp;=&amp;I_1+I_2+I_3+I_4\\
|I_3|&amp;=&amp;\frac{1}{\pi}\left|\int_{x}^{x+\delta}\{f(t)-f(x)\}D_n(t-x)\,dt\right|\\
&amp;=&amp;\frac{1}{\pi}[f(x+\delta)-f(x)]\left|\int_\zeta^{x+\delta}D_n(t-x)\,dt\right|&lt;\varepsilon
K_0\\
|I_2|&amp;=&amp;\frac{1}{\pi}\left|\int_{x-\delta}^xf(t)-f(x)D_n(t-x)\,dt\right|\\
&amp;=&amp;\frac{1}{\pi}[f(x)-f(x-\delta)]\left|\int_{x-\delta}^\xi
D_n(t-x)\,dt\right|&lt;\varepsilon K_1\\
|I_4|&amp;=&amp;\frac{1}{\pi}\left|\int_{x+\delta}^{x+|pi}\frac{f(t)-f(x)}{2\sin\frac{1}{2}(t-x)}.\sin\left(n+\frac{1}{2}\right)(t-x)\,dt\right|
\end{eqnarray*}

$\to0$ uniformly with respect to $x$ as $n\to\infty$. Hence the
result.

\item[Theorem 10]
If $f(t)\in{\cal{L}}(-\pi\ \pi)$ and is periodic $-2\pi$ and if

$$f(t)\sim\frac{1}{2}a_0+\sum_{n=1}^\infty(a_n\cos nt+b_n\sin n)$$

and if $g(x)=\sum_0^x\left\{f(t)-\frac{1}{2}a_0\right\}\,dt$ then
$g(x)$ is periodic $2\pi$ and
$\sum_1^\infty\frac{b_n}{n}+\sum_1^\infty\frac{a_n\sin nx-b_n\cos
nx}{n}$ converges uniformly to $g(x)$ in $[-\pi\ \pi]$.

\item[Proof]

\begin{eqnarray*}
g(x+2\pi)-g(x)&amp;=&amp;\int_x^{x+2\pi}\{f(t)-\frac{1}{2}a_0\}\,dt\\
&amp;=&amp;\int_x^{x+2\pi}f(t)\,dt-\pi a_0\\ &amp;=&amp;\int_{-\pi}^\pi
f(t)\,dt-\pi a_0=0\\ g(x)&amp;\sim&amp;\frac{1}{2}A_0+\sum_1^\infty
A_n\cos nx+B_n\sin nx\\ n\geq1,A_n&amp;=&amp;\frac{1}{\pi}\int_{-\pi}^\pi
g(t)\cos nt\,dt\\ &amp;=&amp;\frac{1}{\pi}\left[g(t)\frac{\sin
nt}{n}\right]_{-\pi}^\pi-\frac{1}{\pi
n}\int_{-\pi}^\pi(f(t)-\frac{1}{2}a_0)\sin nt\,dt\\
&amp;=&amp;-\frac{b_n}{n}
\end{eqnarray*}

Similarly $B_n=+\frac{a_n}{n}$

$$\int_0^x(f(t)-\frac{1}{2}a_0)\,dt=C+\sum_{n=1}^\infty\left(\frac{a_n}{n}\sin
nx-\frac{b_n}{n}\cos nx\right)$$

putting $x=0\ \ C=\sum_{n=1}^\infty\frac{b_n}{n}$.

\item[Theorem 11]
If

\begin{description}

\item[(i)]
$f\in{\cal{L}}(-\pi\ \pi)$ and is periodic - $2\pi$

\item[(ii)]
$|f(t)|\leq M$ in $(x-\delta,x+\delta)\ 0&lt;\delta&lt;\pi$

\end{description}

Then $|S_n(x)|\leq\frac{2}{\pi}M\log m+\frac{2M}{\pi}+O(1)$ as
$m\to\infty$.

\item[Proof]
\begin{eqnarray*}
\frac{\pi}{2}S_n(x)&amp;=&amp;\int_0^\pi\phi_x(t)\frac{\sin
nt}{t}\,dt+O(1)\\ &amp;=&amp;\int_0^\delta+\int_\delta^\pi+O(1)\\
&amp;=&amp;\int_0^\frac{1}{n}+\int_{\frac{1}{n}}^\delta+\int_\delta^\pi+O(1)=I_1+I_2+I_3+O(1)\\
\left|\frac{\sin
nt}{t}\right|&amp;\leq&amp;\left\{\begin{array}{ll}n&amp;t\leq\frac{1}{n}\\\frac{1}{t}&amp;t&gt;\frac{1}{n}\end{array}\right.\\
\textrm{therefore
}|I_1|&amp;\leq&amp;\int_0^\frac{1}{n}|\phi_x(t)|\left|\frac{\sin
nt}{t}\right|\,dt\\ &amp;\leq&amp;M\int_0^\frac{1}{n}n\,dt=M\\
|I_2|&amp;\leq&amp;\int_{\frac{1}{n}}^\delta|\phi_x(t)|\left|\frac{\sin
nt}{t}\right|\,dt\leq M\int_{\frac{1}{n}}^\delta\frac{1}{t}\,dt\\
&amp;=&amp;M(\log n+\log \delta)&lt;M\log n \end{eqnarray*}

choosing $\delta&lt;1\ I_3\to0$ as $n\to\infty$ by Riemann Lebesgue
Theorem therefore

$$|S_n(x)|\leq\frac{2}{\pi}M\log n+\frac{2M}{\pi}+O(1)\textrm{ as
}n\to\infty.$$

\item[Theorem 12]

\begin{description}

\item[(i)]
$f(t)\in{\cal{L}}(-\pi\ \pi)$ and is periodic - $2\pi$

\item[(ii)]
$\phi_x(0+)$ exists

\end{description}

then $|S_n(x)|=O(\log n)$ as $n\to\infty$.

\item[Proof]
$$\frac{\pi}{2} S_n(x)=\int_0^\pi\phi_x(t)\frac{\sin
nt}{t}\,dt+o(1)$$

If $\varepsilon&gt;0\exists \eta$ with
$o&lt;\eta&lt;1||\phi_x(t)-\phi_x(0+)|&lt;\varepsilon$ in $(0,\eta)$.

\begin{eqnarray*}
\frac{\pi}{2}(S_n(x)-\phi_x((0+))&amp;=&amp;\int_0^\pi[\phi_x(t)-\phi_x(0+)]\frac{\sin
nt}{t}+o(1)\\
&amp;=&amp;\int_0^\frac{1}{n}+\int_\frac{1}{n}^\eta+\int_\eta^\pi+o(1)\\
&amp;=&amp;I_1+I_2+I_3+o(1)
\end{eqnarray*}

As in Theorem 11 $|i_1|\leq\varepsilon\ |I_2|\leq\varepsilon(\log
n+\log\eta)&lt;\varepsilon\log n\ |I_3|\to0$ as $n\to\infty$
therefore $\frac{|S_n(X)|}{\log n}\to0$ as $n\to\infty$.

\item[Summability by Ces\`{a}ro's 1st mean, summability (C,1)]
We have the result, due to Cauchy, that If $S_n\to S$ as
$n\to\infty$ then $\frac{S_0+S_1+\ldots+S_n}{n+1}\to S$ as
$n\to\infty$.

If $S_n=\sum_{r=1}^n a_r$ and if
$\frac{S_0+S_1+\ldots+S_n}{n+1}=\sigma_n\to S$ as $n\to\infty$ we
say $\sum_0^\infty a_r$ is summable $(C,1)$ to $S$.

If $\sum_{r=0}^\infty$ converges it is summable $(C,1)$ to $S$,
but the converse is not necessarily true.

\item[Examples]

\begin{enumerate}

\item
$1-1+1-1+\ldots$ is summable $(C,1)$ to $\frac{1}{2}$ but is not
convergent.

\item

\begin{eqnarray*}
S_n&amp;=&amp;\frac{1}{2}+\cos\theta+\cos2\theta+\ldots+\cos n\theta\
0&lt;\theta&lt;2\pi\\
S_n&amp;=&amp;\frac{\sin\left(n+\frac{1}{2}\right)\theta}{2\sin\frac{1}{2}\theta}
\end{eqnarray*}

$$\left|\frac{S_)+S_1+\ldots+S_n}{n+1}\right|=\left|\frac{1-\cos(n+1)\theta}{\left(2\sin\frac{1}{2}\theta\right)^2}\frac{1}{n+1}\right|\leq\frac{1}{n+1}\frac{2}{\left(2\sin\frac{1}{2}\theta\right)^2}$$

$\to0$ as $n\to\infty$. Therefore
$\frac{1}{2}+\sum_1^\infty\cos\nu\theta$ is summable $(C,1)$ to )
for all $\theta|0&lt;\theta&lt;2\pi$.

But suppose $\sin\left(n+\frac{1}{2}\right)\theta\to S$ ads
$n\to\infty$ therefore $\sin\left(n-\frac{1}{2}\right)\theta\to S$
as $n\to\infty$.

Therefore
$\sin\left(n+\frac{1}{2}\right)\theta-\sin\left(n-\frac{1}{2}\right)\theta\to0$
as $n\to\infty$

Therefore $2\cos n\theta\sin\frac{1}{2}\theta\to0$ as
$n\to\infty.\ \sin\frac{1}{2}\theta\neq0$.

Therefore $\cos n\theta\to0$ as $n\to\infty$ therefore
$\cos2n\theta\to0$ as $n\to\infty$ but
$\cos2n\theta=2\cos^2n\theta-1\Rightarrow0=-1$ which is a
contradiction therefore
$\frac{1}{2}\sum_{n=1}^\infty\cos\nu\theta$ does not converge.

\end{enumerate}

\item[Fej\'{e}r's Integral]

\begin{eqnarray*}
S_n(x)&amp;=&amp;\frac{2}{\pi}\int_0^\pi\phi_x(t)D_n(t)\,dt\\
\textrm{therefore }\frac{S_0+\ldots S_n}{n+1}&amp;=&amp;\sigma_n(x)\\
&amp;=&amp;\frac{2}{\pi}\int_0^\pi\phi_x(t)\frac{D_0+\ldots+D_n}{n+1}\,dt\\
\sum_{\nu=0}^n D_\nu
t&amp;=&amp;\sum_{\nu=0}^n\frac{\sin\left(\nu+\frac{1}{2}\right)t}{2\sin\frac{1}{2}t}\\
&amp;=&amp;\sum_{\nu=0}^n\frac{\cos\nu
t-\cos(\nu+1)t}{\left(2\sin\frac{1}{2}\right)^2}\\
&amp;=&amp;\frac{1-\cos(n+1)t}{\left(2\sin\frac{1}{2}\right)^2}\\
\textrm{Fej\'{e}r's Kernel}&amp;\equiv&amp;K_n(t)\\
&amp;=&amp;\frac{D_0+\ldots+D_n}{n+1}\\
&amp;=&amp;\frac{1}{n+1}\frac{1-\cos(n+1)t}{\left(2\sin\frac{1}{2}t\right)^2}\\
&amp;=&amp;\frac{1}{2}\left(\frac{\sin\frac{n+1}{2}t}{\sin\frac{1}{2}t}\right)^2\frac{1}{n+1}\\
\sigma_n&amp;=&amp;\frac{2}{\pi}\int_0^\pi\phi_x(t)K_n(t)\,dt.
\end{eqnarray*}

Now if $f(t)=1$ for all $t$ then $\phi_x(t)=1$ for all t,
$S_n(t)=1$ for all $t$ and $\sigma_n(t)=1$ for all $t$ therefore

$$\sigma_n(x)-S=\frac{2}{\pi}\int_0^\pi(\phi_x(t)_S)K_n(t)\,dt$$

\item[Theorem 13 Fej\'{e}r's Theorem]
If $f\in{\cal{L}}(-\pi\ \pi)$ and is periodic $2\pi$, and if
$\phi_x(0+)$ exists then the Fourier series of $f(t)$ is summable
$(C,1)$ for $t=x$ to $\phi_x(0+)$

\item[Proof]
Let $\varepsilon&gt;0$. Choose $\delta|0&lt;\delta\leq\pi$, and
$|\phi_x(t)-\phi_x(0+)|&lt;\frac{1}{2}\varepsilon$ in $0,\ \delta)$

\begin{eqnarray*}
|I_1|&amp;\leq&amp;\frac{2}{\pi}\int_0^\delta|\phi_x(t)-\phi_x(0+)||K_n(t)|\,dt\\
&amp;\leq&amp;\frac{2}{\pi}\frac{\varepsilon}{2}\int_0^\delta
K_n(t)\,dt\textrm{ as }K_n(t)\geq0\\
&amp;\leq&amp;\frac{2}{\pi}\frac{\varepsilon}{2}\int_0^\pi
K_n(t)\,dt=\frac{\varepsilon}{2}\textrm{ for all }n\geq0\\
|I_2|&amp;\leq&amp;\frac{2}{\pi}\int_\delta^\pi|\phi_x(t)-\phi_x(0+)|
\frac{1-\cos(n+1)t}{(n+1)4\sin^2\frac{1}{2}t}\\
&amp;\leq&amp;\frac{2}{\pi}\frac{1}{n+1}\int_\delta^\pi|
\phi_x(t)-\phi_x(0+)|\frac{2}{4\sin^2\frac{1}{2}t}\,dt\\
&amp;=&amp;\frac{C(\delta)}{n+1}&lt;\frac{\varepsilon}{2}
\end{eqnarray*}

if $n$ is sufficiently large. Hence the result.

\item[Corollary 1]
If $f(x+)$ and $f(x-)$ exist, the Fourier series is summable
$(C,1)$ to $\frac{f(x+)+f(x-)}{2}$.

\item[Corollary 2]
If $f(t)$ is continuous at $x$, the Fourier series is summable
$(C,1)$ to $f(x)$.

\item[Answer to Problem 1]
The Fourier series converges to $\phi(x)$ therefore the Fourier
series is summable $(C,1)$ to $\phi(x)$.

But since $f(t)$ continuous the Fourier series is summable $C,1)$
to $f(x)$ by Fej\'{e}r's theorem therefore $f(x)=\phi(x)$.

\item[Answer to Problem 2]
The Fourier series of $f(x)$ is summable $(C,1)$ to $f(x)$.

The Fourier series of $\phi(x))$ is summable $(C,1)$ to $\phi(x)$.

If $f$ and $\phi$ have the same Fourier series $f(x)=\phi(x)$.

\item[Problem 4]
If $f(t)$ is continuous and its Fourier series is convergent is
its sum $f(t)$.

\item[Theorem 14]
If $f\in{\cal{L}}(-\pi\ \pi)$ and is periodic $2\pi$ then if the
Fourier series converges at $t=x$, and $f(t)$ is continuous at
$t=x$, its sum is $f(x)$.

\item[Proof]
Since the Fourier series is convergent at $t=x$ it has sum $S$
therefore the Fourier series is summable $(C,1)$ to $S$ ate $t=x$
therefore $S=f(x)$.

\item[Uniform Summability]
$\sum a_n(x)$ is summable $(C,1)$ uniformly in $[a\ b]$ to
$\sigma(x)$ if

$$\sigma_n(x)=\frac{S_0(x)+\ldots+S_n(x)}{n+1}\to\sigma(x)$$

uniformly with respect to $x$ in $[a\ b]$ as $n\to\infty$.

\item[Theorem 15 Localisation Property]
If $f(4)\in{\cal{L}}(-\pi\ \pi)$ and is periodic $2\pi$ then

\begin{description}

\item[(i)]
the summability $(C,1)$ of the Fourier series at $t=x$ depends
only on $f(t)$ in $x-\delta,x+\delta)\ \delta&gt;0$.

\item[(ii)]
the uniform summability $(C,1)$ of the Fourier series in $[a\ b]$
depends only on $f(t)$ in $(a-\delta,b+\delta\ \delta&gt;0$.

\end{description}

\item[Proof]

\begin{description}

\item[(i)]

$$\sigma_n(x)=\frac{2}{\pi}\int_0^\pi\phi_x(t)K_n(t)\,dt$$

For $0&lt;\delta&lt;\pi$

\begin{eqnarray*}
\left|\int_\delta^\pi\phi_x(t)K_n(t)\,dt\right|
&amp;=&amp;\frac{1}{n+1}\left|\int_\delta^\pi\phi_x(t)\frac{1-\cos(n+1)t}{
\left(2\sin\frac{1}{2}t\right)^2}\,dt\right|\\
&amp;\leq&amp;\frac{1}{n+1}\int_\delta^\pi|\phi_x(t)|\frac{2}{\left(
2\sin\frac{1}{2}\right)^2}\,dt\\
&amp;=&amp;\frac{C(\delta)}{n+1}\to0\textrm{ as }n\to\infty
\end{eqnarray*}

\item[(ii)]

\begin{eqnarray*}
\sigma_n(x)&amp;=&amp;\frac{2}{\pi}\int_0^\pi\phi_x(t)K_n(t)\,dt\\
&amp;=&amp;\frac{1}{\pi}\int_0^\pi{f(x+t)+f(x-t)}K_n(t)\,dt\\
&amp;=&amp;\frac{1}{\pi}\int_{-\pi}^\pi f(x+t)K_n(t)\,dt\\
&amp;=&amp;\frac{1}{\pi}\int_{x-\pi}^{x+\pi}f(t)K_n(t-x)\,dt\\
&amp;=&amp;\frac{1}{\pi}\int_{x-\pi}^{x-\delta}+\frac{1}{\pi}\int_{x-\delta}^{
x+\delta}+\frac{1}{\pi}\int_{x+\delta}^{x+\delta}\\
&amp;=&amp;I_1+I_2+I_3\\
|I_3|&amp;=&amp;\left|\frac{1}{\pi}\int_{x+\delta}^{x+\pi}f(t)K_)n(t-x)\,dt\right|\\
&amp;\leq&amp;\frac{1}{n+1}\frac{1}{\pi}\int_{x+\delta}^{x+\pi}|f(t)|\frac{2}{\left[
2\sin\frac{1}{2}(t-x)\right]^2}\,dt\\
&amp;\leq&amp;\frac{2}{n+1}\frac{1}{\pi}\frac{1}{\left(2\sin\frac{1}{2}\delta\right)
^2}\int_{x+\delta}^{x+\pi}|f(t)|\,dt\\
&amp;\leq&amp;\frac{2}{n+1}\frac{1}{\pi}\frac{1}{\left(2\sin\frac{1}{2}\delta\right)^2
}\int_{-\pi}^\pi|f(t)|\,dt=\frac{C(\delta)}{n+1}\to0
\end{eqnarray*}

uniformly in any interval and so in the whole range. Similarly
$I_1\to0$ uniformly in the whole range.

\end{description}

\item[Theorem 16 (Modified form of $K_{n-1}(t)$)]
The necessary and sufficient condition for the Fourier series of
$f(t)$ to be summed $(C,1)$ to $S$ is

$$\frac{1}{n}\int_0^\pi(\phi_x(t)-S)\left(\frac{\sin\frac{1}{2}nt}{t}\right)^2\,dt\to0\textrm{
as }n\to\infty$$

\item[Proof]

\begin{eqnarray*}
K_{n-1}(t)&amp;=&amp;\frac{1}{n}\frac{1-\cos
nt}{\left(2\sin\frac{1}{2}t\right)^2}\\
&amp;=&amp;\frac{2}{n}\frac{\sin^2\frac{1}{2}n(t)}{\left(2\sin\frac{1}{2}t\right)^2}\\
I&amp;=&amp;\frac{1}{n}\left|\int_0^\pi[\phi_x(t)-S]
\left[\left(\frac{\sin\frac{1}{2}nt}{2\sin\frac{1}{2}t}\right)^2-
\left(\frac{\sin\frac{1}{2}nt}{t}\right)^2\right]\,dt\right|\\
&amp;\leq&amp;\int_0^\pi|\phi_x(t)-S|\left[\frac{1}{\left(2\sin\frac{1}{2}t\right)^2}
-\frac{1}{t^2}\right]\,dt
\end{eqnarray*}

write
$g(t)=\frac{1}{\left(2\sin\frac{1}{2}t\right)^2}-\frac{1}{t^2}$.

$g(t)$ continuous $0&lt;t\leq\pi$ and tends to a limit as $t\to0$.
Define $g(0)$ by continuity then $g(t)$ bounded in $[0\ \pi]$

$$I\leq\frac{1}{n}A\int_0^\pi|\phi_x(t)-S|\,dt=O\left(\frac{1}{n}\right)$$

Hence the result.

\item[The Lebesgue Set]
Lebesgue showed that if $g(t)\in{\cal{L}}(a\ b)$ and
$\phi(x)=\int_a^xg(t)\,dt\ x\in[a\ b]$ then $\exists\phi'(x)=g(x)$
p.p in $(a\ b)$. He then generalised it to the following result.

\item[Theorem 17]
If $f\in{\cal{L}}(a\ b)\
\lim_{h\to0}\frac{1}{h}\int_x^{x+h}|f(t)-\alpha|\,dt=|f(x)-\alpha|$
for all real $\alpha$, except when $x$ belongs to a set of measure
0 (independent of $\alpha$).

\item[Proof]
For a fixed $\alpha\
\lim_{h\to0}\frac{1}{h}\int_{x}^{x+h}|f(t)-\alpha|\,dt=|f(x)-\alpha|$
for all $x\in[a\ b)$ outside a set $\xi_\alpha$ of measure zero.

Let $\{\alpha_\nu\}$ be an enumeration of the rational numbers.

$$\lim_{h\to0}\frac{1}{h}\int_x^{x+h}|f(t)-\alpha_\nu|\,dt=|f(t)-\alpha_\nu|$$

for all $\alpha_\nu$, and all $x$ outside
$\xi=\cup_{\nu=1}^\infty\xi_{\alpha_nu}$ which is null.

Let $\beta$ be a real number.

$$||f(x)-\beta|-|f(x)-\alpha_\nu||\leq|(f(x)-\beta)
-(f(x)-\alpha_\nu)|=|\beta-\alpha_\nu|$$

Therefore

$$\left|\frac{1}{h}\int_{x}^{x+h}|f(t)-\beta|\,dt
-\frac{1}{h}\int_x^{x+h}|f(t)-\alpha_\nu|\,dt\right|
\leq|\beta-\alpha_\nu|$$

for all $x$ outside $\xi$ we have

\begin{eqnarray*}
&amp;&amp;\left|\frac{1}{h}\int_x^{x+h}|f(t)-\beta|\,dt-|f(t)-\beta|\right|\\
&amp;\leq&amp;\left|\frac{1}{h}\int_x^{x+h}\int_x^{x+h}|f(t)-\beta|\,dt
-\frac{1}{h}\int_x^{x+h}|f(t)-\alpha_\nu|\,dt\right|\\
&amp;&amp;+\left|\frac{1}{h}\int_x^{x+h}|f(t)-\alpha_\nu|\,dt-
|f(t)-\alpha_\nu|\,dt\right|\\
&amp;&amp;+\left||f(t)-\alpha_\nu|-|f(t)-\beta|\right|\\
&amp;\leq&amp;|\beta-\alpha_\nu|+\left|\frac{1}{h}\int_x^{x+h}
|f(t)-\alpha_\nu|\,dt-|f(t)-\alpha_nu|\right|+|\beta-\alpha_\nu|
\end{eqnarray*}

which may be made as small as we please, by choice first of
$\alpha_\nu$ and then of $h$. Hence the result.

\item[Corollary]
$$\int_0^h|f(x+t)-f(x)|\,dt=o(h)\textrm{ as }h\to0$$

for almost all $x$. The set where this holds is called the
Lebesgue set.

\item[Theorem 18 (Fej\'{e}r Lebesgue)]
If $f(t)\in{\cal{L}}(-\pi\ \pi)$ and is periodic $2\pi$ and if
$\int_0^h|\phi_x(t)-S|\,dt=o(h)$ as $h\to0+$ then the Fourier
series of $f(t)$ is summable $(C,1)$ to $S$ at $t=x$.

\item[Proof]

\begin{equation}
\frac{1}{n}\int_0^\pi[\phi_x(t)-S]\left(\frac{\sin\frac{1}{2}nt}{t}\right)^2\,dt
=\sigma_n(x)+o(1)
\end{equation}

Let $\varepsilon&gt;0.\ \exists\delta|0&lt;\delta&lt;\pi$, and

\begin{eqnarray*}
0\leq\Phi(t)&amp;=&amp;\int_0^t|\phi_x(u)-S|\,du&lt;\varepsilon t\textrm{ in
}90\ \delta]\\
\frac{1}{n}\left\{\frac{\sin\frac{1}{2}nt}{t}\right\}^2&amp;\leq&amp;\left\{
\begin{array}{ll}n&amp;0&lt;t\leq\frac{1}{n}\\
\frac{1}{nt^2}&amp;\frac{1}{n}\leq t\leq\pi\end{array}\right.\\
(3)&amp;=&amp;\frac{1}{n}\left\{\int_0^\frac{1}{n}+\int_\frac{1}{n}^\delta
+\in_\delta^\pi\right\}\\ &amp;=&amp;I_1+I_2+I_3\textrm{ taking
}n&gt;\frac{1}{\delta}\\
|I_3|&amp;\leq&amp;\frac{1}{n}\int_\delta^\pi|\phi_x(t)-S|\frac{1}{t^2}\,dt=
\frac{C(\delta)}{n}\\
|I_1|&amp;\leq&amp;n\int_0^\frac{1}{n}|\phi_x(t)-S|\,dt\to0\textrm{ as
}n\to\infty\textrm{ by hypothesis}\\
|I_2|&amp;\leq&amp;\frac{1}{n}\int_{\frac{1}{n}}^\delta|\phi_x(t)-S|\,\frac{dt}{t^2}\\
&amp;=&amp;\frac{1}{n}\left[\frac{\Phi(t)}{t^2}\right]_\frac{1}{n}^\delta+\frac{2}{n}
\int_\frac{1}{n}^\delta\frac{\Phi(t)}{t^3}\,dt\\
&amp;\leq&amp;\frac{1}{n}C(\delta)-n\Phi\left(\frac{1}{n}\right)+
\frac{2\varepsilon}{n}\int_\frac{1}{n}^\delta\frac{dt}{t^2}\\
&amp;\leq&amp;\frac{1}{n}C(\delta)+\frac{2\varepsilon}{n}
\int_\frac{1}{n}^\infty\frac{dt}{t^2}\\
&amp;=&amp;\frac{C(\delta)}{n}+2\varepsilon
\end{eqnarray*}

Hence the result.

\item[Corollary]

$$\frac{1}{h}\int_0^h|\phi_x(t)-f(x)|\,dt\leq\frac{1}{h}
\int_0^h\frac{|f(x+t)-f(x)|}{2}\,dt+\frac{1}{h}\int_0^h
\frac{|f(x-t)-f(x)|}{2}\,dt$$

$\to0$ in the Lebesgue set.

Hence the Fourier series of $f(t)$ is summable $(C,1)$ to $f(x)$
in the Lebesgue set.

\item[Theorem 19]
The necessary and sufficient conditions for $m\leq\sigma_n(x)\leq
M$ for all $n$ and all $x$ is $m\leq f(t)\leq M$ p.p. in $-\pi\
\pi]$

\item[Proof]

\begin{description}

\item[(i)]
Necessity: Since $f(t)\in{\cal{L}}(-\pi\ \pi)$, for all $x$ in the
Lebesgue set $\sigma_n(x)\to f(x)$ as $n\to\infty$.

By hypothesis $m\leq\sigma_n(x)\leq M$ therefore
$m\leq\underline{\lim}\sigma_n(x)\leq\overline{\lim}\sigma_n(x)\leq
M$ therefore for almost all $x,\
\overline{\lim}=\underline{\lim}=f(x)$ therefore $m\leq f(x)\leq
M$ p.p in $[-\pi\ \pi]$.

\item[(ii)]
Sufficiency:

$$\sigma_n(x)-M=\frac{1}{2\pi}\int_0^\pi\frac{\phi_x(t)-M}{n+1}
\left\{\frac{\sin\frac{n+1}{2}t}{\sin\frac{1}{2}t}\right\}^2\,dt$$

Since $f(t)\leq M$ p.p. in $[-\pi\ \pi]\phi_x(t)\leq M$ p.p in
$[0\pi]$ therefore the integrand is $\leq0$p.p therefore the
integral is $\leq 0$ therefore $\sigma_n(x)\leq M$. Similarly
$\sigma_n(x)\geq m$.

\end{description}

\item[Theorem 20 Uniqueness theorem for Fourier Series]
If $f\in{\cal{L}}(-\pi\ \pi)\ g\in{\cal{L}}(-\pi\
\pi)f(t)\sim\frac{1}{2}a_0+\sum(\ )\ \
g(t)\sim\frac{1}{2}a_0+\sum(\ )$ then $f(t)=g(t)$ p.p. in $(-\pi\
\pi)$.

\item[Proof]
The Fourier series of $f(t)$ is summable $(C,1)$ p.p. to $f(t)$.
The Fourier series of $g(t)$ is summable $(C,1)$ to $g(t)$.

\item[Theorem 21]
If $f(t)\in{\cal{L}}(-\pi\ \pi)$ and it's Fourier coefficients are
all zero the $f(t)=0$ p.p in $(-\pi\ \pi)$.

\item[Proof]
From Theorem 20 with $g(t)=0$. Suppose $f(t)$ is even and
$|f(t)||leq1$.

\begin{eqnarray*}
S_n&amp;=&amp;S_n(0)=\frac{2}{\pi}\int_0^\pi
f(t)\frac{\sin\left(n+\frac{1}{2}\right)t}{2\sin\frac{1}{2}t}\,dt\\
|S_n|&amp;\leq&amp;\frac{2}{\pi}\int_0^\pi\frac{|\sin\left(n+\frac{1}{2}\right)t|
}{2\sin\frac{1}{2}t}\,dt
\end{eqnarray*}

If $f(t)=\chi_n(t)=\left\{\begin{array}{ll}1&amp;\textrm{if
}\sin\left(n+\frac{1}{2}\right)t&gt;o\\ -1&amp;\textrm{if
}\sin\left(n+\frac{1}{2}\right)t\leq0\end{array}\right.$ in $[-\
\pi]$

$$S_n(0)=\frac{2}{\pi}\int_0^\pi\frac{|\sin\left(n+\frac{1}{2}\right)|
}{2\sin\frac{1}{2}t}\,dt.$$

Let $\displaystyle
L_n=\frac{2}{\pi}\int_0^\pi\frac{\left|\sin\left(n+\frac{1}{2}\right)t\right|
}{2\sin\frac{1}{2}t}\,dt$

We can show that $|L_n-\frac{4}{\pi^2}\log n|&lt;K$ for all $n$, and
some $K$, and $L_n\sim\frac{4}{\pi^2}\log n$ as $n\to\infty$.

\begin{eqnarray*}
&amp;&amp;\left|\frac{2}{\pi}\int_0^\pi\frac{|\sin\left(n+\frac{1}{2}\right)t|
}{2\sin\frac{1}{2}t}\,dt-\frac{2}{\pi}\int_0^\pi\frac{|\sin
nt|}{2\tan\frac{1}{2}t}\,dt\right|\\
&amp;\leq&amp;\frac{2}{\pi}\int_0^\pi\frac{|\sin\left(n+\frac{1}{2}\right)t-\sin
nt\cos\frac{1}{2}t|}{2\sin\frac{1}{2}t}\,dt\\
&amp;=&amp;\frac{2}{\pi}\int_0^\pi\frac{|\cos
nt|}{2}\leq\frac{2}{\pi}\int_0^\pi\frac{1}{2}\,dt=1
\end{eqnarray*}

Now

\begin{eqnarray*}
&amp;&amp;\left|\frac{2}{\pi}\int_0^\pi\left(\frac{|\sin
nt|}{2\tan\frac{1}{2}t}-\frac{|\sin nt|}{t}\right)\,dt\right|\\
&amp;\leq&amp;\frac{2}{\pi}\int_0^\pi\left|\frac{1}{t}-\frac{1}{2\tan
\frac{1}{2}t}\right|\,dt=G\\ \textrm{ therefore }
&amp;&amp;\left|L_n-\frac{2}{\pi}\int_0^\pi\frac{|\sin
nt|}{t}\,dt\right|\leq G+1
\end{eqnarray*}

\begin{eqnarray*}
\int_0^\pi\frac{|\sin nt|}{t}\,dt&amp;=&amp;\int_0^n\pi\frac{|\sin
u|}{u}\,du\\
&amp;=&amp;\sum_{\nu=0}^{n-1}\int_{\nu\pi}^{(\nu+1)\pi}\frac{|\sin
u|}{u}\,du\\ &amp;=&amp;\sum_{\nu=0}^{n-1}\int_0^\pi\frac{|\sin
u|}{u+\nu\pi}\,du\\ &amp;\geq&amp;\sum_{\nu=0}^{n-1}\int_0^\pi\sin
u\,du\frac{1}{(\nu+1)\pi}\textrm{ as
}\frac{1}{\nu\pi}\geq\frac{1}{u+\nu\pi}\geq\frac{1}{(\nu+1)\pi}\\
&amp;=&amp;\sum_{\nu=0}^{n-1}\frac{2}{(\nu+1)\pi}\\
&amp;\geq&amp;\frac{2}{\pi}\sum_{\nu=0}^{n-1}\sum_{\nu+1}^{\nu+2}\frac{dt}{t}\\
&amp;\geq&amp;\frac{2}{\pi}\log n
\end{eqnarray*}

Similarly

$$\sum_{\nu=0}^{n-1}\int_0^\pi\frac{\sin u}{u+\nu
n}\leq\int_0^\pi\frac{\sin u}{u}\,du+\int_0^\pi\frac{\sin
u}{u+\pi}\,du+\frac{2}{\pi}\log n$$

Therefore

$$\left|L_n-\frac{4}{\pi^2}\log n\right|&lt;K$$

Hence $\overline{\lim}\frac{S_n(x)}{\log n}\leq\frac{4}{\pi^2}$.

\item[Theorem 22]
There is an even continuous periodic function of period $2\pi$
whose Fourier series diverges for $t=0$.

\item[Proof]

$$S_n(0)\geq\frac{2}{\pi}\int_0^\pi f(t)\frac{\sin nt}{t}\,dt-C$$

Write $S_n=\frac{2}{\pi}\int_0^\pi f(t)\frac{\sin nt}{t}\,dt$ then
$S_n(0)\geq\frac{2}{\pi}S_n-C$.

We shall construct $F(t)$ so that $\overline{\lim}S_n=+\infty$.
Consider $a_\nu&gt;0|\sum_1^\infty=1$.

Consider integers $1&lt;n_1&lt;n_2&lt;n_3&lt;\ldots$

$f(t)=\sum_{\nu=1}^\infty a_\nu\sin n_\nu t$ in $[0\ \pi]$, even
and periodic $2\pi$.

$f(t)$ is continuous as the series converges uniformly with
respect to $t$.

\begin{eqnarray*}
S_n&amp;=&amp;\int_0^\pi f(t)\frac{\sin nt}{t}\,dt\\
&amp;=&amp;\int_0^\pi\sum_{\nu=1}^\infty a_\nu\sin n_\nu t.\frac{\sin
nt}{t}\,dt\\ &amp;=&amp;\sum_{\nu=1}^\infty a_\nu\int_0^\pi\frac{\sin
n_\nu t\sin nt}{t}\,dt
\end{eqnarray*}

Choose $n_1|\alpha_21\int_0^\pi\frac{\sin^2 n_1 t}{t}&gt;2$

$$S_{n_1}\geq\alpha_1\int_0^\pi\frac{\sin^2n_1t}{t}\,dt-\sum_{\nu=2}^\infty\alpha_\nu\left|\sum_0^\pi\frac{\sin
n_1t}{t}\sin n_\nu t\,dt\right|$$

Choose $M_1|\left|\int_0^\pi\frac{\sin n_1t}{t}\sin
mt\,dt\right|&lt;1$ for $m\geq M_1$ (possible be Riemann-Lebesgue
Theorem) therefore provided $n_2\geq M_1$

$$\left|\int_0^\pi\frac{\sin n_\nu t\sin n_1t}{t}\,dt\right|&lt;1\
\nu=2,3,\ldots$$

therefore $S_{n_1}\geq2-\sum_{\nu=2}^\infty\alpha_\nu\geq1$.

Choose $n_2\geq M_1|\alpha_2\int_0^\pi\frac{\sin^2
n_2t}{f}\,dt&gt;3$.

$$S_{n_2}\geq\alpha_2\int_0^\pi\frac{\sin^2
n_2t}{t}\,dt-\left(\sum_{\nu&lt;2}+\sum_{\nu&gt;2}\right)\left|\alpha_2\int_0^\pi\frac{\sin
n_\nu t\sin n_2t}{t}\,dt\right|$$

$$\left|\int_0^\pi\frac{\sin n_1t\sin
n_2t}{t}\,dt\right|&lt;1\textrm{ as }n_2\geq M_1.$$

Choose $M_2|\left|\int_0^\pi\frac{\sin n_2t}{t}\sin
mt\,dt\right|&lt;1$ for $m\geq M_2$ therefore provided $n_3\geq M_2$

$$\left|\int_0^\pi\frac{\sin n_\nu t\sin n_2t}{t}\,dt\right|&lt;1\
\nu=3,4,\ldots$$

therefore $S_{n_2}\geq3-\sum_{\nu\neq2}\alpha_\nu.1\geq2$.

Suppose $1\leq n_1&lt;M_1\leq n_2&lt;M_2\leq\ldots\leq
n_{\mu-1}&lt;M_{\mu-1}$ have all been chosen such that
$S_{n_{\mu-1}}&gt;\mu-1$ provided $n_\mu\geq M_{\mu-1}$,

Choose $n_\nu|\alpha_\mu\int_0^\pi\frac{\sin^2n_\mu
t}{t}\,dt&gt;\mu+1$

$$\left|\int_0^\pi\frac{\sin n_\nu t\sin n_\mu
t}{t}\right|&lt;1\textrm{ for $\nu=1,2,\ldots,\mu-1$ as} n_\mu\geq
M_{\mu-1}$$

Choose $M_\mu|\left|\int_0^\pi\frac{\sin n_\mu t\sin
mt}{t}\,dt\right|&lt;1$ provided $m\geq M_\mu$.

Therefore provided $n_{\mu+1}\geq M_\mu$

\begin{eqnarray*}
S_{n_\mu}&amp;\geq&amp;\alpha_\nu\int_0^\pi\frac{\sin^2n_\mu
t}{t}\,dt-\sum_{\nu\neq\mu}\alpha_\nu\left|\int_0^\pi\frac{\sin
n_\nu t\sin n_\mu t}{t}\,dt\right|\\ &amp;\geq&amp;\mu+1-1=\mu
\end{eqnarray*}

\item[Lemma 1]
If $f_N(t)=\sin\left(N+\frac{1}{2}\right)t$ in $[0\ \pi]$ even and
periodic, $N$ a positive integer. Then

\begin{description}

\item[(i)]
$S_n(0)&gt;0$

\item[(ii)]
$S_N(0)&gt;\frac{1}{\pi}\log N-C$

\end{description}


\item[Proof]

\begin{description}

\item[(i)]

$$a_n=\frac{1}{\pi}\left(\frac{1}{N+n+\frac{1}{2}}+
\frac{1}{N-n+\frac{1}{2}}\right)=
\frac{1}{\pi}\frac{2N+\frac{1}{2}}{
\left(N+\frac{1}{2}\right)^2-n^2}$$

$S_n(0)$ increasing and $&gt;0$ in $[0\ N]$ and decreasing in $[N+1\
\infty)$, also $S_\infty(0)=0$ and the function is continuous and
B.V therefore $S_n(0)&gt;0$ for all $n$.

\item[(ii)]
See problems.

\end{description}

\item[Theorem 23]
If $\psi(n)$ is any decreasing sequence tending to zero as
$n\to\infty,\ \exists$ a continuous even periodic function of
period $2\pi$ whose Fourier series

\begin{description}

\item[(i)]
diverges at 0

\item[(ii)]
$s_{n_\nu}(0)&gt;\log n_\nu\psi(n_\nu)$ for a sequence
$n_1&lt;n_2&lt;\ldots$.

\end{description}

\item[Proof]
Let $\alpha_j=\frac{1}{j^2},\
f_n(t)=\sin\left(n+\frac{1}{2}\right)t\ [0\ \pi]$ even and
periodic.

\begin{eqnarray*}
f(t)&amp;=&amp;\sum_{j=1}^\infty\alpha_jf_{n_j}(t)\\
&amp;=&amp;\sum_{j=1}^\infty\frac{\sin\left(n_j+\frac{1}{2}\right)t}{j^2}\\
S_{n_\nu}(0)&amp;=&amp;\sum_{j=1}^\infty\frac{1}{j^2}\frac{2}{\pi}\int_0^\pi\frac{\sin\left(n_j+\frac{1}{2}\right)t\sin\left(n_\nu+\frac{1}{2}\right)t}{2\sin\frac{1}{2}t}\,dt\\
&amp;\geq&amp;\frac{1}{\nu^2}\frac{2}{\pi}\int_0^\pi\frac{\sin^2\left(n_\nu+\frac{1}{2}\right)}{2\sin\frac{1}{2}t}\,dt\\
&amp;&gt;&amp;\frac{1}{\pi}\frac{1}{\nu^2}\log n_\nu-C\\
&amp;&gt;&amp;\frac{1}{2\pi}\frac{1}{\nu^2}\log n_\nu
\end{eqnarray*}

is $\nu$ sufficiently large.

Choose $n_\nu|\psi(n_\nu)&lt;K\frac{1}{2\pi}.\frac{1}{\nu^2}$
(possible as $\psi\to0$).

\item[General Trigonometrical Series]

\item[Theorem 24 Cantor's Lemma]
If $a_n\cos nx+b_n\sin nx\to0$ for all $x$ in a set $E$ of
positive measure, then $a_n\to0,b_n\to0$ as $n\to\infty$.

\item[Proof]
Suppose without loss of generality $\exists\subset[-\pi\ \pi]$

Assume $a_n^2+b_n^2\not\to0\ \exists\delta&gt;0$ and
$\{n_\nu\}|a_{n_\nu}^2+b_{n_\nu}^2\geq\delta$.

Let $\displaystyle g_\nu(x)=\frac{(a_{n_\nu}\cos n_\nu
x+b_{n_\nu}\sin n_\nu x)^2}{a_{n_\nu}^2+b_{n_\nu}^2}$

$g_\nu(x)\to0$ boundedly in $E$ for

\begin{eqnarray*}
|g_nu(x)|&amp;\leq&amp;\frac{a_{n_\nu}\cos n_\nu x+b_{n_\nu}\sin n_\nu
x)^2}{\delta}\to0,\ x\in E\\
|g_\nu(x)|&amp;=&amp;\left(\frac{a_{n_\nu}}{\sqrt{a_{n_\nu}^2+b_{n_\nu}^2}}\cos
n_\nu x+\frac{b_{n_\nu}}{\sqrt{a_{n_\nu}^2+b_{n_\nu}^2}}\sin n_nu
x\right)^2\\ &amp;=&amp;\sin^2(n_\nu x+\alpha_\nu\leq1
\end{eqnarray*}

Therefore $\int_e g_\nu(x)\,dx\to0$ as $\nu\to\infty$ by the Dom.
Cgce. Theorem.

\begin{eqnarray*}
\int_Eg_\nu(x)\,dx&amp;=&amp;\int_E\frac{(a_{n_\nu}\cos n_\nu
x+b_{n_\nu}\sin n_\nu x)^2}{a_{n_\nu}^2+b_{n_\nu}^2}\,dx\\
&amp;=&amp;\int_E\frac{a_{n_\nu}^2\cos^2n_\nu x+2a_\nu b_\nu\sin n_\nu
x\cos n_\nu x+b_{n_\nu}^2\sin^2n_\nu
x}{a_{n_\nu}^2+b_{n_\nu}^2}\,dx\\
&amp;=&amp;\frac{1}{2}\int_E\frac{a_{n_\nu}^2+b_{n_\nu}^2}{a_{n_\nu}^2+b_{n_\nu}^2}+C\int_E\cos2n_\nu
x\,dx+C'\int_E\sin2n_\nu x\,dx\\ &amp;=&amp;I_1+I_2+I_3
\end{eqnarray*}

$I_2,I_3\to0$ as $n\to\infty$ by Riemann-Lebesgue theorem,
therefore $\int_Eg_\nu(x)\,dx\to\frac{1}{2}mE$ as $n\to\infty$
which is a contradiction.

\item[Theorem 25]
If $\ds\frac{1}{2}a_0+\sum_{n=1}^\infty a_n\cos nx+bb_n\sin nx$
converges to $S$ and if $\ds
F(t)=\frac{1}{4}a_0t^2-\sum_1^\infty\frac{a_n}{n^2}\cos
nt+\frac{b_n}{n^2}\sin nt$ converges in $|t-x|&lt;\delta$.

$$\lim_{h\to0}\frac{F(x+h)-2F(x)+F(x-h)}{h^2}=S$$

\item[Proof]
If $0&lt;h&lt;\delta$ then we have

\begin{eqnarray*}
&amp;&amp;\frac{F(x+h)-2F(x)+F(x-h)}{h^2}\\
&amp;=&amp;\frac{1}{4}a_0^2\frac{(x+h)^2-2x^2+(x-h)^2}{h^2}\\
&amp;&amp;-\frac{1}{h^2}\sum_{n=1}^\infty\left[\frac{a_n}{n^2}\left\{\cos
n(x+h)-2\cos nx+\cos n(x-h)\right\}\right.\\
&amp;&amp;\left.+\frac{b_n}{n^2}\left\{\sin n(x+h)-2\sin nx+\sin
n(x-h)\right\}\right]\\
&amp;=&amp;\frac{1}{2}a_0+\sum_{n=1}^\infty\left(a_n\cos nx+b_n\sin
nx\right)\left(\frac{\sin\frac{1}{2}nh}{\frac{1}{2}nh}\right)^2
\end{eqnarray*}

$\to S$ as $h\to0$.

\item[Corollary Riemann's 1st Lemma]
If $a_n\to0$ and $b_n\to0$, and if
$\ds\frac{1}{2}a_0+\sum_{n=1}^\infty a_n\cos nx+b_n\sin nx$
converges to $S$, $\ds
F(t)=\frac{1}{4}a_0t^2-\sum_{n=1}^\infty\frac{a_n}{n^2}\cos
nt+\frac{b_n}{n^2}\sin nt$ exists for all $t$ and $\exists
\ds\lim_{h\to0}\frac{F(x+h)-2F(x)+f(x-h)}{h^2}=S$.

\item[Generalised Derivatives]

\begin{description}

\item[I]
First derivative $\ds f'(x)=\lim_{h\to0}frac{f(x+h)-f(x)}{h}$

Symmetric 1st derivative $\ds
D'f(x)=\lim_{h\to0}\frac{f(x+h)-f(x-h)}{2h}$

[If $\exists f'(x)\Rightarrow D'f(x)=f'(x)$]

\item[II]
Generalised 2nd derivative.

$$f_{(2)}(x)=\lim_{h\to
0}\frac{f(x+h)-f(x)-f'(x).h}{\frac{1}{2}h^2}$$

[If $\exists f''(x)\Rightarrow f_{(2)}(x)=f''(x)$]

\item[III]
Symmetric generalised 2nd derivative.

$$D^2f(x)=\lim_{h\to0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}$$

[If $\exists f''(x)\Rightarrow D^2f(x)=f''(x)$]

\end{description}

\item[Schwarz's Lemma]
If $D^2F(x)=0$ in $(a\ b)$ and $F$ is continuous in $[a\
b]\Rightarrow f(x)=Ax+B$ in $[a\ b]$.

\item[Proof]
We prove: If $D^2F(x)\geq0$ in $(a\ b)$, then

$$F(x)\leq\frac{F(a)(b-x)+F(b)(x-a)}{(b-a)}$$

in $[a\ b]$.

Let $\ds L(x)=\frac{F(a)(b-x)+F(b)(x-a)}{b-a}$

$L''(x)=0$ therefore $D^2L(x)=0$.

Let $\phi(x)=F(x)-L(x)$.

$\phi(x)$ is continuous in $[a\ b]$ and $D^2\phi(x)=D^2F(x)\geq0$.

$\phi(a)=\phi(b)=0$ R.T.P. $\phi(c)\leq0\ a&lt;c&lt;b$.

Suppose $\phi(c)&gt;0\ a&lt;c&lt;b$ therefore the upper bound of $\phi$ in
$(a\ b)$ is $&gt;0$ and attained at $\xi$ say.

Let $g(x)=-\frac{1}{2}\varepsilon (x-a)(b-x)\ \
g''(x)=\varepsilon$.

Let $\psi(x)=\phi(x)+g(x)$.

$\psi(x)$ is continuous in $[a\ b]$ and

$$D^2\psi(x)=D^2\phi(x)+D^2g(x)\geq\varepsilon.$$

Choose $\varepsilon|\psi(c)&gt;0$ therefore the upper bound of $\psi$
in $(a\ b)$ is $&gt;0$ and attained at $\zeta$ say.

$a&lt;\zeta&lt;b$ and $\psi(\zeta)\geq\psi(x)\ x\in[a\ b]$ therefore

$$\frac{\psi(\zeta+h)-2\psi(\zeta)+\psi(\zeta-h)}{h^2}\leq0$$

therefore $D^2\psi(\zeta)\leq0$ which is a contradiction.

Therefore $D^2F(x)\geq0$ in $(a\ b)$

$$\Rightarrow F(x)\leq\frac{F(a)(b-x)+F(b)(x-a)}{(b-a)}$$

Therefore $D^2F(x)=0$ in $(a\ b)$

$$\Rightarrow F(x)=\frac{F(a)(b-x)+F(b)(x-a)}{(b-a)}$$

\item[Theorem 26 1st uniqueness theorem for Trig. Series]
If a Trigonometric series $\ds\frac{1}{2}a_0+\sum_{n=1}^\infty
a_n\cos nx+b_n\sin nx$ converges to zero for all $x$ then
$a_0=a_n=b_n=0$.

\item[Proof]

\begin{enumerate}

\item
Since the trigonometric series converges for all $x$ $(a_n\cos
nx+b_n\sin nx)\to0$ as $n\to\infty$ for all $x$ in a set of
positive measure.

\item
By Cantor's Lemma $a_n\to0\ b_n\to0$ as $n\to\infty$

\item
Therefore $\ds F(t)=\frac{1}{4}a_0t^2-\sim_{n=1}^\infty
\frac{a_n}{n^2}\cos nt+\frac{b_n}{n^2}\sin nt$ converges uniformly
with respect to $t$ in any finite interval therefore $F(t)$ is
continuous everywhere.

\item
For each $x\ \frac{1}{2}a_0+\sum_1^\infty a_n\cos nx+b_n\sin nx$
converges to zero therefore by Riemann's 1st Lemma $D^2F(x)$
exists and =0 for every $x$.

\item
By Schwarz's Lemma, in an arbitrary finite interval $[a\ b]$,
since $F$ is continuous, it follows that $F(x)$ is linear in $(a\
b)$.

\item
Since $F(x)$ is linear in every interval $[a\ b),\ F(x)$ in linear
in $-\infty,\ \infty)$

\item

$$\frac{1}{4}a_0x^2-\sum_{n=1}^\infty\frac{a_n}{n^2}\cos
nx+\frac{b_n}{n^2}\sin nx=Ax+B$$

therefore

$$\sum_{bn=1}^\infty\frac{a_n}{n^2}\cos nx+\frac{b_n}{n^2}\sin
nx=\frac{1}{4}a_0x^2-Ax-B\ \ -\infty&lt;x&lt;\infty$$

But LHS is periodic - $2\pi$ and continuous in $[-\pi\ \pi]$ and
therefore is bounded in $(-\infty\ \infty)$. Therefore $a_0=)\
A=0$.

\item
Therefore

\begin{eqnarray*}
B&amp;=&amp;\sum_1^\infty\frac{a_n}{n^2}\cos nx+\frac{b_n}{n^2}\sin nx\\
\frac{a_n}{n^2}&amp;=&amp;\frac{1}{\pi}\int_{-\pi}^\pi-B\cos nx\,dx\\
\frac{b_n}{n^2}=\frac{1}{\pi}\int_{-\pi}^\pi-B\sin nx\,dx
\end{eqnarray*}

Therefore $a_n=b_n=0$.

\end{enumerate}

\item[Theorem 27 2nd Uniqueness theorem for trig series]
If for all $x$ with a finite number of exceptions
$\ds\frac{1}{2}a_0+\sum_{n=1}^infty a_n\cos nx+b_n\sin nx$
converges to zero then $a_0=0\ a_n=b_n=0\ n\geq1$

\item[Lemma Riemann's 2nd Lemma]
If $a_n$ and $b_n\to0$ as $n\to\infty$ and
$F(t)=\frac{1}{4}a_0t^2-\sum_{n=1}^\infty\frac{a_n}{n^2}\cos
nt+\frac{b_n}{n^2}\sin nt$ then
$\ds\lim_{h\to0}\frac{F(x+h)-2F(x)+F(x-h)}{h}=0$

\item[Proof]

\begin{eqnarray*}
&amp;&amp;\frac{F(x+h)-2F(x)+F(x-h)}{h}\\
&amp;=&amp;\frac{1}{2}a_0h+h\sum_{n=1}^infty (a_n\cos nx+b_n\sin
nx)\left(\frac{\sin\frac{1}{2}nh}{\frac{1}{2}nh}\right)^2
\end{eqnarray*}

If $\varepsilon&gt;0\exists N||a_n\cos nx+b_n\sin nx|&lt;\varepsilon
n&gt;N$.

Also $\sin^2\frac{1}{2}nh\leq\left(\frac{1}{2}nh\right)^2$ for
$n\leq\frac{1}{h}$.

Let $A$ be the upper bound of $\left\{\frac{1}{2}|a_0||a_n\cos
nx+b_n\sin nx|\right\}$.

Then

\begin{eqnarray*}
&amp;&amp;\left|\frac{F(x+h)-2F(x)+F(x-h)}{h}\right|\\
&amp;\leq&amp;Ah+\sum_{n=1}^N
Ah+h\sum_{N&lt;n\leq\frac{1}{h}}\varepsilon+4\sum\frac{\varepsilon}{n^2h}\\
&amp;\leq&amp;A(N+1)h+\varepsilon+\frac{4\varepsilon}{h}\int_\frac{1}{h}^\infty\frac{du}{(u-1)^2}\\
&amp;=&amp;A(N+1)h+\varepsilon+\frac{4\varepsilon}{1-h}
\end{eqnarray*}

this may be made as small as we please first by choice of
$\varepsilon$, then by choice of $h$.

\item[Proof of Theorem 27]
Theorem 26 tells us that $F(t)$ is linear in the interval between
any two of the exceptional points, and since $F$ is continuous the
straight lines in adjacent intervals join. Applying Riemann's 2nd
Lemma at the exceptional point tells us that the slopes on both
sides of this point are the same. Hence $F(t)$ is linear
throughout $(-\infty\ \infty)$ and the result follows as in
theorem 26.

\item[Theorem 28]
If a trigonometric series $\frac{1}{2}a_0+\sum(a_n\cos nx+b_n\sin
nx)$ converges for all $x$ to $f(x)$, and $f$ is bounded then the
trigonometric series is the Fourier series of $f(x)$.

\item[Lemma 1]
$F(x)$ continuous in $[a\ b]$ and $D^2F(x)\geq0$ in $[a\
b]\Rightarrow F(x)$ convex in $[a\ b]$ i.e. if
$a\leq\alpha&lt;\gamma&lt;\beta\leq b$

$$F(\gamma)\leq\frac{F(\alpha)(\beta-\gamma)+F(\beta)(\gamma-\alpha)}{\beta-\alpha}$$

In particular $\beta=x+h\ \alpha=x-h\ \gamma=x$ gives

$$F(x)\leq\frac{F(x+h)+F(x-h)}{2}$$

Therefore

$$\Delta_h^2=F(x+h)+F(x-h)-2F(x)\geq0$$

i.e. as $\ds D^2F(x)=\lim_{h\to0}\frac{\Delta_h^2F(x)}{h^2}\geq0$
in $(a\ b)\Rightarrow\frac{1}{h^2}\Delta_h^2F(x)\geq0\ a&lt;x&lt;b$ if
$h$ is sufficiently small.

\item[Lemma 2]
If $F(x)$ is continuous in $[a\ b]$ and if $m\leq D^2F(x)\leq M$
in $(a\ b)$ and if $m\leq D^2F(x)\leq M$ in $(a\ b)$ then $\ds
m\leq\frac{\Delta_h^2F(x)}{h^2}\leq M\ \ D^2F(x)-m\geq0$

Therefore

$$D^2\left(F(x)-\frac{1}{2}mx^2\right)\geq0\Rightarrow\frac{
\Delta_h^2\left(F(x)-\frac{1}{2}mx^2\right)}{h^2}\geq0
\Rightarrow\frac{\Delta_h^2F(x)}{h^2}\geq m.$$

Similarly $\frac{\Delta_h^2F(x)}{h^2}\leq M$

\item[Proof of Theorem 28]

\begin{enumerate}

\item
$a_n\cos nx+b_n\sin nx\to0$ for all $x$ as $n\to\infty\Rightarrow
a_n\to0\ b_n\to0$ as $n\to\infty$

\item
$$F(t)=\frac{1}{4}a_0t^2-\sum_{n=1}^\infty
\left(\frac{a_n}{n^2}\cos nt+\frac{b_n}{n^2}\sin nt\right)$$

is uniformly convergent in every finite interval and so $F(t)$ is
continuous everywhere.

\item
For every given $x$

$$\frac{1}{2}a_0+sum a_n\cos nx+b_n\sin nx$$

converges to $f(x)$, say therefore $\exists D^2F(x)=f(x)$
therefore $\frac{\Delta_h^2F(x)}{h^2}\to f(x)$ as $h\to0$

\item
$$|F(x)|\leq
M\Rightarrow\left|\frac{\Delta_h^2F(x)}{h^2}\right|\leq M$$

for all $x$ in $[-\pi\ \pi]$ and all $h$ sufficiently small.

\item

\begin{eqnarray*}
\frac{\Delta_h^2F(x)}{h^2}&amp;=&amp;\frac{1}{2}a_0+\sum_1^\infty(a_n\cos
nx+b_n\sin
nx)\left(\frac{\sin\frac{1}{2}nh}{\frac{1}{2}nh}\right)^2\\
&amp;=&amp;\frac{1}{2}a_0+\sum_{n=1}^\infty A_n\cos nx+B_n\sin nx
\end{eqnarray*}

where
$A_n=a_n\left(\frac{\sin\frac{1}{2}nh}{\frac{1}{2}nh}\right)^2\ \
B_n=b_n\left(\frac{\sin\frac{1}{2}nh}{\frac{1}{2}nh}\right)^2$

\item
For any fixed $h$

$$|A_n|=|a_n|\left(\frac{\sin\frac{1}{2}nh}{
\frac{1}{2}nh}\right)^2\leq\frac{4}{h^2}\frac{|A_n|}{n^2}.\ \
|B_n|\leq\frac{4}{h^2}\frac{|b_n|}{n^2}$$

therefore $\sum_1^\infty A_n\cos nx+B_n\sin nx$ converges
uniformly with respect to $x$ in $[-\pi\ \pi]$.

Therefore

\begin{eqnarray*}
A_m=\frac{1}{\pi}\int_{-\pi}^\pi\frac{\Delta_h^2F(x)}{h^2}\cos
nx\,dx\\
B_m&amp;=&amp;\frac{1}{\pi}\int_{-\pi}^\pi\frac{\Delta_h^2F(x)}{h^2}\sin
nx\,dx
\end{eqnarray*}

Therefore

$$\lim_{h\to0}a_m\left(\frac{\sin\frac{1}{2}nh}{\frac{1}{2}nh}\right)^2
=\lim_{h\to0}\int_{-\pi}^\pi\frac{\Delta_h^2F(x)}{h^2}\sin
nx\,dx$$

$\frac{\Delta_h^2F(x)}{h^2}\to f(x)$ boundedly for $-\pi\leq x\leq
\pi$ as $h\to0$ therefore $a_m=\frac{1}{\pi}\int_{-\pi}^\pi
f(x)\sin mx\,dx$.

Hence the result.

\end{enumerate}

\end{description}

\end{document}
</pre></body></html>