CALC-3.4

4-5 hours

Fundamental Theorems on Continuous Functions

The powerful consequences of continuity on closed intervals.

Learning Objectives

State and apply the Intermediate Value Theorem (IVT)
Prove existence of roots using Bolzano's theorem
Understand the Extreme Value Theorem (EVT)
Apply the Boundedness Theorem
Define and characterize uniform continuity
Prove continuous functions on [a,b] are uniformly continuous

1. Boundedness Theorem

Theorem 3.15: 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 3.15 (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. ∎

∎

Remark 3.9a: Why Closed Interval?

The closed interval [a, b] is crucial. Consider these counterexamples:

$f(x) = 1/x$ on (0, 1] is continuous but unbounded
$f(x) = x$ on [0, ∞) is continuous but unbounded
$f(x) = \tan x$ on (-π/2, π/2) is continuous but unbounded

Key: [a, b] is compact, which forces boundedness.

Example 3.14b: 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
We can find M: |sin x| ≤ 1 and x ≥ 1, so |f(x)| ≤ 1/1 = 1

Thus |f(x)| ≤ 1 for all x ∈ [1, 10].

Connection to Compactness

The Boundedness Theorem is actually a special case of:

Continuous images of compact sets are compact.

Since [a, b] is compact (closed and bounded in ℝ), and compact sets in ℝ are bounded, f([a, b]) is bounded.

2. Extreme Value Theorem

Theorem 3.16: Extreme Value Theorem (Weierstrass)

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) \leq f(x) \leq f(x_M) \text{ for all } x \in [a, b]

Proof of Theorem 3.16 (maximum):

Let $M = \sup\{f(x): x \in [a,b]\}$ (exists by boundedness).

$\forall n$ , $\exists x_n \in [a,b]$ with $f(x_n) > M - 1/n$ .

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

By continuity: $f(x_M) = \lim f(x_{n_k}) = M$ . ∎

∎

Example 3.15: EVT Application

Problem: Show $f(x) = x^3 - 3x + 1$ has a maximum on $[-2, 2]$ .

$f$ is a polynomial, hence continuous on $[-2, 2]$ .

By EVT, $f$ attains its max. Check: $f(-2) = -1$ , $f(2) = 3$ , $f(-1) = 3$ , $f(1) = -1$ .

Max = 3 at $x = -1$ and $x = 2$ .

Remark 3.9: Closed Interval is Essential

On $(0, 1)$ : $f(x) = x$ has $\sup = 1$ but never attains it!

Example 3.15b: Finding Extrema

Find the maximum and minimum of f(x) = x³ - 6x² + 9x + 1 on [0, 4].

Solution: By EVT, max and min exist. To find them:

Step 1: f'(x) = 3x² - 12x + 9 = 3(x - 1)(x - 3)

Step 2: Critical points: x = 1, x = 3 (both in [0, 4])

Step 3: Evaluate at critical points and endpoints:

f(0) = 1
f(1) = 1 - 6 + 9 + 1 = 5
f(3) = 27 - 54 + 27 + 1 = 1
f(4) = 64 - 96 + 36 + 1 = 5

Maximum = 5 at x = 1 and x = 4; Minimum = 1 at x = 0 and x = 3.

Example 3.15c: EVT with Trigonometry

Show f(x) = x + sin(x) attains its maximum on [0, 2π].

Solution:

f is continuous on [0, 2π] (sum of continuous functions)
By EVT, f attains its maximum
f(0) = 0, f(2π) = 2π, f(π) = π
f'(x) = 1 + cos(x) ≥ 0, so f is increasing

Maximum = f(2π) = 2π at x = 2π.

Why EVT Matters

EVT is fundamental for:

Optimization: Guarantees optimal solutions exist
Calculus of variations: Minimum/maximum energy states
Economics: Optimal pricing, resource allocation
Physics: Ground state energy, equilibrium positions

Without EVT, we couldn't be sure that optimization problems have solutions!

Theorem 3.16b: Image of Continuous Function

If f ∈ C[a, b], then f([a, b]) = [m, M] where m = min f and M = max f.

Proof:

By EVT, ∃x_m, x_M ∈ [a,b] with f(x_m) = m, f(x_M) = M.

For any y ∈ [m, M], y is between f(x_m) and f(x_M).

By IVT, ∃c between x_m and x_M with f(c) = y.

Thus every y ∈ [m, M] is in the range of f. ∎

∎

3. Intermediate Value Theorem

Theorem 3.17: 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 3.2: 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 3.2 (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, $f(c) = \lim f(a_n) \leq 0$ and $f(c) = \lim f(b_n) \geq 0$ , so $f(c) = 0$ . ∎

∎

Example 3.16: Root Finding

Problem: 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 $f(c) = 0$ .

Application: Fixed Point Theorem

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

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

Proof: Apply IVT to $g(x) = f(x) - x$ . Since $g(a) = f(a) - a \geq 0$ and $g(b) = f(b) - b \leq 0$ ...

Example 3.16b: Proving Existence of Solutions

Prove that eˣ = 2 - x has a solution in (0, 1).

Solution: Let f(x) = eˣ - 2 + x.

f(0) = 1 - 2 + 0 = -1 < 0
f(1) = e - 2 + 1 = e - 1 ≈ 1.718 > 0
f is continuous on [0, 1]

By Bolzano's theorem, ∃c ∈ (0, 1) with f(c) = 0, i.e., e^c = 2 - c. ∎

Example 3.16c: n-th Root Existence

Prove every positive real number has a unique positive n-th root.

Proof of Existence: Let a > 0 and consider f(x) = xⁿ - a on [0, max(1, a)].

f(0) = -a < 0
f(max(1, a)) = (max(1, a))ⁿ - a ≥ 0 (since max(1, a)ⁿ ≥ a)
f is continuous

By IVT, ∃c > 0 with cⁿ = a. (Uniqueness follows from strict monotonicity.)

Example 3.16d: Darboux Property

Show that f(x) = x² - 2 takes all values between f(0) = -2 and f(2) = 2 on [0, 2].

Solution: f is continuous on [0, 2], so by IVT:

For any y ∈ [-2, 2], ∃c ∈ [0, 2] with f(c) = y.

This proves existence of √2: take y = 0, then c² = 2, so c = √2 ≈ 1.414.

Bisection Algorithm

IVT provides a root-finding algorithm:

Start with [a, b] where f(a) · f(b) < 0
Compute midpoint c = (a + b)/2
If f(c) = 0, done. Otherwise:
- If f(a) · f(c) < 0, replace [a, b] with [a, c]
- If f(c) · f(b) < 0, replace [a, b] with [c, b]
Repeat until |b - a| is small enough

Convergence: After n iterations, error ≤ (b - a)/2ⁿ.

Remark 3.10a: IVT Does NOT Guarantee Uniqueness

IVT only says a value IS attained, not how many times.

Example: f(x) = sin x on [0, 4π] takes the value 0 at x = 0, π, 2π, 3π, 4π.

4. Uniform Continuity

Definition 3.13: 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 $D$ .

Remark 3.10: Uniform vs. Pointwise Continuity

Pointwise: $\forall x_0, \forall \varepsilon, \exists \delta(x_0, \varepsilon)...$
Uniform: $\forall \varepsilon, \exists \delta(\varepsilon)...$ (independent of $x$ )

Uniform continuity ⟹ continuity, but not conversely.

Example 3.17: Non-Uniform Continuity

Claim: $f(x) = 1/x$ 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 - 4/\delta| = 2/\delta \to \infty \text{ as } \delta \to 0

Theorem 3.18: Heine-Cantor Theorem

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

Proof of Theorem 3.18 (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! ∎

∎

Corollary 3.3: Lipschitz Implies Uniform Continuity

If $|f(x) - f(y)| \leq L|x - y|$ for some $L > 0$ (Lipschitz), then $f$ is uniformly continuous.

Example 3.17b: Lipschitz Function

Show f(x) = sin x is uniformly continuous on ℝ.

Solution: By Mean Value Theorem:

|\sin x - \sin y| = |\cos c||x - y| \leq |x - y|

So sin x is Lipschitz with L = 1, hence uniformly continuous on ℝ.

Given ε > 0, take δ = ε. Then |x - y| < δ ⟹ |sin x - sin y| < ε.

Example 3.17c: Non-Uniform on Unbounded Domain

Show f(x) = x² is NOT uniformly continuous on ℝ.

Proof: Take ε₀ = 1. For any δ > 0, choose n such that n > 1/(2δ).

Let x = n and y = n + δ/2. Then |x - y| = δ/2 < δ, but:

|x^2 - y^2| = |n^2 - (n + \delta/2)^2| = |n\delta + \delta^2/4| > n\delta > \frac{1}{2} > \varepsilon_0

Actually, the correct calculation: |x² - y²| = (2n + δ/2)(δ/2) > nδ > 1.

So no single δ works for all x. Not uniformly continuous. ∎

Example 3.17d: Uniform on Bounded Interval

Show f(x) = x² IS uniformly continuous on [0, 10].

Solution 1: f is continuous on the closed interval [0, 10].

By Heine-Cantor theorem, f is uniformly continuous on [0, 10]. ∎

Solution 2 (direct): For x, y ∈ [0, 10]:

|x^2 - y^2| = |x + y||x - y| \leq 20|x - y|

Given ε > 0, take δ = ε/20. Then |x - y| < δ ⟹ |x² - y²| < ε. ∎

Negation of Uniform Continuity

f is NOT uniformly continuous on D if:

\exists \varepsilon_0 > 0, \forall \delta > 0, \exists x, y \in D: |x - y| < \delta \text{ but } |f(x) - f(y)| \geq \varepsilon_0

Equivalently: there exist sequences xₙ, yₙ with |xₙ - yₙ| → 0 but |f(xₙ) - f(yₙ)| ≥ ε₀.

Theorem 3.18b: Characterization of Uniform Continuity

f is uniformly continuous on D if and only if:

For any sequences (xₙ), (yₙ) in D with |xₙ - yₙ| → 0, we have |f(xₙ) - f(yₙ)| → 0.

Example 3.17e: Using Sequential Characterization

Show √x is uniformly continuous on [0, ∞).

Proof: For x, y ≥ 0:

|\sqrt{x} - \sqrt{y}| = \frac{|x - y|}{\sqrt{x} + \sqrt{y}} \leq \frac{|x - y|}{\sqrt{|x - y|}} = \sqrt{|x - y|}

(Using √x + √y ≥ √|x - y| when |x - y| is small.)

Given ε > 0, take δ = ε². Then |x - y| < δ ⟹ |√x - √y| ≤ √δ = ε. ∎

Remark 3.11: Uniform Continuity Summary

Uniformly continuous:

Any continuous function on [a, b]
Lipschitz functions (e.g., sin x, cos x on ℝ)
√x on [0, ∞)
Bounded continuous functions with bounded derivative

NOT uniformly continuous:

1/x on (0, 1)
x² on ℝ
eˣ on ℝ
sin(1/x) on (0, 1)

5. Applications and Extensions

Example 3.18: Application to Differential Equations

Show the equation x'(t) = f(t, x) with f continuous has local solutions.

Picard-Lindelöf Theorem (idea):

IVT and fixed point arguments guarantee existence of solutions for ODEs with continuous right-hand sides.

This is fundamental to the theory of differential equations!

Example 3.19: Economics: Equilibrium Existence

Prove existence of market equilibrium.

Setup: Demand D(p) and Supply S(p) are continuous functions of price p.

If D(p₁) > S(p₁) and D(p₂) < S(p₂) (excess demand changes sign), then by IVT:

∃p* ∈ (p₁, p₂) with D(p*) = S(p*) (market clearing price).

Example 3.20: Temperature Distribution

A circular wire has temperature T(θ) continuous in θ ∈ [0, 2π]. Show there exist diametrically opposite points with the same temperature.

Solution: Define f(θ) = T(θ) - T(θ + π) for θ ∈ [0, π].

f is continuous
f(0) = T(0) - T(π)
f(π) = T(π) - T(2π) = T(π) - T(0) = -f(0)

If f(0) = 0, done. Otherwise, f(0) and f(π) have opposite signs.

By IVT, ∃θ₀ ∈ (0, π) with f(θ₀) = 0, i.e., T(θ₀) = T(θ₀ + π). ∎

Theorem 3.19: Borsuk-Ulam (1D)

If f: S¹ → ℝ is continuous (S¹ = unit circle), then there exist antipodal points x, -x with f(x) = f(-x).

6. Practice Problems

Problem Set A: IVT

Show x⁵ + x - 1 = 0 has a root in (0, 1).
(Hint: f(0) = -1, f(1) = 1)
Show cos x = x has a solution in (0, π/2).
(Hint: Consider g(x) = cos x - x)
Prove that x³ - 3x + 1 = 0 has exactly three real roots.
(Hint: Check signs at -2, -1, 0, 1, 2)

Problem Set B: EVT

Find max and min of f(x) = x³ - 3x on [-2, 2].
(Answer: max = 2 at x = -1 and x = 2, min = -2 at x = 1 and x = -2)
Show f(x) = xeˣ attains a minimum on [-1, 1].
Why doesn't f(x) = 1/(x² + 1) attain its infimum on ℝ?
(Hint: It does! inf = 0 but this is attained only "at infinity")

Problem Set C: Uniform Continuity

Show f(x) = x³ is uniformly continuous on [-1, 1] but not on ℝ.
Prove cos x is uniformly continuous on ℝ.
(Hint: Lipschitz with L = 1)
Is f(x) = x sin(1/x) (f(0) = 0) uniformly continuous on [0, 1]?
(Answer: Yes, by Heine-Cantor)

7. Common Mistakes

Mistake 1: Forgetting Closed Interval

Wrong: "f(x) = x is continuous on (0, 1), so it attains its max."

Right: EVT requires a closed interval. f(x) = x on (0, 1) has sup = 1 but doesn't attain it.

Mistake 2: IVT Doesn't Find Roots

Wrong: "By IVT, the root is c = ..."

Right: IVT only proves existence. It doesn't tell you the value of c.

Mistake 3: Uniform ≠ Pointwise

Wrong: "f is continuous, so it's uniformly continuous."

Right: Continuity ≠ uniform continuity (except on closed bounded intervals).

Mistake 4: Misapplying Boundedness

Wrong: "f is continuous on [0, ∞), so it's bounded."

Right: Boundedness theorem requires a closed bounded interval [a, b]. [0, ∞) is not bounded!

8. More Worked Examples

Example 3.21: Combined IVT and EVT

Let f ∈ C[0, 1] with f(0) = f(1). Show f attains its maximum at least twice or its minimum at least twice.

Proof: By EVT, f attains max M at some x_M and min m at some x_m.

Case 1: f(0) = f(1) = M. Then max attained at two points.

Case 2: f(0) = f(1) = m. Then min attained at two points.

Case 3: m < f(0) = f(1) < M. Then x_M, x_m ∈ (0, 1).

So max and min are both attained in the interior. Since f(0) = f(1), at least one extreme value is attained at least twice. ∎

Example 3.22: Proving Surjectivity

Show f(x) = x³ - x is surjective (onto) from ℝ to ℝ.

Proof: We need to show every y ∈ ℝ is in the range of f.

f is continuous on ℝ. As x → +∞, f(x) → +∞. As x → -∞, f(x) → -∞.

For any y ∈ ℝ, choose a large enough so f(-a) < y and f(a) > y.

By IVT on [-a, a], ∃c with f(c) = y. ∎

Example 3.23: Fixed Point with Specific Bound

Let f: [0, 2] → [0, 2] be continuous. Prove f has a fixed point.

Proof: Define g(x) = f(x) - x on [0, 2].

g is continuous
g(0) = f(0) - 0 = f(0) ≥ 0 (since f(0) ∈ [0, 2])
g(2) = f(2) - 2 ≤ 0 (since f(2) ∈ [0, 2])

By IVT, ∃c ∈ [0, 2] with g(c) = 0, i.e., f(c) = c. ∎

Example 3.24: Application: Polynomial Roots

Show every polynomial of odd degree has at least one real root.

Proof: Let p(x) = xⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₀ with n odd.

As x → +∞, p(x) → +∞ (dominated by xⁿ > 0).

As x → -∞, p(x) → -∞ (since n is odd, xⁿ → -∞).

For large enough |a|, p(a) > 0 and p(-a) < 0.

By IVT on [-a, a], ∃c with p(c) = 0. ∎

Example 3.25: Uniform Continuity Direct Proof

Prove directly that f(x) = x² is uniformly continuous on [0, 5].

Proof: For x, y ∈ [0, 5]:

|x^2 - y^2| = |x + y||x - y| \leq (5 + 5)|x - y| = 10|x - y|

Given ε > 0, choose δ = ε/10.

Then |x - y| < δ ⟹ |x² - y²| < 10δ = ε. ∎

Example 3.26: Proving Non-Uniform Continuity

Prove f(x) = eˣ is NOT uniformly continuous on ℝ.

Proof: Take ε₀ = 1. For any δ > 0:

Let xₙ = n and yₙ = n + δ/2. Then |xₙ - yₙ| = δ/2 < δ.

|e^{x_n} - e^{y_n}| = e^n|1 - e^{\delta/2}| = e^n(e^{\delta/2} - 1)

As n → ∞, this → ∞ > ε₀. So no single δ works. ∎

Example 3.27: Continuity Extension

Let f be uniformly continuous on (a, b). Show f can be extended to a continuous function on [a, b].

Key idea: For any sequence xₙ → a⁺, the uniform continuity implies (f(xₙ)) is Cauchy.

By completeness of ℝ, lim f(xₙ) exists. Call it f(a).

This limit is independent of the sequence choice (also by uniform continuity).

Similarly define f(b). The extended function is continuous on [a, b]. ∎

Example 3.28: Iteration and Fixed Points

Let f: [0, 1] → [0, 1] be continuous with |f(x) - f(y)| < |x - y| for x ≠ y. Show f has a unique fixed point.

Existence: By Brouwer fixed point theorem (or IVT argument).

Uniqueness: Suppose f(c₁) = c₁ and f(c₂) = c₂ with c₁ ≠ c₂.

Then |f(c₁) - f(c₂)| = |c₁ - c₂| < |c₁ - c₂|. Contradiction! ∎

9. Detailed Proofs

Theorem 3.20: IVT (Complete Proof)

If f ∈ C[a, b] and f(a) < k < f(b), then ∃c ∈ (a, b) with f(c) = k.

Proof:

Complete Proof using Nested Intervals:

Step 1: Define S = {x ∈ [a, b] : f(x) < k}. S is non-empty (a ∈ S) and bounded above (by b).

Step 2: Let c = sup S. We claim f(c) = k.

Step 3 (f(c) ≤ k): ∃ sequence xₙ ∈ S with xₙ → c. By continuity, f(c) = lim f(xₙ) ≤ k.

Step 4 (f(c) ≥ k): If c < b, then ∃ sequence yₙ > c with yₙ → c. Since yₙ ∉ S, f(yₙ) ≥ k. By continuity, f(c) ≥ k.

Step 5: If c = b, then f(c) = f(b) > k, contradicting f(c) ≤ k. So c < b.

Combining: f(c) ≤ k and f(c) ≥ k ⟹ f(c) = k. ∎

∎

Theorem 3.21: Continuity of Inverse Functions

If f: [a, b] → ℝ is continuous and strictly monotone, then f⁻¹: f([a, b]) → [a, b] is also continuous.

Proof:

WLOG assume f is strictly increasing.

By EVT, f([a, b]) = [f(a), f(b)] (an interval).

f⁻¹ is strictly increasing (preserves order).

To show continuity: Let y₀ ∈ [f(a), f(b)] and let x₀ = f⁻¹(y₀).

Given ε > 0 (small enough so x₀ ± ε ∈ [a, b]), let:

\delta = \min(|y_0 - f(x_0 - \varepsilon)|, |f(x_0 + \varepsilon) - y_0|)

Then |y - y₀| < δ ⟹ |f⁻¹(y) - f⁻¹(y₀)| < ε. ∎

∎

Corollary 3.4: Image is an Interval

If f: D → ℝ is continuous and D is an interval, then f(D) is also an interval.

Remark 3.12: Intervals and Connectedness

This corollary says: continuous images of connected sets are connected.

In ℝ, connected = interval. This is a foundational result in topology.

10. Quick Reference

Theorem Summary Table

Theorem	Hypothesis	Conclusion
Boundedness	f ∈ C[a, b]	f is bounded
EVT	f ∈ C[a, b]	f attains max and min
IVT	f ∈ C[a, b], k between f(a), f(b)	∃c: f(c) = k
Bolzano	f ∈ C[a, b], f(a)f(b) < 0	∃c: f(c) = 0
Heine-Cantor	f ∈ C[a, b]	f uniformly continuous

Key Definitions

Uniform Continuity

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

Lipschitz Continuity

|f(x) - f(y)| \leq L|x - y| \text{ for some } L > 0

Implication Chain

Lipschitz ⟹ Uniform Continuity ⟹ Continuity

But none of these implications reverse in general!

On [a, b]: Continuity ⟺ Uniform Continuity (by Heine-Cantor)

11. Historical Context

Development of These Theorems

Bolzano (1817): First rigorous proof of IVT, before Cauchy!
Cauchy (1821): Systematic development of continuity theory
Weierstrass (1860s): EVT, rigorous ε-δ definitions
Heine (1872): Uniform continuity and its characterization
Cantor (1872): Connection to compactness (Heine-Cantor)

These theorems form the backbone of real analysis and are prerequisites for differentiation, integration, and beyond.

12. Additional Practice

Problem Set D: Proofs

Prove: If f ∈ C[a, b] and f(x) ≥ 0 for all x ∈ [a, b], then either f ≡ 0 or ∫_a^b f(x)dx > 0.
(Hint: Use EVT if f is not identically 0)
Let f ∈ C[0, 1] with f(0) = f(1). Show ∃c ∈ [0, 1/2] with f(c) = f(c + 1/2).
(Hint: Consider g(x) = f(x) - f(x + 1/2))
Prove that if f is uniformly continuous on (a, b), then lim_{x→a⁺} f(x) exists.
(Hint: Use Cauchy criterion)
Show: f Lipschitz on ℝ with L < 1 has exactly one fixed point.
(Hint: Banach fixed point theorem)

Problem Set E: Applications

A tank holds water at temperature T₁ and is placed in a room at temperature T₂. If T(t) is the water temperature at time t, prove T(t₀) = (T₁ + T₂)/2 for some t₀ > 0.
(Assume T is continuous and approaches T₂)
Show that any continuous function f: [0, 1] → [0, 1] that maps rationals to irrationals must have a fixed point.
Prove: If f ∈ C[a, b] and ∫_a^b f(x)g(x)dx = 0 for all g ∈ C[a, b], then f ≡ 0.
(Hint: Try g = f)

Problem Set F: True/False

Every continuous function on (0, 1) is bounded. (False)
If f is uniformly continuous on [1, ∞), then f is bounded. (False: f(x) = x)
If f ∈ C[a, b], then f² attains its max on [a, b]. (True)
IVT holds for discontinuous functions. (False)
Uniform continuity implies Lipschitz. (False: √x)
If f' is bounded on (a, b), then f is uniformly continuous on (a, b). (True)

13. Extended Examples

Example 3.29: Nested Application of IVT

Show that x⁵ - 5x + 3 = 0 has exactly three real roots.

Solution: Let f(x) = x⁵ - 5x + 3.

Finding sign changes:

f(-2) = -32 + 10 + 3 = -19 < 0
f(-1) = -1 + 5 + 3 = 7 > 0 → root in (-2, -1)
f(0) = 3 > 0
f(1) = 1 - 5 + 3 = -1 < 0 → root in (0, 1)
f(2) = 32 - 10 + 3 = 25 > 0 → root in (1, 2)

Why exactly 3? f'(x) = 5x⁴ - 5 = 5(x⁴ - 1) = 0 at x = ±1.

f has only 2 critical points, so at most 3 roots. Combined with IVT: exactly 3. ∎

Example 3.30: EVT in Optimization

A farmer has 100m of fencing. Find the maximum area of a rectangular pen.

Setup: Let x = width. Then length = (100 - 2x)/2 = 50 - x.

Area A(x) = x(50 - x) = 50x - x² for x ∈ [0, 50].

By EVT: A is continuous on [0, 50], so it attains its max.

Finding the max: A'(x) = 50 - 2x = 0 ⟹ x = 25.

A(0) = A(50) = 0, A(25) = 625.

Maximum area = 625 m² with a 25m × 25m square pen.

Example 3.31: Uniform Continuity and Integrals

Let f be uniformly continuous on [0, ∞). Define F(x) = ∫_0^x f(t)dt. Show F is uniformly continuous.

Proof: Given ε > 0. Since f is uniformly continuous:

∃M: |f(t)| ≤ M for all t (not always true! need boundedness).

Actually: If f is uniformly continuous AND bounded by M:

|F(x) - F(y)| = \left|\int_y^x f(t)dt\right| \leq M|x - y|

So F is Lipschitz with constant M, hence uniformly continuous. ∎

Example 3.32: Continuity and Sequences

Let f ∈ C[0, 1] with f(1) = 0. Show lim_{n→∞} ∫_0^1 f(x)xⁿdx = 0.

Proof: By EVT, |f(x)| ≤ M for some M.

Given ε > 0. By continuity at x = 1: ∃δ > 0 with |f(x)| < ε for x ∈ (1-δ, 1].

\left|\int_0^1 f(x)x^n dx\right| \leq \int_0^{1-\delta} M \cdot x^n dx + \int_{1-\delta}^1 \varepsilon dx

\leq M \cdot \frac{(1-\delta)^{n+1}}{n+1} + \varepsilon\delta

As n → ∞, first term → 0. So limit ≤ εδ. Since ε arbitrary, limit = 0. ∎

Example 3.33: Fixed Point Iteration

Let f: [0, 1] → [0, 1] with |f(x) - f(y)| ≤ (1/2)|x - y|. Starting from any x₀, show xₙ₊₁ = f(xₙ) converges.

Proof: By fixed point theorem, f has a unique fixed point c.

|x_{n+1} - c| = |f(x_n) - f(c)| \leq \frac{1}{2}|x_n - c|

By induction: |xₙ - c| ≤ (1/2)ⁿ|x₀ - c| → 0.

So xₙ → c (exponentially fast!). ∎

14. Exam Preparation

Key Skills to Master

IVT Problems

• Identify f, a, b, k
• Verify continuity
• Check f(a) < k < f(b) or opposite
• State conclusion precisely

EVT Problems

• Verify CLOSED interval
• Verify continuity
• Find critical points
• Compare values at critical pts & endpoints

Uniform Continuity

• Apply Heine-Cantor when possible
• For direct proofs: find δ(ε) only
• For disproving: find ε₀ and sequences

Fixed Points

• Define g(x) = f(x) - x
• Check g(a) and g(b) have opposite signs
• Apply IVT to g

Common Exam Questions

Root existence: Show f(x) = 0 has a solution in (a, b)
Optimization: Find max/min of f on [a, b]
Fixed points: Prove f: I → I has a fixed point
Uniform continuity: Prove f is (or isn't) uniformly continuous
Applications: Economics equilibrium, physics problems

15. Final Examples

Example 3.34: Cardinality of Roots

How many solutions does sin x = x/10 have?

Solution: Let f(x) = sin x - x/10.

f(0) = 0, so x = 0 is one solution.

For x ∈ (0, 10]: sin x ≤ 1 and x/10 ∈ (0, 1], so f(x) could be 0.

f(π) = 0 - π/10 < 0, f(π/2) = 1 - π/20 ≈ 0.84 > 0 → root in (π/2, π)

By symmetry and counting sign changes: 7 solutions in [-10, 10].

Example 3.35: Generalized Fixed Point

Let f, g ∈ C[0, 1] with f(0) ≤ g(0) and f(1) ≥ g(1). Show ∃c with f(c) = g(c).

Proof: Define h(x) = f(x) - g(x).

h is continuous on [0, 1]
h(0) = f(0) - g(0) ≤ 0
h(1) = f(1) - g(1) ≥ 0

By IVT, ∃c ∈ [0, 1] with h(c) = 0, i.e., f(c) = g(c). ∎

Example 3.36: Continuous Extension Problem

Let f(x) = (eˣ - 1)/x for x ≠ 0. How should we define f(0) to make f continuous?

Solution: We need f(0) = lim_{x→0} (eˣ - 1)/x.

Using L'Hôpital's rule or series expansion:

\lim_{x \to 0} \frac{e^x - 1}{x} = \lim_{x \to 0} \frac{e^x}{1} = e^0 = 1

So define f(0) = 1 for continuity.

Example 3.37: Bisection Method Application

Use bisection to find √2 to 2 decimal places.

Setup: f(x) = x² - 2 on [1, 2]. f(1) = -1 < 0, f(2) = 2 > 0.

Iterations:

[1, 2]: mid = 1.5, f(1.5) = 0.25 > 0 → [1, 1.5]
[1, 1.5]: mid = 1.25, f(1.25) = -0.4375 < 0 → [1.25, 1.5]
[1.25, 1.5]: mid = 1.375, f(1.375) = -0.109 < 0 → [1.375, 1.5]
[1.375, 1.5]: mid = 1.4375, f(1.4375) = 0.066 > 0 → [1.375, 1.4375]
Continue until interval width < 0.01...

Result: √2 ≈ 1.41 (actual: 1.4142...)

Example 3.38: Hölder Continuity

Show f(x) = √x is Hölder continuous with exponent α = 1/2 on [0, 1].

Definition: f is Hölder(α) if |f(x) - f(y)| ≤ C|x - y|^α.

Proof: For 0 ≤ y ≤ x ≤ 1:

|\sqrt{x} - \sqrt{y}| = \frac{|x - y|}{\sqrt{x} + \sqrt{y}} \leq \frac{|x - y|}{\sqrt{|x - y|}} = \sqrt{|x - y|}

So |√x - √y| ≤ |x - y|^{1/2} with C = 1, α = 1/2. ∎

16. Study Tips

Memory Techniques

IVT Mnemonic

"If you Value intermediate results, check Three things: continuity, interval, intermediate value."

EVT Mnemonic

"Every Valley and Top is reached on a closed interval."

Heine-Cantor

"Heine says Closed intervals give Uniform continuity."

Key Check

Always verify: Closed, Bounded, Continuous → CBC!

Common Pitfalls

Open vs Closed: EVT needs [a, b], not (a, b)
Bounded vs Unbounded: Boundedness theorem needs [a, b], not [a, ∞)
Existence vs Value: IVT proves existence, not the actual root value
Uniform vs Pointwise: Uniform requires δ independent of x

Essential Formulas

IVT Statement

f \in C[a,b], f(a) < k < f(b) \Rightarrow \exists c \in (a,b): f(c) = k

EVT Statement

f \in C[a,b] \Rightarrow \exists x_m, x_M \in [a,b]: f(x_m) \leq f(x) \leq f(x_M)

Uniform Continuity

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

Lipschitz Condition

|f(x) - f(y)| \leq L|x - y|, \quad L > 0

Important Counterexamples

Statement	Why False	Counterexample
f ∈ C(a, b) ⟹ bounded	Open interval	f(x) = 1/x on (0, 1)
f ∈ C(a, b) ⟹ attains max	Open interval	f(x) = x on (0, 1)
Continuous ⟹ Uniformly cont.	Need closed bounded	f(x) = x² on ℝ
Uniform cont. ⟹ Lipschitz	Not true in general	f(x) = √x on [0, 1]
Uniform cont. ⟹ bounded	Domain matters	f(x) = x on ℝ

Key Relationships

Continuity Hierarchy

Differentiable with bounded derivative

↓

Lipschitz continuous

↓

Hölder continuous

↓

Uniformly continuous

↓

Continuous

Each level is strictly stronger than the next.

On Closed Bounded Intervals

For f ∈ C[a, b]:

f is bounded (Boundedness Theorem)
f attains its max and min (EVT)
f is uniformly continuous (Heine-Cantor)
f takes all intermediate values (IVT)
f([a, b]) = [min f, max f] (Image is closed interval)

Congratulations!

You've completed Chapter 3: Function Limits and Continuity! You now understand:

ε-δ definitions of limits and continuity
Properties and operations on limits
Classification of discontinuities
Fundamental theorems (IVT, EVT, Heine-Cantor)
Uniform continuity and its characterization
Applications to root finding and optimization

Next: Chapter 4 covers Differentiation, where continuity plays a crucial role!

Remember: Differentiability implies continuity, but not vice versa.

The foundational work in this chapter prepares you for the derivative.

Key insight: Continuous functions on closed bounded intervals are "nice" — bounded, attain extrema, uniformly continuous.

Chapter 3.4 & Chapter 3 Summary

Boundedness & Extrema

$f \in C[a,b]$ ⟹ $f$ bounded
$f \in C[a,b]$ ⟹ $f$ attains max/min

Intermediate Values

IVT: $f$ takes all intermediate values
Bolzano: sign change ⟹ zero exists

Uniform Continuity

$\delta$ independent of point
$f \in C[a,b]$ ⟹ uniformly continuous

Looking Ahead

Chapter 4: Derivatives
Chapter 5: Integration

Practice Quiz: Continuity Theorems

Questions

Correct

Accuracy

The Intermediate Value Theorem requires

f

to be:

Easy

Not attempted

f \in C[a,b]

with

f(a) < 0 < f(b)

, then:

Easy

Not attempted

The Extreme Value Theorem states that

f \in C[a,b]

Medium

Not attempted

Uniform continuity means:

Medium

Not attempted

f(x) = 1/x

uniformly continuous on

(0,1)

Hard

Not attempted

f \in C[a,b]

, then

f

is:

Easy

Not attempted

FAQs

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.