Continuity

Continuity

I tell my calculus students that the continuous functions are exactly those whose limits are easy to compute - just plug it in! That is, $f$ should be continuous at $x=c$ if $$\lim_{x\to c} f(x) = f(c).$$

That fits pretty well with the intuitiont that you should be able to draw the graph without lifting your pen and it translates directly to an epsilon-delta definition.

Definition

Def 4.3.1 A function $f: A\to \mathbb R$ is continuous at a point $c\in A$ if, for all $\varepsilon>0$, there exists a $\delta>0$ such that whenever $|x−c|<\delta$ (and $x\in A$) it follows that $|f(x) − f(c)| < \varepsilon$.

Comment: Note that a key distinction between this definition the definition of limit is that we have no $0<|x-c|<\delta$; that is, we do care about the value of the function at $x=c$.

Example

Claim: Any affine function is continuous. That is, if $m,b\in\mathbb R$, then $f:\mathbb R \to \mathbb R$ defined by $$f(x) = mx+b \text{ is continuous at all } c\in\mathbb R.$$

Algebraic properties

The obvious algebraic properties follow directly from the relationship with the limit.

Thm 4.3.4: Let $f$ and $g$ be functions defined on a domain $A\subset\mathbb R$, and assume that both are continuous at some point $c$ of $A$. Then,

1. $kf$ is continuous at $c$ for all $k\in\mathbb R$,
2. $\displaystyle f(x)+g(x)$ is continuous at $c$,
3. $\displaystyle f(x)g(x)$ is continuous at $c$, and
4. $\displaystyle f(x)/g(x)$ is continuous at $c$ provided $g(c)\neq0$.

Sequential characterization

Thm 4.3.2 (iii): If $f(x_n)\to f(c)$ for every sequence $(x_n)$ in the domain of $f$ that converges to $c$, then $f$ is continuous at $c$.

Cor 4.3.3: Let $f : A \to \mathbb R$, and let $c\in A$ be a limit point of $A$. If there exists a sequence $(x_n)\subset A$ with $x_n\to c$ but such that $f(x_n)$ does not converge to $f(c)$, we may conclude that $f$ is not continuous at $c$.

Example

Here's an example where the sequential criterion is pretty convenient, particularly for showing discontinuity.

Claim: Let $p\in\mathbb R$ and let $$f_p(x) = \begin{cases} x^p \sin(1/x) & x \neq 0 \\ 0 & x=0.\end{cases}$$ Then, $f_p$ is continuous at the origin iff $p>0$.

The graph

Uniform continuity

Let's start with a proof of the fact that $f(x)=x^2$ is continuous on $\mathbb R$. To this end, let $c\in\mathbb R$ and note that if $\delta\in(0,1)$ and $|x-c|<\delta$, then $$|x+c| \leq |x|+|c| < |c|+1+|c| = 2|c|+1.$$ Next, let $\varepsilon>0$ and choose $\delta>0$ such that $\delta<\min(\varepsilon/(2|c|+1),1).$ Then, for $|x-c|<\delta$, $$\begin{aligned} |x^2 - c^2| = |x+c||x-c| <(2|c|+1)\frac{\varepsilon}{2|c|+1} = \varepsilon. \end{aligned}$$

Comment

A key difference bewteen this proof and the last is that the choice of $\delta$ depends on $c$, as well as on $\varepsilon$.

Functions where the this choice does not depend on $c$ are called uniformly continuous.

Definition

Def 4.4.4: A function $f : A \to \mathbb R$ is uniformly continuous on $A$ if for every $\varepsilon > 0$ there exists a $\delta > 0$ such that for all $x, y \in A$, $|x−y|<\delta$ implies $|f(x)−f(y)|<\varepsilon$.

Note that this definition applies over an entire set, rather than just at a point.

It should be pretty clear that, if $f$ is uniformly contnuous on $A$, then it's continuous at every point of $A$.

Compactness

A key reason that $f(x) = x^2$ is not uniformly continuous on $\mathbb R$ is because it's unbounded. The following is then a reasonable conjecture:

Thm 4.4.7: A function that is continuous on a compact set $K$ is uniformly continuous on $K$.

EVT and IVT

The extreme value theorem and the intermediate value theorem are two theorems involving continuous functions that are of tremendous importance in applied mathematics.

That's because they are both examples of existence theorems - that is, the assert the existence of numbers that satisfy certain conditions.

The extreme value theorem

The extreme value theorem involves continuity and compactness.

Thm 4.4.2 (EVT): If $f : K \to \mathbb R$ is continuous on a compact set $K \subset \mathbb R$, then $f$ attains a maximum and minimum value.

Ultimately, this is beause the image of a compact set is again compact.

Preservation of compact sets

Thm 4.4.1: Let $f : A \to \mathbb R$ be continuous on $A$. If $K \subset A$ is compact, then $f(K)=\{f(x):x\in K\}$ is compact as well.

Idea behind the proof: We need to show that $f(K)$ is closed and bounded. So, suppose that $x_n\in K$ with $x_n\to x$, which must also be in $K$. Then, by continuity, $f(x_n) \to f(x)\in f(K)$.

The intermediate value theorem

Thm 4.5.1: Let $f : [a,b] \to \mathbb R$ be continuous. If $L$ is a real number satisfying $$f(a) < L < f(b) \text{ or } f(a) > L > f(b).$$ Then there exists a point $c \in (a,b)$ where $f(c) = L$.