# Math4121 Lecture 13

## New book Chapter 2

Riemann's motivation: Fourier series

$$
F(x) = \frac{a_0}{2} + \sum_{k=1}^{\infty} a_k \cos(kx) + b_k \sin(kx)
$$

$a_k = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(kx) dx$

$b_k = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(kx) dx$

To study the convergence of the Fourier series, we need to study the convergence of the sequence of partial sums. (Riemann integration)

Why Riemann integration?

Let

$$
((x)) = \begin{cases} 
x-\lfloor x \rfloor & x \in [\lfloor x \rfloor, \lfloor x \rfloor + \frac{1}{2}) \\
0 & x=\lfloor x \rfloor + \frac{1}{2}\\
x-\lfloor x \rfloor - 1 & x \in (\lfloor x \rfloor + \frac{1}{2}, \lfloor x \rfloor + 1] \end{cases}
$$

We define

$$
f(x) = \sum_{n=1}^{\infty} \frac{((nx))}{n^2}=\lim_{N\to\infty}\sum_{n=1}^{N} \frac{((nx))}{n^2}
$$

(i) The series converges uniformly over $x\in[0,1]$.

$$
\left|f(x)-\sum_{n=1}^{N} \frac{((nx))}{n^2}\right|\leq \sum_{n=N+1}^{\infty}\frac{|((nx))|}{n^2}\leq \sum_{n=N+1}^{\infty} \frac{1}{n^2}<\epsilon
$$

As a consequence, $f(x)\in \mathscr{R}$.

(ii) $f$ has a discontinuity at every rational number with even denominator.

$$
\begin{aligned}
\lim_{h\to 0^+}f(\frac{a}{2b}+h)-f(\frac{a}{2b})&=\lim_{h\to 0^+}\sum_{n=1}^{\infty}\frac{((\frac{na}{2b}+h))}{n^2}-\sum_{n=1}^{\infty}\frac{((\frac{na}{2b}))}{n^2}\\
&=\lim_{h\to 0^+}\sum_{n=1}^{\infty}\frac{((\frac{na}{2b}+h))-((\frac{na}{2b}))}{n^2}\\
&=\sum_{n=1}^{\infty}\lim_{h\to 0^+}\frac{((\frac{na}{2b}+h))-((\frac{na}{2b}))}{n^2}\\
&>0
\end{aligned}
$$

### Back to the fundamental theorem of calculus

Suppose $f$ is integrable on $[a,b]$, then

$$
F(x)=\int_a^x f(t)dt
$$

$F$ is continuous on $[a,b]$.

if $f$ is continuous at $x_0$, then $F$ is differentiable at $x_0$ and $F'(x_0)=f(x_0)$.

#### Theorem (Darboux's theorem)

If $\lim_{x\to a^-}f(x)=L^-$, then $\lim_{h\to 0} \frac{f(a+h)-f(a)}{h}=L^-$.

Proof:

$$
h\sup_{x\in [0,h]}f(x)\geq F(a+h)\geq \inf_{x\in [0,h]}f(x)h
$$

Consequently,

$$
f(x)=\sum_{n=1}^{\infty} \frac{((nx))}{n^2}
$$

then

$$
F(x)=\int_0^x f(t)dt
$$

is continuous on $[0,1]$.

However, since $\lim_{x\to 0^+}f(x)\neq \lim_{x\to 0^-}f(x)$ holds for all the rational numbers with even denominator, $F$ is not differentiable at all the rational numbers with even denominator.

Moral: There exists a continuous function on $[0,1]$ that is not differentiable at any rational number with even denominator. (Dense set)

#### Weierstrass function

$$
g(x)=\sum_{n=0}^{\infty} a^n \cos(b^n \pi x)
$$

where $0<a<1$ and $ab>1+\frac{3}{2}\pi$.

$g(x)$ is continuous on $\mathbb{R}$ but nowhere differentiable.

_If we change our integral, will be differentiable at most points?_