Calculus Some Theorems

Calculus Some Theorems September 28, 2018

Definitions

(a) A set S of real numbers is called bounded from above if there is a real number k such that k ≥ s for all s in S.

The element k is called an upper bound of S.

(b) A set S of real numbers is called bounded from below if there is a real number l such that s ≥ l for all s in S.

The element l is called a lower bound of S.

(c) A set S of real numbers is called bounded if there is a real number m such that

|s| ≤ m for all s in S.

Definitions

(a) b is called a least upper bound of S, denoted by b = l.u.b. S, if (i) b is an upper bound of S, i.e. b ≥ s for all s in S.

(ii) if k is another upper bound of S then k ≥ b, i.e. ∀  > 0, ∃ s ∈ S such that s > b − .

(b) l is called a greatest lower bound of S, denoted by l = g.l.b. S, if (i) l is a lower bound of S, i.e. s ≥ l for all s in S.

(ii) if m is another lower bound of S then m ≤ l, i.e. ∀  > 0, ∃ s ∈ S such that s < l + .

(c) Completeness Axiom of R. Every nonempty, bounded from above set of real numbers has a least upper bound (or supremum) in the set of real numbers. Equivalently, every nonempty, bounded from below set of real numbers has a greatest lower bound (or infimum) in the set of real numbers.

Remarks

(a) Let S be a nonempty, bounded from above subset of R and let b = l.u.b. S. By taking

_n = 1

n, for n = 1, 2, . . . ,

there exists xn∈ S such that b − 1

n = b − n < xn ≤ b for all n ∈ N.

This implies that

n→∞lim xn = l.u.b. S by the squeeze theorem.

(b) Let S be a nonempty, bounded from below subset of R and let l = g.l.b. S. By taking n = 1 n, for n = 1, 2, . . . ,

there exists y_n∈ S such that l ≤ y_n< l + _n = l + 1

n for all n ∈ N.

This imples that

n→∞lim y_n= g.l.b. S by the squeeze theorem.

Calculus Some Theorems (Continued) September 28, 2018 (c) Note that the existence of l.u.b. S does not guarantee it belongs to S. For example, if

S = {x ∈ Q | x² < 2} then

l.u.b. S = √

2 /∈ S and g.l.b. S = −√ 2 /∈ S.

Lemma Let f be continuous on [a, b]. If

either f (a) < 0 < f (b), or f (b) < 0 < f (a) then there exists c ∈ (a, b) such that f (c) = 0.

Proof. Suppose that f (a) < 0 < f (b).

The continuity of f at x = a and |f (a)| > 0 implies that there is a δ > 0 such that f (x) < f (a)

2 < 0 for all x ∈ [a, a + δ).

This implies that the set S defined by

S = {ξ | f (x) < 0, for all x ∈ [a, ξ)}

is a nonempty subset of R and it is bounded from above by b since f (b) > 0, b /∈ S. By the completeness of R, there exists a real number a < c < b such that

c = l.u.b. {ξ | f (x) < 0, for all x ∈ [a, ξ)}.

If f (c) > 0 =⇒ ∃ δ > 0 such that f (x) > 0 ∀ x ∈ (c − δ, c + δ) by the continuity of f at c

=⇒ f (c − δ 2) > 0

=⇒ c = l.u.b. {ξ | f (x) < 0, for all x ∈ [a, ξ)} ≤ c − δ

2 which is a contradiction.

If f (c) < 0 =⇒ ∃ δ > 0 such that f (x) < 0 ∀ x ∈ (c − δ, c + δ) by the continuity of f at c

=⇒ f (x) < 0 ∀ x ∈ [a, c + δ)

=⇒ c = l.u.b. {ξ | f (x) < 0, for all x ∈ [a, ξ)} ≥ c + δ which is a contradiction.

Hence, f (c) = 0.

The Intermediate Value Theorem Suppose that f is continuous on the closed interval [a, b]

and let N be any number between f (a) and f (b), where f (a) 6= f (b). Then there exists c ∈ (a, b) such that f (c) = N.

Proof. Suppose that f (a) < N < f (b).

The function g defined by

g(x) = f (x) − N is continuous on [a, b]. Since

g(a) = f (a) − N < 0 and g(b) = f (b) − N > 0 there is a c ∈ (a, b) such that g(c) = 0, i.e. f (c) = N.

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Calculus Some Theorems (Continued) September 28, 2018

Lemma If f is continuous on [a, b], then f is bounded on [a, b].

Proof. Consider the set A defined by

A = {x | x ∈ [a, b] and f is bounded on [a, x] } ⊂ R

Then A is bounded above by b and, since f is continuous at a, f is bounded on [a, a], we have a ∈ A and A 6= ∅. By the completeness of R, there exists c ≤ b such that

c = l.u.b. {x | x ∈ [a, b] and f is bounded on [a, x] }.

If c < b =⇒ ∃ δ > 0 such that |f (x) − f (c)| < 1 ∀ x ∈ (c − δ, c + δ) by the continuity of f at c

=⇒ |f (x)| < |f (c)| + 1 ∀ x ∈ (c − δ, c + δ) i.e. f is bounded on (c − δ, c + δ)

=⇒ f is bounded on [a, c +δ

2] = [a, c − δ

2] ∪ [c − δ 2, c + δ

2]

=⇒ c = l.u.b. {x | x ∈ [a, b] and f is bounded on [a, x] } ≥ c +δ

2 which is a contradiction.

Hence, c = b, i.e. f is bounded on [a, b].

Extreme Value Theorem Let f be continuous on [a, b]. Then there exist x₁, x₂ ∈ [a, b] such that

f (x₁) = max{f (x) | x ∈ [a, b]}

f (x₂) = min{f (x) | x ∈ [a, b]}

Proof. Since f is continuous on [a, b], f is bounded on [a, b].

Set M = l.u.b. {f (x) | x ∈ [a, b]}.

Suppose that there does not exist any x ∈ [a, b] such that f (x) = M, the function g defined by

g(x) = 1

M − f (x) is continuous on [a, b]

and g(x) is bounded on [a, b]. This is a contradiction since there exists a (converging) sequence of points {z_n} ⊂ [a, b] such that

n→∞lim f (z_n) = M ⇔ lim

n→∞|g(z_n)| = ∞.

Hence, there exists x₁ ∈ [a, b] such that f (x₁) = max{f (x) | x ∈ [a, b]}.

By setting

h(x) = −f (x) ∀ x ∈ [a, b], there exists x₂ ∈ [a, b] such that

−f (x₂) = h(x₂) = max{h(x) | x ∈ [a, b]} = max{−f (x) | x ∈ [a, b]} = − min{f (x) | x ∈ [a, b]}.

This implies that

f (x₂) = min{f (x) | x ∈ [a, b]}.

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