Calculus (I)

Calculus (I)

W^EN-C^HING L^IEN

Department of Mathematics National Cheng Kung University

2008

2-5: Continuity (II)-exercises

Example1:

f(x) =

1

q , if x = ^p_q ∈(0,1)∩Q,(p,q) =1 0 , if x ∈(0,1)∩(R\Q)

Show that f(x)is continuous on(0,1)∩(R\Q)and discontinuous on (0,1)∩Q

(Hint: ∀ǫ >0 , choose q0∈Ns.t. _q¹

0 < ǫ, then

#{^p_q ∈(0,1)∩Q:0<q<q0}is a finite number.)

Example2:

IF|a|<1 , then lim

n→∞

aⁿ =0

proof: |a| <1 and a6=0⇒ ∃b >0 s.t. |a|= _1+b¹

⇒0≤ |aⁿ−0| = _(1+b)¹ n ≤ _1+nb¹ < _nb¹ Sandwich theorem ⇒ lim

n→∞

aⁿ =0 2

2-5: Continuity (II)-exercises

Example1:

f(x) =

1

q , if x = ^p_q ∈(0,1)∩Q,(p,q) =1 0 , if x ∈(0,1)∩(R\Q)

Show that f(x)is continuous on(0,1)∩(R\Q)and discontinuous on (0,1)∩Q

(Hint: ∀ǫ >0 , choose q0∈Ns.t. _q¹

0 < ǫ, then

#{^p_q ∈(0,1)∩Q:0<q<q0}is a finite number.)

Example2:

IF|a|<1 , then lim

n→∞

aⁿ =0

proof: |a| <1 and a6=0⇒ ∃b >0 s.t. |a|= _1+b¹

⇒0≤ |aⁿ−0| = _(1+b)¹ n ≤ _1+nb¹ < _nb¹ Sandwich theorem ⇒ lim

n→∞

aⁿ =0 2

2-5: Continuity (II)-exercises

Example1:

f(x) =

1

q , if x = ^p_q ∈(0,1)∩Q,(p,q) =1 0 , if x ∈(0,1)∩(R\Q)

Show that f(x)is continuous on(0,1)∩(R\Q)and discontinuous on (0,1)∩Q

(Hint: ∀ǫ >0 , choose q0∈Ns.t. _q¹

0 < ǫ, then

#{^p_q ∈(0,1)∩Q:0<q<q0}is a finite number.)

Example2:

IF|a|<1 , then lim

n→∞

aⁿ =0

proof: |a| <1 and a6=0⇒ ∃b >0 s.t. |a|= _1+b¹

⇒0≤ |aⁿ−0| = _(1+b)¹ n ≤ _1+nb¹ < _nb¹ Sandwich theorem ⇒ lim

n→∞

aⁿ =0 2

2-5: Continuity (II)-exercises

Example1:

f(x) =

1

q , if x = ^p_q ∈(0,1)∩Q,(p,q) =1 0 , if x ∈(0,1)∩(R\Q)

Show that f(x)is continuous on(0,1)∩(R\Q)and discontinuous on (0,1)∩Q

(Hint: ∀ǫ >0 , choose q0∈Ns.t. _q¹

0 < ǫ, then

#{^p_q ∈(0,1)∩Q:0<q<q0}is a finite number.)

Example2:

IF|a|<1 , then lim

n→∞

aⁿ =0

proof: |a| <1 and a6=0⇒ ∃b >0 s.t. |a|= _1+b¹

⇒0≤ |aⁿ−0| = _(1+b)¹ n ≤ _1+nb¹ < _nb¹ Sandwich theorem ⇒ lim

n→∞

aⁿ =0 2

2-5: Continuity (II)-exercises

Example1:

f(x) =

1

q , if x = ^p_q ∈(0,1)∩Q,(p,q) =1 0 , if x ∈(0,1)∩(R\Q)

Show that f(x)is continuous on(0,1)∩(R\Q)and discontinuous on (0,1)∩Q

(Hint: ∀ǫ >0 , choose q0∈Ns.t. _q¹

0 < ǫ, then

#{^p_q ∈(0,1)∩Q:0<q<q0}is a finite number.)

Example2:

IF|a|<1 , then lim

n→∞

aⁿ =0

proof: |a| <1 and a6=0⇒ ∃b >0 s.t. |a|= _1+b¹

⇒0≤ |aⁿ−0| = _(1+b)¹ n ≤ _1+nb¹ < _nb¹ Sandwich theorem ⇒ lim

n→∞

aⁿ =0 2

2-5: Continuity (II)-exercises

Example1:

f(x) =

1

q , if x = ^p_q ∈(0,1)∩Q,(p,q) =1 0 , if x ∈(0,1)∩(R\Q)

Show that f(x)is continuous on(0,1)∩(R\Q)and discontinuous on (0,1)∩Q

(Hint: ∀ǫ >0 , choose q0∈Ns.t. _q¹

0 < ǫ, then

#{^p_q ∈(0,1)∩Q:0<q<q0}is a finite number.)

Example2:

IF|a|<1 , then lim

n→∞

aⁿ =0

proof: |a| <1 and a6=0⇒ ∃b >0 s.t. |a|= _1+b¹

⇒0≤ |aⁿ−0| = _(1+b)¹ n ≤ _1+nb¹ < _nb¹ Sandwich theorem ⇒ lim

n→∞

aⁿ =0 2

2-5: Continuity (II)-exercises

Example1:

f(x) =

1

q , if x = ^p_q ∈(0,1)∩Q,(p,q) =1 0 , if x ∈(0,1)∩(R\Q)

Show that f(x)is continuous on(0,1)∩(R\Q)and discontinuous on (0,1)∩Q

(Hint: ∀ǫ >0 , choose q0∈Ns.t. _q¹

0 < ǫ, then

#{^p_q ∈(0,1)∩Q:0<q<q0}is a finite number.)

Example2:

IF|a|<1 , then lim

n→∞

aⁿ =0

proof: |a| <1 and a6=0⇒ ∃b >0 s.t. |a|= _1+b¹

⇒0≤ |aⁿ−0| = _(1+b)¹ n ≤ _1+nb¹ < _nb¹ Sandwich theorem ⇒ lim

n→∞

aⁿ =0 2

2-5: Continuity (II)-exercises

Example1:

f(x) =

1

q , if x = ^p_q ∈(0,1)∩Q,(p,q) =1 0 , if x ∈(0,1)∩(R\Q)

Show that f(x)is continuous on(0,1)∩(R\Q)and discontinuous on (0,1)∩Q

(Hint: ∀ǫ >0 , choose q0∈Ns.t. _q¹

0 < ǫ, then

#{^p_q ∈(0,1)∩Q:0<q<q0}is a finite number.)

Example2:

IF|a|<1 , then lim

n→∞

aⁿ =0

proof: |a| <1 and a6=0⇒ ∃b >0 s.t. |a|= _1+b¹

⇒0≤ |aⁿ−0| = _(1+b)¹ n ≤ _1+nb¹ < _nb¹ Sandwich theorem ⇒ lim

n→∞

aⁿ =0 2

2-5: Continuity (II)-exercises

Example1:

f(x) =

1

q , if x = ^p_q ∈(0,1)∩Q,(p,q) =1 0 , if x ∈(0,1)∩(R\Q)

Show that f(x)is continuous on(0,1)∩(R\Q)and discontinuous on (0,1)∩Q

(Hint: ∀ǫ >0 , choose q0∈Ns.t. _q¹

0 < ǫ, then

#{^p_q ∈(0,1)∩Q:0<q<q0}is a finite number.)

Example2:

IF|a|<1 , then lim

n→∞

aⁿ =0

proof: |a| <1 and a6=0⇒ ∃b >0 s.t. |a|= _1+b¹

⇒0≤ |aⁿ−0| = _(1+b)¹ n ≤ _1+nb¹ < _nb¹ Sandwich theorem ⇒ lim

n→∞

aⁿ =0 2

Example3:

If f satisfies the following

|f(x1)−f(x2)| ≤2|x1−x2|¹³ ,∀x1,x2∈R, Show that f is continuous on R.

Remark:

(1) If f(x)is continuous at x =ξthen

nlim→∞

f(xn) =f(ξ), for every sequence xn that converges toξ i.e. lim

n→∞

f(xn) =f(lim

n→∞

xn).

(2) f(x)is uniformly continuous if∀ǫ >0 ,∃δ(ǫ)>0 such that|f(x)−f(ξ)|< ǫwhenever|x −ξ|< δ.

Remark:

(1) If f(x)is continuous at x =ξthen

nlim→∞

f(xn) =f(ξ), for every sequence xn that converges toξ i.e. lim

n→∞

f(xn) =f(lim

n→∞

xn).

(2) f(x)is uniformly continuous if∀ǫ >0 ,∃δ(ǫ)>0 such that|f(x)−f(ξ)|< ǫwhenever|x −ξ|< δ.

Remark:

(1) If f(x)is continuous at x =ξthen

nlim→∞

f(xn) =f(ξ), for every sequence xn that converges toξ i.e. lim

n→∞

f(xn) =f(lim

n→∞

xn).

(2) f(x)is uniformly continuous if∀ǫ >0 ,∃δ(ǫ)>0 such that|f(x)−f(ξ)|< ǫwhenever|x −ξ|< δ.

Theorem

Every function continuous in a closed interval[a,b]is uniformly continuous in that interval.

Ex:

Let f(x)be continuous for 0≤x ≤1. Suppose further that f(x)assumes rational values only and that f(x) = ¹₂ when x = ¹₂.

Prove that f(x) = ¹₂ everywhere.

Example:

f(x) = ¹_x , x ∈(0,1].

Prove that f is not uniformly continuous on (0,1].

Proof.

Letǫ= ¹₂

For anyδ >0, we can find n∈Ns.t. ¹_n < δ, However,

|f(1

n)−f( 1

n+1)|=1> ǫ, for|1

n − 1

n+1|= 1

n(n+1) < δ.

∴f(x)is not uniformly continuous.

(i.e. ∃ǫ >0 s.t. ∀δ >0 ,∃x,y ∈Domain(f)

|f(x)−f(y)| ≥ǫand|x −y|< δ.)

Example:

f(x) = ¹_x , x ∈(0,1].

Prove that f is not uniformly continuous on (0,1].

Proof.

Letǫ= ¹₂

For anyδ >0, we can find n∈Ns.t. ¹_n < δ, However,

|f(1

n)−f( 1

n+1)|=1> ǫ, for|1

n − 1

n+1|= 1

n(n+1) < δ.

∴f(x)is not uniformly continuous.

(i.e. ∃ǫ >0 s.t. ∀δ >0 ,∃x,y ∈Domain(f)

|f(x)−f(y)| ≥ǫand|x −y|< δ.)

Example:

f(x) = ¹_x , x ∈(0,1].

Prove that f is not uniformly continuous on (0,1].

Proof.

Letǫ= ¹₂

For anyδ >0, we can find n∈Ns.t. ¹_n < δ, However,

|f(1

n)−f( 1

n+1)|=1> ǫ, for|1

n − 1

n+1|= 1

n(n+1) < δ.

∴f(x)is not uniformly continuous.

(i.e. ∃ǫ >0 s.t. ∀δ >0 ,∃x,y ∈Domain(f)

|f(x)−f(y)| ≥ǫand|x −y|< δ.)

Example:

f(x) = ¹_x , x ∈(0,1].

Prove that f is not uniformly continuous on (0,1].

Proof.

Letǫ= ¹₂

For anyδ >0, we can find n∈Ns.t. ¹_n < δ, However,

|f(1

n)−f( 1

n+1)|=1> ǫ, for|1

n − 1

n+1|= 1

n(n+1) < δ.

∴f(x)is not uniformly continuous.

(i.e. ∃ǫ >0 s.t. ∀δ >0 ,∃x,y ∈Domain(f)

|f(x)−f(y)| ≥ǫand|x −y|< δ.)

Example:

f(x) = ¹_x , x ∈(0,1].

Prove that f is not uniformly continuous on (0,1].

Proof.

Letǫ= ¹₂

For anyδ >0, we can find n∈Ns.t. ¹_n < δ, However,

|f(1

n)−f( 1

n+1)|=1> ǫ, for|1

n − 1

n+1|= 1

n(n+1) < δ.

∴f(x)is not uniformly continuous.

(i.e. ∃ǫ >0 s.t. ∀δ >0 ,∃x,y ∈Domain(f)

|f(x)−f(y)| ≥ǫand|x −y|< δ.)

Example:

f(x) = ¹_x , x ∈(0,1].

Prove that f is not uniformly continuous on (0,1].

Proof.

Letǫ= ¹₂

For anyδ >0, we can find n∈Ns.t. ¹_n < δ, However,

|f(1

n)−f( 1

n+1)|=1> ǫ, for|1

n − 1

n+1|= 1

n(n+1) < δ.

∴f(x)is not uniformly continuous.

(i.e. ∃ǫ >0 s.t. ∀δ >0 ,∃x,y ∈Domain(f)

|f(x)−f(y)| ≥ǫand|x −y|< δ.)

Example:

f(x) = ¹_x , x ∈(0,1].

Prove that f is not uniformly continuous on (0,1].

Proof.

Letǫ= ¹₂

For anyδ >0, we can find n∈Ns.t. ¹_n < δ, However,

|f(1

n)−f( 1

n+1)|=1> ǫ, for|1

n − 1

n+1|= 1

n(n+1) < δ.

∴f(x)is not uniformly continuous.

(i.e. ∃ǫ >0 s.t. ∀δ >0 ,∃x,y ∈Domain(f)

|f(x)−f(y)| ≥ǫand|x −y|< δ.)

Lemma

If f : [a,b]−→Ris continuous, then for anyǫ >0, we can divide[a,b]into finite subintervals so that for any two points x,y in the same subinterval,

|f(x)−f(y)|< ǫ 2

Proof of theorem.

Givenǫ >0, we can find a=x0<x1<· · ·<xn =b s.t.

∀x,y ∈[xi−1,xi],|f(x)−f(y)|< ǫ 2 Now chooseδ =min{xi −xi−1:i =1,· · · ,n}.

If x,y ∈[a,b]and|x −y|< δ,

either (i) x,y ∈[xi,x_i+1]for some i ⇒ |f(x)−f(y)|< ₂^ǫ or (ii) x ∈[xi,x_i+1], y ∈[xi+1,x_i+2] for some i

⇒ |f(x)−f(y)| ≤ |f(x)−f(xi+1)|+|f(xi+1)−f(y)|

≤ ₂^ǫ +₂^ǫ =ǫ

Thus, f(x)is uniformly continuous on[a,b].

Proof of theorem.

Givenǫ >0,we can find a=x0<x1<· · ·<xn =b s.t.

∀x,y ∈[xi−1,xi],|f(x)−f(y)|< ǫ 2 Now chooseδ =min{xi −xi−1:i =1,· · · ,n}.

If x,y ∈[a,b]and|x −y|< δ,

either (i) x,y ∈[xi,x_i+1]for some i ⇒ |f(x)−f(y)|< ₂^ǫ or (ii) x ∈[xi,x_i+1], y ∈[xi+1,x_i+2] for some i

⇒ |f(x)−f(y)| ≤ |f(x)−f(xi+1)|+|f(xi+1)−f(y)|

≤ ₂^ǫ +₂^ǫ =ǫ

Thus, f(x)is uniformly continuous on[a,b].

Proof of theorem.

Givenǫ >0, we can find a=x0<x1<· · ·<xn =b s.t.

∀x,y ∈[xi−1,xi],|f(x)−f(y)|< ǫ 2 Now chooseδ =min{xi −xi−1:i =1,· · · ,n}.

If x,y ∈[a,b]and|x −y|< δ,

either (i) x,y ∈[xi,x_i+1]for some i ⇒ |f(x)−f(y)|< ₂^ǫ or (ii) x ∈[xi,x_i+1], y ∈[xi+1,x_i+2] for some i

⇒ |f(x)−f(y)| ≤ |f(x)−f(xi+1)|+|f(xi+1)−f(y)|

≤ ₂^ǫ +₂^ǫ =ǫ

Thus, f(x)is uniformly continuous on[a,b].

Proof of theorem.

Givenǫ >0, we can find a=x0<x1<· · ·<xn =b s.t.

∀x,y ∈[xi−1,xi],|f(x)−f(y)|< ǫ 2 Now chooseδ =min{xi −xi−1:i =1,· · · ,n}.

If x,y ∈[a,b]and|x −y|< δ,

either (i) x,y ∈[xi,x_i+1]for some i ⇒ |f(x)−f(y)|< ₂^ǫ or (ii) x ∈[xi,x_i+1], y ∈[xi+1,x_i+2] for some i

⇒ |f(x)−f(y)| ≤ |f(x)−f(xi+1)|+|f(xi+1)−f(y)|

≤ ₂^ǫ +₂^ǫ =ǫ

Thus, f(x)is uniformly continuous on[a,b].

Proof of theorem.

Givenǫ >0, we can find a=x0<x1<· · ·<xn =b s.t.

∀x,y ∈[xi−1,xi],|f(x)−f(y)|< ǫ 2 Now chooseδ =min{xi −xi−1:i =1,· · · ,n}.

If x,y ∈[a,b]and|x −y|< δ,

either (i) x,y ∈[xi,x_i+1]for some i ⇒ |f(x)−f(y)|< ₂^ǫ or (ii) x ∈[xi,x_i+1], y ∈[xi+1,x_i+2] for some i

⇒ |f(x)−f(y)| ≤ |f(x)−f(xi+1)|+|f(xi+1)−f(y)|

≤ ₂^ǫ +₂^ǫ =ǫ

Thus, f(x)is uniformly continuous on[a,b].

Proof of theorem.

Givenǫ >0, we can find a=x0<x1<· · ·<xn =b s.t.

∀x,y ∈[xi−1,xi],|f(x)−f(y)|< ǫ 2 Now chooseδ =min{xi −xi−1:i =1,· · · ,n}.

If x,y ∈[a,b]and|x −y|< δ,

either (i) x,y ∈[xi,x_i+1]for some i ⇒ |f(x)−f(y)|< ₂^ǫ or (ii) x ∈[xi,x_i+1], y ∈[xi+1,x_i+2] for some i

⇒ |f(x)−f(y)| ≤ |f(x)−f(xi+1)|+|f(xi+1)−f(y)|

≤ ₂^ǫ +₂^ǫ =ǫ

Thus, f(x)is uniformly continuous on[a,b].

Proof of theorem.

Givenǫ >0, we can find a=x0<x1<· · ·<xn =b s.t.

∀x,y ∈[xi−1,xi],|f(x)−f(y)|< ǫ 2 Now chooseδ =min{xi −xi−1:i =1,· · · ,n}.

If x,y ∈[a,b]and|x −y|< δ,

either (i) x,y ∈[xi,x_i+1]for some i ⇒ |f(x)−f(y)|< ₂^ǫ or (ii) x ∈[xi,x_i+1], y ∈[xi+1,x_i+2] for some i

⇒ |f(x)−f(y)| ≤ |f(x)−f(xi+1)|+|f(xi+1)−f(y)|

≤ ₂^ǫ +₂^ǫ =ǫ

Thus, f(x)is uniformly continuous on[a,b].

Proof of theorem.

Givenǫ >0, we can find a=x0<x1<· · ·<xn =b s.t.

∀x,y ∈[xi−1,xi],|f(x)−f(y)|< ǫ 2 Now chooseδ =min{xi −xi−1:i =1,· · · ,n}.

If x,y ∈[a,b]and|x −y|< δ,

either (i) x,y ∈[xi,x_i+1]for some i ⇒ |f(x)−f(y)|< ₂^ǫ or (ii) x ∈[xi,x_i+1], y ∈[xi+1,x_i+2] for some i

⇒ |f(x)−f(y)| ≤ |f(x)−f(xi+1)|+|f(xi+1)−f(y)|

≤ ₂^ǫ +₂^ǫ =ǫ

Thus, f(x)is uniformly continuous on[a,b].

Ex:f(x)is continuous at x =c and f(c)>0.

Then,∃δ >0 s.t. f(x)>0 , for x ∈(c−δ,c+δ).

Remark:

Definition (1) Sis a set ofR

B - upper bound ofS: x ≤B, ∀x ∈S A - lower bound ofS: A≥x, ∀x ∈S

Definition (2)

supremum = least upper bound infimum = greatest lower bound

Remark:

Definition (1) Sis a set ofR

B - upper bound ofS: x ≤B, ∀x ∈S A - lower bound ofS: A≥x, ∀x ∈S

Definition (2)

supremum = least upper bound infimum = greatest lower bound

Thank you.