Fourier Analysis 1

The first Fourier analysis lecture on 15/01/2024 began with a general introduction to the concepts of the topic.

A Review of Riemann Integration

Let $-\infty < a < b < + \infty$ , then $I = [(a,b)]$ is an interval $I \subset \mathbb{R}$ , open, closed or semi-closed. If $f : I \rightarrow \mathbb{R}$ or ( $f : I \rightarrow \mathbb{C}$ ) is a function then we are interested in defining the integral $S = \int_a^b f(x)dx$ .

Definition [Riemann Integration]:

A partition of $I$ is $\Delta : a = t_0 < t_1 < ... < t_n = b$ , of mesh (or norm):

|\Delta| \coloneq \max*{0 \leq j \leq n-1}|t*{j+1} - t_j|.

The Riemann sum of $f$ corresponding to $\Delta$ is:

S(f;\Delta) \coloneq \sum_{j=0}^{n-1}f(t_j)\cdot (t_{j+1}-t_j).

We say $f$ is integrable on $I$ , with value of integral $s \coloneq \int_a^b f(x)dx$ , if for all $\varepsilon > 0$ there exists a $\delta > 0$ such that for every partition $\Delta$ of $I$ of mesh size $|\Delta|< \delta$ we have:

|S(f,\Delta) - s| < \varepsilon

That is:

\int_a^b f(x)dx \coloneq \lim_{|\Delta| \rightarrow 0}S(f,\Delta).

The following facts on Riemann integration were established as part of a standard analysis course:

If a function $f$ is not bounded on $I$ , then $f$ is not integrable on $I$ .
The upper (resp. lower) Riemann sums:

U(f;\Delta) \coloneq \sum_{j=0}^{n-1}\left(\sup_{t \in [t_j,t_{j+1}]}f(t)\right)\cdot (t_{j+1}-t_j)

resp.:

L(f;\Delta) \coloneq \sum_{j=0}^{n-1}\left(\inf_{t \in [t_j,t_{j+1}]}f(t)\right)\cdot (t_{j+1}-t_j).

Then $f$ is integrable on $I$ if and only if for every $\varepsilon > 0$ there exists a partition $\Delta$ such that:

0 \leq U(f,\Delta) - L(f,\Delta) < \varepsilon .