# 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.