Advanced Calculus (II)

Advanced Calculus (II)

W^EN-C^HINGL^IEN

Department of Mathematics National Cheng Kung University

2009

Ch9: Convergence in R

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

φ_k(t)dt = Z b

a



k →∞lim φ_k(t)

 dt.

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 {I_j}_j∈Nthat covers E such that

∞

X

j=1

|I_j| ≤ 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.

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.

Proof.

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

∞

X

j=1

|I_j^{(k )}| < ε 2^k.

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

∞

X

k =1

∞

X

j=1

|I_j^{(k )}| ≤

∞

X

j=1

ε 2^k = ε.

Consequently, E is of measure zero.

Proof.

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

∞

X

j=1

|I_j^{(k )}| < ε 2^k.

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

∞

X

k =1

∞

X

j=1

|I_j^{(k )}| ≤

∞

X

j=1

ε 2^k = ε.

Consequently, E is of measure zero.

Proof.

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

∞

X

j=1

|I_j^{(k )}| < ε 2^k.

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

∞

X

k =1

∞

X

j=1

|I_j^{(k )}| ≤

∞

X

j=1

ε 2^k = ε.

Consequently, E is of measure zero.

Proof.

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

∞

X

j=1

|I_j^{(k )}| < ε 2^k.

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

∞

X

k =1

∞

X

j=1

|I_j^{(k )}| ≤

∞

X

j=1

ε 2^k = ε.

Consequently, E is of measure zero.

Proof.

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

∞

X

j=1

|I_j^{(k )}| < ε 2^k.

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

∞

X

k =1

∞

X

j=1

|I_j^{(k )}| ≤

∞

X

j=1

ε 2^k = ε.

Consequently, E is of measure zero.

Proof.

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

∞

X

j=1

|I_j^{(k )}| < ε 2^k.

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

∞

X

k =1

∞

X

j=1

|I_j^{(k )}| ≤

∞

X

j=1

ε 2^k = ε.

Consequently, E is of measure zero.

Proof.

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

∞

X

j=1

|I_j^{(k )}| < ε 2^k.

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

∞

X

k =1

∞

X

j=1

|I_j^{(k )}| ≤

∞

X

j=1

ε 2^k = ε.

Consequently, E is of measure zero.

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 (x_k) →y , as k → ∞. Hence, x ∈ I and since f is continuous, f (xk) →f (x ). In particular, the graph of f is closed.

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 (x_k) →y , as k → ∞. Hence, x ∈ I and since f is continuous, f (xk) →f (x ). In particular, the graph of f is closed.

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 (x_k) →y , as k → ∞. Hence, x ∈ I and since f is continuous, f (xk) →f (x ). In particular, the graph of f is closed.

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 (x_k) →y , as k → ∞. Hence, x ∈ I and since f is continuous, f (xk) →f (x ). In particular, the graph of f is closed.

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 (x_k) →y , as k → ∞. Hence, x ∈ I and since f is continuous, f (xk) →f (x ). In particular, the graph of f is closed.

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 (x_k) →y , as k → ∞. Hence, x ∈ I and since f is continuous, f (xk) →f (x ). In particular, the graph of f is closed.

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 (x_k) →y , as k → ∞.Hence, x ∈ I and since f is continuous, f (xk) →f (x ). In particular, the graph of f is closed.

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 (x_k) →y , as k → ∞. Hence, x ∈ I and since f is continuous, f (xk) →f (x ). In particular, the graph of f is closed.

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 (x_k) →y , as k → ∞.Hence, x ∈ I and since f is continuous, f (xk) →f (x ). In particular, the graph of f is closed.

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 (x_k) →y , as k → ∞. Hence, x ∈ I and since f is continuous, f (xk) →f (x ). In particular, the graph of f is closed.

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.

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.

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.

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.

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.

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.

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.

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

f (xk_j) > 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.

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

f (xk_j) > 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.

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

f (xk_j) > 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.

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

f (xk_j) > 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.

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

f (xk_j) > 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.

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

f (xk_j) > 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.

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

f (xk_j) > 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.

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

f (xk_j) > 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.

Thank you.