Real Analysis

Sequences of Functions

Sequences of Functions

Pointwise Convergence

Consider \(\{f_n(x)\}\). For every point \(x\), \(f(x) = \lim\limits_{n \rightarrow \infty} f_n(x)\).

For example,

  1. \(f_n(x) = \frac{x}{n} \xrightarrow{\text{pointwise}} f(x) = 0\).
  2. Let the domain \(E\) of \(f_n\) be \([0, 1]\). \[\begin{align} f_n(x) = x^n \xrightarrow{\text{pointwise}} f(x) = \begin{cases} 0 & \text{if } x \neq 1 \\ 1 & \text{otherwise} \end{cases} \end{align}\]
  3. \(f_n(x) = \frac{1}{n}\sin(n^2x) \xrightarrow{\text{pointwise}} f(x) = 0\).

This notion of limit does not preserve continuity, derivation, and integration.

Uniform Convergence

Let \(\|f\| = \sup\limits_{x \in E} |f(x)|\). Say \(f_n \xrightarrow{\text{uniform}} f\) (\(f_n\) converges uniformly to \(f\) on \(E\)) if for every \(\epsilon > 0\), there exists \(\mathcal{N}\) such that \(n \geq \mathcal{N}\) implies \(\|f_n - f\| < \epsilon\).

Informally, we can draw an \(\epsilon\)-ribbon about the limit \(f\) and \(f_n\) eventually stays in the ribbon. The same \(\mathcal{N}\) works for every \(x \in E\).

This is the usual convergence in metric space \(\mathcal{C}_b(E)\), the continuous bounded functions on \(E\), where \(d(f, g) = \|f - g\|\).

Theorem. \(\mathcal{C}_b(E)\) is complete when the image is in \(\mathbb{R}^n\).

So we have the Cauchy criterion.

Theorem. \(f_n \xrightarrow{\text{uniform}} f\) on \(E\) if and only if for every \(\epsilon > 0\), there exists \(\mathcal{N}\) such that for all \(m, n > \mathcal{N}\) and \(x \in E\), \(|f_n(x) - f_m(x)| < \epsilon\).

Theorem. If \(f_n \xrightarrow{\text{uniform}} f\) on \(E\) and \(f_n\) is continuous then \(f\) is continuous.

Proof. Clearly, \(|f(x) - f(y)| \leq |f(x) - f_n(x)| + |f_n(x) - f_n(y)| + |f_n(y) - f(y)|\). Fix \(x \in E\). For all \(\epsilon > 0\), choose \(f_n\) so that \(\|f_n - f\| < \frac{\epsilon}{3}\). Thus, \(|f(x) - f_n(x)| < \frac{\epsilon}{3}\) and \(|f_n(y) - f(y)| < \frac{\epsilon}{3}\). Since \(f_n\) is continuous, there exists \(\delta > 0\) such that \(|x - y| < \delta\) implies \(|f_n(x) - f_n(y)| < \frac{\epsilon}{3}\). So, for every \(\epsilon > 0\), we found \(\delta > 0\) such that \(|f(x) - f(y)| < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon\). QED.

Theorem. There exists \(f : [0, 1] \rightarrow [0, 1]^2\) that is space filling.

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