Advanced Calculus (I)
W EN -C HING L IEN
Department of Mathematics
National Cheng Kung University
W EN -C HING L IEN
Advanced Calculus (I)
4.4 Monotone Function and The Inverse
Function Theorem
Definition
Let E be a nonempty subset of R and f : E → R.
(i)
f is said to be increasing (respectively, strictly increasing)
on E if and only if x1 , x2 ∈ E and x1 < x2 imply
f (x1 ) ≤ f (x2 ) (respectively, f (x1 ) < f (x2 )).
(ii)
f is said to be dereasing (respectively, strictly decreasing)
on E if and only if x1 , x2 ∈ E and x1 < x2 imply
f (x1 ) ≥ f (x2 ) (respectively, f (x1 ) > f (x2 )).
W EN -C HING L IEN
Advanced Calculus (I)
4.4 Monotone Function and The Inverse
Function Theorem
Definition
Let E be a nonempty subset of R and f : E → R.
(i)
f is said to be increasing (respectively, strictly increasing)
on E if and only if x1 , x2 ∈ E and x1 < x2 imply
f (x1 ) ≤ f (x2 ) (respectively, f (x1 ) < f (x2 )).
(ii)
f is said to be dereasing (respectively, strictly decreasing)
on E if and only if x1 , x2 ∈ E and x1 < x2 imply
f (x1 ) ≥ f (x2 ) (respectively, f (x1 ) > f (x2 )).
W EN -C HING L IEN
Advanced Calculus (I)
4.4 Monotone Function and The Inverse
Function Theorem
Definition
Let E be a nonempty subset of R and f : E → R.
(i)
f is said to be increasing (respectively, strictly increasing)
on E if and only if x1 , x2 ∈ E and x1 < x2 imply
f (x1 ) ≤ f (x2 ) (respectively, f (x1 ) < f (x2 )).
(ii)
f is said to be dereasing (respectively, strictly decreasing)
on E if and only if x1 , x2 ∈ E and x1 < x2 imply
f (x1 ) ≥ f (x2 ) (respectively, f (x1 ) > f (x2 )).
W EN -C HING L IEN
Advanced Calculus (I)
4.4 Monotone Function and The Inverse
Function Theorem
Definition
Let E be a nonempty subset of R and f : E → R.
(i)
f is said to be increasing (respectively, strictly increasing)
on E if and only if x1 , x2 ∈ E and x1 < x2 imply
f (x1 ) ≤ f (x2 ) (respectively, f (x1 ) < f (x2 )).
(ii)
f is said to be dereasing (respectively, strictly decreasing)
on E if and only if x1 , x2 ∈ E and x1 < x2 imply
f (x1 ) ≥ f (x2 ) (respectively, f (x1 ) > f (x2 )).
W EN -C HING L IEN
Advanced Calculus (I)
4.4 Monotone Function and The Inverse
Function Theorem
Definition
Let E be a nonempty subset of R and f : E → R.
(i)
f is said to be increasing (respectively, strictly increasing)
on E if and only if x1 , x2 ∈ E and x1 < x2 imply
f (x1 ) ≤ f (x2 ) (respectively, f (x1 ) < f (x2 )).
(ii)
f is said to be dereasing (respectively, strictly decreasing)
on E if and only if x1 , x2 ∈ E and x1 < x2 imply
f (x1 ) ≥ f (x2 ) (respectively, f (x1 ) > f (x2 )).
W EN -C HING L IEN
Advanced Calculus (I)
4.4 Monotone Function and The Inverse
Function Theorem
Definition
Let E be a nonempty subset of R and f : E → R.
(i)
f is said to be increasing (respectively, strictly increasing)
on E if and only if x1 , x2 ∈ E and x1 < x2 imply
f (x1 ) ≤ f (x2 ) (respectively, f (x1 ) < f (x2 )).
(ii)
f is said to be dereasing (respectively, strictly decreasing)
on E if and only if x1 , x2 ∈ E and x1 < x2 imply
f (x1 ) ≥ f (x2 ) (respectively, f (x1 ) > f (x2 )).
W EN -C HING L IEN
Advanced Calculus (I)
Definition
(iii)
f is said to be monotone (respectively, strictly monotone)
on E if and only if f is either decreasing or increasing
(respectively, either strictly decreasing or strictly
increasing) on E.
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Advanced Calculus (I)
Definition
(iii)
f is said to be monotone (respectively, strictly monotone)
on E if and only if f is either decreasing or increasing
(respectively, either strictly decreasing or strictly
increasing) on E.
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Advanced Calculus (I)
Definition
(iii)
f is said to be monotone (respectively, strictly monotone)
on E if and only if f is either decreasing or increasing
(respectively, either strictly decreasing or strictly
increasing) on E.
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Advanced Calculus (I)
Theorem
Suppose that a, b ∈ R, with a 6= b, that f is continuous on
[a,b], and that f is differentiable on (a,b).
(i)
If f 0 (x) > 0 (respectively, f’(x)¡0) for all x ∈ (a, b), then f is
strictly increasing (respectively, strictly decreasing) on
[a,b].
(ii)
If f 0 (x) = 0 for all x ∈ (a, b), then f is constant on [a,b].
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Advanced Calculus (I)
Theorem
Suppose that a, b ∈ R, with a 6= b, that f is continuous on
[a,b], and that f is differentiable on (a,b).
(i)
If f 0 (x) > 0 (respectively, f’(x)¡0) for all x ∈ (a, b), then f is
strictly increasing (respectively, strictly decreasing) on
[a,b].
(ii)
If f 0 (x) = 0 for all x ∈ (a, b), then f is constant on [a,b].
W EN -C HING L IEN
Advanced Calculus (I)
Theorem
Suppose that a, b ∈ R, with a 6= b, that f is continuous on
[a,b], and that f is differentiable on (a,b).
(i)
If f 0 (x) > 0 (respectively, f’(x)¡0) for all x ∈ (a, b), then f is
strictly increasing (respectively, strictly decreasing) on
[a,b].
(ii)
If f 0 (x) = 0 for all x ∈ (a, b), then f is constant on [a,b].
W EN -C HING L IEN
Advanced Calculus (I)
Theorem
Suppose that a, b ∈ R, with a 6= b, that f is continuous on
[a,b], and that f is differentiable on (a,b).
(i)
If f 0 (x) > 0 (respectively, f’(x)¡0) for all x ∈ (a, b), then f is
strictly increasing (respectively, strictly decreasing) on
[a,b].
(ii)
If f 0 (x) = 0 for all x ∈ (a, b), then f is constant on [a,b].
W EN -C HING L IEN
Advanced Calculus (I)
Theorem
Suppose that a, b ∈ R, with a 6= b, that f is continuous on
[a,b], and that f is differentiable on (a,b).
(i)
If f 0 (x) > 0 (respectively, f’(x)¡0) for all x ∈ (a, b), then f is
strictly increasing (respectively, strictly decreasing) on
[a,b].
(ii)
If f 0 (x) = 0 for all x ∈ (a, b), then f is constant on [a,b].
W EN -C HING L IEN
Advanced Calculus (I)
Theorem
Suppose that a, b ∈ R, with a 6= b, that f is continuous on
[a,b], and that f is differentiable on (a,b).
(i)
If f 0 (x) > 0 (respectively, f’(x)¡0) for all x ∈ (a, b), then f is
strictly increasing (respectively, strictly decreasing) on
[a,b].
(ii)
If f 0 (x) = 0 for all x ∈ (a, b), then f is constant on [a,b].
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Let a ≤ x1 < x2 ≤ b. By the Mean Value Theorem, there
is a c ∈ (a, b) such that f (x2 ) − f (x1 ) = f 0 (c)(x2 − x1 ).
Thus, f (x2 ) > f (x1 ) when f 0 (c) > 0 and f (x2 ) < f (x1 ) when
f 0 (c) < 0. This proves part (i).
To prove part (ii), let a ≤ x ≤ b. By the Mean Value
Theorem and hypothesis there is a c ∈ (a, b) such that
f (x) − f (a) = f 0 (c)(x − a) = 0.
Thus, f (x) = f (a) for all x ∈ [a, b]. 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Let a ≤ x1 < x2 ≤ b. By the Mean Value Theorem, there
is a c ∈ (a, b) such that f (x2 ) − f (x1 ) = f 0 (c)(x2 − x1 ).
Thus, f (x2 ) > f (x1 ) when f 0 (c) > 0 and f (x2 ) < f (x1 ) when
f 0 (c) < 0. This proves part (i).
To prove part (ii), let a ≤ x ≤ b. By the Mean Value
Theorem and hypothesis there is a c ∈ (a, b) such that
f (x) − f (a) = f 0 (c)(x − a) = 0.
Thus, f (x) = f (a) for all x ∈ [a, b]. 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Let a ≤ x1 < x2 ≤ b. By the Mean Value Theorem, there
is a c ∈ (a, b) such that f (x2 ) − f (x1 ) = f 0 (c)(x2 − x1 ).
Thus, f (x2 ) > f (x1 ) when f 0 (c) > 0 and f (x2 ) < f (x1 ) when
f 0 (c) < 0. This proves part (i).
To prove part (ii), let a ≤ x ≤ b. By the Mean Value
Theorem and hypothesis there is a c ∈ (a, b) such that
f (x) − f (a) = f 0 (c)(x − a) = 0.
Thus, f (x) = f (a) for all x ∈ [a, b]. 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Let a ≤ x1 < x2 ≤ b. By the Mean Value Theorem, there
is a c ∈ (a, b) such that f (x2 ) − f (x1 ) = f 0 (c)(x2 − x1 ).
Thus, f (x2 ) > f (x1 ) when f 0 (c) > 0 and f (x2 ) < f (x1 ) when
f 0 (c) < 0. This proves part (i).
To prove part (ii), let a ≤ x ≤ b. By the Mean Value
Theorem and hypothesis there is a c ∈ (a, b) such that
f (x) − f (a) = f 0 (c)(x − a) = 0.
Thus, f (x) = f (a) for all x ∈ [a, b]. 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Let a ≤ x1 < x2 ≤ b. By the Mean Value Theorem, there
is a c ∈ (a, b) such that f (x2 ) − f (x1 ) = f 0 (c)(x2 − x1 ).
Thus, f (x2 ) > f (x1 ) when f 0 (c) > 0 and f (x2 ) < f (x1 ) when
f 0 (c) < 0. This proves part (i).
To prove part (ii), let a ≤ x ≤ b. By the Mean Value
Theorem and hypothesis there is a c ∈ (a, b) such that
f (x) − f (a) = f 0 (c)(x − a) = 0.
Thus, f (x) = f (a) for all x ∈ [a, b]. 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Let a ≤ x1 < x2 ≤ b. By the Mean Value Theorem, there
is a c ∈ (a, b) such that f (x2 ) − f (x1 ) = f 0 (c)(x2 − x1 ).
Thus, f (x2 ) > f (x1 ) when f 0 (c) > 0 and f (x2 ) < f (x1 ) when
f 0 (c) < 0. This proves part (i).
To prove part (ii), let a ≤ x ≤ b. By the Mean Value
Theorem and hypothesis there is a c ∈ (a, b) such that
f (x) − f (a) = f 0 (c)(x − a) = 0.
Thus, f (x) = f (a) for all x ∈ [a, b]. 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Let a ≤ x1 < x2 ≤ b. By the Mean Value Theorem, there
is a c ∈ (a, b) such that f (x2 ) − f (x1 ) = f 0 (c)(x2 − x1 ).
Thus, f (x2 ) > f (x1 ) when f 0 (c) > 0 and f (x2 ) < f (x1 ) when
f 0 (c) < 0. This proves part (i).
To prove part (ii), let a ≤ x ≤ b. By the Mean Value
Theorem and hypothesis there is a c ∈ (a, b) such that
f (x) − f (a) = f 0 (c)(x − a) = 0.
Thus, f (x) = f (a) for all x ∈ [a, b]. 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Let a ≤ x1 < x2 ≤ b. By the Mean Value Theorem, there
is a c ∈ (a, b) such that f (x2 ) − f (x1 ) = f 0 (c)(x2 − x1 ).
Thus, f (x2 ) > f (x1 ) when f 0 (c) > 0 and f (x2 ) < f (x1 ) when
f 0 (c) < 0. This proves part (i).
To prove part (ii), let a ≤ x ≤ b. By the Mean Value
Theorem and hypothesis there is a c ∈ (a, b) such that
f (x) − f (a) = f 0 (c)(x − a) = 0.
Thus, f (x) = f (a) for all x ∈ [a, b]. 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Let a ≤ x1 < x2 ≤ b. By the Mean Value Theorem, there
is a c ∈ (a, b) such that f (x2 ) − f (x1 ) = f 0 (c)(x2 − x1 ).
Thus, f (x2 ) > f (x1 ) when f 0 (c) > 0 and f (x2 ) < f (x1 ) when
f 0 (c) < 0. This proves part (i).
To prove part (ii), let a ≤ x ≤ b. By the Mean Value
Theorem and hypothesis there is a c ∈ (a, b) such that
f (x) − f (a) = f 0 (c)(x − a) = 0.
Thus, f (x) = f (a) for all x ∈ [a, b]. 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Let a ≤ x1 < x2 ≤ b. By the Mean Value Theorem, there
is a c ∈ (a, b) such that f (x2 ) − f (x1 ) = f 0 (c)(x2 − x1 ).
Thus, f (x2 ) > f (x1 ) when f 0 (c) > 0 and f (x2 ) < f (x1 ) when
f 0 (c) < 0. This proves part (i).
To prove part (ii), let a ≤ x ≤ b. By the Mean Value
Theorem and hypothesis there is a c ∈ (a, b) such that
f (x) − f (a) = f 0 (c)(x − a) = 0.
Thus, f (x) = f (a) for all x ∈ [a, b]. 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Let a ≤ x1 < x2 ≤ b. By the Mean Value Theorem, there
is a c ∈ (a, b) such that f (x2 ) − f (x1 ) = f 0 (c)(x2 − x1 ).
Thus, f (x2 ) > f (x1 ) when f 0 (c) > 0 and f (x2 ) < f (x1 ) when
f 0 (c) < 0. This proves part (i).
To prove part (ii), let a ≤ x ≤ b. By the Mean Value
Theorem and hypothesis there is a c ∈ (a, b) such that
f (x) − f (a) = f 0 (c)(x − a) = 0.
Thus, f (x) = f (a) for all x ∈ [a, b]. 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Let a ≤ x1 < x2 ≤ b. By the Mean Value Theorem, there
is a c ∈ (a, b) such that f (x2 ) − f (x1 ) = f 0 (c)(x2 − x1 ).
Thus, f (x2 ) > f (x1 ) when f 0 (c) > 0 and f (x2 ) < f (x1 ) when
f 0 (c) < 0. This proves part (i).
To prove part (ii), let a ≤ x ≤ b. By the Mean Value
Theorem and hypothesis there is a c ∈ (a, b) such that
f (x) − f (a) = f 0 (c)(x − a) = 0.
Thus, f (x) = f (a) for all x ∈ [a, b]. 2
W EN -C HING L IEN
Advanced Calculus (I)
Theorem
If f is 1-1 and continuous on an interval I, then f is strictly
monotone on I and f −1 is continuous and strictly
monotone on f(I).
W EN -C HING L IEN
Advanced Calculus (I)
Theorem
If f is 1-1 and continuous on an interval I, then f is strictly
monotone on I and f −1 is continuous and strictly
monotone on f(I).
W EN -C HING L IEN
Advanced Calculus (I)
Theorem (Inverse Function Theorem)
Let f be 1-1 and continuous on an open interval I. If
a ∈ f (I) and if f 0 (f −1 (a)) exists and is nonzero, then f −1 is
differentiable at a and
(f −1 )0 (a) =
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1
f 0 (f −1 (a))
.
Advanced Calculus (I)
Theorem (Inverse Function Theorem)
Let f be 1-1 and continuous on an open interval I. If
a ∈ f (I) and if f 0 (f −1 (a)) exists and is nonzero, then f −1 is
differentiable at a and
(f −1 )0 (a) =
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1
f 0 (f −1 (a))
.
Advanced Calculus (I)
Lemma:
Suppose that f is increasing on [a,b].
(i) If x0 ∈ [a, b), then f (x0 +) exists and f (x0 ) ≤ f (x0 +).
(ii) If x0 ∈ (a, b], then f (x0 −) exists and f (x0 −) ≤ f (x0 ).
W EN -C HING L IEN
Advanced Calculus (I)
Lemma:
Suppose that f is increasing on [a,b].
(i) If x0 ∈ [a, b), then f (x0 +) exists and f (x0 ) ≤ f (x0 +).
(ii) If x0 ∈ (a, b], then f (x0 −) exists and f (x0 −) ≤ f (x0 ).
W EN -C HING L IEN
Advanced Calculus (I)
Lemma:
Suppose that f is increasing on [a,b].
(i) If x0 ∈ [a, b), then f (x0 +) exists and f (x0 ) ≤ f (x0 +).
(ii) If x0 ∈ (a, b], then f (x0 −) exists and f (x0 −) ≤ f (x0 ).
W EN -C HING L IEN
Advanced Calculus (I)
Lemma:
Suppose that f is increasing on [a,b].
(i) If x0 ∈ [a, b), then f (x0 +) exists and f (x0 ) ≤ f (x0 +).
(ii) If x0 ∈ (a, b], then f (x0 −) exists and f (x0 −) ≤ f (x0 ).
W EN -C HING L IEN
Advanced Calculus (I)
Lemma:
Suppose that f is increasing on [a,b].
(i) If x0 ∈ [a, b), then f (x0 +) exists and f (x0 ) ≤ f (x0 +).
(ii) If x0 ∈ (a, b], then f (x0 −) exists and f (x0 −) ≤ f (x0 ).
W EN -C HING L IEN
Advanced Calculus (I)
Lemma:
Suppose that f is increasing on [a,b].
(i) If x0 ∈ [a, b), then f (x0 +) exists and f (x0 ) ≤ f (x0 +).
(ii) If x0 ∈ (a, b], then f (x0 −) exists and f (x0 −) ≤ f (x0 ).
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Fix x0 ∈ (a, b]. By symmetry it suffices to show that f (x0 −)
exists and satisfies f (x0 −) ≤ f (x0 ). Set
E = {f (x) : a < x < x0 } and s = sup E. Since f is
increasing, f (x0 ) is an upper bound of E. Hence, s is a
finite real number that satisfies s ≤ f (x0 ). Given > 0,
choose by the Approximation Property an x1 ∈ (a, x0 )
such that s − < f (x1 ) ≤ s. Since f is increasing,
s − < f (x1 ) ≤ f (x) ≤ s
for all x1 < x < x0 . Therefore, f (x0 −) exists and satisfies
f (x0 −) = s ≤ f (x0 ). 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Fix x0 ∈ (a, b]. By symmetry it suffices to show that f (x0 −)
exists and satisfies f (x0 −) ≤ f (x0 ). Set
E = {f (x) : a < x < x0 } and s = sup E. Since f is
increasing, f (x0 ) is an upper bound of E. Hence, s is a
finite real number that satisfies s ≤ f (x0 ). Given > 0,
choose by the Approximation Property an x1 ∈ (a, x0 )
such that s − < f (x1 ) ≤ s. Since f is increasing,
s − < f (x1 ) ≤ f (x) ≤ s
for all x1 < x < x0 . Therefore, f (x0 −) exists and satisfies
f (x0 −) = s ≤ f (x0 ). 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Fix x0 ∈ (a, b]. By symmetry it suffices to show that f (x0 −)
exists and satisfies f (x0 −) ≤ f (x0 ). Set
E = {f (x) : a < x < x0 } and s = sup E. Since f is
increasing, f (x0 ) is an upper bound of E. Hence, s is a
finite real number that satisfies s ≤ f (x0 ). Given > 0,
choose by the Approximation Property an x1 ∈ (a, x0 )
such that s − < f (x1 ) ≤ s. Since f is increasing,
s − < f (x1 ) ≤ f (x) ≤ s
for all x1 < x < x0 . Therefore, f (x0 −) exists and satisfies
f (x0 −) = s ≤ f (x0 ). 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Fix x0 ∈ (a, b]. By symmetry it suffices to show that f (x0 −)
exists and satisfies f (x0 −) ≤ f (x0 ). Set
E = {f (x) : a < x < x0 } and s = sup E. Since f is
increasing, f (x0 ) is an upper bound of E. Hence, s is a
finite real number that satisfies s ≤ f (x0 ). Given > 0,
choose by the Approximation Property an x1 ∈ (a, x0 )
such that s − < f (x1 ) ≤ s. Since f is increasing,
s − < f (x1 ) ≤ f (x) ≤ s
for all x1 < x < x0 . Therefore, f (x0 −) exists and satisfies
f (x0 −) = s ≤ f (x0 ). 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Fix x0 ∈ (a, b]. By symmetry it suffices to show that f (x0 −)
exists and satisfies f (x0 −) ≤ f (x0 ). Set
E = {f (x) : a < x < x0 } and s = sup E. Since f is
increasing, f (x0 ) is an upper bound of E. Hence, s is a
finite real number that satisfies s ≤ f (x0 ). Given > 0,
choose by the Approximation Property an x1 ∈ (a, x0 )
such that s − < f (x1 ) ≤ s. Since f is increasing,
s − < f (x1 ) ≤ f (x) ≤ s
for all x1 < x < x0 . Therefore, f (x0 −) exists and satisfies
f (x0 −) = s ≤ f (x0 ). 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Fix x0 ∈ (a, b]. By symmetry it suffices to show that f (x0 −)
exists and satisfies f (x0 −) ≤ f (x0 ). Set
E = {f (x) : a < x < x0 } and s = sup E. Since f is
increasing, f (x0 ) is an upper bound of E. Hence, s is a
finite real number that satisfies s ≤ f (x0 ). Given > 0,
choose by the Approximation Property an x1 ∈ (a, x0 )
such that s − < f (x1 ) ≤ s. Since f is increasing,
s − < f (x1 ) ≤ f (x) ≤ s
for all x1 < x < x0 . Therefore, f (x0 −) exists and satisfies
f (x0 −) = s ≤ f (x0 ). 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Fix x0 ∈ (a, b]. By symmetry it suffices to show that f (x0 −)
exists and satisfies f (x0 −) ≤ f (x0 ). Set
E = {f (x) : a < x < x0 } and s = sup E. Since f is
increasing, f (x0 ) is an upper bound of E. Hence, s is a
finite real number that satisfies s ≤ f (x0 ). Given > 0,
choose by the Approximation Property an x1 ∈ (a, x0 )
such that s − < f (x1 ) ≤ s. Since f is increasing,
s − < f (x1 ) ≤ f (x) ≤ s
for all x1 < x < x0 . Therefore, f (x0 −) exists and satisfies
f (x0 −) = s ≤ f (x0 ). 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Fix x0 ∈ (a, b]. By symmetry it suffices to show that f (x0 −)
exists and satisfies f (x0 −) ≤ f (x0 ). Set
E = {f (x) : a < x < x0 } and s = sup E. Since f is
increasing, f (x0 ) is an upper bound of E. Hence, s is a
finite real number that satisfies s ≤ f (x0 ). Given > 0,
choose by the Approximation Property an x1 ∈ (a, x0 )
such that s − < f (x1 ) ≤ s. Since f is increasing,
s − < f (x1 ) ≤ f (x) ≤ s
for all x1 < x < x0 . Therefore, f (x0 −) exists and satisfies
f (x0 −) = s ≤ f (x0 ). 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Fix x0 ∈ (a, b]. By symmetry it suffices to show that f (x0 −)
exists and satisfies f (x0 −) ≤ f (x0 ). Set
E = {f (x) : a < x < x0 } and s = sup E. Since f is
increasing, f (x0 ) is an upper bound of E. Hence, s is a
finite real number that satisfies s ≤ f (x0 ). Given > 0,
choose by the Approximation Property an x1 ∈ (a, x0 )
such that s − < f (x1 ) ≤ s. Since f is increasing,
s − < f (x1 ) ≤ f (x) ≤ s
for all x1 < x < x0 . Therefore, f (x0 −) exists and satisfies
f (x0 −) = s ≤ f (x0 ). 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Fix x0 ∈ (a, b]. By symmetry it suffices to show that f (x0 −)
exists and satisfies f (x0 −) ≤ f (x0 ). Set
E = {f (x) : a < x < x0 } and s = sup E. Since f is
increasing, f (x0 ) is an upper bound of E. Hence, s is a
finite real number that satisfies s ≤ f (x0 ). Given > 0,
choose by the Approximation Property an x1 ∈ (a, x0 )
such that s − < f (x1 ) ≤ s. Since f is increasing,
s − < f (x1 ) ≤ f (x) ≤ s
for all x1 < x < x0 . Therefore, f (x0 −) exists and satisfies
f (x0 −) = s ≤ f (x0 ). 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Fix x0 ∈ (a, b]. By symmetry it suffices to show that f (x0 −)
exists and satisfies f (x0 −) ≤ f (x0 ). Set
E = {f (x) : a < x < x0 } and s = sup E. Since f is
increasing, f (x0 ) is an upper bound of E. Hence, s is a
finite real number that satisfies s ≤ f (x0 ). Given > 0,
choose by the Approximation Property an x1 ∈ (a, x0 )
such that s − < f (x1 ) ≤ s. Since f is increasing,
s − < f (x1 ) ≤ f (x) ≤ s
for all x1 < x < x0 . Therefore, f (x0 −) exists and satisfies
f (x0 −) = s ≤ f (x0 ). 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Fix x0 ∈ (a, b]. By symmetry it suffices to show that f (x0 −)
exists and satisfies f (x0 −) ≤ f (x0 ). Set
E = {f (x) : a < x < x0 } and s = sup E. Since f is
increasing, f (x0 ) is an upper bound of E. Hence, s is a
finite real number that satisfies s ≤ f (x0 ). Given > 0,
choose by the Approximation Property an x1 ∈ (a, x0 )
such that s − < f (x1 ) ≤ s. Since f is increasing,
s − < f (x1 ) ≤ f (x) ≤ s
for all x1 < x < x0 . Therefore, f (x0 −) exists and satisfies
f (x0 −) = s ≤ f (x0 ). 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Fix x0 ∈ (a, b]. By symmetry it suffices to show that f (x0 −)
exists and satisfies f (x0 −) ≤ f (x0 ). Set
E = {f (x) : a < x < x0 } and s = sup E. Since f is
increasing, f (x0 ) is an upper bound of E. Hence, s is a
finite real number that satisfies s ≤ f (x0 ). Given > 0,
choose by the Approximation Property an x1 ∈ (a, x0 )
such that s − < f (x1 ) ≤ s. Since f is increasing,
s − < f (x1 ) ≤ f (x) ≤ s
for all x1 < x < x0 . Therefore, f (x0 −) exists and satisfies
f (x0 −) = s ≤ f (x0 ). 2
W EN -C HING L IEN
Advanced Calculus (I)
Theorem
If f is monotone on an open interval I, then f has at most
countably many points of discontinuity on I.
W EN -C HING L IEN
Advanced Calculus (I)
Theorem
If f is monotone on an open interval I, then f has at most
countably many points of discontinuity on I.
W EN -C HING L IEN
Advanced Calculus (I)
Theorem (Intermediate Value Theorem For Derivatives)
Suppose that f is differentiable on [a,b] with f 0 (a) 6= f 0 (b).
If y0 is a real number that lies between f 0 (a) and f 0 (b),
then there is an x0 ∈ (a, b) such that f 0 (x0 ) = y0 .
W EN -C HING L IEN
Advanced Calculus (I)
Theorem (Intermediate Value Theorem For Derivatives)
Suppose that f is differentiable on [a,b] with f 0 (a) 6= f 0 (b).
If y0 is a real number that lies between f 0 (a) and f 0 (b),
then there is an x0 ∈ (a, b) such that f 0 (x0 ) = y0 .
W EN -C HING L IEN
Advanced Calculus (I)
Thank you.
W EN -C HING L IEN
Advanced Calculus (I)