# Modes of Convergence

In this section we discuss four basic types of convergence for series with functional coefficients.

Introduction to Linear Algebra with Mathematica

# Modes of convergence

Suppose { fn } is a sequence of continuous functions on [0, 1] (it can be any finite interval). We assume that $$\lim_{n \to \onfty} f(x) = f(x)$$ exists for every x, and inquire as to the nature of the limiting function f(x).

If we suppose that the convergence is uniform, matters are straight-forward and f is then everywhere continuous. However, once we drop the assumption of uniform convergence, things may change radically and the issues that arise can be quite subtle. An example of this is given by the fact that one can construct a sequence of continuous functions { fn } converging everywhere to f so that it is not Riemann integrable. So the relation

$\int_0^1 f(x)\,{\text d}x = \lim_{n\to \infty} \int_0^1 f_n (x)\,{\text d}x$
holds with Lebesgue integration, but not the Riemann one. For example, let
$f_n (x) = \begin{cases} n, & \ \mbox{ if } \ 0 < x < 1/n , \\ 0 , & \ \mbox{ otherwise. } \end{cases}$
Then fn(x) → 0 for all x, yet $$\int f_n (x)\,{\text d}x = 1$$ for all n ∈ ℤ+.
For every periodic function f ∈ 𝔏²(-π,π) with period 2π, its Fourier series converges in 𝔏²(-π,π), i.e.,
$\lim_{N\to \infty} \frac{1}{2\pi} \int_{-\pi}^{\pi} \left\vert f(x) - \sum_{|n|\le N} c_n (f) \,e^{{\bf j} nx} \right\vert^2 {\text d} x = 0 ,$
and Parseval’s equality
$\frac{1}{2\pi} \int_{-\pi}^{\pi} \left\vert f(x) \right\vert^2 {\text d} x = \sum_{n=-\infty}^{\infty} \left\vert c_n (f) \right\vert^2$
holds.
Riesz--Fischer Theorem: Suppose { bn }n = -∞ is a sequence of complex numbers with $$\sum_{n=-\infty}^{\infty} \left\vert b_n \right\vert^2 < \infty .$$ Then there is a unique periodic function f ∈ 𝔏²(-π,π) such that $$b_n = c_n (f) = \frac{1}{2\pi} \,\int_{-\pi}^{\pi} f(x)\, e^{-{\bf j} nx} \,{\text d}x .$$
$\alpha_n (f) = \frac{1}{2\ell} \int_{-\ell}^{\ell} f(x) \,e^{-n{\bf j} \pi x/\ell} {\text d} x = O \left( \frac{1}{|n|^m} \right) \qquad\mbox{as} \quad |n| \to \infty .$