• 沒有找到結果。

N/A
N/A
Protected

Copied!
37
0
0

(1)

WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

2009

(2)

## Ch9: Convergence in R

n

### 9.5: Applications

Theorem (9.40 Dini)

Suppose that H is a compact subset ofRnand fk :H →R is a pointwise monotone sequence of continuous

functions. If fk → f pointwise on H as k → ∞ and f is continuous on H, then fk → f uniformly on H. In particular, if φk is a pointwise monotone sequence of functions continuous on an interval [a,b] that converges pointwise to a continuous function, then

k →∞lim Z b

a

φk(t)dt = Z b

a



k →∞lim φk(t)

 dt.

(3)

Definition (9.41)

(i) A set E ⊆R is said to be of measure zero if and only if for every ε > 0 there is a countable collection of intervals {Ij}j∈Nthat covers E such that

X

j=1

|Ij| ≤ ε.

(ii) A function f : [a, b] →R is said to be almost

everywhere continuous on [a, b] if and only if the set of points x ∈ [a, b] where f is discontinuous is a set of measure zero.

(4)

Remark (9.43)

If E1,E2, . . . is a sequence of sets of measure zero, then

E =

[

k =1

Ek

is also a set of measure zero.

(5)

Proof.

Let ε > 0.By hypothesis, given k ∈N we can choose a collection of intervals {Ij(k )}j∈Nthat covers Ek such that

X

j=1

|Ij(k )| < ε 2k.

then the collection {Ij(k )}k ,j∈Nis countable, covers E , and

X

k =1

X

j=1

|Ij(k )| ≤

X

j=1

ε 2k = ε.

Consequently, E is of measure zero.

(6)

Proof.

Let ε > 0. By hypothesis, given k ∈Nwe can choose a collection of intervals {Ij(k )}j∈Nthat covers Ek such that

X

j=1

|Ij(k )| < ε 2k.

then the collection {Ij(k )}k ,j∈Nis countable, covers E , and

X

k =1

X

j=1

|Ij(k )| ≤

X

j=1

ε 2k = ε.

Consequently, E is of measure zero.

(7)

Proof.

Let ε > 0. By hypothesis, given k ∈N we can choose a collection of intervals {Ij(k )}j∈Nthat covers Ek such that

X

j=1

|Ij(k )| < ε 2k.

then the collection {Ij(k )}k ,j∈Nis countable, covers E ,and

X

k =1

X

j=1

|Ij(k )| ≤

X

j=1

ε 2k = ε.

Consequently, E is of measure zero.

(8)

Proof.

Let ε > 0. By hypothesis, given k ∈N we can choose a collection of intervals {Ij(k )}j∈Nthat covers Ek such that

X

j=1

|Ij(k )| < ε 2k.

then the collection {Ij(k )}k ,j∈Nis countable, covers E , and

X

k =1

X

j=1

|Ij(k )| ≤

X

j=1

ε 2k = ε.

Consequently, E is of measure zero.

(9)

Proof.

Let ε > 0. By hypothesis, given k ∈N we can choose a collection of intervals {Ij(k )}j∈Nthat covers Ek such that

X

j=1

|Ij(k )| < ε 2k.

then the collection {Ij(k )}k ,j∈Nis countable, covers E ,and

X

k =1

X

j=1

|Ij(k )| ≤

X

j=1

ε 2k = ε.

Consequently, E is of measure zero.

(10)

Proof.

Let ε > 0. By hypothesis, given k ∈N we can choose a collection of intervals {Ij(k )}j∈Nthat covers Ek such that

X

j=1

|Ij(k )| < ε 2k.

then the collection {Ij(k )}k ,j∈Nis countable, covers E , and

X

k =1

X

j=1

|Ij(k )| ≤

X

j=1

ε 2k = ε.

Consequently, E is of measure zero.

(11)

Proof.

Let ε > 0. By hypothesis, given k ∈N we can choose a collection of intervals {Ij(k )}j∈Nthat covers Ek such that

X

j=1

|Ij(k )| < ε 2k.

then the collection {Ij(k )}k ,j∈Nis countable, covers E , and

X

k =1

X

j=1

|Ij(k )| ≤

X

j=1

ε 2k = ε.

Consequently, E is of measure zero.

(12)

Theorem (9.51 Closed Graph Theorem)

Let I be a closed interval and f : I →R. Then f is

continuous on I if and only if the graph of f is closed and connected inR2.

Proof.

For any interval J ⊆ I, let G(J) represent the graph of y = f (x ) for x ∈ J.Suppose that f is continuous on I. The function x 7→ (x , f (x )) is continuous from I intoR2, and I is connected inR. Thus G(I) is connected in R2by Theorem 9.30. To prove that G(I) is closed, we shall use Theorem 9.8. Let xk ∈ I and (xk,f (xk)) → (x , y ) as k → ∞. Then xk → x and f (xk) →y , as k → ∞. Hence, x ∈ I and since f is continuous, f (xk) →f (x ). In particular, the graph of f is closed.

(13)

Theorem (9.51 Closed Graph Theorem)

Let I be a closed interval and f : I →R. Then f is

continuous on I if and only if the graph of f is closed and connected inR2.

Proof.

For any interval J ⊆ I,let G(J) represent the graph of y = f (x ) for x ∈ J. Suppose that f is continuous on I.The function x 7→ (x , f (x )) is continuous from I intoR2, and I is connected inR. Thus G(I) is connected in R2by Theorem 9.30. To prove that G(I) is closed, we shall use Theorem 9.8. Let xk ∈ I and (xk,f (xk)) → (x , y ) as k → ∞. Then xk → x and f (xk) →y , as k → ∞. Hence, x ∈ I and since f is continuous, f (xk) →f (x ). In particular, the graph of f is closed.

(14)

Theorem (9.51 Closed Graph Theorem)

Let I be a closed interval and f : I →R. Then f is

continuous on I if and only if the graph of f is closed and connected inR2.

Proof.

For any interval J ⊆ I, let G(J) represent the graph of y = f (x ) for x ∈ J.Suppose that f is continuous on I. The function x 7→ (x , f (x )) is continuous from I intoR2, and I is connected inR.Thus G(I) is connected inR2by Theorem 9.30. To prove that G(I) is closed, we shall use Theorem 9.8. Let xk ∈ I and (xk,f (xk)) → (x , y ) as k → ∞. Then xk → x and f (xk) →y , as k → ∞. Hence, x ∈ I and since f is continuous, f (xk) →f (x ). In particular, the graph of f is closed.

(15)

Theorem (9.51 Closed Graph Theorem)

Let I be a closed interval and f : I →R. Then f is

continuous on I if and only if the graph of f is closed and connected inR2.

Proof.

For any interval J ⊆ I, let G(J) represent the graph of y = f (x ) for x ∈ J. Suppose that f is continuous on I.The function x 7→ (x , f (x )) is continuous from I intoR2, and I is connected inR. Thus G(I) is connected in R2by Theorem 9.30.To prove that G(I) is closed, we shall use Theorem 9.8. Let xk ∈ I and (xk,f (xk)) → (x , y ) as k → ∞. Then xk → x and f (xk) →y , as k → ∞. Hence, x ∈ I and since f is continuous, f (xk) →f (x ). In particular, the graph of f is closed.

(16)

Theorem (9.51 Closed Graph Theorem)

Let I be a closed interval and f : I →R. Then f is

continuous on I if and only if the graph of f is closed and connected inR2.

Proof.

For any interval J ⊆ I, let G(J) represent the graph of y = f (x ) for x ∈ J. Suppose that f is continuous on I. The function x 7→ (x , f (x )) is continuous from I intoR2, and I is connected inR.Thus G(I) is connected inR2by Theorem 9.30. To prove that G(I) is closed, we shall use Theorem 9.8.Let xk ∈ I and (xk,f (xk)) → (x , y ) as k → ∞. Then xk → x and f (xk) →y , as k → ∞. Hence, x ∈ I and since f is continuous, f (xk) →f (x ). In particular, the graph of f is closed.

(17)

Theorem (9.51 Closed Graph Theorem)

Let I be a closed interval and f : I →R. Then f is

continuous on I if and only if the graph of f is closed and connected inR2.

Proof.

For any interval J ⊆ I, let G(J) represent the graph of y = f (x ) for x ∈ J. Suppose that f is continuous on I. The function x 7→ (x , f (x )) is continuous from I intoR2, and I is connected inR. Thus G(I) is connected in R2by Theorem 9.30.To prove that G(I) is closed, we shall use Theorem 9.8. Let xk ∈ I and (xk,f (xk)) → (x , y ) as k → ∞.Then xk → x and f (xk) →y , as k → ∞. Hence, x ∈ I and since f is continuous, f (xk) →f (x ). In particular, the graph of f is closed.

(18)

Theorem (9.51 Closed Graph Theorem)

Let I be a closed interval and f : I →R. Then f is

continuous on I if and only if the graph of f is closed and connected inR2.

Proof.

For any interval J ⊆ I, let G(J) represent the graph of y = f (x ) for x ∈ J. Suppose that f is continuous on I. The function x 7→ (x , f (x )) is continuous from I intoR2, and I is connected inR. Thus G(I) is connected in R2by Theorem 9.30. To prove that G(I) is closed, we shall use Theorem 9.8.Let xk ∈ I and (xk,f (xk)) → (x , y ) as k → ∞. Then xk → x and f (xk) →y , as k → ∞.Hence, x ∈ I and since f is continuous, f (xk) →f (x ). In particular, the graph of f is closed.

(19)

Theorem (9.51 Closed Graph Theorem)

Let I be a closed interval and f : I →R. Then f is

continuous on I if and only if the graph of f is closed and connected inR2.

Proof.

For any interval J ⊆ I, let G(J) represent the graph of y = f (x ) for x ∈ J. Suppose that f is continuous on I. The function x 7→ (x , f (x )) is continuous from I intoR2, and I is connected inR. Thus G(I) is connected in R2by Theorem 9.30. To prove that G(I) is closed, we shall use Theorem 9.8. Let xk ∈ I and (xk,f (xk)) → (x , y ) as k → ∞.Then xk → x and f (xk) →y , as k → ∞. Hence, x ∈ I and since f is continuous, f (xk) →f (x ). In particular, the graph of f is closed.

(20)

Theorem (9.51 Closed Graph Theorem)

Let I be a closed interval and f : I →R. Then f is

continuous on I if and only if the graph of f is closed and connected inR2.

Proof.

For any interval J ⊆ I, let G(J) represent the graph of y = f (x ) for x ∈ J. Suppose that f is continuous on I. The function x 7→ (x , f (x )) is continuous from I intoR2, and I is connected inR. Thus G(I) is connected in R2by Theorem 9.30. To prove that G(I) is closed, we shall use Theorem 9.8. Let xk ∈ I and (xk,f (xk)) → (x , y ) as k → ∞. Then xk → x and f (xk) →y , as k → ∞.Hence, x ∈ I and since f is continuous, f (xk) →f (x ). In particular, the graph of f is closed.

(21)

Theorem (9.51 Closed Graph Theorem)

Let I be a closed interval and f : I →R. Then f is

continuous on I if and only if the graph of f is closed and connected inR2.

Proof.

For any interval J ⊆ I, let G(J) represent the graph of y = f (x ) for x ∈ J. Suppose that f is continuous on I. The function x 7→ (x , f (x )) is continuous from I intoR2, and I is connected inR. Thus G(I) is connected in R2by Theorem 9.30. To prove that G(I) is closed, we shall use Theorem 9.8. Let xk ∈ I and (xk,f (xk)) → (x , y ) as k → ∞. Then xk → x and f (xk) →y , as k → ∞. Hence, x ∈ I and since f is continuous, f (xk) →f (x ). In particular, the graph of f is closed.

(22)

Proof.

Conversely, suppose that the graph of f is closed and connected inR2.We first show that f satisfies the

Intermediate Value Theorem on I. Indeed, suppose to the contrary that there exist x1<x2in I with f (x1) 6=f (x2)and a value y0between f (x1)and f (x2)such that f (t) 6= y0for all t ∈ [x1,x2].Suppose for simplicity that f (x1) <f (x2).

Since f (t) 6= y0for any t ∈ [x1,x2], the open sets U = {(x , y ) : x < x1} ∪ {(x, y ) : x < x2,y < y0},

V = {(x , y ) : x > x2} ∪ {(x, y ) : x > x1,y > y0} separate G(I), a contradiction. Therefore, f satisfies the Intermediate Value Theorem on I.

(23)

Proof.

Conversely, suppose that the graph of f is closed and connected inR2. We first show that f satisfies the

Intermediate Value Theorem on I.Indeed, suppose to the contrary that there exist x1<x2in I with f (x1) 6=f (x2)and a value y0between f (x1)and f (x2)such that f (t) 6= y0for all t ∈ [x1,x2]. Suppose for simplicity that f (x1) <f (x2).

Since f (t) 6= y0for any t ∈ [x1,x2], the open sets U = {(x , y ) : x < x1} ∪ {(x, y ) : x < x2,y < y0},

V = {(x , y ) : x > x2} ∪ {(x, y ) : x > x1,y > y0} separate G(I), a contradiction. Therefore, f satisfies the Intermediate Value Theorem on I.

(24)

Proof.

Conversely, suppose that the graph of f is closed and connected inR2. We first show that f satisfies the

Intermediate Value Theorem on I. Indeed, suppose to the contrary that there exist x1<x2in I with f (x1) 6=f (x2)and a value y0between f (x1)and f (x2)such that f (t) 6= y0for all t ∈ [x1,x2].Suppose for simplicity that f (x1) <f (x2).

Since f (t) 6= y0for any t ∈ [x1,x2],the open sets U = {(x , y ) : x < x1} ∪ {(x, y ) : x < x2,y < y0},

V = {(x , y ) : x > x2} ∪ {(x, y ) : x > x1,y > y0} separate G(I), a contradiction. Therefore, f satisfies the Intermediate Value Theorem on I.

(25)

Proof.

Conversely, suppose that the graph of f is closed and connected inR2. We first show that f satisfies the

Intermediate Value Theorem on I. Indeed, suppose to the contrary that there exist x1<x2in I with f (x1) 6=f (x2)and a value y0between f (x1)and f (x2)such that f (t) 6= y0for all t ∈ [x1,x2]. Suppose for simplicity that f (x1) <f (x2).

Since f (t) 6= y0for any t ∈ [x1,x2], the open sets U = {(x , y ) : x < x1} ∪ {(x, y ) : x < x2,y < y0},

V = {(x , y ) : x > x2} ∪ {(x, y ) : x > x1,y > y0} separate G(I), a contradiction.Therefore, f satisfies the Intermediate Value Theorem on I.

(26)

Proof.

Conversely, suppose that the graph of f is closed and connected inR2. We first show that f satisfies the

Intermediate Value Theorem on I. Indeed, suppose to the contrary that there exist x1<x2in I with f (x1) 6=f (x2)and a value y0between f (x1)and f (x2)such that f (t) 6= y0for all t ∈ [x1,x2]. Suppose for simplicity that f (x1) <f (x2).

Since f (t) 6= y0for any t ∈ [x1,x2],the open sets U = {(x , y ) : x < x1} ∪ {(x, y ) : x < x2,y < y0},

V = {(x , y ) : x > x2} ∪ {(x, y ) : x > x1,y > y0} separate G(I), a contradiction. Therefore, f satisfies the Intermediate Value Theorem on I.

(27)

Proof.

Conversely, suppose that the graph of f is closed and connected inR2. We first show that f satisfies the

Intermediate Value Theorem on I. Indeed, suppose to the contrary that there exist x1<x2in I with f (x1) 6=f (x2)and a value y0between f (x1)and f (x2)such that f (t) 6= y0for all t ∈ [x1,x2]. Suppose for simplicity that f (x1) <f (x2).

Since f (t) 6= y0for any t ∈ [x1,x2], the open sets U = {(x , y ) : x < x1} ∪ {(x, y ) : x < x2,y < y0},

V = {(x , y ) : x > x2} ∪ {(x, y ) : x > x1,y > y0} separate G(I), a contradiction.Therefore, f satisfies the Intermediate Value Theorem on I.

(28)

Proof.

Conversely, suppose that the graph of f is closed and connected inR2. We first show that f satisfies the

Intermediate Value Theorem on I. Indeed, suppose to the contrary that there exist x1<x2in I with f (x1) 6=f (x2)and a value y0between f (x1)and f (x2)such that f (t) 6= y0for all t ∈ [x1,x2]. Suppose for simplicity that f (x1) <f (x2).

Since f (t) 6= y0for any t ∈ [x1,x2], the open sets U = {(x , y ) : x < x1} ∪ {(x, y ) : x < x2,y < y0},

V = {(x , y ) : x > x2} ∪ {(x, y ) : x > x1,y > y0} separate G(I), a contradiction. Therefore, f satisfies the Intermediate Value Theorem on I.

(29)

Proof.

If f is not continuous on I,then there exist numbers x0 ∈ I, ε0>0, and xk ∈ I such that xk → x0 and

|f (xk) −f (x0)| > ε0. By symmetry, we may suppose that f (xk) >f (x0) + ε0 for infinitely many k ’s,say

f (xkj) > f (x0) + ε0>f (x0), j ∈ N.

By the Intermediate Value Theorem, choose cj between xkj and x0such that f (cj) = f (x0) + ε0. By construction, (cj,f (cj)) → (x0,f (x0) + ε0)and cj → x0as j → ∞. Hence, the graph of f on I is not closed.

(30)

Proof.

If f is not continuous on I, then there exist numbers x0 ∈ I, ε0>0, and xk ∈ I such that xk → x0 and

|f (xk) −f (x0)| > ε0. By symmetry, we may suppose that f (xk) >f (x0) + ε0 for infinitely many k ’s, say

f (xkj) > f (x0) + ε0>f (x0), j ∈ N.

By the Intermediate Value Theorem, choose cj between xkj and x0such that f (cj) = f (x0) + ε0. By construction, (cj,f (cj)) → (x0,f (x0) + ε0)and cj → x0as j → ∞. Hence, the graph of f on I is not closed.

(31)

Proof.

If f is not continuous on I, then there exist numbers x0 ∈ I, ε0>0, and xk ∈ I such that xk → x0 and

|f (xk) −f (x0)| > ε0. By symmetry, we may suppose that f (xk) >f (x0) + ε0 for infinitely many k ’s,say

f (xkj) > f (x0) + ε0>f (x0), j ∈ N.

By the Intermediate Value Theorem,choose cj between xkj and x0such that f (cj) = f (x0) + ε0. By construction, (cj,f (cj)) → (x0,f (x0) + ε0)and cj → x0as j → ∞. Hence, the graph of f on I is not closed.

(32)

Proof.

If f is not continuous on I, then there exist numbers x0 ∈ I, ε0>0, and xk ∈ I such that xk → x0 and

|f (xk) −f (x0)| > ε0. By symmetry, we may suppose that f (xk) >f (x0) + ε0 for infinitely many k ’s, say

f (xkj) > f (x0) + ε0>f (x0), j ∈ N.

By the Intermediate Value Theorem, choose cj between xkj and x0such that f (cj) = f (x0) + ε0.By construction, (cj,f (cj)) → (x0,f (x0) + ε0)and cj → x0as j → ∞. Hence, the graph of f on I is not closed.

(33)

Proof.

If f is not continuous on I, then there exist numbers x0 ∈ I, ε0>0, and xk ∈ I such that xk → x0 and

|f (xk) −f (x0)| > ε0. By symmetry, we may suppose that f (xk) >f (x0) + ε0 for infinitely many k ’s, say

f (xkj) > f (x0) + ε0>f (x0), j ∈ N.

By the Intermediate Value Theorem,choose cj between xkj and x0such that f (cj) = f (x0) + ε0. By construction, (cj,f (cj)) → (x0,f (x0) + ε0)and cj → x0as j → ∞.Hence, the graph of f on I is not closed.

(34)

Proof.

If f is not continuous on I, then there exist numbers x0 ∈ I, ε0>0, and xk ∈ I such that xk → x0 and

|f (xk) −f (x0)| > ε0. By symmetry, we may suppose that f (xk) >f (x0) + ε0 for infinitely many k ’s, say

f (xkj) > f (x0) + ε0>f (x0), j ∈ N.

By the Intermediate Value Theorem, choose cj between xkj and x0such that f (cj) = f (x0) + ε0.By construction, (cj,f (cj)) → (x0,f (x0) + ε0)and cj → x0as j → ∞. Hence, the graph of f on I is not closed.

(35)

Proof.

If f is not continuous on I, then there exist numbers x0 ∈ I, ε0>0, and xk ∈ I such that xk → x0 and

|f (xk) −f (x0)| > ε0. By symmetry, we may suppose that f (xk) >f (x0) + ε0 for infinitely many k ’s, say

f (xkj) > f (x0) + ε0>f (x0), j ∈ N.

By the Intermediate Value Theorem, choose cj between xkj and x0such that f (cj) = f (x0) + ε0. By construction, (cj,f (cj)) → (x0,f (x0) + ε0)and cj → x0as j → ∞.Hence, the graph of f on I is not closed.

(36)

Proof.

If f is not continuous on I, then there exist numbers x0 ∈ I, ε0>0, and xk ∈ I such that xk → x0 and

|f (xk) −f (x0)| > ε0. By symmetry, we may suppose that f (xk) >f (x0) + ε0 for infinitely many k ’s, say

f (xkj) > f (x0) + ε0>f (x0), j ∈ N.

By the Intermediate Value Theorem, choose cj between xkj and x0such that f (cj) = f (x0) + ε0. By construction, (cj,f (cj)) → (x0,f (x0) + ε0)and cj → x0as j → ∞. Hence, the graph of f on I is not closed.

(37)

## Thank you.

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung