# Wednesday, January 16, 2013 Uniform Continuity

## 1Image as an interval¶

In Analysis 1 we touched on characterization of intervals (Theorem 2.5.1 in the book). It's stated as follows:

If $S$ is a subset of $\mathbb{R}$ that contains at least two points and has the property:
(1) $x, y \in S$ and $x < y$ then $[x, y] \subseteq S$
Then $S$ is an interval.

We'll use this to prove the following:

Theorem: Let $I$ be an interval and $f : I \to \mathbb{R}$ be continuous on $I$, then $f(I)$ is an interval

Proof: If $f(I) = {a}$ then the corresponding interval is $[a, a]$. Otherwise let $a,b \in f(I)$ be such that $a < b$. then there exists $\alpha, \beta \in I$ such that $f(\alpha) = a$ and $f(\beta) = b$. Either $\alpha < \beta$ or $\alpha > \beta$ - WLOG assume $\alpha < \beta$. If $k$ is such that $a < k < b$ then since $f$ is continuous on $[\alpha, \beta]$ by Bolzano's IVT there exists $c \in (\alpha, \beta)$ such that $f(c) = k$, thus $[a, b] \subseteq f(I)$. By characterization of intervals, this qualifies $f(I)$ as an interval. $\blacksquare$

## 2Uniform Continuity¶

Recall the definition of continuity. The $\delta$ that plays an important part in general depends not only on $\epsilon$ but also on the point $c$ under consideration. Consider $1 / x$ for example: the closer to 0 $c$ is, the smaller $\delta$ can be. With this in mind, we provide the following definition:

Definition: Let $A \subseteq \mathbb{R}$, $f : A \to \mathbb{R}$. We say that $f$ is uniformly continuous on $A$ if for a given $\epsilon > 0$, there exists a $\delta > 0$ such that if $|x - y| < \delta$ then $|f(x) - f(y)| < \epsilon$ $\forall x, y \in A$

Clearly the set of uniformly continuous functions is a subset of the set of all continuous functions (which is just a fancy way of saying that a uniformly continuous function is also continuous). Now let's cover some useful equivalencies we can use when working with uniform continuity:

Theorem: Let $A \subseteq \mathbb{R}$, $f : A \to \mathbb{R}$. The following are equivalent (TFAE):
(i) $f$ is not uniformly continuous
(ii) There exists some $\epsilon > 0$ such that for any $\delta > 0$ there exists $x_\delta, y_\delta \in A$ such that $|x_\delta - y_\delta| < \delta$, but $|f(x_\delta) - f(y_\delta)| \geq \epsilon$
(iii) There exist $\epsilon > 0$ and sequences $(x_n), (y_n) \in A$ such that $\lim x_n - y_n = 0$ but $|f(x_n) - f(y_n)| \geq \epsilon$ for all $n \in \mathbb{N}$

Let's look at example of showing that something isn't uniformly continuous. Let's show that $g(x) = 1/x$ is not uniformly continuous on $(0,1)$:

Proof: Let $\epsilon = \frac{1}{2}$. Suppose there exists $\delta > 0$ such that if $|x-y|<\delta$ then $|f(x) - f(y)| < \epsilon$. As $\frac{1}{n} \to 0$ there exists $N \in \mathbb{N}$ such that if $n \geq N$ then $\frac{1}{n} < \frac{\delta}{2}$. Thus let $x = \frac{1}{N}, y = \frac{1}{N+1}$. Then $|x-y| \leq |x|+|y| = \frac{1}{N} + \frac{1}{N+1} < \frac{\delta}{2} + \frac{\delta}{2} = \delta$, but $|f(x) - f(y)| = |1 / \frac{1}{N} - 1 / \frac{1}{N+1}| = 1 \geq \frac{1}{2} = \epsilon$ $\blacksquare$

We've covered a number of interesting properties so far about continuous functions on closed, bounded intervals, so how about one more:

Theorem: If $I = [a, b]$ is a closed and bounded interval and $f : I \to \mathbb{R}$ is a continuous function on $I$, then $f$ is uniformly continuous on $I$.

Proof: We'll tackle this one with proof by contradiction. Suppose $f$ is not uniformly continuous on $I$. Then by (iii) above, there exist sequences $(x_n), (y_n) \in I$ and $\epsilon > 0$ such that $|x_n - y_n| < \frac{1}{n}$, but $|f(x_n) - f(y_n)| \geq \epsilon$. By Bolzano-Weierstrass, there exists a convergent subsequence $(x_{n_k}) \to x \in I$. Observe that $(y_{n_k}) \to x$ as well, since:

$$|y_{n_k} - x| = |y_{n_k} - x_{n_k} + x_{n_k} - x| \leq |y_{n_k} - x_{n_k}| + |x_{n_k} - x|$$

And both terms in the rightmost side can be made arbitrarily small. So we know $(x_{n_k}) \to x$ and $(y_{n_k}) \to x$, but $f$ is continuous on $I$, so $f(x_{n_k}) \to f(x)$ and $f(y_{n_k}) \to f(x)$. Thus:

$$|f(x_{n_k}) - f(y_{n_k})| = |f(x_{n_k}) - f(x) + f(x) - f(y_{n_k})| \leq |f(x_{n_k}) - f(x)| + |f(y_{n_k}) - f(x)|$$

For any given $\epsilon$ there exists $N \in \mathbb{N}$ such that if $k \geq N$, $|f(x_{n_k}) - f(x)| < \frac{\epsilon}{2}$ and there exists $M \in \mathbb{N}$ such that if $k \geq N$, $|f(y_{n_k}) - f(x)| < \frac{\epsilon}{2}$. Thus for $k \geq \sup\{N, M\}$, $|f(x_{n_k}) - f(y_{n_k})| < \epsilon$. But by our assumption that $f$ is not uniformly continuous, $|f(x_{n_k}) - f(y_{n_k})| \geq \epsilon$. This is a contradiction, completing our proof. $\blacksquare$