Honors Calculus I - 2.3 Continuity

Continuous Functions

$f$ is said to be continuous at $a$ if $\lim_{x\to a}f(x)=f(a)$

$\varepsilon-\delta$ definition:

For every $\varepsilon >0$, there exists a $\delta >0$ such that $0<|x-a|<\delta$ satisfy $|f(x)-f(a)|<\varepsilon$

• Left-continuous : $\lim_{x\to a^{-}}f(x)=f(a)$
• Right-continuous: $\lim_{x\to a^{+}}f(x)=f(a)$

A function $f$ is continuous at $a$ iff it is both left-continuous and right-continuous at $a$.

• If $f$ is continuous at every point in $(a,b)$, then $f$ is said to be continuous on $(a,b)$
• If $f$ is continuous at every point in $(a,b)$, and is right-continuous at $a$ and left-continuous at $b$, then $f$ is said to be continuous on $[a,b]$

Operations on continuous functions

• Suppose $f$ and $g$ are continuous functions on $(a,b)$, then $f+g,f-g,fg,f/g$ are continuous on $(a,b)$ (for quotient, the denominator is not zero)
• If $g$ is continuous at a point $p\in (a,b)$ and $f$ is continuous at the point $g(p)$, then the composite of $h = f\circ g$ is continuous at $p$

Properties of Continuous Functions

Extreme Value Theorem

Suppose $f$ is a continuous functions on a bounded closed interval $[a,b]$, then exists $p,q\in [a,b]$ such that:

$$f(p)=\inf_{x\in[a,b]}f(x)=\min_{x\in[a,b]}f(x)\\ f(q)=\sup_{x\in[a,b]}f(x)=\max_{x\in[a,b]}f(x)$$

Intermediate Value Theorem

Suppose $f$ is continuous on $[a,b]$, for any value $\xi$ between $f(a)$ and $f(b)$, there exists at least one point $c\in[a,b]$ such that $f(c)=\xi$.

One can modify the Intermediate Value Theorem to:

Any real-valued continuous function $f$ on $[a,b]$ attains all intermediate values between $\min_{x\in[a,b]}f(x)$ and $\max_{x\in[a,b]}f(x)$.

Bolzano’s Theorem:

Any real continuous function $f$ on $[a,b]$ has at least one root if $f(a)\cdot f(b)<0$.

Continuity of the inverse function

What is inverse? It must be bijective.

For a continuous interval:

A continuous function on an interval is invertible iff it is strictly monotonic.

If $f:[a,b]\to \R$ is strictly increasing and continuous, then $f$ is invertible, and its inverse $f^{-1}$ is also strictly increasing and continuous. (The same holds if $f$ is strictly decreasing and continuous)

Uniform Continuity

$f$ is uniformly continuity on $E$ if:

$$\forall \varepsilon >0, \;\exist \delta>0,\;\forall |x-y|<\delta\\ |f(x)-f(y)|<\varepsilon$$

If a function is uniformly continuity, is must be continuity (but the inverse is not true)

(Continuity is a pointwise property, whereas uniform continuity pertains to a whole set)

Note that the $\delta$ only depends on $\varepsilon$, not $x$ or $y$.

Theorem: Relationship between continuity and uniform continuity

Every continuous functions defined on a bounded closed interval $[a,b]$ is uniformly continuous on that interval.