Advanced Calculus (II)

Advanced Calculus (II)

W^EN-C^HING L^IEN

Department of Mathematics National Cheng Kung University

2009

Ch9: Convergence in R

9.3: Continuous functions

Definition (9.22)

Let E be a nonempty subset of Rⁿ and let f : E →R^m. (i) f is said to be continuous at a∈E if and only if for every ε >0 there is aδ > 0 (which in general depends on ε, f , and a) such that

(3) kx−ak < δand x∈E imply kf(x) −f(a)k < ε.

(ii) f is said to be continuous on E (notation: f :E →R^m is continuous) if and only if f is continuous at every x ∈E.

Let E be a nonempty subset of Rⁿ and let f : E →R^m. Then f is said to be uniformly continuous on E (notation:

f :E →R^m is uniformly continuous) if and only if for every ε >0 there is aδ >0 such that

kx−ak < δ and x,a∈E imply kf(x) −f(a)k < ε.

Let E be a nonempty compact subset of Rⁿ.If f is continuous on E, then f is uniformly continuous on E.

Suppose that f is continuous on E.Given ε >0 and a∈E, chooseδ(a) >0 such that

x∈Bδ(a)(a)and x∈E imply kf(x) −f(a)k < ε 2. Sinceδ(a)/2 is positive for all a ∈E, we can choose finitely many points aj ∈E and numbersδj := δ(aj)/2 such that

(4) E ⊂

[N

j=1

Bδ_j(aj).

Setδ :=min{δ1, . . . , δN}.

Suppose that f is continuous on E.Givenε >0 and a∈E, chooseδ(a) >0 such that

x∈Bδ(a)(a)and x∈E imply kf(x) −f(a)k < ε 2. Sinceδ(a)/2 is positive for all a ∈E, we can choose finitely many points aj ∈E and numbersδj := δ(aj)/2 such that

(4) E ⊂

[N

j=1

Bδ_j(aj).

Setδ :=min{δ1, . . . , δN}.

Suppose that f is continuous on E.Given ε >0 and a∈E, chooseδ(a) >0 such that

x∈Bδ(a)(a)and x∈E imply kf(x) −f(a)k < ε 2. Sinceδ(a)/2 is positive for all a ∈E,we can choose finitely many points aj ∈E and numbersδj := δ(aj)/2 such that

(4) E ⊂

[N

j=1

Bδ_j(aj).

Setδ :=min{δ1, . . . , δN}.

Suppose that f is continuous on E.Given ε >0 and a∈E, chooseδ(a) >0 such that

x∈Bδ(a)(a)and x∈E imply kf(x) −f(a)k < ε 2. Sinceδ(a)/2 is positive for all a ∈E, we can choose finitely many points aj ∈E and numbersδj := δ(aj)/2 such that

(4) E ⊂

[N

j=1

Bδ_j(aj).

Setδ :=min{δ1, . . . , δN}.

Suppose that f is continuous on E.Given ε >0 and a∈E, chooseδ(a) >0 such that

x∈Bδ(a)(a)and x∈E imply kf(x) −f(a)k < ε 2. Sinceδ(a)/2 is positive for all a ∈E,we can choose finitely many points aj ∈E and numbersδj := δ(aj)/2 such that

(4) E ⊂

[N

j=1

Bδ_j(aj).

Setδ :=min{δ1, . . . , δN}.

Suppose that f is continuous on E.Given ε >0 and a∈E, chooseδ(a) >0 such that

x∈Bδ(a)(a)and x∈E imply kf(x) −f(a)k < ε 2. Sinceδ(a)/2 is positive for all a ∈E, we can choose finitely many points aj ∈E and numbersδj := δ(aj)/2 such that

(4) E ⊂

[N

j=1

Bδ_j(aj).

Setδ :=min{δ1, . . . , δN}.

Suppose that f is continuous on E.Given ε >0 and a∈E, chooseδ(a) >0 such that

x∈Bδ(a)(a)and x∈E imply kf(x) −f(a)k < ε 2. Sinceδ(a)/2 is positive for all a ∈E, we can choose finitely many points aj ∈E and numbersδj := δ(aj)/2 such that

(4) E ⊂

[N

j=1

Bδ_j(aj).

Setδ :=min{δ1, . . . , δN}.

Suppose that x,a∈E and kx−ak < δ.By (4), x belongs to Bδ_j(aj)for some 1≤j ≤N.Hence,

ka−ajk ≤ ka−xk + kx−ajk< δj+ δj =2δj = δ(aj), i.e., a also belongs to Bδ(aj)(aj).It follows, therefore, from the choice ofδ(aj)that

(5)

kf(x) −f(a)k ≤ kf(x) −f(aj)k+kf(aj) −f(a)k < ε 2+ε

2 = ε.

This proves that f is uniformly continuous on E.

Suppose that x,a∈E and kx−ak < δ.By (4), x belongs to Bδ_j(aj)for some 1≤j ≤N.Hence,

ka−ajk ≤ ka−xk + kx−ajk< δj+ δj =2δj = δ(aj), i.e., a also belongs to Bδ(aj)(aj).It follows, therefore, from the choice ofδ(aj)that

(5)

kf(x) −f(a)k ≤ kf(x) −f(aj)k+kf(aj) −f(a)k < ε 2+ε

2 = ε.

This proves that f is uniformly continuous on E.

Suppose that x,a∈E and kx−ak < δ.By (4), x belongs to Bδ_j(aj)for some 1≤j ≤N.Hence,

ka−ajk ≤ ka−xk + kx−ajk< δj+ δj =2δj = δ(aj), i.e., a also belongs to Bδ(aj)(aj).It follows, therefore, from the choice ofδ(aj)that

(5)

kf(x) −f(a)k ≤ kf(x) −f(aj)k+kf(aj) −f(a)k < ε 2+ε

2 = ε.

This proves that f is uniformly continuous on E.

Suppose that x,a∈E and kx−ak < δ.By (4), x belongs to Bδ_j(aj)for some 1≤j ≤N.Hence,

ka−ajk ≤ ka−xk + kx−ajk< δj+ δj =2δj = δ(aj), i.e., a also belongs to Bδ(aj)(aj).It follows, therefore, from the choice ofδ(aj)that

(5)

kf(x) −f(a)k ≤ kf(x) −f(aj)k+kf(aj) −f(a)k < ε 2+ε

2 = ε.

This proves that f is uniformly continuous on E.

Suppose that x,a∈E and kx−ak < δ.By (4), x belongs to Bδ_j(aj)for some 1≤j ≤N.Hence,

ka−ajk ≤ ka−xk + kx−ajk< δj+ δj =2δj = δ(aj),i.e., a also belongs to Bδ(aj)(aj).It follows, therefore, from the choice ofδ(aj)that

(5)

kf(x) −f(a)k ≤ kf(x) −f(aj)k+kf(aj) −f(a)k < ε 2+ε

2 = ε.

This proves that f is uniformly continuous on E.

Suppose that x,a∈E and kx−ak < δ.By (4), x belongs to Bδ_j(aj)for some 1≤j ≤N.Hence,

ka−ajk ≤ ka−xk + kx−ajk< δj+ δj =2δj = δ(aj), i.e., a also belongs to Bδ(aj)(aj).It follows, therefore, from the choice ofδ(aj)that

(5)

kf(x) −f(a)k ≤ kf(x) −f(aj)k+kf(aj) −f(a)k < ε 2+ε

2 = ε.

This proves that f is uniformly continuous on E.

Suppose that x,a∈E and kx−ak < δ.By (4), x belongs to Bδ_j(aj)for some 1≤j ≤N.Hence,

ka−ajk ≤ ka−xk + kx−ajk< δj+ δj =2δj = δ(aj),i.e., a also belongs to Bδ(aj)(aj).It follows, therefore, from the choice ofδ(aj)that

(5)

kf(x) −f(a)k ≤ kf(x) −f(aj)k+kf(aj) −f(a)k < ε 2+ε

2 = ε.

This proves that f is uniformly continuous on E.

Suppose that x,a∈E and kx−ak < δ.By (4), x belongs to Bδ_j(aj)for some 1≤j ≤N.Hence,

ka−ajk ≤ ka−xk + kx−ajk< δj+ δj =2δj = δ(aj), i.e., a also belongs to Bδ(aj)(aj).It follows, therefore, from the choice ofδ(aj)that

(5)

kf(x) −f(a)k ≤ kf(x) −f(aj)k+kf(aj) −f(a)k < ε 2+ε

2 = ε.

This proves that f is uniformly continuous on E.

Suppose that x,a∈E and kx−ak < δ.By (4), x belongs to Bδ_j(aj)for some 1≤j ≤N.Hence,

ka−ajk ≤ ka−xk + kx−ajk< δj+ δj =2δj = δ(aj), i.e., a also belongs to Bδ(aj)(aj).It follows, therefore, from the choice ofδ(aj)that

(5)

kf(x) −f(a)k ≤ kf(x) −f(aj)k+kf(aj) −f(a)k < ε 2+ε

2 = ε.

This proves that f is uniformly continuous on E.

Suppose that x,a∈E and kx−ak < δ.By (4), x belongs to Bδ_j(aj)for some 1≤j ≤N.Hence,

ka−ajk ≤ ka−xk + kx−ajk< δj+ δj =2δj = δ(aj), i.e., a also belongs to Bδ(aj)(aj).It follows, therefore, from the choice ofδ(aj)that

(5)

kf(x) −f(a)k ≤ kf(x) −f(aj)k+kf(aj) −f(a)k < ε 2+ε

2 = ε.

This proves that f is uniformly continuous on E.

Suppose that x,a∈E and kx−ak < δ.By (4), x belongs to Bδ_j(aj)for some 1≤j ≤N.Hence,

ka−ajk ≤ ka−xk + kx−ajk< δj+ δj =2δj = δ(aj), i.e., a also belongs to Bδ(aj)(aj).It follows, therefore, from the choice ofδ(aj)that

(5)

kf(x) −f(a)k ≤ kf(x) −f(aj)k+kf(aj) −f(a)k < ε 2+ε

2 = ε.

This proves that f is uniformly continuous on E.

Suppose that x,a∈E and kx−ak < δ.By (4), x belongs to Bδ_j(aj)for some 1≤j ≤N.Hence,

ka−ajk ≤ ka−xk + kx−ajk< δj+ δj =2δj = δ(aj), i.e., a also belongs to Bδ(aj)(aj).It follows, therefore, from the choice ofδ(aj)that

(5)

kf(x) −f(a)k ≤ kf(x) −f(aj)k+kf(aj) −f(a)k < ε 2+ε

2 = ε.

This proves that f is uniformly continuous on E.

Let n,m∈N and f :Rⁿ →R^m.Then the following three conditions are equivalent.

(i) f is continuous on Rⁿ.

(ii) f⁻¹(V)is open in Rⁿfor every open subset V of R^m. (iii) f⁻¹(E)is closed in Rⁿ for every closed subset E of R^m.

Let n,m∈N and f :Rⁿ →R^m.Then the following three conditions are equivalent.

(i) f is continuous on Rⁿ.

(ii) f⁻¹(V)is open in Rⁿfor every open subset V of R^m. (iii) f⁻¹(E)is closed in Rⁿ for every closed subset E of R^m.

Let n,m∈N and f :Rⁿ →R^m.Then the following three conditions are equivalent.

(i) f is continuous on Rⁿ.

(ii) f⁻¹(V)is open in Rⁿfor every open subset V of R^m. (iii) f⁻¹(E)is closed in Rⁿ for every closed subset E of R^m.

Let n,m∈N and f :Rⁿ →R^m.Then the following three conditions are equivalent.

(i) f is continuous on Rⁿ.

(ii) f⁻¹(V)is open in Rⁿfor every open subset V of R^m. (iii) f⁻¹(E)is closed in Rⁿ for every closed subset E of R^m.

Let n,m∈N, let E be open in Rⁿ, and suppose that f :E →R^m. Then f is continuous on E if and only if f⁻¹(V)is open in E for every open set V in R^m.

(i) If f(x) = _x2¹+1 and E = (0,1], then f is continuous on R and E is bounded, but f⁻¹(E) = (−∞, ∞)is not bounded.

(ii) If f(x) = x² and E = (1,4), then f is continuous on R and E is connected, but f⁻¹(E) = (−2, −1) ∪ (1,2)is not connected.

Let n,m∈N. If H is compact in Rⁿand f :H →R^m is continuous on H, then f(H)is compact in R^m.

Let n,m∈N. If E is connected in Rⁿ and f :E →R^m is continuous on E, then f(E)is connected in R^m.

The graph y =f(x)of a continuous real function f on an interval[a,b]is compact and connected.

Suppose that H is a nonempty subset of Rⁿand

f :H →R. If H is compact, and f is continuous on H, then M :=sup{f(x) :x∈H} and m :=inf{f(x) :x∈H}

are finite real numbers. Moreover, there exist points xM,xm ∈H such that M =f(xM)and m=f(xm).

Let n,m∈N. If H is a compact subset of Rⁿand

f :H →R^m is 1-1 and continuous, then f⁻¹ is continuous on f(H).

Thank you.