## 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 R**^{n}*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 R**^{n}*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 R*** ^{n}*.

*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**∈*B*δ(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 a**

*j*∈

*E and numbers*δ

*j*:= δ(a

*j*)/2 such that

(4) *E* ⊂

[*N*

*j=1*

*B*δ* _{j}*(a

*j*).

Setδ :=*min{δ*1, . . . , δ*N*}.

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

**x**∈*B*δ(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 a**

*j*∈

*E and numbers*δ

*j*:= δ(a

*j*)/2 such that

(4) *E* ⊂

[*N*

*j=1*

*B*δ* _{j}*(a

*j*).

Setδ :=*min{δ*1, . . . , δ*N*}.

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

**x**∈*B*δ(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 a**

*j*∈

*E and numbers*δ

*j*:= δ(a

*j*)/2 such that

(4) *E* ⊂

[*N*

*j=1*

*B*δ* _{j}*(a

*j*).

Setδ :=*min{δ*1, . . . , δ*N*}.

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

**x**∈*B*δ(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 a**

*j*∈

*E and numbers*δ

*j*:= δ(a

*j*)/2 such that

(4) *E* ⊂

[*N*

*j=1*

*B*δ* _{j}*(a

*j*).

Setδ :=*min{δ*1, . . . , δ*N*}.

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

**x**∈*B*δ(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 a**

*j*∈

*E and numbers*δ

*j*:= δ(a

*j*)/2 such that

(4) *E* ⊂

[*N*

*j=1*

*B*δ* _{j}*(a

*j*).

Setδ :=*min{δ*1, . . . , δ*N*}.

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

**x**∈*B*δ(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 a**

*j*∈

*E and numbers*δ

*j*:= δ(a

*j*)/2 such that

(4) *E* ⊂

[*N*

*j=1*

*B*δ* _{j}*(a

*j*).

Setδ :=*min{δ*1, . . . , δ*N*}.

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

**x**∈*B*δ(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 a**

*j*∈

*E and numbers*δ

*j*:= δ(a

*j*)/2 such that

(4) *E* ⊂

[*N*

*j=1*

*B*δ* _{j}*(a

*j*).

Setδ :=*min{δ*1, . . . , δ*N*}.

**Suppose that x,a**∈*E and* kx−**ak < δ.By (4), x belongs**
*to B*δ* _{j}*(a

*j*)for some 1≤

*j*≤

*N.*Hence,

ka−**a***j*k ≤ ka−**xk + kx**−**a***j*k< δ*j*+ δ*j* =2δ*j* = δ(a*j*), i.e.,
* a also belongs to B*δ(a

*j*)(a

*j*).It follows, therefore, from the choice ofδ(a

*j*)that

(5)

kf(x) −*f*(a)k ≤ kf(x) −*f*(a*j*)k+kf(a*j*) −*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 B*δ* _{j}*(a

*j*)for some 1≤

*j*≤

*N.*Hence,

ka−**a***j*k ≤ ka−**xk + kx**−**a***j*k< δ*j*+ δ*j* =2δ*j* = δ(a*j*), i.e.,
* a also belongs to B*δ(a

*j*)(a

*j*).It follows, therefore, from the choice ofδ(a

*j*)that

(5)

kf(x) −*f*(a)k ≤ kf(x) −*f*(a*j*)k+kf(a*j*) −*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 B*δ* _{j}*(a

*j*)for some 1≤

*j*≤

*N.*Hence,

ka−**a***j*k ≤ ka−**xk + kx**−**a***j*k< δ*j*+ δ*j* =2δ*j* = δ(a*j*), i.e.,
* a also belongs to B*δ(a

*j*)(a

*j*).It follows, therefore, from the choice ofδ(a

*j*)that

(5)

kf(x) −*f*(a)k ≤ kf(x) −*f*(a*j*)k+kf(a*j*) −*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 B*δ* _{j}*(a

*j*)for some 1≤

*j*≤

*N.*Hence,

**a***j*k ≤ ka−**xk + kx**−**a***j*k< δ*j*+ δ*j* =2δ*j* = δ(a*j*), i.e.,
* a also belongs to B*δ(a

*j*)(a

*j*).It follows, therefore, from the choice ofδ(a

*j*)that

(5)

kf(x) −*f*(a)k ≤ kf(x) −*f*(a*j*)k+kf(a*j*) −*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 B*δ* _{j}*(a

*j*)for some 1≤

*j*≤

*N.*Hence,

ka−**a***j*k ≤ ka−**xk + kx**−**a***j*k< δ*j*+ δ*j* =2δ*j* = δ(a*j*),i.e.,
* a also belongs to B*δ(a

*j*)(a

*j*).It follows, therefore, from the choice ofδ(a

*j*)that

(5)

kf(x) −*f*(a)k ≤ kf(x) −*f*(a*j*)k+kf(a*j*) −*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 B*δ* _{j}*(a

*j*)for some 1≤

*j*≤

*N.*Hence,

**a***j*k ≤ ka−**xk + kx**−**a***j*k< δ*j*+ δ*j* =2δ*j* = δ(a*j*), i.e.,
* a also belongs to B*δ(a

*j*)(a

*j*).It follows, therefore, from the choice ofδ(a

*j*)that

(5)

kf(x) −*f*(a)k ≤ kf(x) −*f*(a*j*)k+kf(a*j*) −*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 B*δ* _{j}*(a

*j*)for some 1≤

*j*≤

*N.*Hence,

ka−**a***j*k ≤ ka−**xk + kx**−**a***j*k< δ*j*+ δ*j* =2δ*j* = δ(a*j*),i.e.,
* a also belongs to B*δ(a

*j*)(a

*j*).It follows, therefore, from the choice ofδ(a

*j*)that

(5)

kf(x) −*f*(a)k ≤ kf(x) −*f*(a*j*)k+kf(a*j*) −*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 B*δ* _{j}*(a

*j*)for some 1≤

*j*≤

*N.*Hence,

**a***j*k ≤ ka−**xk + kx**−**a***j*k< δ*j*+ δ*j* =2δ*j* = δ(a*j*), i.e.,
* a also belongs to B*δ(a

*j*)(a

*j*).It follows, therefore, from the choice ofδ(a

*j*)that

(5)

kf(x) −*f*(a)k ≤ kf(x) −*f*(a*j*)k+kf(a*j*) −*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 B*δ* _{j}*(a

*j*)for some 1≤

*j*≤

*N.*Hence,

**a***j*k ≤ ka−**xk + kx**−**a***j*k< δ*j*+ δ*j* =2δ*j* = δ(a*j*), i.e.,
* a also belongs to B*δ(a

*j*)(a

*j*).It follows, therefore, from the choice ofδ(a

*j*)that

(5)

kf(x) −*f*(a)k ≤ kf(x) −*f*(a*j*)k+kf(a*j*) −*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 B*δ* _{j}*(a

*j*)for some 1≤

*j*≤

*N.*Hence,

**a***j*k ≤ ka−**xk + kx**−**a***j*k< δ*j*+ δ*j* =2δ*j* = δ(a*j*), i.e.,
* a also belongs to B*δ(a

*j*)(a

*j*).It follows, therefore, from the choice ofδ(a

*j*)that

(5)

kf(x) −*f*(a)k ≤ kf(x) −*f*(a*j*)k+kf(a*j*) −*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 B*δ* _{j}*(a

*j*)for some 1≤

*j*≤

*N.*Hence,

**a***j*k ≤ ka−**xk + kx**−**a***j*k< δ*j*+ δ*j* =2δ*j* = δ(a*j*), i.e.,
* a also belongs to B*δ(a

*j*)(a

*j*).It follows, therefore, from the choice ofδ(a

*j*)that

(5)

kf(x) −*f*(a)k ≤ kf(x) −*f*(a*j*)k+kf(a*j*) −*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 B*δ* _{j}*(a

*j*)for some 1≤

*j*≤

*N.*Hence,

**a***j*k ≤ ka−**xk + kx**−**a***j*k< δ*j*+ δ*j* =2δ*j* = δ(a*j*), i.e.,
* a also belongs to B*δ(a

*j*)(a

*j*).It follows, therefore, from the choice ofδ(a

*j*)that

(5)

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

2 = ε.

*This proves that f is uniformly continuous on E.*

*Let n,m*∈*N and f* :**R*** ^{n}* →

**R**

*.*

^{m}*Then the following three*

*conditions are equivalent.*

**(i) f is continuous on R**^{n}*.*

*(ii) f*^{−1}(V)**is open in R**^{n}**for every open subset V of R**^{m}*.*
*(iii) f*^{−1}(E)**is closed in R**^{n}**for every closed subset E of R**^{m}*.*

*Let n,m*∈*N and f* :**R*** ^{n}* →

**R**

*.*

^{m}*Then the following three*

*conditions are equivalent.*

**(i) f is continuous on R**^{n}*.*

*(ii) f*^{−1}(V)**is open in R**^{n}**for every open subset V of R**^{m}*.*
*(iii) f*^{−1}(E)**is closed in R**^{n}**for every closed subset E of R**^{m}*.*

*Let n,m*∈*N and f* :**R*** ^{n}* →

**R**

*.*

^{m}*Then the following three*

*conditions are equivalent.*

**(i) f is continuous on R**^{n}*.*

*(ii) f*^{−1}(V)**is open in R**^{n}**for every open subset V of R**^{m}*.*
*(iii) f*^{−1}(E)**is closed in R**^{n}**for every closed subset E of R**^{m}*.*

*Let n,m*∈*N and f* :**R*** ^{n}* →

**R**

*.*

^{m}*Then the following three*

*conditions are equivalent.*

**(i) f is continuous on R**^{n}*.*

*(ii) f*^{−1}(V)**is open in R**^{n}**for every open subset V of R**^{m}*.*
*(iii) f*^{−1}(E)**is closed in R**^{n}**for every closed subset E of R**^{m}*.*

*Let n,m*∈**N, let E be open in R**^{n}*, 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) = * _{x}*2

^{1}+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) = *x*^{2} *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 R**^{n}*and 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 R**^{n}*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 R**^{n}*and*

*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*
**x***M*,**x***m* ∈*H such that M* =*f*(x*M*)*and m*=*f*(x*m*).

*Let n,m*∈**N. If H is a compact subset of R**^{n}*and*

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