Advanced Calculus (I)
W EN -C HING L IEN
Department of Mathematics
National Cheng Kung University
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Advanced Calculus (I)
3.3 Continuity
Definition
Let E be a nonempty subset of R and f : E → R.
(i)
f is said to be continuous at a point a ∈ E if and only if
given ǫ > 0 there is a δ > 0 (which in general depends on
ǫ, f, and a) such that
|x − a| < δ and x ∈ E imply |f (x) − f (a)| < ǫ.
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Advanced Calculus (I)
3.3 Continuity
Definition
Let E be a nonempty subset of R and f : E → R.
(i)
f is said to be continuous at a point a ∈ E if and only if
given ǫ > 0 there is a δ > 0 (which in general depends on
ǫ, f, and a) such that
|x − a| < δ and x ∈ E imply |f (x) − f (a)| < ǫ.
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Advanced Calculus (I)
3.3 Continuity
Definition
Let E be a nonempty subset of R and f : E → R.
(i)
f is said to be continuous at a point a ∈ E if and only if
given ǫ > 0 there is a δ > 0 (which in general depends on
ǫ, f, and a) such that
|x − a| < δ and x ∈ E imply |f (x) − f (a)| < ǫ.
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Advanced Calculus (I)
3.3 Continuity
Definition
Let E be a nonempty subset of R and f : E → R.
(i)
f is said to be continuous at a point a ∈ E if and only if
given ǫ > 0 there is a δ > 0 (which in general depends on
ǫ, f, and a) such that
|x − a| < δ and x ∈ E imply |f (x) − f (a)| < ǫ.
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Advanced Calculus (I)
Definition
(ii)
f is said to be continuous on E (notation: f : E → R is
continuous) if and only if f is continuous at every x ∈ E.
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Advanced Calculus (I)
Definition
(ii)
f is said to be continuous on E (notation: f : E → R is
continuous) if and only if f is continuous at every x ∈ E.
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Advanced Calculus (I)
Definition
(ii)
f is said to be continuous on E (notation: f : E → R is
continuous) if and only if f is continuous at every x ∈ E.
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Advanced Calculus (I)
Remark:
Let I be an open interval that contains a point a and
f : I → R. Then f is continuous at a ∈ I if and only if
f (a) = lim f (x).
x→a
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Advanced Calculus (I)
Remark:
Let I be an open interval that contains a point a and
f : I → R. Then f is continuous at a ∈ I if and only if
f (a) = lim f (x).
x→a
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Advanced Calculus (I)
Example:
1. f (x) = x 2 , x ∈ [0, 1].
2. f (x) = x 2 , x ∈ R.
1
3. f (x) = , x ∈ (0, 1].
x
√
4. f (x) = x, x ∈ R + .
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Advanced Calculus (I)
Theorem
Suppose that E is a nonempty subset of R, a ∈ E, and
f : E → R. Then the following statements are equivalent:
(i) f is continuous at a ∈ E
(ii) If xn converges to a and xn ∈ E, then f (xn ) → f (a) as
n → ∞.
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Advanced Calculus (I)
Theorem
Suppose that E is a nonempty subset of R, a ∈ E, and
f : E → R. Then the following statements are equivalent:
(i) f is continuous at a ∈ E
(ii) If xn converges to a and xn ∈ E, then f (xn ) → f (a) as
n → ∞.
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Advanced Calculus (I)
Theorem
Suppose that E is a nonempty subset of R, a ∈ E, and
f : E → R. Then the following statements are equivalent:
(i) f is continuous at a ∈ E
(ii) If xn converges to a and xn ∈ E, then f (xn ) → f (a) as
n → ∞.
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Advanced Calculus (I)
Theorem
Suppose that E is a nonempty subset of R, a ∈ E, and
f : E → R. Then the following statements are equivalent:
(i) f is continuous at a ∈ E
(ii) If xn converges to a and xn ∈ E, then f (xn ) → f (a) as
n → ∞.
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Advanced Calculus (I)
Theorem
Suppose that A and B are subsets of R and that f : A → R
and g : B → R with f (A) ⊆ B.
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Advanced Calculus (I)
Theorem
Suppose that A and B are subsets of R and that f : A → R
and g : B → R with f (A) ⊆ B.
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Advanced Calculus (I)
Theorem
(i)
If A := I \ {a}, where I is a nondegenerate interval that
either contains a or has a as one of its endpoint if
L := lim f (x)
x→a
x∈I
exists and belongs to B, and if g is continuous at L ∈ B,
then
!
lim (g ◦ f )(x) = g
x→a
x∈I
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lim f (x) .
x→a
x∈I
Advanced Calculus (I)
Theorem
(i)
If A := I \ {a}, where I is a nondegenerate interval that
either contains a or has a as one of its endpoint if
L := lim f (x)
x→a
x∈I
exists and belongs to B, and if g is continuous at L ∈ B,
then
!
lim (g ◦ f )(x) = g
x→a
x∈I
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lim f (x) .
x→a
x∈I
Advanced Calculus (I)
Theorem
(ii)
If f is continuous at a ∈ A and g is continuous at f (a) ∈ B,
then g ◦ f is continuous at a ∈ A.
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Advanced Calculus (I)
Theorem
(ii)
If f is continuous at a ∈ A and g is continuous at f (a) ∈ B,
then g ◦ f is continuous at a ∈ A.
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Advanced Calculus (I)
Proof:
Suppose that xn ∈ I \ {a} and xn → a as n → ∞. Since
f (A) ⊆ B, f (xn ) ∈ B. Also, by the Sequential
Characterization of Limits (Theorem 3.17), f (xn ) → L as
n → ∞. Since g is continuous at L ∈ B, it follows from
Theorem 3.21 that g ◦ f (xn ) := g(f (xn )) → g(L) as n → ∞.
Hence by Theorem 3.17, g ◦ f (x) → g(L) as x → a in I.
This prove (i). A similar proof establishes part (ii). 2
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Advanced Calculus (I)
Proof:
Suppose that xn ∈ I \ {a} and xn → a as n → ∞. Since
f (A) ⊆ B, f (xn ) ∈ B. Also, by the Sequential
Characterization of Limits (Theorem 3.17), f (xn ) → L as
n → ∞. Since g is continuous at L ∈ B, it follows from
Theorem 3.21 that g ◦ f (xn ) := g(f (xn )) → g(L) as n → ∞.
Hence by Theorem 3.17, g ◦ f (x) → g(L) as x → a in I.
This prove (i). A similar proof establishes part (ii). 2
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Advanced Calculus (I)
Proof:
Suppose that xn ∈ I \ {a} and xn → a as n → ∞. Since
f (A) ⊆ B, f (xn ) ∈ B. Also, by the Sequential
Characterization of Limits (Theorem 3.17), f (xn ) → L as
n → ∞. Since g is continuous at L ∈ B, it follows from
Theorem 3.21 that g ◦ f (xn ) := g(f (xn )) → g(L) as n → ∞.
Hence by Theorem 3.17, g ◦ f (x) → g(L) as x → a in I.
This prove (i). A similar proof establishes part (ii). 2
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Advanced Calculus (I)
Proof:
Suppose that xn ∈ I \ {a} and xn → a as n → ∞. Since
f (A) ⊆ B, f (xn ) ∈ B. Also, by the Sequential
Characterization of Limits (Theorem 3.17), f (xn ) → L as
n → ∞. Since g is continuous at L ∈ B, it follows from
Theorem 3.21 that g ◦ f (xn ) := g(f (xn )) → g(L) as n → ∞.
Hence by Theorem 3.17, g ◦ f (x) → g(L) as x → a in I.
This prove (i). A similar proof establishes part (ii). 2
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Advanced Calculus (I)
Proof:
Suppose that xn ∈ I \ {a} and xn → a as n → ∞. Since
f (A) ⊆ B, f (xn ) ∈ B. Also, by the Sequential
Characterization of Limits (Theorem 3.17), f (xn ) → L as
n → ∞. Since g is continuous at L ∈ B, it follows from
Theorem 3.21 that g ◦ f (xn ) := g(f (xn )) → g(L) as n → ∞.
Hence by Theorem 3.17, g ◦ f (x) → g(L) as x → a in I.
This prove (i). A similar proof establishes part (ii). 2
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Advanced Calculus (I)
Proof:
Suppose that xn ∈ I \ {a} and xn → a as n → ∞. Since
f (A) ⊆ B, f (xn ) ∈ B. Also, by the Sequential
Characterization of Limits (Theorem 3.17), f (xn ) → L as
n → ∞. Since g is continuous at L ∈ B, it follows from
Theorem 3.21 that g ◦ f (xn ) := g(f (xn )) → g(L) as n → ∞.
Hence by Theorem 3.17, g ◦ f (x) → g(L) as x → a in I.
This prove (i). A similar proof establishes part (ii). 2
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Advanced Calculus (I)
Proof:
Suppose that xn ∈ I \ {a} and xn → a as n → ∞. Since
f (A) ⊆ B, f (xn ) ∈ B. Also, by the Sequential
Characterization of Limits (Theorem 3.17), f (xn ) → L as
n → ∞. Since g is continuous at L ∈ B, it follows from
Theorem 3.21 that g ◦ f (xn ) := g(f (xn )) → g(L) as n → ∞.
Hence by Theorem 3.17, g ◦ f (x) → g(L) as x → a in I.
This prove (i). A similar proof establishes part (ii). 2
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Advanced Calculus (I)
Proof:
Suppose that xn ∈ I \ {a} and xn → a as n → ∞. Since
f (A) ⊆ B, f (xn ) ∈ B. Also, by the Sequential
Characterization of Limits (Theorem 3.17), f (xn ) → L as
n → ∞. Since g is continuous at L ∈ B, it follows from
Theorem 3.21 that g ◦ f (xn ) := g(f (xn )) → g(L) as n → ∞.
Hence by Theorem 3.17, g ◦ f (x) → g(L) as x → a in I.
This prove (i). A similar proof establishes part (ii). 2
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Advanced Calculus (I)
Proof:
Suppose that xn ∈ I \ {a} and xn → a as n → ∞. Since
f (A) ⊆ B, f (xn ) ∈ B. Also, by the Sequential
Characterization of Limits (Theorem 3.17), f (xn ) → L as
n → ∞. Since g is continuous at L ∈ B, it follows from
Theorem 3.21 that g ◦ f (xn ) := g(f (xn )) → g(L) as n → ∞.
Hence by Theorem 3.17, g ◦ f (x) → g(L) as x → a in I.
This prove (i). A similar proof establishes part (ii). 2
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Advanced Calculus (I)
Proof:
Suppose that xn ∈ I \ {a} and xn → a as n → ∞. Since
f (A) ⊆ B, f (xn ) ∈ B. Also, by the Sequential
Characterization of Limits (Theorem 3.17), f (xn ) → L as
n → ∞. Since g is continuous at L ∈ B, it follows from
Theorem 3.21 that g ◦ f (xn ) := g(f (xn )) → g(L) as n → ∞.
Hence by Theorem 3.17, g ◦ f (x) → g(L) as x → a in I.
This prove (i). A similar proof establishes part (ii). 2
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Advanced Calculus (I)
Proof:
Suppose that xn ∈ I \ {a} and xn → a as n → ∞. Since
f (A) ⊆ B, f (xn ) ∈ B. Also, by the Sequential
Characterization of Limits (Theorem 3.17), f (xn ) → L as
n → ∞. Since g is continuous at L ∈ B, it follows from
Theorem 3.21 that g ◦ f (xn ) := g(f (xn )) → g(L) as n → ∞.
Hence by Theorem 3.17, g ◦ f (x) → g(L) as x → a in I.
This prove (i). A similar proof establishes part (ii). 2
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Advanced Calculus (I)
Proof:
Suppose that xn ∈ I \ {a} and xn → a as n → ∞. Since
f (A) ⊆ B, f (xn ) ∈ B. Also, by the Sequential
Characterization of Limits (Theorem 3.17), f (xn ) → L as
n → ∞. Since g is continuous at L ∈ B, it follows from
Theorem 3.21 that g ◦ f (xn ) := g(f (xn )) → g(L) as n → ∞.
Hence by Theorem 3.17, g ◦ f (x) → g(L) as x → a in I.
This prove (i). A similar proof establishes part (ii). 2
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Advanced Calculus (I)
Definition
Let E be a nonempty subset of R. A function f : E → R is
said to be bounded on E if and only if there is an M ∈ R
such that |f (x)| ≤ M for all x ∈ E.
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Advanced Calculus (I)
Definition
Let E be a nonempty subset of R. A function f : E → R is
said to be bounded on E if and only if there is an M ∈ R
such that |f (x)| ≤ M for all x ∈ E.
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Advanced Calculus (I)
Theorem (Extreme Value Theorem)
If I is a closed, bounded interval and f : I → R is
continuous on I, then f is bounded on I. Moreover, if
M = sup f (x) and m = inf f (x),
x∈I
x∈I
then there exist points xm , xM ∈ I such that
(6)
f (xM ) = M and f (xm ) = m.
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Advanced Calculus (I)
Theorem (Extreme Value Theorem)
If I is a closed, bounded interval and f : I → R is
continuous on I, then f is bounded on I. Moreover, if
M = sup f (x) and m = inf f (x),
x∈I
x∈I
then there exist points xm , xM ∈ I such that
(6)
f (xM ) = M and f (xm ) = m.
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Advanced Calculus (I)
Proof:
Suppose first that f is not bounded on I. Then there exist
xn ∈ I such that
(7)
|f (xn )| > n,
n ∈ N.
Since I is bounded, we know (by the Bolzano-Weierstrass
Theorem) that {xn } has a convergent subsequence, say
xnk → a as k → ∞. Since I is closed, we also know (by the
Comparison Theorem) that a ∈ I. In particular, f (a) ∈ R.
On the other hand, substituting nk for n in (7) and taking
the limit of this inequility as k → ∞, we have |f (a)| = ∞, a
contradiction. Hence, the function f is bounded on I.
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Advanced Calculus (I)
Proof:
Suppose first that f is not bounded on I. Then there exist
xn ∈ I such that
(7)
|f (xn )| > n,
n ∈ N.
Since I is bounded, we know (by the Bolzano-Weierstrass
Theorem) that {xn } has a convergent subsequence, say
xnk → a as k → ∞. Since I is closed, we also know (by the
Comparison Theorem) that a ∈ I. In particular, f (a) ∈ R.
On the other hand, substituting nk for n in (7) and taking
the limit of this inequility as k → ∞, we have |f (a)| = ∞, a
contradiction. Hence, the function f is bounded on I.
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Advanced Calculus (I)
Proof:
Suppose first that f is not bounded on I. Then there exist
xn ∈ I such that
(7)
|f (xn )| > n,
n ∈ N.
Since I is bounded, we know (by the Bolzano-Weierstrass
Theorem) that {xn } has a convergent subsequence, say
xnk → a as k → ∞. Since I is closed, we also know (by the
Comparison Theorem) that a ∈ I. In particular, f (a) ∈ R.
On the other hand, substituting nk for n in (7) and taking
the limit of this inequility as k → ∞, we have |f (a)| = ∞, a
contradiction. Hence, the function f is bounded on I.
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Suppose first that f is not bounded on I. Then there exist
xn ∈ I such that
(7)
|f (xn )| > n,
n ∈ N.
Since I is bounded, we know (by the Bolzano-Weierstrass
Theorem) that {xn } has a convergent subsequence, say
xnk → a as k → ∞. Since I is closed, we also know (by the
Comparison Theorem) that a ∈ I. In particular, f (a) ∈ R.
On the other hand, substituting nk for n in (7) and taking
the limit of this inequility as k → ∞, we have |f (a)| = ∞, a
contradiction. Hence, the function f is bounded on I.
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Suppose first that f is not bounded on I. Then there exist
xn ∈ I such that
(7)
|f (xn )| > n,
n ∈ N.
Since I is bounded, we know (by the Bolzano-Weierstrass
Theorem) that {xn } has a convergent subsequence, say
xnk → a as k → ∞. Since I is closed, we also know (by the
Comparison Theorem) that a ∈ I. In particular, f (a) ∈ R.
On the other hand, substituting nk for n in (7) and taking
the limit of this inequility as k → ∞, we have |f (a)| = ∞, a
contradiction. Hence, the function f is bounded on I.
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Suppose first that f is not bounded on I. Then there exist
xn ∈ I such that
(7)
|f (xn )| > n,
n ∈ N.
Since I is bounded, we know (by the Bolzano-Weierstrass
Theorem) that {xn } has a convergent subsequence, say
xnk → a as k → ∞. Since I is closed, we also know (by the
Comparison Theorem) that a ∈ I. In particular, f (a) ∈ R.
On the other hand, substituting nk for n in (7) and taking
the limit of this inequility as k → ∞, we have |f (a)| = ∞, a
contradiction. Hence, the function f is bounded on I.
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Suppose first that f is not bounded on I. Then there exist
xn ∈ I such that
(7)
|f (xn )| > n,
n ∈ N.
Since I is bounded, we know (by the Bolzano-Weierstrass
Theorem) that {xn } has a convergent subsequence, say
xnk → a as k → ∞. Since I is closed, we also know (by the
Comparison Theorem) that a ∈ I. In particular, f (a) ∈ R.
On the other hand, substituting nk for n in (7) and taking
the limit of this inequility as k → ∞, we have |f (a)| = ∞, a
contradiction. Hence, the function f is bounded on I.
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Suppose first that f is not bounded on I. Then there exist
xn ∈ I such that
(7)
|f (xn )| > n,
n ∈ N.
Since I is bounded, we know (by the Bolzano-Weierstrass
Theorem) that {xn } has a convergent subsequence, say
xnk → a as k → ∞. Since I is closed, we also know (by the
Comparison Theorem) that a ∈ I. In particular, f (a) ∈ R.
On the other hand, substituting nk for n in (7) and taking
the limit of this inequility as k → ∞, we have |f (a)| = ∞, a
contradiction. Hence, the function f is bounded on I.
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Suppose first that f is not bounded on I. Then there exist
xn ∈ I such that
(7)
|f (xn )| > n,
n ∈ N.
Since I is bounded, we know (by the Bolzano-Weierstrass
Theorem) that {xn } has a convergent subsequence, say
xnk → a as k → ∞. Since I is closed, we also know (by the
Comparison Theorem) that a ∈ I. In particular, f (a) ∈ R.
On the other hand, substituting nk for n in (7) and taking
the limit of this inequility as k → ∞, we have |f (a)| = ∞, a
contradiction. Hence, the function f is bounded on I.
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Suppose first that f is not bounded on I. Then there exist
xn ∈ I such that
(7)
|f (xn )| > n,
n ∈ N.
Since I is bounded, we know (by the Bolzano-Weierstrass
Theorem) that {xn } has a convergent subsequence, say
xnk → a as k → ∞. Since I is closed, we also know (by the
Comparison Theorem) that a ∈ I. In particular, f (a) ∈ R.
On the other hand, substituting nk for n in (7) and taking
the limit of this inequility as k → ∞, we have |f (a)| = ∞, a
contradiction. Hence, the function f is bounded on I.
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Suppose first that f is not bounded on I. Then there exist
xn ∈ I such that
(7)
|f (xn )| > n,
n ∈ N.
Since I is bounded, we know (by the Bolzano-Weierstrass
Theorem) that {xn } has a convergent subsequence, say
xnk → a as k → ∞. Since I is closed, we also know (by the
Comparison Theorem) that a ∈ I. In particular, f (a) ∈ R.
On the other hand, substituting nk for n in (7) and taking
the limit of this inequility as k → ∞, we have |f (a)| = ∞, a
contradiction. Hence, the function f is bounded on I.
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Suppose first that f is not bounded on I. Then there exist
xn ∈ I such that
(7)
|f (xn )| > n,
n ∈ N.
Since I is bounded, we know (by the Bolzano-Weierstrass
Theorem) that {xn } has a convergent subsequence, say
xnk → a as k → ∞. Since I is closed, we also know (by the
Comparison Theorem) that a ∈ I. In particular, f (a) ∈ R.
On the other hand, substituting nk for n in (7) and taking
the limit of this inequility as k → ∞, we have |f (a)| = ∞, a
contradiction. Hence, the function f is bounded on I.
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Suppose first that f is not bounded on I. Then there exist
xn ∈ I such that
(7)
|f (xn )| > n,
n ∈ N.
Since I is bounded, we know (by the Bolzano-Weierstrass
Theorem) that {xn } has a convergent subsequence, say
xnk → a as k → ∞. Since I is closed, we also know (by the
Comparison Theorem) that a ∈ I. In particular, f (a) ∈ R.
On the other hand, substituting nk for n in (7) and taking
the limit of this inequility as k → ∞, we have |f (a)| = ∞, a
contradiction. Hence, the function f is bounded on I.
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Suppose first that f is not bounded on I. Then there exist
xn ∈ I such that
(7)
|f (xn )| > n,
n ∈ N.
Since I is bounded, we know (by the Bolzano-Weierstrass
Theorem) that {xn } has a convergent subsequence, say
xnk → a as k → ∞. Since I is closed, we also know (by the
Comparison Theorem) that a ∈ I. In particular, f (a) ∈ R.
On the other hand, substituting nk for n in (7) and taking
the limit of this inequility as k → ∞, we have |f (a)| = ∞, a
contradiction. Hence, the function f is bounded on I.
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Suppose first that f is not bounded on I. Then there exist
xn ∈ I such that
(7)
|f (xn )| > n,
n ∈ N.
Since I is bounded, we know (by the Bolzano-Weierstrass
Theorem) that {xn } has a convergent subsequence, say
xnk → a as k → ∞. Since I is closed, we also know (by the
Comparison Theorem) that a ∈ I. In particular, f (a) ∈ R.
On the other hand, substituting nk for n in (7) and taking
the limit of this inequility as k → ∞, we have |f (a)| = ∞, a
contradiction. Hence, the function f is bounded on I.
W EN -C HING L IEN
Advanced Calculus (I)
We have proved that both M and m are finite real
numbers. To show that there is an xM ∈ I such that
f (xM ) = M, suppose to the contrary that f (x) < M for all
x ∈ I. Then the function
g(x) =
1
M − f (x)
is continuous, hence, bounded on I.
W EN -C HING L IEN
Advanced Calculus (I)
We have proved that both M and m are finite real
numbers. To show that there is an xM ∈ I such that
f (xM ) = M, suppose to the contrary that f (x) < M for all
x ∈ I. Then the function
g(x) =
1
M − f (x)
is continuous, hence, bounded on I.
W EN -C HING L IEN
Advanced Calculus (I)
We have proved that both M and m are finite real
numbers. To show that there is an xM ∈ I such that
f (xM ) = M, suppose to the contrary that f (x) < M for all
x ∈ I. Then the function
g(x) =
1
M − f (x)
is continuous, hence, bounded on I.
W EN -C HING L IEN
Advanced Calculus (I)
We have proved that both M and m are finite real
numbers. To show that there is an xM ∈ I such that
f (xM ) = M, suppose to the contrary that f (x) < M for all
x ∈ I. Then the function
g(x) =
1
M − f (x)
is continuous, hence, bounded on I.
W EN -C HING L IEN
Advanced Calculus (I)
We have proved that both M and m are finite real
numbers. To show that there is an xM ∈ I such that
f (xM ) = M, suppose to the contrary that f (x) < M for all
x ∈ I. Then the function
g(x) =
1
M − f (x)
is continuous, hence, bounded on I.
W EN -C HING L IEN
Advanced Calculus (I)
We have proved that both M and m are finite real
numbers. To show that there is an xM ∈ I such that
f (xM ) = M, suppose to the contrary that f (x) < M for all
x ∈ I. Then the function
g(x) =
1
M − f (x)
is continuous, hence, bounded on I.
W EN -C HING L IEN
Advanced Calculus (I)
We have proved that both M and m are finite real
numbers. To show that there is an xM ∈ I such that
f (xM ) = M, suppose to the contrary that f (x) < M for all
x ∈ I. Then the function
g(x) =
1
M − f (x)
is continuous, hence, bounded on I.
W EN -C HING L IEN
Advanced Calculus (I)
In particular, there is a C > 0 such that |g(x)| = g(x) ≤ C.
It follows that
(8)
f (x) ≤ M −
1
c
for all x ∈ I. Taking the supremum of (8) over all x ∈ I, we
obtain M ≤ M − 1/C < M, a contradiction. Hence, there
is an xM ∈ I, such that f (xM ) = M. A similar argument
proves that there is an xm ∈ I such that f (xm ) = m. 2
W EN -C HING L IEN
Advanced Calculus (I)
In particular, there is a C > 0 such that |g(x)| = g(x) ≤ C.
It follows that
(8)
f (x) ≤ M −
1
c
for all x ∈ I. Taking the supremum of (8) over all x ∈ I, we
obtain M ≤ M − 1/C < M, a contradiction. Hence, there
is an xM ∈ I, such that f (xM ) = M. A similar argument
proves that there is an xm ∈ I such that f (xm ) = m. 2
W EN -C HING L IEN
Advanced Calculus (I)
In particular, there is a C > 0 such that |g(x)| = g(x) ≤ C.
It follows that
(8)
f (x) ≤ M −
1
c
for all x ∈ I. Taking the supremum of (8) over all x ∈ I, we
obtain M ≤ M − 1/C < M, a contradiction. Hence, there
is an xM ∈ I, such that f (xM ) = M. A similar argument
proves that there is an xm ∈ I such that f (xm ) = m. 2
W EN -C HING L IEN
Advanced Calculus (I)
In particular, there is a C > 0 such that |g(x)| = g(x) ≤ C.
It follows that
(8)
f (x) ≤ M −
1
c
for all x ∈ I. Taking the supremum of (8) over all x ∈ I, we
obtain M ≤ M − 1/C < M, a contradiction. Hence, there
is an xM ∈ I, such that f (xM ) = M. A similar argument
proves that there is an xm ∈ I such that f (xm ) = m. 2
W EN -C HING L IEN
Advanced Calculus (I)
In particular, there is a C > 0 such that |g(x)| = g(x) ≤ C.
It follows that
(8)
f (x) ≤ M −
1
c
for all x ∈ I. Taking the supremum of (8) over all x ∈ I, we
obtain M ≤ M − 1/C < M, a contradiction. Hence, there
is an xM ∈ I, such that f (xM ) = M. A similar argument
proves that there is an xm ∈ I such that f (xm ) = m. 2
W EN -C HING L IEN
Advanced Calculus (I)
In particular, there is a C > 0 such that |g(x)| = g(x) ≤ C.
It follows that
(8)
f (x) ≤ M −
1
c
for all x ∈ I. Taking the supremum of (8) over all x ∈ I, we
obtain M ≤ M − 1/C < M, a contradiction. Hence, there
is an xM ∈ I, such that f (xM ) = M. A similar argument
proves that there is an xm ∈ I such that f (xm ) = m. 2
W EN -C HING L IEN
Advanced Calculus (I)
In particular, there is a C > 0 such that |g(x)| = g(x) ≤ C.
It follows that
(8)
f (x) ≤ M −
1
c
for all x ∈ I. Taking the supremum of (8) over all x ∈ I, we
obtain M ≤ M − 1/C < M, a contradiction. Hence, there
is an xM ∈ I, such that f (xM ) = M. A similar argument
proves that there is an xm ∈ I such that f (xm ) = m. 2
W EN -C HING L IEN
Advanced Calculus (I)
In particular, there is a C > 0 such that |g(x)| = g(x) ≤ C.
It follows that
(8)
f (x) ≤ M −
1
c
for all x ∈ I. Taking the supremum of (8) over all x ∈ I, we
obtain M ≤ M − 1/C < M, a contradiction. Hence, there
is an xM ∈ I, such that f (xM ) = M. A similar argument
proves that there is an xm ∈ I such that f (xm ) = m. 2
W EN -C HING L IEN
Advanced Calculus (I)
In particular, there is a C > 0 such that |g(x)| = g(x) ≤ C.
It follows that
(8)
f (x) ≤ M −
1
c
for all x ∈ I. Taking the supremum of (8) over all x ∈ I, we
obtain M ≤ M − 1/C < M, a contradiction. Hence, there
is an xM ∈ I, such that f (xM ) = M. A similar argument
proves that there is an xm ∈ I such that f (xm ) = m. 2
W EN -C HING L IEN
Advanced Calculus (I)
In particular, there is a C > 0 such that |g(x)| = g(x) ≤ C.
It follows that
(8)
f (x) ≤ M −
1
c
for all x ∈ I. Taking the supremum of (8) over all x ∈ I, we
obtain M ≤ M − 1/C < M, a contradiction. Hence, there
is an xM ∈ I, such that f (xM ) = M. A similar argument
proves that there is an xm ∈ I such that f (xm ) = m. 2
W EN -C HING L IEN
Advanced Calculus (I)
Lemma: [Sign-Preserving Property].
Let f : I → R where I is an open, nondegenerate interval.
If f is continuous at a point x0 ∈ I and f (x0 ) > 0, then there
are positive number ǫ and δ such that |x − x0 | < δ implies
f (x) > ǫ.
W EN -C HING L IEN
Advanced Calculus (I)
Lemma: [Sign-Preserving Property].
Let f : I → R where I is an open, nondegenerate interval.
If f is continuous at a point x0 ∈ I and f (x0 ) > 0, then there
are positive number ǫ and δ such that |x − x0 | < δ implies
f (x) > ǫ.
W EN -C HING L IEN
Advanced Calculus (I)
Theorem (Intermediate Value Theorem)
Let I be a nondegenerate interval and f : I → R be
continuous. If a, b ∈ I with a < b, and if y0 lies between
f (a) and f (b), then there is an x0 ∈ (a, b) such that
f (x0 ) = y0 .
W EN -C HING L IEN
Advanced Calculus (I)
Theorem (Intermediate Value Theorem)
Let I be a nondegenerate interval and f : I → R be
continuous. If a, b ∈ I with a < b, and if y0 lies between
f (a) and f (b), then there is an x0 ∈ (a, b) such that
f (x0 ) = y0 .
W EN -C HING L IEN
Advanced Calculus (I)
Example:
Assuming that sin x is continuous on R, prove that
(
1
sin
x 6= 0
f (x) =
x
1
x =0
is continuous on (−∞, 0) and (0, ∞), discontinuous at 0,
and neither f (0+) nor f (0−) exists. (see Figure 3.1 on
p.61.)
W EN -C HING L IEN
Advanced Calculus (I)
Example:
Assuming that sin x is continuous on R, prove that
(
1
x 6= 0
sin
f (x) =
x
1
x =0
is continuous on (−∞, 0) and (0, ∞), discontinuous at 0,
and neither f (0+) nor f (0−) exists. (see Figure 3.1 on
p.61.)
W EN -C HING L IEN
Advanced Calculus (I)
Example:
The Dirchlet function is defined on R by
1 x ∈Q
f (x) :=
0 x ∈Q
Prove that every point x ∈ R is a point of discontinuity of f.
(Such functions are called nowhere continuous.)
W EN -C HING L IEN
Advanced Calculus (I)
Example:
The Dirchlet function is defined on R by
1 x ∈Q
f (x) :=
0 x ∈Q
Prove that every point x ∈ R is a point of discontinuity of f.
(Such functions are called nowhere continuous.)
W EN -C HING L IEN
Advanced Calculus (I)
Example:
Prove that the function

p
 1
x = ∈ Q (in reduced form)
f (x) =
q
q
 0 x ∈ Q.
is continuous at every irrational in the interval (0, 1) but
discontinuous at every rational in (0, 1).
W EN -C HING L IEN
Advanced Calculus (I)
Example:
Prove that the function

p
 1
x = ∈ Q (in reduced form)
f (x) =
q
q
 0 x ∈ Q.
is continuous at every irrational in the interval (0, 1) but
discontinuous at every rational in (0, 1).
W EN -C HING L IEN
Advanced Calculus (I)
Thank you.
W EN -C HING L IEN
Advanced Calculus (I)