Continuity Theorems | Calculus Course

Section 1: Boundedness Theorem

Theorem 1.1: Boundedness Theorem

If $f \in C[a, b]$ , then $f$ is bounded on $[a, b]$ :

\exists M > 0: |f(x)| \leq M \text{ for all } x \in [a, b]

Proof of Theorem 1.1 (by contradiction):

Suppose $f$ is unbounded. Then $\forall n \in \mathbb{N}$ , $\exists x_n \in [a,b]$ with $|f(x_n)| > n$ .

By Bolzano-Weierstrass, $\{x_n\}$ has a convergent subsequence $x_{n_k} \to c \in [a,b]$ .

Since $f$ is continuous at $c$ : $f(x_{n_k}) \to f(c)$ .

But $|f(x_{n_k})| > n_k \to \infty$ , contradiction. ∎

∎

Example 1.1: Verifying Boundedness

Show f(x) = sin(x)/x on [1, 10] is bounded.

Solution:

f is continuous on [1, 10] (ratio of continuous functions, denominator ≠ 0).

By Boundedness Theorem, f is bounded. Since |sin x| ≤ 1 and x ≥ 1, we have |f(x)| ≤ 1.

Section 2: Extreme Value Theorem

Theorem 2.1: Extreme Value Theorem (EVT)

If $f \in C[a, b]$ , then $f$ attains its maximum and minimum on $[a, b]$ :

\exists x_m, x_M \in [a, b]: f(x_m) = \min_{x \in [a,b]} f(x), \quad f(x_M) = \max_{x \in [a,b]} f(x)

Proof of Theorem 2.1:

By Boundedness Theorem, $M = \sup_{x \in [a,b]} f(x)$ exists and is finite.

By supremum property, $\exists x_n \in [a,b]$ with $f(x_n) \to M$ .

By Bolzano-Weierstrass, $x_{n_k} \to x_M \in [a,b]$ .

By continuity, $f(x_M) = \lim f(x_{n_k}) = M$ . Similarly for minimum. ∎

∎

Example 2.1: Finding Extrema

Find max and min of f(x) = x³ - 3x on [-2, 2].

Solution:

f is continuous on [-2, 2], so by EVT, max and min exist.

f'(x) = 3x² - 3 = 3(x² - 1) = 0 when x = ±1.

Check: f(-2) = -8 + 6 = -2, f(-1) = -1 + 3 = 2, f(1) = 1 - 3 = -2, f(2) = 8 - 6 = 2.

Max = 2 at x = -1 and x = 2, Min = -2 at x = -2 and x = 1.

Section 3: Intermediate Value Theorem

Theorem 3.1: Intermediate Value Theorem (IVT)

If $f \in C[a, b]$ and $k$ is between $f(a)$ and $f(b)$ , then:

\exists c \in (a, b): f(c) = k

Corollary 3.1: Bolzano's Theorem (Zero Theorem)

If $f \in C[a, b]$ and $f(a) \cdot f(b) < 0$ (opposite signs), then:

\exists c \in (a, b): f(c) = 0

Proof of Corollary 3.1 (Bisection):

WLOG $f(a) < 0 < f(b)$ . Let $c = (a+b)/2$ .

If $f(c) = 0$ , done. If $f(c) > 0$ , apply to $[a, c]$ . If $f(c) < 0$ , apply to $[c, b]$ .

Get nested intervals $[a_n, b_n]$ with $b_n - a_n = (b-a)/2^n \to 0$ .

By completeness, $a_n, b_n \to c$ . By continuity, $f(c) = 0$ . ∎

∎

Example 3.1: Root Finding

Show: $x^3 + x - 1 = 0$ has a root in $(0, 1)$ .

Let $f(x) = x^3 + x - 1$ . Then $f(0) = -1 < 0$ and $f(1) = 1 > 0$ .

By Bolzano's theorem, $\exists c \in (0, 1)$ with $f(c) = 0$ .

Section 4: Uniform Continuity

Definition 4.1: Uniform Continuity

$f$ is uniformly continuous on $D$ if:

\forall \varepsilon > 0, \exists \delta > 0: \forall x, y \in D, |x - y| < \delta \Rightarrow |f(x) - f(y)| < \varepsilon

Key: the same $\delta$ works for ALL $x, y$ in $D$ .

Theorem 4.1: Heine-Cantor Theorem

If $f \in C[a, b]$ , then $f$ is uniformly continuous on $[a, b]$ .

Proof of Theorem 4.1 (by contradiction):

Suppose not. Then $\exists \varepsilon_0 > 0$ such that $\forall \delta > 0$ , $\exists x, y$ with $|x - y| < \delta$ but $|f(x) - f(y)| \geq \varepsilon_0$ .

Take $\delta = 1/n$ . Get sequences $x_n, y_n$ with $|x_n - y_n| < 1/n$ but $|f(x_n) - f(y_n)| \geq \varepsilon_0$ .

By Bolzano-Weierstrass, $x_{n_k} \to c \in [a,b]$ . Then $y_{n_k} \to c$ too.

By continuity: $f(x_{n_k}) \to f(c)$ and $f(y_{n_k}) \to f(c)$ , so $|f(x_{n_k}) - f(y_{n_k})| \to 0$ . Contradiction! ∎

∎

Example 4.1: Non-Uniform Continuity

Show: $f(x) = 1/x$ is NOT uniformly continuous on $(0, 1)$ .

Take $\varepsilon = 1$ . For any $\delta > 0$ , let $x = \delta/2$ , $y = \delta/4$ .

Then $|x - y| = \delta/4 < \delta$ , but $|f(x) - f(y)| = 2/\delta \to \infty$ as $\delta \to 0$ .

Section 5: Lipschitz Continuity

Definition 5.1: Lipschitz Function

A function $f: D \to \mathbb{R}$ is Lipschitz continuous if there exists $L > 0$ such that:

|f(x) - f(y)| \leq L|x - y| \quad \forall x, y \in D

The smallest such $L$ is called the Lipschitz constant.

Theorem 5.1: Lipschitz Implies Uniform Continuity

If $f$ is Lipschitz continuous, then $f$ is uniformly continuous.

Proof:

Given $\varepsilon > 0$ , take $\delta = \varepsilon/L$ .

Then $|x - y| < \delta \Rightarrow |f(x) - f(y)| \leq L|x - y| < L \cdot \frac{\varepsilon}{L} = \varepsilon$ .

∎

Example 5.1: Lipschitz Examples

$f(x) = x$ is Lipschitz with $L = 1$
$f(x) = |x|$ is Lipschitz with $L = 1$
$f(x) = \sin x$ is Lipschitz with $L = 1$ (by Mean Value Theorem)
$f(x) = \sqrt{x}$ on $[1, \infty)$ is NOT Lipschitz

Section 6: Fixed Point Theorem

Definition 6.1: Fixed Point

A point $c$ is a fixed point of $f$ if $f(c) = c$ .

Theorem 6.1: Brouwer Fixed Point Theorem (1D)

If $f: [a, b] \to [a, b]$ is continuous, then $f$ has at least one fixed point.

Proof:

Consider $g(x) = f(x) - x$ . Then $g(a) = f(a) - a \geq 0$ and $g(b) = f(b) - b \leq 0$ .

By Intermediate Value Theorem, $\exists c \in [a, b]$ with $g(c) = 0$ , i.e., $f(c) = c$ .

∎

Example 6.1: Application

The function $f(x) = \cos x$ on $[0, 1]$ maps into $[\cos 1, 1] \subset [0, 1]$ , so it has a fixed point.

Section 7: Continuity and Limits

Theorem 7.1: Continuity via Limits

A function $f$ is continuous at $x_0$ if and only if:

\lim_{x \to x_0} f(x) = f(x_0)

Theorem 7.2: Composition of Continuous Functions

If $f$ is continuous at $x_0$ and $g$ is continuous at $f(x_0)$ , then $g \circ f$ is continuous at $x_0$ .

Example 7.1: Composition

Since $f(x) = x^2$ and $g(x) = \sin x$ are continuous, $(g \circ f)(x) = \sin(x^2)$ is continuous.

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Practice Quiz: Continuity Theorems

10

Questions

0

Correct

0%

Accuracy

1

The Intermediate Value Theorem requires

f

to be:

Easy

Not attempted

2

If

f \in C[a,b]

with

f(a) < 0 < f(b)

, then:

Easy

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3

The Extreme Value Theorem states that

f \in C[a,b]

:

Medium

Not attempted

4

Uniform continuity means:

Medium

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5

Is

f(x) = 1/x

uniformly continuous on

(0,1)

?

Hard

Not attempted

6

If

f \in C[a,b]

, then

f

is:

Easy

Not attempted

7

The Boundedness Theorem requires:

Easy

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8

If

f

is Lipschitz continuous, then:

Medium

Not attempted

9

Bolzano's theorem is a special case of:

Medium

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10

On which interval is

f(x) = x^2

uniformly continuous?

Hard

Not attempted

Frequently Asked Questions

Why must the interval be closed for EVT?

On open intervals, f can 'escape to infinity' near endpoints. Example: f(x) = 1/x on (0,1) is continuous but has no max.

What's the difference between continuity and uniform continuity?

Continuity: for each x₀ and ε, find δ (may depend on x₀). Uniform: for each ε, find δ that works for ALL x simultaneously.

How is IVT used to find roots?

If f(a) and f(b) have opposite signs and f is continuous, there's a root in (a,b). Bisection method repeatedly applies this.

Does IVT guarantee uniqueness of roots?

No! IVT only proves existence. A function can have multiple roots. Example: sin x = 0 has infinitely many solutions.

What is the Heine-Cantor theorem?

If f is continuous on a closed bounded interval [a,b], then f is uniformly continuous on [a,b]. This is a fundamental result connecting continuity and uniform continuity.