Two problems of Chapter 2

Fourier Analysis

Problem 1

  One can construct Riemann integrable functions on $[0,1]$ that have a dense set of discontinuities as follows.

 (a) Let $f(x)=0$, when $x<0$, and $f(x)=1$ if $x\geq0$. Choose a countable dense sequence $\{r_n\}$ in $[0,1]$. Then, show that function

$F(x)=\sum_{n=1}^{\infty} \dfrac{1}{n^2} f(x-r_n)$

is integrable and has discontinuities at all points of the sequence $\{r_n\}$

Proof:

 Step one: Note that $F(x)$ is monotonic and bounded for the definition of $f$,and $F(x) \leq$ Euler’s series, so that it has countable discontinuities at most. Hence, $F(x)$ is integrable on $[0,1]$.

  Step two: We consider that $F(r_n)$. $\{r_n\}$ is dense in $[0,1]$ means $\forall \varepsilon >0,$ , $\exists n \in \mathbb{N}$ such that $|x-r_n|<\varepsilon$. Or $\forall x_1,x_2 \in [0,1]$, $\exists n \in \mathbb{N}$ such that $r_n$ is between $x_1$ and $x_2$. Fixed $x_2 = r_n$, $n=1,2,3,...$ and consider that:

$|F(x)-F(r_n)|$

$\forall \delta > 0$, when $|x-r_n| < \delta$, we can find $r_m$ is between $x$ and $r_n$. Due to $F(x)$ is monotonic, we have

Where $C_{n,m}$ is a constant(If we fix $m$ and $n$). That is $F(x)$ is discontinuous at all points $x=r_n$.

$(\#)$

  (b)Consider next

$F(x)=\sum_{n=1}^{\infty} 3^{-n}g(x-r_n)$

where $g(x)=\sin{\dfrac{1}{x}}$ when $x\neq0$, and $g(0)=0$. Then $F$ is integrable, discontinuous at each $x=r_n$, and fails to be monotonic in any subinterval of $[0,1]$.

Proof:

  Note that $g(x)$ is bounded, and the geometric series $\sum 3^{-n}$ is convergence. So $F(x)$ is bounded.

Next, we discuss the integrability of $F(x)$. Denote

$\mathcal{U}=\sum_{j=1}^{N}{\left[\sup_{x_{j-1} \leq x \leq x_j}\left\{\sum_{n=1}^{\infty} 3^{-n}g(x-r_n)\right\}\right](x_j-x_{j-1})}$

$\mathcal{L}=\sum_{j=1}^{N}{\left[\inf_{x_{j-1} \leq x \leq x_j}\left\{\sum_{n=1}^{\infty} 3^{-n}g(x-r_n)\right\}\right](x_j-x_{j-1})}$

Then,

$\mathcal{U-L}=\sum_{j=1}^{N}{\omega_j(x)(x_j-x_{j-1})}$

Where

So, we prove that there is at most countable $\omega_j(x) > 0$ as $N\to \infty$.

 © The original example of Riemann is the function

$F(x)=\sum_{n=1}^{\infty} \dfrac{(nx)}{n^2}$

where $(x)=x$ for $x \in (-\dfrac{1}{2},\dfrac{1}{2}]$, and $(x)$ is continued to $\mathbb{R}$ by periodicity, that is, $(x+1)=(x)$. It can be shown that $F$ is discontinuous whenever $x=\dfrac{m}{2n}$, where $m,n\in \mathbb{Z}$ with $m$ odd and $n \neq 0$.

Proof:

Problem 2

  Let $D_N$ denote the Dirichlet kernel

$D_N(\theta)=\sum_{k=-N}^{N}e^{ik\theta}=\dfrac{\sin\left(N+\dfrac{1}{2} \theta\right)}{\sin\dfrac{\theta}{2}}$

and define

$L_N=\dfrac{1}{2\pi}\int_{-\pi}^{\pi}\left|D_N(\theta)\right|d\theta$

  (a)Prove that

$L_N \geq c\log N$

for some constant $c>0$.

  Proof:

  We calculate $L_N$ straitly:

The last inequality is used Euler’s formula

$\sum_{k=1}^{n}{\dfrac{1}{k}}=\log n + \gamma_n$

Where $\gamma_n$ is Euler constant.

  A more careful estimate gives

$L_N \ge \dfrac{4}{\pi^2} \log N +O(1)$

  (b)Prove the following as consequence: for each $n\geq 1$, there exists a continuous function $f_n$ such that $\left|f_n \right|<1$ and $\left| S_n(f_n)(0)\right| \geq c' \log n$.