Supremum Norm Complete

Why

We want a complete norm on the vector space of continuous functions.

Result

The supremum norm is complete.

Let $R$ denote the real numbers. Let $\seq{f}$ be an egoprox sequence in $C[a, b]$.

Candidate. $\seq{f}$ is egoprox means $\forall \epsilon > 0, \exists N$ so that

\[ m, n > N \implies \supnorm{f_n - f_m} < \epsilon . \]

Since $\supnorm{f_n - f_m} < \epsilon \implies \abs{f_n(x) - f_m(x)} < \epsilon $ for all $x \in [a, b]$, the sequence of real numbers $\set{f_n(x)}_n$ is egoprox for each $x \in [a, b]$. Since the metric space $(R, \abs{\cdot })$ is complete, there is a limit $l_x \in R$ such that $f_n(x) \goesto l_x$ as $n \goesto \infty$, for each $x \in [a, b]$. Define $f: [a, b] \to R$ by $f(x) = l_x$ for each $x \in [a, b]$.

Candidate is Limit. First, we argue that $\supnorm{f_n - f} \goesto 0$ as $n \goesto \infty$. Since $\seq{f}$ is an egoprox sequence, there exists $n_0$ so that

\[ n,m \geq n_0 \implies \supnorm{f_n - f_m} < \epsilon /2. \]

So for all $x \in [a, b]$,

\[ n,m \geq n_0 \implies \abs{f_n(x) - f_m(x)} < \epsilon /2. \]

For all $x \in [a, b]$, and $n \geq n_0$,

\[ \lim_{m \to \infty} \abs{f_n(x) - f_m(x)} \leq \epsilon /2 < \epsilon . \]

The sequence $\set{f_k(x)}_{k = m}^{\infty}$ is a final part of $\set{f_k(x)}_{k = 1}^{\infty}$, and so has the same limit, $f(x)$. Therefore, using continuity of subtraction and the absolute value,

\[ \lim_{m \to \infty} \abs{f_n(x) - f_m(x)} = \abs{f_n(x) - f(x)}. \]

We conclude that for $n \geq n_0$, $x \in [a, b]$, $\abs{f_n(x) - f(x)} < \epsilon $, from which we deduce $\supnorm{f_n - f} < \epsilon $. Thus $f_n \goesto f$ as $n \goesto \infty$.

Limit is Continuous. Next, we argue that $f$ is continuous. Let $x_0 \in [a, b]$. Let $\epsilon > 0$. Since $f_n \goesto f$ there exists $n_0$ so that

\[ \supnorm{f_{n_0} - f} < \epsilon /3. \]

By the triangle inequality,

\[ \begin{aligned} \abs{f(x_0) - f(x)} \leq \abs{f(x_0) - f_{n_0}(x_0)} + \abs{f_{n_0}(x_0) - f(x)}, \end{aligned} \]

for all $x \in [a, b]$. Using $\abs{f(x_0) - f_{n_0}(x_0)} < \epsilon /3$,

\[ \begin{aligned} \abs{f(x_0) - f(x)} &< \epsilon /3 + \abs{f_{n_0}(x_0) - f(x)}, \end{aligned} \]

for all $x \in [a, b]$. Using the triangle inequality,

\[ \abs{f(x_0) - f(x)} < \epsilon /3 + \abs{f_{n_0}(x_0) - f_{n_0}(x)} + \abs{f_{n_0}(x) - f(x)} \\ \]

for all $x \in [a, b]$. Using $\abs{f_{n_0}(x_0) - f(x)} < \epsilon /3$

\[ \begin{aligned} \abs{f(x_0) - f(x)} < \epsilon /3 + \abs{f_{n_0}(x_0) - f_{n_0}(x)} + \epsilon /3 \\ \end{aligned} \]

for all $x \in [a, b]$. Since $f_{n_0}$ is continuous, there exists $\delta > 0$ so that

\[ \abs{x_0 - x} < \delta \implies \abs{f_{n_0}(x_0) - f_{n_0}(x)} < \epsilon /3, \]

for $x \in [a, b]$. In this case,

\[ \begin{aligned} \abs{f(x_0) - f(x)} &< \epsilon /3 + \epsilon /3 + \epsilon /3 = \epsilon . \end{aligned} \]

Since $\epsilon $ was arbitrary, $f$ is continuous at $x_0$. Since $x_0$ was arbitrary, $f$ is continuous everywhere. Some call the above the three epsilon argument.