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\begin{document}

\begin{center}
 \textbf{Introductory Course: Fourier Analysis and its many uses}
 \end{center}
\begin{center}
 \textbf{Solutions - Exercises 16.1-16.5, 16.9, 16.12, 16.13, 16.15, 16.16, 16.17, 16.20, 16.21, 16.23 from ``A First Look at Fourier Analysis" by T.W. K\"{o}rner}
 \end{center}
\begin{center}
 (prepared by Mihai Stoiciu)
\end{center}

\medskip
\medskip

 \textbf{Exercise 16.1.} i) Take $P(t)=\frac{f^{(n)}(0)}{n\,!}\,t^n
 + \frac{f^{(n-1)}(0)}{(n-1)\,!}\,t^{n-1} + \cdots + f(0).$

 ii) Note first that $g(0)=g'(0)= \cdots = g^{(n)}(0)=0$. Since $g(0)=g(t)=0$ it follows from Rolle's theorem that
 there exists a $c_1 \in (0,t)$ such that $g'(c_1)=0$. Observe now
 that $g'(0)=g'(c_1)=0$ so Rolle's theorem implies that there exists a $c_2 \in (0, c_1)$ such
 that $g''(c_2)=0$. Repeating this procedure $(n+1)$ times we get
 that there exists $c=c_{n+1} \in (0,t)$ such that $g^{(n+1)}(c) =
 0$. This immediately implies
 $f(t)=P(t)+\frac{f^{(n+1)}(c)}{(n+1)!}\, t^{n+1}.$

 \medskip

\textbf{Exercise 16.2.} A polynomial $P$ of degree at most
$(2n+1)$ with prescribed values for $P(0),\, P'(0),\, ...\,
P^{(n)}(0),\, P(1),\,P'(1),\, ...\, P^{(n)}(1)$ can be obtained by
taking a linear combination of the polynomials $\frac{1}{k!}x^k
(x-1)^{n+1}$, $0 \leq k \leq n$ and $\frac{1}{k!}x^{n+1}(x-1)^k$,
$0 \leq k \leq n$.

We prove now that $P$ is unique. Let $Q$ be a polynomial of degree
at most $(2n+1)$ such that $P(0)=Q(0), P'(0)=Q'(0), ...
P^{(n)}(0)=Q^{(n)}(0), P(1)=Q(1), P'(1)=Q'(1), ...\,
P^{(n)}(1)=Q^{(n)}(1)$. $R=P-Q$ is a polynomial of degree at most
$2n+1$ and $R(0)=R'(0)= \cdots =R^{(n)}(0)=R(1)=R'(1)= \cdots
=R^{(n)}(1)=0$. Therefore the polynomials $X^{n+1}$ and
$(X-1)^{n+1}$ divide $R$ and since $\deg R \leq 2n+1$ we obtain
$R=0$.

Let $P$ be the unique polynomial of degree at most $(2n+1)$ such
that $P(0)=f(0), P'(0)=f'(0), ...\, P^{(n)}(0)=f^{(n)}(0),
P(1)=f(1), P'(1)=f'(1), ...\, P^{(n)}(1)=f^{(n)}(1)$. For a fixed
$y \in (0,1)$ let
$$
g(x)=f(x)-P(x)-\frac{f(y)-P(y)}{y^{n+1}(y-1)^{n+1}}\ x^{n+1}
(x-1)^{n+1}
$$

Let's observe that $g(0)=g'(0)= \cdots = g^{(n)}(0)=0$ and
$g(1)=g'(1)= \cdots = g^{(n)}(1)=0$. Furthermore
$g(0)=g(y)=g(1)=0$. Therefore, from Rolle's theorem it follows
that there exists two points $c_1^1 \in (0,y)$ and $c_2^1 \in
(y,1)$ such that $g'(c_1^1)=g'(c_2^1)=0$. Since $g'(0)=g'(1)=0$ we
can apply Rolle's theorem again and we get three points $c_1^2 \in
(0,c_1^1),\ c_2^2 \in (c_1^1,c_2^1)$ and $c_3^2 \in (c_2^1,1)$
such that $g''(c_1^2)=g''(c_2^2)=g''(c_3^2)=0$. Repeating this
procedure $(n+1)$ times we get $(n+2)$ points
$c_1^{n+1},c_2^{n+1},...\,c_{n+2}^{n+1} \in (0,1)$ such that
$g^{(n+1)}(c_1^{n+1})=g^{(n+1)}(c_2^{n+1})=\cdots=g^{(n+1)}(c_{n+2}^{n+1})=0$.


Now we can apply Rolle's theorem again and we get $(n+1)$ points
$$
c_1^{n+2} \in (c_1^{n+1},c_2^{n+1}),\ c_2^{n+2} \in
(c_2^{n+1},c_3^{n+1}),\ ...\ c_{n+1}^{n+2} \in
(c_{n+1}^{n+1},c_{n+2}^{n+1})
$$ such that
$g^{(n+2)}(c_1^{n+2})=g^{(n+2)}(c_2^{n+2})= \cdots =
g^{(n+2)}(c_{n+1}^{n+2})=0$. Repeating this procedure $(n+1)$
times we get that there exists a point $c=c_1^{2n+2} \in (0,1)$
such that $g^{(2n+2)}(c)=0$.

Since $g^{(2n+2)}(x)= f^{(2n+2)}(x) -
\frac{f(y)-P(y)}{y^{n+1}(y-1)^{n+1}}\,(2n+2)\,!\ $ we immediately
get
$$E(y) = f(y)-P(y) = \frac{f^{(2n+2)}(c)}{(2n+2)\,!}\ y^{n+1}
(y-1)^{n+1}$$

\medskip

\textbf{Exercise 16.3.} Taking imaginary parts in the de Moivre
formula it is easy to see that
$$
U_n(t) = \sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}
\begin{pmatrix}
n+1 \\
2k+1
\end{pmatrix} (t^2 -1)^k\ t^{n-2k}
$$
Taking $\frac{\partial}{\partial \theta}$ in $T_n(\cos \theta) =
\cos n \theta\ $ we get $\ T_n'(\cos \theta)\ \sin \theta = n
(\sin n\theta)$ which implies that $T_n'(x) = n U_{n-1}(x)$.

\medskip

\textbf{Exercise 16.4.} For any $t \in \bbR$ with $|t| < 1$ and
for any $\theta \in \bbR$ we have $|t e^{i \theta}| < 1$ so
$$
\sum_{n=0}^{\infty} t^n e^{in \theta} = \frac{1}{1-t e^{i \theta}}
$$

Since $\textrm{Re} \left[ t^n e^{in \theta}\right] = t^n \cos(n
\theta)$ and $\textrm{Re} \left[\frac{1}{1-t e^{i \theta}}\right]
= \frac{1- t \cos \theta}{1 - 2 t \cos \theta + t^2}$ we get that
$\frac{1 - t \cos \theta}{1 - 2 t \cos \theta + t^2} =
\sum_{n=0}^{\infty} T_n(\cos \theta)\ t^n$ for any $|t| < 1$ and
any $\theta \in \bbR$, which implies $$\frac{1 - tx}{1 - 2 tx +
t^2} = \sum_{n=0}^{\infty} T_n(x)\ t^n$$ for any $|t| <1$ and $|x|
\leq 1$.

\medskip

\textbf{Exercise 16.5.} Let's observe first that for any $r \in
(0,1)$ we have $P_r(f, \theta) = (f * P_r) (\theta)$ where $P_r:
[-\pi, \pi] \rightarrow \mathbb{R} ,\ P_r(t) =
\sum_{n=-\infty}^{\infty} r^{|n|} e^{i n t} = \frac{1 - r^2}{1 +
r^2 - 2 r \cos t}$. Furthermore $P_r(t) > 0$ for any $t \in [-\pi,
\pi]$ and:
$$
\frac{1}{2 \pi} \int_{-\pi}^{\pi} P_r(t)\,dt = \frac{1}{2 \pi}
\sum_{n=-\infty}^{\infty} \int_{-\pi}^{\pi} r^{|n|} e^{i n t} dt =
1
$$

Let $\varepsilon > 0$. For any $r \in (0,1)$ we have:
\begin{align*}
|P_r(f, \theta) - f(\theta)| &= \left|(f*P_r) (\theta) - f(\theta)
\frac{1}{2 \pi} \int_{-\pi}^{\pi} P_r(t)\,dt \right| =
\left|\frac{1}{2 \pi} \int_{-\pi}^{\pi} \left[f(\theta -
t)-f(\theta)\right] P_r(t)\,dt \right|\\
&\leq \frac{1}{2 \pi} \int_{|t| < \delta} \left|f(\theta -
t)-f(\theta)\right| P_r(t)\,dt + \frac{1}{2 \pi} \int_{|t| \geq
\delta} \left|f(\theta - t)-f(\theta)\right| P_r(t)\,dt
\end{align*}

Since $f \in \mathrm{C}(\mathbb{T})$ it follows that $f$ is
bounded and uniformly continuous. Let $M > 0$ such that $|f(s)| <
M$ for any $s \in [-\pi, \pi]$ and let $\delta_{\varepsilon} > 0$
such that $|t-t'| < \delta_{\varepsilon}$ implies $|f(t) - f(t')|
< \frac{\varepsilon}{2}$.

For any $t$ with $|t| > \delta_{\varepsilon}$ we have $\frac{1 -
r^2}{1 + r^2 - 2 r \cos t} < \frac{1 - r^2}{1 + r^2 - 2 r \cos
\delta_{\varepsilon}} < \frac{1-r^2}{1 - (\cos
\delta_{\varepsilon})^2}$ and therefore

$$
\int_{|t| \geq \delta_{\varepsilon}} P_r(t)\,dt\ <\ \frac{1-r^2}{1
- (\cos \delta_{\varepsilon})^2}
$$

Let $r_{\varepsilon} > 0$ such that $r \in (r_{\varepsilon},1)$
implies $\frac{1-r^2}{1 - (\cos \delta_{\varepsilon})^2} < \frac{2
\pi \varepsilon}{4 M}$. For any $r \in (r_{\varepsilon},1)$ we
have:
$$
\frac{1}{2 \pi} \int_{|t| < \delta_{\varepsilon}} \left|f(\theta -
t)-f(\theta)\right| P_r(t)\, dt < \frac{\varepsilon}{2}
$$and$$\frac{1}{2 \pi} \int_{|t| \geq
\delta_{\varepsilon}} \left|f(\theta - t)-f(\theta)\right|
P_r(t)\, dt < \frac{1}{2 \pi}\, 2M\, \frac{2 \pi \varepsilon}{4 M}
= \frac{\varepsilon}{2}
$$

We can conclude now that $|P_r(f, \theta) - f(\theta)| <
\varepsilon$ for any $r \in (r_{\varepsilon}, 1)$ and any $\theta
\in [-\pi, \pi]$. Therefore $P_r(f, \cdot) \rightarrow f$
uniformly as $r \rightarrow 1, \ r < 1$.

\medskip

\textbf{Exercise 16.9.} i) Let $\varepsilon > 0$. Since $f$ has
real values it follows from Theorem 8.1 ii) that there exists a
real trigonometric polynomial
$$
Q(t)= \sum_{n=0}^{N} \left(a_n \sin n t + b_n \cos n t\right)
$$
\noindent such that $\| f - Q \|_{\infty} < \varepsilon$. Let
$Q_1(t)=Q(-t)$. Since $f(t) = f(-t)$ it follows that $\| f - Q_1
\|_{\infty} < \varepsilon$. For $R(t) = \frac{1}{2} (Q(t) +
Q_1(t))$ we have $\| f - R \|_{\infty} < \varepsilon$. Furthermore
a straightforward computation shows that:
$$
R(t) = \sum_{n=0}^{N} b_n \cos n t
$$

ii) Let $g : [-\pi, \pi] \rightarrow \bbR$ defined by $g(t) =
F(|\cos t|)$. Let $\varepsilon > 0$. Since $g \in
\mathrm{C}(\mathbb{T})$ and is real valued it follows from i) that
there exists a real trigonometric polynomial $ R(t) =
\sum_{n=0}^{N} c_n \cos n t\ $ such that $\| f - R \|_{\infty} <
\varepsilon$. Let's observe now that
$$
R(t) = \sum_{n=0}^{N} c_n \cos n t = \sum_{n=0}^{N} c_n\, T_n
(\cos t) = Q(\cos t)
$$ \noindent
where $T_n$ is the $n$-th Chebyshev polynomial of the first kind
and $Q \in \mathbb{R}[\mathrm{X}]$.

Therefore $|g(t) - R(t)| = \left|F(|\cos t|) - Q(\cos t)\right| <
\varepsilon$ for any $t \in [-\pi, \pi]$ which immediately implies
$\| F - Q \|_{\infty} < \varepsilon$.

iii) Since $\int_0^1 g(t)\,t^n\,dt = 0$ for any $n$ it follows
that $\int_0^1 g(t)\,P(t)\,dt = 0$ for any polynomial $P \in
\mathbb{R}[\mathrm{X}]$. Let $P_n$ be a sequence of polynomials
such that $P_n \rightarrow \overline{g}$ uniformly. Since
$$
\lim_{n \rightarrow \infty} \int_0^1 g(t) P_n(t) dt = \int_0^1
g(t)\ \overline{g(x)}\ dt
$$
\noindent it follows that $\int_0^1 g(t)\ \overline{g(x)}\ dt =
\int_0^1 |g(t)|^2 \ dt = 0\ $ so $\ g = 0$.

\medskip

\textbf{Exercise 16.12.} i) Since $f$ is continuous it follows
immediately that $\int_{-\pi}^{\pi} |f(t)|^2 dt < \infty$ and
Theorem 8.2. implies that $\sum_{n=-\infty}^{\infty}
|\widehat{f}(n)|^2 < \infty$. We get $\lim_{n \rightarrow \infty}
\widehat{f}(n) =0$ and $\lim_{n \rightarrow -\infty}
\widehat{f}(n) =0$.

ii) We can write $\sin t = \frac{1}{2 i} \left(e^{it} -
e^{-it}\right)$ and therefore $f_1(t) = \frac{1}{2 i} \left(g_1(t)
e^{it} - g_1(t) e^{-it}\right)$. This implies $\widehat{f_1}(j) =
\frac{1}{2 i} \left(\widehat{g_1}(j-1)- \widehat{g_1}(j+1)\right)$
so for any $n \geq 1$
$$
S_n(f_1,0) = \sum_{j=-n}^{n} \widehat{f_1}(j) = \widehat{f_1}(n) +
\widehat{f_1}(n+1) - \widehat{f_1}(-n) - \widehat{f_1}(-n-1)
$$
Using part i) we get $\lim_{n \rightarrow \infty} S_n(f_1,0) = 0$.

iii) For any $t \neq n \pi$ we have $g_2 (t) = \frac{1}{\sin
t}f_2(t).$ We must prove that $\frac{1}{\sin t}f_2(t), t \neq n
\pi$ can be extended to a continuous function on $\mathbb{T}$. For
any $t \in (-\pi, \pi), t \neq 0$

$$
\frac{1}{\sin t}\,f_2(t) = \frac{t}{\sin t}\ \frac{f_2(t)}{t} =
\frac{t}{\sin t}\ \frac{f_2(t) - f_2(0)}{t-0}
$$

Therefore $\lim_{t \rightarrow 0} \frac{1}{\sin t}f_2(t) =
f_2'(0)$, so the function $\frac{1}{\sin t}f_2(t)$ can be extended
by continuity at $t=0$. Similarly
$$\lim_{t \rightarrow \pi}
\frac{1}{\sin t}\,f_2(t) = \lim_{t \rightarrow \pi} \frac{1}{\sin
(\pi - t)}\,f_2(t) = \lim_{t \rightarrow \pi} \frac{t - \pi}{\sin
(\pi - t)}\,\frac{f_2(t) - f_2(\pi)}{t- \pi} = - f_2'(\pi)$$

$$\lim_{t \rightarrow -\pi}
\frac{1}{\sin t}\,f_2(t) = \lim_{t \rightarrow -\pi}
\frac{-1}{\sin (\pi + t)}\,f_2(t) = \lim_{t \rightarrow -\pi}
\frac{-(t + \pi)}{\sin (\pi + t)}\,\frac{f_2(t) - f_2(-\pi)}{t +
\pi} = - f_2'(\pi)$$

\noindent so the function $\frac{1}{\sin t}f_2(t)$ can be extended
to a continuous function $g_2 : \mathbb{T} \rightarrow
\mathbb{C}$. Using ii) we get $\lim_{n \rightarrow \infty}
S_n(f_2,0) = 0$.

iv) It is easy to see that $f_4$ is continuous and differentiable
at $0$. Furthermore for any $j=2 n, n \in \mathbb{Z}$ we have:
$$
\widehat{f_4}(j) = \widehat{f_4}(2n) = \frac{1}{2 \pi}
\int_{-\pi}^{\pi} f_3(2t) \ e^{-2int} dt = \frac{1}{4 \pi} \int_{-
2 \pi}^{2 \pi} f_3(s)\ e^{-i n s} ds = \widehat{f_3}(n)
$$

>From part iii) we get that there exists a function $g \in
\mathrm{C}(\mathbb{T})$ such that $f_4(t) = g(t)\,\sin t$ and from
ii) we get that $\widehat{f_4} (2n) = \frac{1}{2 i}
(\widehat{g}(2n-1) - \widehat{g}(2n+1))$. We can now conclude that
$$
S_N(f_3,0) = \sum_{n=-N}^{N} \widehat{f_3} (n) = \sum_{n=-N}^{N}
\widehat{f_4} (2n) = \frac{1}{2 i} \left(\widehat{g}(-2n-1) -
\widehat{g}(2n+1)\right)
$$

Therefore $\lim_{N \rightarrow \infty} S_N(f_3,0) = 0$.

 v) Suppose $f$ is continuous and differentiable at $x_0 \in [-\pi, \pi]$. Let $g : [-\pi, \pi] \rightarrow
 \mathbb{C},\ g(x) = f(x+x_0)-f(x_0)$. Clearly $g(0) = 0$ and $g$ is differentiable at $0$. Using part iv) we
 get $\lim_{N \rightarrow \infty} S_N(g,0) = 0$. But
\begin{align*}
 S_N(g,0) &= \sum_{n=-N}^{n=N} \widehat{g}(n)\ =\ \widehat{g}(0) + \sum_{n=-N\ n\neq
 0}^{n=N} \widehat{g} (n) = -f(x_0) + \sum_{n=-N}^{N} \frac{1}{2 \pi}
 \int_{-\pi}^{\pi} f(x+x_0) e^{-i n x} dx\\ &= -f(x_0) + \sum_{n=-N}^{N}  e^{i n x_0} \frac{1}{2 \pi}
 \int_{-\pi}^{\pi} f(x+x_0) e^{-i n (x + x_0)} dx = -f(x_0)
 + \sum_{n=-N}^{N}  e^{i n x_0} \widehat{f}(n) \\ &= -f(x_0)+ S_N(f,x_0)
\end{align*}

 \noindent which implies $\lim_{N \rightarrow \infty} S_N(f,x_0) = f(x_0)$.
\medskip

\textbf{Exercise 16.13. - Reformulated}

i) Prove that for any $\varepsilon > 0$ and any $K > 0$ we can
find a continuous function $f \in \mathrm{C}(\mathbb{T})$ and an
integer $M > 0$ such that $\|f\|_{\infty} < \varepsilon$ and
$|S_M(f,0)| > K$.

\medskip

ii) Prove that the function $f$ from part i) can be chosen to be a
trigonometric polynomial.

\medskip

iii) Prove that for any $\varepsilon > 0$, any $K > 0$ and any
positive integer $m > 0$ we can find a trigonometric polynomial
$P$ and a positive integer $M$ such that $\|P\|_{\infty} <
\varepsilon$, $\widehat{P}(r) = 0$ for any integer $|r| \leq m$
and $|S_M(P,0)|
> K$.

\medskip

iv) For any nonzero trigonometric polynomial $P(t) =
\sum_{n=-N}^{N} a_n e^{int}$ we denote by $\deg P = \max$ \{$n$,
$\ n$ nonnegative integer s.t. $a_n \neq 0$ or $a_{-n} \neq 0$\}.
Prove that we can find a sequence of trigonometric polynomials
$\{P_n\}_{n \geq 1}$ such that the sequence $m(n) = \deg
(P_{n-1})$ is increasing and:

a) $\|P_n\|_{\infty} < 2^{-n}$

b) $\widehat{P_n}(r)=0\ $ if \ $|r| \leq m(n)$ or $|r| > m(n+1)$

c) $|S_{\deg P_n}(P_n,0)| \geq 2^n + \sum_{k=1}^{n-1}|S_{\deg
P_k}(P_k, 0)|$

\medskip

v) Prove that $\sum_{n=1}^{\infty} P_n$ is uniformly convergent to
some continuous function $f$ and that for any $n$ and for any
integer $r$ such that $m(n)+1 \leq r \leq m(n+1)$ we have
$\widehat{f}(r) = \widehat{P_n}(r)$.

\medskip

vi) Deduce that $|S_{m(n+1)}(f,0)| \geq 2^n$ for any $n$ and
therefore $\{S_{N}(f,0)\}$ diverges.

\medskip

\textbf{Solution} i) Let $M$ large enough so that $\varepsilon B
\log M > K$. From Lemma 6.5. there exists a function $g \in
\mathrm{C}(\mathbb{T})$ such that $\| g\|_{\infty} < 1$ and
$|S_M(g,0)| \geq B \log M
> \frac{K}{\varepsilon}$. Let $f = \varepsilon g$. We have $\| f
\|_{\infty} < \varepsilon$ and $|S_M (f,0)| = \varepsilon |S_M
(g,0)| > K$.

ii) Using  part i) we can find a function $f \in
\mathrm{C}(\mathbb{T})$ and a positive integer $M$ such that $\| f
\|_{\infty} < \frac{\varepsilon}{2}$ and $|S_M (f,0)| > (K+1)$.
Since the trigonometric polynomials are dense in
$\mathrm{C}(\mathbb{T})$ we can find a trigonometric polynomial
$P$ such that $\|P-f\|_{\infty} < \min \{\frac{1}{2M+1} ,
\frac{\varepsilon}{2}\}$. Clearly $\|P\|_{\infty} < \varepsilon$.
Furthermore:
$$
|S_M (f,0)-S_M (P,0)| = |S_M (f-P,0)| \leq \sum_{n=-M}^{M}
\frac{1}{2 \pi} \left|\int_{-\pi}^{\pi} (f-P)(t) e^{-int} dt
\right| \ < 1
$$

\noindent which shows that $|S_M (P,0)| > K$.

iii) For any nonzero trigonometric polynomial $P(t) =
\sum_{n=-N}^{N} a_n e^{int}$ we denote by $\deg P = \max$ \{n
nonnegative integer s.t. $a_n \neq 0$ or $a_{-n} \neq 0$\}. Let
$Q$ be a trigonometric polynomial such that $\|Q\|_{\infty} <
\varepsilon$ and $|S_{\deg Q}(Q,0)| > K$. Let $P(t) = e^{(\deg Q +
m + 1) i t} Q(t)$. Clearly $\|P\|_{\infty} < \varepsilon$ and
$\widehat{P}(r) =0$ for any integer $r \leq m$. Let $M = 2 \deg P
+ m +1$. Then $S_M(P,0) = S_{\deg Q}(Q,0)$ so $|S_M(P,0)| > K$.

iv) We will repeatedly use iii) to get the sequence of
trigonometric polynomials $\{P_n\}$. Let $P_1$ be a trigonometric
polynomial such that $\|P_1\|_{\infty} \leq 2^{-1}$ and $|S_{\deg
P_1}(P_1, 0)| \geq 2$. Let $m(2)=\deg P_2$. Let now $P_2$ be a
trigonometric polynomial such that $\|P_2\|_{\infty} \leq 2^{-2}$,
$|S_{\deg P_2}(P_2, 0)| \geq 2^2 + |S_{\deg P_1}(P_1, 0)|$ and
$\widehat{P_2}(r)=0$ for all integers $r,\ -m(2) \leq r \leq
m(2)$. Let $m(3)=\deg P_2$. Repeating this procedure we get a
sequence of trigonometric polynomials $\{P_n\}$ and an increasing
sequence of integers $\{m(n)\}$, $m(n)= \deg P_{n-1}$ for any $n
\geq 2$, such that $\|P_n\|_{\infty} \leq 2^{-n}$, $|S_{\deg
P_n}(P_n, 0)| \geq 2^n + \sum_{k=1}^{n-1}|S_{\deg P_k}(P_k, 0)| $
and $\widehat{P_n}(r)=0$ for any $r$, $|r| \leq m(n)$ or $|r| >
m(n+1)$.

v) Since $\|P_n\|_{\infty} < 2^{-n}$ and the series
$\sum_{n=1}^{\infty} 2^{-n}$ is convergent it follows that
$\sum_{n=1}^{\infty} P_n$ converges uniformly to a continuous
function $f$. For any integer $r$, $m(n)+1 \leq |r| \leq m(n+1)$
\begin{align*}
\widehat{f}(r) &= \frac{1}{2 \pi} \int_{-\pi}^{\pi} f(t) e^{irt}
dt = \frac{1}{2 \pi} \int_{-\pi}^{\pi} \left( \sum_{n=1}^{\infty}
P_n(t) \right) e^{irt} dt \\&= \frac{1}{2 \pi} \sum_{n=1}^{\infty}
\int_{-\pi}^{\pi} P_n(t) e^{irt} dt =\sum_{k=1}^{\infty}
\widehat{P_k}(r) = \widehat{P_n}(r)
\end{align*}

Therefore for any integer $n$ we have that
\begin{align*}
S_{m(n+1)}(f,0)&=\sum_{k=-m(n+1)}^{m(n+1)}\widehat{f}(k) =
\sum_{k=-m(2)}^{m(2)}\widehat{f}(k) + \sum_{l=2}^{n}
\left[\sum_{m(l)+1 \leq |k| \leq m(l+1)}\widehat{f}(k)\right]\\
&=\sum_{k=-m(2)}^{m(2)}\widehat{P_1}(k) + \sum_{l=2}^{n}
\left[\sum_{m(l)+1 \leq |k| \leq m(l+1)}\widehat{P_l}(k)\right] =
S_{m(2)}(P_1,0) + \sum_{l=2}^{n}S_{m(l+1)}(P_l,0)\\
&=S_{\deg P_n}(P_n,0) + \sum_{l=1}^{n-1} S_{\deg P_l}(P_l,0)
\end{align*}

Since $|S_{\deg P_n}(P_n, 0)| \geq 2^n + \sum_{k=1}^{n-1}|S_{\deg
P_k}(P_k, 0)|$ we get $|S_{m(n+1)}(f,0)| \geq 2^n$ for any
positive integer $n$ which shows that $\overline{\lim}_{N
\rightarrow \infty} |S_{N}(f,0)| = \infty$, so $\ S_{N}(f,0)
\nrightarrow f(0)$.

The function $f$ constructed before has complex values. Since for
any positive integer $N$ we have $S_{N}(f,0) =
S_{N}(\mathrm{Re}(f),0) + i S_{N}(\mathrm{Im}(f),0)$ it follows
that at least one of $\overline{\lim}_{N \rightarrow \infty}
|S_{N}(\mathrm{Re}(f),0)|$ and $\overline{\lim}_{N \rightarrow
\infty} |S_{N}(\mathrm{Im}(f),0)|$ is $\infty$ so at least one of
$\mathrm{Re}(f)$ and $\mathrm{Im}(f)$ is a real valued continuous
function for which the Fourier series diverges at 0.

\medskip

\textbf{Exercise 16.15.} Let's observe that $\langle \log_{10} n
\rangle \in [0, 1/2]$ if and only if there exists a $p \in
\mathbb{N}$ such that $10^p \leq n \leq 10^{p+1/2}$. Let
$$
a_N = \frac{\mathrm{card}\{1 \leq n \leq N,\ \langle \log_{10} n
\rangle \in [0, 1/2]\}}{N}
$$

Let $S(k) = 10^k -1$ and $T(k) = 3 \cdot 10^k$. Since $10^{1/2} <
4$ we clearly have
$$\limsup_{k \rightarrow \infty}\,
a_{S(k)} < 4/10
$$
Also $$\liminf_{k \rightarrow \infty}\, a_{T(k)} > \lim_{k
\rightarrow \infty} \frac{1}{3 \cdot 10^k} (3 \cdot 10^k - 10^k) =
2/3$$ \noindent which shows that $\{a_N\}$ does not have a limit
as $N \rightarrow \infty$. In particular $\langle \log_{10} n
\rangle$ are not equidistributed in $[0,1]$.

For $\varepsilon > 0$, $x \in [0,1]$ and $n \in \mathbb{N}\ n > 0\
$, $|\langle \log_{10} n \rangle - x| < \varepsilon$ if and only
if there exists a $p \in \mathbb{N}$ such that $n \in [10^{p +
x-\varepsilon}, 10^{p + x+\varepsilon}]$. Since $\lim_{p
\rightarrow \infty} (10^{p + x+\varepsilon} - 10^{p +
x-\varepsilon}) = \infty$ we can pick a $p \in \mathbb{N}$ large
enough such that the set $\ M = [10^{p + x-\varepsilon}, 10^{p +
x+\varepsilon}] \bigcap \mathbb{N}\ $ is not empty. Any $n \in M$
will satisfy $|\langle \log_{10} n \rangle - x| < \varepsilon$.

\medskip

\textbf{Exercise 16.16.} i) This is a different proof for the
`Riemann-Lebesgue' lemma. Obviously if $f$ is a trigonometric
polynomial then $\widehat{f}(n) = 0$ if $|n|$ is large enough.

Let $\varepsilon > 0$ and $f \in \mathrm{C}(\mathbb{T})$. Let
$P(t) = \sum_{k=-N}^{N} a_k e^{ikt}$ be a trigonometric polynomial
such that $\|f-P\|_{\infty} < \varepsilon$. Then for any $n \in
\mathbb{Z}$ such that $|n| > N$, we have
$$
\widehat{f}(n) =  \frac{1}{2 \pi} \int_{-\pi}^{\pi} (f(s) -
P(s))\, e^{ins} ds = \frac{1}{2 \pi} \int_{-\pi}^{\pi} (f(s) -
P(s))\, e^{ins} ds + \frac{1}{2 \pi} \int_{-\pi}^{\pi} P(s)\,
e^{ins} ds
$$ so $$|\widehat{f}(n)| \leq \frac{1}{2 \pi} \int_{-\pi}^{\pi} \left|(f(s) -
P(s))\, e^{ins}\right| ds\ <\ \varepsilon $$ which shows that
$\lim_{n \rightarrow \infty} \widehat{f}(n) = 0$ and $\lim_{n
\rightarrow -\infty} \widehat{f}(n) = 0$.

Let $f(t) = 1$. Then
\begin{align*}
\int_{0}^{2 \pi} |\sin nt|\,dt &= \frac{1}{n} \int_{0}^{2 n \pi}
|\sin u|\,du = \frac{1}{n} \sum_{k=0}^{n-1} \int_{2 k \pi}^{2 k
\pi
+ 2 \pi} |\sin u|\,du = \frac{1}{n}\, 4n = 4\\
&=\frac{4}{2 \pi} \int_0^{2 \pi} 1\,dt
\end{align*}

Let now $f(t) = e^{imt}, m \in \mathbb{Z}, m \neq 0$ and $n > m$.
Then, integrating by parts twice, we get:
\begin{align*}
I_{m,n} &= \int_{0}^{2 \pi} e^{imt} |\sin nt|\, dt =
\sum_{k=0}^{2n-1} (-1)^k \int_{\frac{k \pi}{n}}^{\frac{(k+1)
\pi}{n}} e^{imx} \sin nx \, dx\\
&=\sum_{k=0}^{2n-1} (-1)^k \int_{\frac{k \pi}{n}}^{\frac{(k+1)
\pi}{n}} \left(\frac{e^{imx}}{im}\right)' \sin nx \, dx =
\frac{n}{im} \sum_{k=0}^{2n - 1} (-1)^{k+1} \int_{\frac{k
\pi}{n}}^{\frac{(k+1) \pi}{n}} e^{imx} \cos nx \, dx\\
&=\frac{n}{(im)^2} \sum_{k=0}^{2n - 1} (-1)^{k+1}
\left[\left(e^{imx} \cos nx \right)\Big|_{x=\frac{k
\pi}{n}}^{x=\frac{(k+1) \pi}{n}} + n \int_{0}^{2 \pi} e^{imx}
\sin nx \, dx\right]\\
&=\frac{n}{(im)^2} \sum_{k=0}^{2n - 1} \left(e^{im \frac{(k+1)
\pi}{n}} + e^{im \frac{k \pi}{n}} \right) + \frac{n^2}{(im)^2}
\sum_{k=0}^{2n-1} (-1)^{k+1} \int_{\frac{k \pi}{n}}^{\frac{(k+1)
\pi}{n}} e^{imx} \sin nx\,dx\\
&= \frac{n}{(im)^2} (e^{im \frac{\pi}{n}}\, +\, 1) \sum_{k=0}^{2n
- 1} \left(e^{\frac{im \pi}{n}}\right)^{k} \ +\  \frac{n^2}{m^2}
I_{m,n} = \frac{n^2}{m^2} I_{m,n}
\end{align*}

Therefore $(1 - \frac{n^2}{m^2}) I_{m,n} = 0$ so $I_{m,n} = 0$.
Since $\int_{0}^{2 \pi} e^{imx} dx = 0$ for any $m \neq 0$ we can
conclude now that
$$
\lim_{n \rightarrow \infty} \int_{0}^{2 \pi} P(t) |\sin nt|\, dt =
\frac{4}{2 \pi} \int_{0}^{2 \pi} P(t)\, dt
$$ for any trigonometric polynomial $P(t) = \sum_{n=-N}^{N} a_n
e^{int}$.

We can now finish the proof. Let $f: \mathbb{T} \rightarrow
\mathbb{R}$ continuous and let $\varepsilon > 0$. Let $P(t) =
\sum_{n=-N}^{N} a_n e^{int}$ be a trigonometric polynomial such
that $\| f - P \|_{\infty} < \frac{\varepsilon}{6 \pi}$. Let $n_0$
large enough such that for any $n \geq n_0$ we have
$$
\left| \int_{0}^{2 \pi} P(t) |\sin nt|\, dt - \frac{4}{2 \pi}
\int_{0}^{2 \pi} P(t)\, dt \right| < \frac{\varepsilon}{3}
$$

Then for any $n \geq n_0$ we have
\begin{align*}
\left| \int_{0}^{2 \pi} f(t)\, |\sin nt|\, dt - \frac{4}{2 \pi}
\int_{0}^{2 \pi} f(t)\, dt \right|\ <\ \int_{0}^{2 \pi} |f(t) -
P(t)|\, |\sin nt|\, dt\\
+ \left| \int_{0}^{2 \pi} P(t)\, |\sin nt|\, dt - \frac{4}{2 \pi}
\int_{0}^{2 \pi} P(t)\, dt \right| + \frac{4}{2 \pi} \int_{0}^{2
\pi} \left|f(t) - P(t) \right|\, dt \ < \ \varepsilon
\end{align*} and therefore
$$
\lim_{n \rightarrow \infty} \int_{0}^{2 \pi} f(t)\, |\sin nt|\, dt
= \frac{4}{2 \pi} \int_{0}^{2 \pi}\, f(t)\, dt
$$

\medskip

\textbf{Exercise 16.17.} A simple computation shows that
$\int_{\alpha}^{\beta} e^{2 \pi i t} dt = 0$ if and only if
$(\alpha - \beta) \in \mathbb{Z}$. Without loss of generality we
can assume that $R = [0,a] \times [0,b]$. For any integer $j,\ 1
\leq j \leq k$ we have $R(j) = [x_j, x_j + a_j] \times [y_j, y_j +
b_j]$. Furthermore:
$$
\int\int_{R_j} e^{2 \pi i (x+y)}\ dx\,dy = \left(\int_{x_j}^{x_j +
a_j} e^{2 \pi i x}\ dx\right)\left(\int_{y_j}^{y_j + b_j} e^{2 \pi
i y}\ dy\right) = 0
$$ since at least one of $a_j$ and $b_j$ is an integer. Therefore
$$
\left(\int_0^a e^{2 \pi i x}\ dx\right)\left(\int_0^b e^{2 \pi i
y}\ dy\right) = \int\int_R e^{2 \pi i (x+y)}\ dx\,dy =
\sum_{j=1}^k \int\int_{R_j} e^{2 \pi i (x+y)}\ dx\,dy  = 0
$$ which shows that at least one of $a$ and $b$ is an integer.

\medskip

\textbf{Exercise 16.20.} i) This is the Cauchy Schwarz inequality
for $a=(a_{-N},a_{-N+1},...a_{N-1},a_{N})$ and
$b=(b_{-N},b_{-N+1},...b_{N-1},b_{N})$ in $\mathbb{C}^{2N+1}$.

ii) For any $N \geq 0$
$$
\sum_{j=-N}^{N} |a_j b_j| \leq \left(\sum_{j=-N}^{N}
|a_j|^2\right)^{1/2} \left(\sum_{j=-N}^{N} |b_j|^2\right)^{1/2}
\leq \left(\sum_{j=-\infty}^{\infty} |a_j|^2\right)^{1/2}
\left(\sum_{j=-\infty}^{\infty} |b_j|^2\right)^{1/2} $$ which
implies that $\{\sum_{j=-N}^{N} |a_j b_j|\}_{N \in \mathbb{N}}$
converges and
$$
\sum_{j=-\infty}^{\infty} |a_j b_j|\ \leq\
\left(\sum_{j=-\infty}^{\infty} |a_j|^2\right)^{1/2}
\left(\sum_{j=-\infty}^{\infty} |b_j|^2\right)^{1/2}
$$

iii) For any $f \in \mathrm{C}(\mathbb{T})$ and any $j \in
\mathbb{Z}$ we get, using integration by parts:
$$\widehat{f'}(j)\, =\, \frac{1}{2 \pi} \int_{-\pi}^{\pi} f'(t)
\,e^{-i j t}\,dt\, =\, ij\, \frac{1}{2 \pi} \int_{-\pi}^{\pi}
f(t)\, e^{-i j t}\,dt\, =\, ij\, \widehat{f}(j)
$$

Since by Plancherel theorem (Theorem 8.2. (i)) we have:
$$
\sum_{j=-\infty}^{\infty}|\widehat{f'}(j)|^2 = \frac{1}{2 \pi}
\int_{-\pi}^{\pi} f'(t) dt
$$ we immediately get
$$\sum_{j=-\infty}^{\infty} j^2 |\widehat{f}(j)|^2 = \frac{1}{2
\pi} \int_{\mathbb{T}}|f'(t)|^2 dt$$

iv) Using ii) we get \begin{align*} \sum_{j=-\infty}^{\infty}
|\widehat{f}(j)|\ &=\ |\widehat{f}(0)| + \sum_{|j| \geq 1}
|\widehat{f}(j)|\,j\,\frac{1}{j}\, \leq \,|\widehat{f}(0)| +
\left( \sum_{|j| \geq 1}
|\widehat{f}(j)|^2\,j^2\right)^{1/2}\left( \sum_{|j| \geq 1}
\frac{1}{j^2}\right)^{1/2}\\ &\leq |\widehat{f}(0)|
+\left(\frac{1}{2\pi} \int_{-\pi}^{\pi}|f'(t)|^2\,dt\right)^{1/2}
\left(\frac{\pi^2}{3}\right)^{1/2}
\end{align*} which shows that
$\sum_{j=-\infty}^{\infty}|\widehat{f}(j)|$ converges. Therefore
$\sum_{j=-\infty}^{\infty} \widehat{f}(j)\, e^{ijt}$ converges
uniformly to a continuous function $g$. It remains to show that
$g=f$. But $\widehat{g}(j) = \widehat{f}(j)$ for any $j \in
\mathbb{Z}$. Therefore, from Theorem 7.4 we get $g=f$.

\medskip

\textbf{Exercise 16.21. (Wirtinger's inequality)\ } i) Since by
hypothesis $\frac{1}{2 \pi}\int_{\mathbb{T}} u(t)\,dt = 0$, we
have $\widehat{u}(0) = 0$. From Plancherel's theorem and part iii)
of the previous problem we have:
\begin{align*}
\frac{1}{2\pi}\int_{-\pi}^{\pi} (u(t))^2\,dt =
\sum_{j=-\infty}^{\infty} |\widehat{u}(j)|^2\ = \sum_{|j| \geq 1}
|\widehat{u}(j)|^2 \leq \sum_{|j| \geq 1} j^2 |\widehat{u}(j)|^2 =
\frac{1}{2 \pi} \int_{-\pi}^{\pi} (u'(t))^2 dt
\end{align*}

Note that in the previous relation we have equality if and only if
$\widehat{u} (j) = 0$ for any $|j| \geq 2$. Therefore $u(t) =
\widehat{u}(1) e^{it} + \widehat{u}(-1) e^{-it}$. Since $u$ is
real valued we must have $u(t) = a \cos(t) + b \sin(t),\ a,b \in
\mathbb{R}$. If $C^2 = a^2 + b^2 \neq 0$ we have
$$
u(t) = \sqrt{a^2 + b^2} \left(\frac{a}{\sqrt{a^2 + b^2}} \cos t +
\frac{b}{\sqrt{a^2 + b^2}} \sin t \right) = C \cos(t + \varphi)
$$ where $\varphi \in [0, 2 \pi)$ is chosen such that $\ \sin
\varphi = -\frac{b}{\sqrt{a^2 + b^2}}\ $ and $\ \cos \varphi =
\frac{a}{\sqrt{a^2 + b^2}}$.

ii) Let $v \in \mathrm{C}^1([0, \frac{\pi}{2}])$ be a real valued
function with $v(0) = 0$ and $v'(\frac{\pi}{2}) = 0$. Let $f :
[-\pi, \pi] \rightarrow \mathbb{R}$ defined by:
$$
f(t) = \begin{cases} v(t) \qquad &\mathrm{if\ \ } t \in \left[0,
\frac{\pi}{2}\right)\\
v(\pi - t) \qquad &\mathrm{if\ \ } t \in \left[\frac{\pi}{2},
\pi\right)\\
-v(-t) \qquad &\mathrm{if\ \ } t \in \left[-\frac{\pi}{2},0 \right)\\
-v(\pi + t) \qquad &\mathrm{if\ \ } t \in \left[-\pi, -\frac{\pi}{2} \right)\\
\end{cases}
$$

It is easy to see that $f \in \mathrm{C}(\mathbb{T})$. For any $\
x_0 \in \mathbb{T}\ $ let's denote by $f'_{-}(x_0)$ the left-hand
derivative of $f$ at $x_0$ and by $f'_{+}(x_0)$ the right-hand
derivative of $f$ at $x_0$. Simple computations show that
$f'_{+}(0) = f'_{-}(0) = v'(0),\ \ f'_{+}(\frac{\pi}{2}) =
f'_{-}(\frac{\pi}{2}) = 0,\ \ f'_{-}(\pi) = f'_{+}(-\pi)=-v'(0)\ $
and $f'_{+}(-\frac{\pi}{2}) = f'_{-}(-\frac{\pi}{2}) = 0$.
Therefore $f \in \mathrm{C}^1(\mathbb{T})$. From part i) we get
$$
\int_{-\pi}^{\pi} (f(t))^2\, dt \, \leq \, \int_{-\pi}^{\pi}
(f'(t))^2\, dt
$$
Since $\ \int_{-\pi}^{\pi} (f(t))^2 dt = 4\,
\int_{0}^{\frac{\pi}{2}} (v(t))^2 dt\ $ and $\ \int_{-\pi}^{\pi}
(f'(t))^2 dt = 4\, \int_{0}^{\frac{\pi}{2}} (v'(t))^2 dt\ $ we get
$$
\int_{0}^{\frac{\pi}{2}} (v(t))^2\, dt \, \leq \,
\int_{0}^{\frac{\pi}{2}} (v'(t))^2\, dt
$$
for any $v \in \mathrm{C}^1(\mathbb{T})$.

The previous inequality becomes equality if and only if there
exist $C \in \mathbb{R}$ and $\varphi \in [0, 2\pi)$ such that
$f(t) = C \cos(t+\varphi)$. When $C \neq 0$, since
$f(0)=f'(\frac{\pi}{2})=0$, it follows that $\cos \varphi =0$ and
$\sin(\frac{\pi}{2}+\varphi) = 0$. Therefore $\varphi =
\frac{\pi}{2}$ or $\varphi = \frac{3 \pi}{2}$. In both cases we
can conclude that $f(t) = C_1 \sin t\ $, so $\ v(t) = C_1 \sin t\
$ for a constant $C_1 \in \mathbb{R}$.

iii) Let $w \in \mathrm{C}^1([0, \frac{\pi}{2}])$ with $w(0) =0$.
Let $M
> 0$ such that $|w(t)| < M$ and $|w'(t)| < M$ for any $t \in [0,
\frac{\pi}{2}]$.

Let $\varepsilon > 0$, $\varepsilon < 7 M^2$ and let $t_0 =
\frac{\pi}{2} - \frac{\varepsilon}{7 M^2}$. Define $v: [0,
\frac{\pi}{2}] \rightarrow \mathbb{R}$ by
$$
v(t)= \begin{cases} w(t) \qquad & \mathrm{if\ \ \ } t \in [0,
t_0]\\
w(t_0) + \frac{1}{2}\, w'(t_0) \left(\frac{\pi}{2} -t_0\right) -
\frac{1}{2}\, w'(t_0) \left(\frac{\pi}{2}
-t\right)^2 \left(\frac{\pi}{2}-t_0\right)^{-1} \qquad & \mathrm{if\ \ \ } t \in (t_0, \frac{\pi}{2}]\\
\end{cases}
$$

It is easy to see that $v \in C^1([0,\frac{\pi}{2}]), v(0)=w(0)=0$
and $v'(\frac{\pi}{2})=0$. Therefore, from part ii) we get:
$$
\int_{0}^{\frac{\pi}{2}} (v(t))^2\, dt \, \leq \,
\int_{0}^{\frac{\pi}{2}} (v'(t))^2\, dt \qquad (1)
$$

It is also easy to see that $\| v \|_{\infty} < 2 M$ and $\| v'
\|_{\infty} < M$. Therefore
\begin{align*}
&\left| \int_{0}^{\frac{\pi}{2}} (v(t))^2\, dt -
\int_{0}^{\frac{\pi}{2}} (w(t))^2\, dt \right|\, = \, \left|
\int_{t_0}^{\frac{\pi}{2}} (v(t))^2\, dt -
\int_{t_0}^{\frac{\pi}{2}} (w(t))^2\, dt \right|\\
&\leq \int_{t_0}^{\frac{\pi}{2}} (v(t))^2\, dt +
\int_{t_0}^{\frac{\pi}{2}} (w(t))^2\, dt \ \leq\  5 M^2 \frac{\varepsilon}{7 M^2} = \frac{5\, \varepsilon}{7} \qquad (2)\\
\end{align*} and
\begin{align*}
&\left|\int_{0}^{\frac{\pi}{2}} (v'(t))^2\, dt -
\int_{0}^{\frac{\pi}{2}} (w'(t))^2\, dt \right|\, = \,
\left|\int_{t_0}^{\frac{\pi}{2}} (v'(t))^2\, dt -
\int_{t_0}^{\frac{\pi}{2}} (w'(t))^2\, dt \right|\\
&\leq \int_{t_0}^{\frac{\pi}{2}} (v'(t))^2\, dt +
\int_{t_0}^{\frac{\pi}{2}} (w'(t))^2\, dt\ <\ 2 M^2
\frac{\varepsilon}{7 M^2} = \frac{2\,\varepsilon}{7} \qquad (3)
\end{align*}

Using (1), (2) and (3) we obtain
$$
\int_{0}^{\frac{\pi}{2}} (w(t))^2\, dt \, \leq \,
\int_{0}^{\frac{\pi}{2}} (w'(t))^2\, dt\, + \, \varepsilon
$$

Since $\varepsilon$ is arbitrarily small we can conclude that

$$
\int_{0}^{\frac{\pi}{2}} (w(t))^2\, dt \, \leq \,
\int_{0}^{\frac{\pi}{2}} (w'(t))^2\, dt
$$
\medskip

\textbf{Exercise 16.23. - (The Gibbs Phenomenon)} i) Let $\lambda
= \frac{1}{2 \pi} (f_+(0) - f_-(0))$ and $g : [-\pi, \pi]
\rightarrow \mathbb{R}$ defined by
$$ g(t) =
\begin{cases} f(t) - \lambda F(t) \qquad &\mathrm{if\ } t\neq 0\\
\frac{1}{2} \left(f_+(0) + f_-(0)\right) \qquad &\mathrm{if\ } t =
0
\end{cases}$$ It is easy to see that $g_-(0) = g_+(0) = \frac{1}{2}
(f_+(0) + f_-(0))$ and therefore $g \in
\mathrm{C}(\mathbb{\mathbb{T}})$. Furthermore, $g$ is continuously
differentiable on $\mathbb{T} \backslash \{ 0 \}$. We clearly have
$f = g + \lambda F$.

ii) For $r \in \mathbb{Z}, r \neq 0$ we have
\begin{align*}
\widehat{F}(r) &= \frac{1}{2 \pi} \int_{-\pi}^{\pi} F(t)\,
e^{-irt}\, dt =  \frac{1}{2 \pi} \int_{-\pi}^{0} (-\pi -t)\,
e^{-irt}\, dt +
\frac{1}{2 \pi} \int_{0}^{\pi} (\pi - t)\, e^{-irt}\, dt\\
&= \frac{1}{2 i r} \int_{-\pi}^{0} (e^{-irt})'\, dt - \frac{1}{2 i
r} \int_{0}^{\pi} (e^{-irt})'\, dt + \frac{1}{2 \pi i r}
\int_{-\pi}^{\pi} t (e^{-irt})'\, dt\\
&= \frac{1}{2 i r} \left(-2 (\cos r \pi) + 2\right)\, +\,
\frac{1}{2 \pi i r} 2 \pi \cos r \pi = \frac{1}{ir}
\end{align*}

Since obviously $\widehat{F} (0) = 0$ we have
$$
S_n(F,t) = \sum_{r=-n}^{n} \widehat{F}(r)\, e^{irt} =
\sum_{r=1}^{n} \left(-\frac{1}{ir}\, e^{-irt} + \frac{1}{ir}\,
e^{irt}\right) = 2 \sum_{r=1}^{n} \frac{\sin rt}{r}
$$

iii) For any $\tau > 0$ we have that $\mathcal{P}_n = \{ r
\frac{\tau}{n}, \ \ 1 \leq r \leq n \}$ is a partition of $[0,
\tau]$ in $n$ intervals of length $\frac{\tau}{n}$. Therefore
$$
S_n(F, \frac{\tau}{n}) = 2 \,\frac{\tau}{n}\, \sum_{r=1}^{n}
\frac{1}{\frac{r \tau}{n}}\, \sin \frac{r \tau}{n}
$$ is the Riemann sum associated to the function $\varphi : [0, \tau]
\rightarrow \mathbb{R},\ \  \varphi(x) = 2\, \frac{\sin x}{x}$,
the partition $\mathcal{P}_n$ of $[0, \tau]$ and the set of
intermediate points obtained by taking the right-hand endpoint
from each interval $[\frac{r-1}{n}, \frac{r}{n}]$. Since the
function $\varphi$ is Riemann integrable on $[0, \tau]$ we get
$$
\lim_{n \rightarrow \infty} S_n(F, \frac{\tau}{n}) = 2
\int_0^{\tau} \frac{\sin x}{x} dx
$$

iv) The graph of the function $G(\tau) = \int_{0}^{\tau}
\frac{\sin x}{x}\,dx$ is

\begin{center}
\includegraphics[width=4in,scale=1,angle=0]{graph1.eps}
\end{center}

This suggest that $\lim_{\tau \rightarrow \infty} \int_0^{\tau}
\frac{\sin x}{x} dx$ exists. (Actually $\lim_{\tau \rightarrow
\infty} \int_0^{\tau} \frac{\sin x}{x} dx =\frac{\pi}{2}$)

v) When $\tau$ is small and $n$ is large we have that $t=\tau/n$
is small. The graph of $S_{50}(F,t)$ is

\begin{center}
\includegraphics[width=5.5in,scale=1,angle=0]{graph2.eps}
\end{center}

We can observe the two ``bumps" near the origin which give a point
of maximum to the right of 0 and a point of minimum to the left of
0.

Let's also observe that the graph of the function $h: (0,\infty)
\rightarrow \mathbb{R}$ defined by $$h(t)= S_n(F,t) - F(t)$$ is
(for $n=50$)

\begin{center}
\includegraphics[width=5.5in,scale=1,angle=0]{graph3.eps}
\end{center}

We can see that this graph is similar to the graph of the function
$G$ from part iv).
\end{document}