Advanced Calculus (I)
WEN-CHING LIEN
Department of Mathematics National Cheng Kung University
WEN-CHINGLIEN 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 aδ >0 (which in general depends on ǫ, f, and a) such that
|x−a| < δ and x ∈E imply |f(x) −f(a)| < ǫ.
WEN-CHINGLIEN 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 aδ >0 (which in general depends on ǫ, f, and a) such that
|x−a| < δ and x ∈E imply |f(x) −f(a)| < ǫ.
WEN-CHINGLIEN 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 aδ >0 (which in general depends on ǫ, f, and a) such that
|x−a| < δ and x ∈E imply |f(x) −f(a)| < ǫ.
WEN-CHINGLIEN 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 aδ >0 (which in general depends on ǫ, f, and a) such that
|x−a| < δ and x ∈E imply |f(x) −f(a)| < ǫ.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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
x→af(x).
WEN-CHINGLIEN 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
x→af(x).
WEN-CHINGLIEN Advanced Calculus (I)
Example:
1. f(x) =x2, x ∈ [0,1].
2. f(x) =x2, x ∈R.
3. f(x) = 1
x, x ∈ (0,1].
4. f(x) =√
x, x ∈R+.
WEN-CHINGLIEN 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 → ∞.
WEN-CHINGLIEN 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 → ∞.
WEN-CHINGLIEN 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 → ∞.
WEN-CHINGLIEN 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 → ∞.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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
x→a x∈I
f(x)
exists and belongs to B, and if g is continuous at L∈B, then
xlim→a x∈I
(g◦f)(x) =g lim
x→a x∈I
f(x)
! .
WEN-CHINGLIEN 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
x→a x∈I
f(x)
exists and belongs to B, and if g is continuous at L∈B, then
xlim→a x∈I
(g◦f)(x) =g lim
x→a x∈I
f(x)
! .
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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
WEN-CHINGLIEN 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
WEN-CHINGLIEN 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
WEN-CHINGLIEN 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
WEN-CHINGLIEN 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
WEN-CHINGLIEN 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
WEN-CHINGLIEN 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
WEN-CHINGLIEN 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
WEN-CHINGLIEN 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
WEN-CHINGLIEN 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
WEN-CHINGLIEN 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
WEN-CHINGLIEN 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
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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
x∈I
f(x) and m=inf
x∈If(x), then there exist points xm,xM ∈I such that (6)
f(xM) =M and f(xm) =m.
WEN-CHINGLIEN 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
x∈I
f(x) and m=inf
x∈If(x), then there exist points xm,xM ∈I such that (6)
f(xM) =M and f(xm) =m.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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.
WEN-CHINGLIEN 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
WEN-CHINGLIEN 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
WEN-CHINGLIEN 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
WEN-CHINGLIEN 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
WEN-CHINGLIEN 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
WEN-CHINGLIEN 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
WEN-CHINGLIEN 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
WEN-CHINGLIEN 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
WEN-CHINGLIEN 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
WEN-CHINGLIEN 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
WEN-CHINGLIEN 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 numberǫandδ such that|x−x0| < δ implies f(x) > ǫ.
WEN-CHINGLIEN 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 numberǫandδ such that|x−x0| < δ implies f(x) > ǫ.
WEN-CHINGLIEN 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 y0lies between f(a)and f(b), then there is an x0∈ (a,b)such that f(x0) =y0.
WEN-CHINGLIEN 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 y0lies between f(a)and f(b), then there is an x0∈ (a,b)such that f(x0) =y0.
WEN-CHINGLIEN Advanced Calculus (I)
Example:
Assuming that sin x is continuous on R, prove that
f(x) = (
sin 1
x x 6=0
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.)
WEN-CHINGLIEN Advanced Calculus (I)
Example:
Assuming that sin x is continuous on R, prove that
f(x) = (
sin 1
x x 6=0
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.)
WEN-CHINGLIEN Advanced Calculus (I)
Example:
The Dirchlet function is defined on R by
f(x) :=
1 x ∈Q 0 x ∈Q
Prove that every point x ∈R is a point of discontinuity of f.
(Such functions are called nowhere continuous.)
WEN-CHINGLIEN Advanced Calculus (I)
Example:
The Dirchlet function is defined on R by f(x) :=
1 x ∈Q 0 x ∈Q
Prove that every point x ∈R is a point of discontinuity of f.
(Such functions are called nowhere continuous.)
WEN-CHINGLIEN Advanced Calculus (I)
Example:
Prove that the function
f(x) =
1
q x = p
q ∈Q (in reduced form) 0 x ∈Q.
is continuous at every irrational in the interval (0,1)but discontinuous at every rational in (0,1).
WEN-CHINGLIEN Advanced Calculus (I)
Example:
Prove that the function
f(x) =
1
q x = p
q ∈Q (in reduced form) 0 x ∈Q.
is continuous at every irrational in the interval (0,1)but discontinuous at every rational in (0,1).
WEN-CHINGLIEN Advanced Calculus (I)
Thank you.
WEN-CHINGLIEN Advanced Calculus (I)