Boundedness theorem

Calculus is the most powerful weapon of thought yet devised by the wit of man, Wallace B. Smith

Recall

The derivative of a function at a chosen input value, when it exists, is the slope of the tangent line to the graph of the function at that point. It is the instantaneous rate of change, the ratio of the instantaneous change in the dependent variable to that of the independent variable.

Definition. A function f(x) is differentiable at a point “a” of its domain, if its domain contains an open interval containing “a”, and the limit $\lim _{h \to 0}{\frac {f(a+h)-f(a)}{h}}$ exists, f’(a) = L = $\lim _{h \to 0}{\frac {f(a+h)-f(a)}{h}}$. More formally, for every positive real number ε, there exists a positive real number δ, such that for every h satisfying 0 < |h| < δ, then |L-$\frac {f(a+h)-f(a)}{h}$|< ε.

The Bolzano–Weierstrass theorem states that each infinite bounded sequence in ℝn has a convergent subsequence.

Theorem. If $\{x_n\}_{n=1}^\infty$ is a convergent sequence and a ≤ xn ≤ b ∀n ∈ ℕ, then a ≤ $\lim_{n \to ∞} x_n = c$ ≤ b.. In other words, the limit of $\{x_n\}_{n=1}^\infty$ must lie between a and b.

Proof.

For the sake of contradiction, assume that c ∉ [a, b]. Without loss of generality, suppose that c < a.

Select ε = $\frac{a-c}{2}, \lim_{n \to ∞} x_n = c$ ⇒ ∃N: ∀n ≥ N, |xn - c| < ε ↭ -ε < xn - c < ε. We only need xn -c < ε = $\frac{a-c}{2}$.

However xn ≥ a, xn -c ≥ a - c = 2ε >[ε> 0] ε

Boundedness theorem

Definition. A function is bounded if for all x in its domain there is a real number M such that |f(x)| ≤ M.

Boundedness theorem. Let I = [a, b] be a closed bounded interval, let f: I → R be continuous on I. Then, f is bounded on I.

Proof. Suppose for the sake of contradiction that the function is not bounded on [a,b] ⇒ for any n ∈ ℝ, ∃xn ∈ [a,b] (we are abusing notation for the sake of brevity) such that |f(xn)| > n.

Because [a,b] is obviously bounded, $\{x_n\}_{n=1}^\infty$ is bounded ⇒

[By the Bolzano–Weierstrass theorem] it implies there exists a convergent subsequence {xnk} of $\{x_n\}_{n=1}^\infty$ with {xnk} → x0.

Theorem. If $\{x_n\}_{n=1}^\infty$ is a convergent sequence and if a ≤ xn ≤ b ∀n ∈ ℕ, then a ≤ $\lim_{n \to ∞} x_n$ ≤ b. ⇒ As [a,b] is closed, it contains x0.

Since f is continuous, f is continuos at x0, and {xnk} → x0 ⇒[Theorem. If f: ℝ → ℝ is continuos at a point x ∈ ℝ and {xn} is a real sequence converging to x, the sequence {f(xn)} converges to f(x)] {f(xnk)} → f(x0) and every convergent sequence is bounded.

However, |f(xn)| > n ⇒ [and this obviously applies to every subsequence] |f(xnk)| > nk, which implies {f(xnk)} diverges ⊥

Examples

• Counterexample. f needs to be continuous on [a, b], $f(x) = \begin{cases} \frac{1}{x}, &0 < x < 1 \\ 0, &x = 0 \end{cases}$

f(x) is not bounded on (0, 1) -Figure 1.g.- because f is not continuos at x = 0, hence f manages to escape to infinity as 0 is not there to keep f in check.

Therefore, we can deduce that the conclusion of the Boundedness Theorem does not necessarily follow.

• f: [0, 4] → ℝ, defined by f(x) = x2. f is continuous on all real numbers ⇒ f is also continuous on the closed bounded interval [0, 4] ⇒ f is is bounded. Furthermore, f’(x) = 2x > 0 on [0, 4] ⇒ f is an increasing function ⇒ |f(x)| ≤ f(4) = 42 = 16.

• f: [$\frac{-π}{2}, \frac{π}{2}$] → ℝ, defined by f(x) = arctan(x). f is continuous on all real numbers ⇒ f is also continuous on the closed bounded interval [$\frac{-π}{2}, \frac{π}{2}$] ⇒ f is is bounded. Furthermore, f’(x) = $\frac{1}{x^2+1}$ > 0 ⇒ f is an increasing function ⇒ |f(x)| ≤ $arctan(\frac{π}{2})$ ≈ 1.003. Besides, arctan(x) is bounded with $\frac{-π}{2} < y < \frac{π}{2}$, $\lim_{x \to -∞} arctan(x) = \frac{-π}{2}, \lim_{x \to ∞} arctan(x) = \frac{π}{2}.$

• f: [-2π, 2π] → ℝ, defined by f(x) = sin(x). f is continuous on all real numbers ⇒ f is also continuous on the closed interval [0, 4] ⇒ f is bounded. Furthermore, f is bounded since |sin(x)| ≤ 1 ∀x ∈ ℝ.

• f: [0, ∞) → ℝ defined by f(x) = 7x + 4 does not satisfy the conditions of the boundedness theorem because f is continuous, but the interval [0, ∞) is not bounded. $\lim_{x \to ∞} 7x + 4 = +∞$ ⇒ f is not bounded.

• f: (-1, 1) → ℝ defined by $\frac{2x}{1-x^2}$, (-1, 1) is not a closed interval ⇒ the conclusion of the Boundedness Theorem does not necessarily follow. Futhermore, $\lim_{x \to 1⁻} \frac{2x}{1-x^2} = \frac{2}{0⁺} = ∞, \lim_{x \to -1⁺} \frac{2x}{1-x^2} = \frac{-2}{0⁺} = -∞$ ⇒ f is not bounded, it has two vertical asymptotes at x = -1 and x = 1.

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