5.1 – Functional Iteration

Sol. Non-linear Fun.

³⁷

Sol. Non-linear Fun.

³⁷

5.1 – Functional Iteration

Fixed-point iteration or functional iteration: Given a continuous function

f

^,

choose an initial point

x

₀ and generate

{x

^k

}

^{k≥0 by}

x

_k+1

= f (x

_k

), k ≥ 0.

{x

^k

}

may not converge, e.g.,

f (x) = 3x

^.

However, when the sequence converges, say,

k→∞

lim x

_k

= x

^∗

,

then, since

f

is continuous,

f (x

^∗

) = f ( lim

k→∞

x

_k

) = lim

k→∞

f (x

_k

) = lim

k→∞

x

_k+1

= x

^∗

.

That is,

x

^∗ is a fixed point of

f

^.

Note that Newton’s method for solving

g(x) = 0 x

_k+1

= x

_k

− g(x

_k

)

g

⁰

(x

_k

)

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

³⁷

5.1 – Functional Iteration

Fixed-point iteration or functional iteration: Given a continuous function

f

^,

choose an initial point

x

₀ and generate

{x

^k

}

^{k≥0 by}

x

_k+1

= f (x

_k

), k ≥ 0.

{x

^k

}

may not converge, e.g.,

f (x) = 3x

. However, when the sequence converges, say,

k→∞

lim x

_k

= x

^∗

,

then, since

f

is continuous,

f (x

^∗

) = f ( lim

k→∞

x

_k

) = lim

k→∞

f (x

_k

) = lim

k→∞

x

_k+1

= x

^∗

.

That is,

x

^∗ is a fixed point of

f

^.

Note that Newton’s method for solving

g(x) = 0 x

_k+1

= x

_k

− g(x

_k

)

g

⁰

(x

_k

)

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

³⁷

5.1 – Functional Iteration

Fixed-point iteration or functional iteration: Given a continuous function

f

^,

choose an initial point

x

₀ and generate

{x

^k

}

^{k≥0 by}

x

_k+1

= f (x

_k

), k ≥ 0.

{x

^k

}

may not converge, e.g.,

f (x) = 3x

. However, when the sequence converges, say,

k→∞

lim x

_k

= x

^∗

,

then, since

f

is continuous,

f (x

^∗

) = f ( lim

k→∞

x

_k

) = lim

k→∞

f (x

_k

) = lim

k→∞

x

_k+1

= x

^∗

.

That is,

x

^∗ is a fixed point of

f

^.

Note that Newton’s method for solving

g(x) = 0 x

_k+1

= x

_k

− g(x

_k

)

g

⁰

(x

_k

)

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

³⁷

5.1 – Functional Iteration

Fixed-point iteration or functional iteration: Given a continuous function

f

^,

choose an initial point

x

₀ and generate

{x

^k

}

^{k≥0 by}

x

_k+1

= f (x

_k

), k ≥ 0.

{x

^k

}

may not converge, e.g.,

f (x) = 3x

. However, when the sequence converges, say,

k→∞

lim x

_k

= x

^∗

,

then, since

f

is continuous,

f (x

^∗

) = f ( lim

k→∞

x

_k

) = lim

k→∞

f (x

_k

) = lim

k→∞

x

_k+1

= x

^∗

.

That is,

x

^∗ is a fixed point of

f

^.

Note that Newton’s method for solving

g(x) = 0 x

_k+1

= x

_k

− g(x

_k

)

g

⁰

(x

_k

)

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

³⁷

5.1 – Functional Iteration

Fixed-point iteration or functional iteration: Given a continuous function

f

^,

choose an initial point

x

₀ and generate

{x

^k

}

^{k≥0 by}

x

_k+1

= f (x

_k

), k ≥ 0.

{x

^k

}

may not converge, e.g.,

f (x) = 3x

. However, when the sequence converges, say,

k→∞

lim x

_k

= x

^∗

,

then, since

f

is continuous,

f (x

^∗

) = f ( lim

k→∞

x

_k

) = lim

k→∞

f (x

_k

) = lim

k→∞

x

_k+1

= x

^∗

.

That is,

x

^∗ is a fixed point of

f

^.

Note that Newton’s method for solving

g(x) = 0 x

_k+1

= x

_k

− g(x

_k

)

g

⁰

(x

_k

)

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

³⁸

is just a special case of functional iteration in which

f (x) = x − g(x) g

⁰

(x) .

Definition 5 A function (mapping)

f

is said to be contractive if there exists a constant

0 ≤ λ < 1

^{such that}

|f(x) − f(y)| ≤ λ|x − y|

for all

x, y

in the domain of

f

^.

Theorem 4 (Contractive Mapping Theorem) Suppose

f : D → D

^{, where}

D ⊆ R

is a closed set, is a contractive mapping. Then

f

^{has a} unique fixed point in

D

. Moreover, this fixed point is the limit of every sequence obtained by

x

_k+1

= f (x

_k

)

with any initial point

x

₀^.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

³⁸

is just a special case of functional iteration in which

f (x) = x − g(x) g

⁰

(x) .

Definition 5 A function (mapping)

f

is said to be contractive if there exists a constant

0 ≤ λ < 1

^{such that}

|f(x) − f(y)| ≤ λ|x − y|

for all

x, y

in the domain of

f

^.

Theorem 4 (Contractive Mapping Theorem) Suppose

f : D → D

^{, where}

D ⊆ R

is a closed set, is a contractive mapping. Then

f

^{has a} unique fixed point in

D

. Moreover, this fixed point is the limit of every sequence obtained by

x

_k+1

= f (x

_k

)

with any initial point

x

₀^.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

³⁸

is just a special case of functional iteration in which

f (x) = x − g(x) g

⁰

(x) .

Definition 5 A function (mapping)

f

is said to be contractive if there exists a constant

0 ≤ λ < 1

^{such that}

|f(x) − f(y)| ≤ λ|x − y|

for all

x, y

in the domain of

f

^.

Theorem 4 (Contractive Mapping Theorem) Suppose

f : D → D

^{, where}

D ⊆ R

is a closed set, is a contractive mapping. Then

f

^{has a} unique fixed point in

D

. Moreover, this fixed point is the limit of every sequence obtained by

x

_k+1

= f (x

_k

)

with any initial point

x

₀^.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

³⁹

Proof: We first show that

lim

k→∞

x

_k ^exists.

Since

x

_k

= x

₀

+(x

₁

−x

⁰

)+(x

₂

−x

¹

)+· · ·+(x

^k

−x

^k−1

) = x

₀

+

k

X

i=1

(x

_i

−x

ⁱ⁻¹

), {x

^k

}

^k≥0 converges if and only if

P

∞

i=1

(x

_i

− x

ⁱ⁻¹

)

converges and it is sufficient to show

P

∞

i=1

|x

ⁱ

− x

ⁱ⁻¹

|

^converges.

Since

f

is contractive, we have

|x

ⁱ

− x

ⁱ⁻¹

| = |f(x

ⁱ⁻¹

) − f(x

ⁱ⁻²

)|

≤ λ|x

ⁱ⁻¹

− x

ⁱ⁻²

|

≤ λ

²

|x

ⁱ⁻²

− x

ⁱ⁻³

|

.. .

≤ λ

ⁱ⁻¹

|x

¹

− x

⁰

|.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

³⁹

Proof: We first show that

lim

k→∞

x

_k exists. Since

x

_k

= x

₀

+(x

₁

−x

⁰

)+(x

₂

−x

¹

)+· · ·+(x

^k

−x

^k−1

) = x

₀

+

k

X

i=1

(x

_i

−x

ⁱ⁻¹

),

{x

^k

}

^k≥0 converges if and only if

P

∞

i=1

(x

_i

− x

ⁱ⁻¹

)

converges and it is sufficient to show

P

∞

i=1

|x

ⁱ

− x

ⁱ⁻¹

|

^converges.

Since

f

is contractive, we have

|x

ⁱ

− x

ⁱ⁻¹

| = |f(x

ⁱ⁻¹

) − f(x

ⁱ⁻²

)|

≤ λ|x

ⁱ⁻¹

− x

ⁱ⁻²

|

≤ λ

²

|x

ⁱ⁻²

− x

ⁱ⁻³

|

.. .

≤ λ

ⁱ⁻¹

|x

¹

− x

⁰

|.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

³⁹

Proof: We first show that

lim

k→∞

x

_k exists. Since

x

_k

= x

₀

+(x

₁

−x

⁰

)+(x

₂

−x

¹

)+· · ·+(x

^k

−x

^k−1

) = x

₀

+

k

X

i=1

(x

_i

−x

ⁱ⁻¹

), {x

^k

}

^k≥0 converges if and only if

P

∞

i=1

(x

_i

− x

ⁱ⁻¹

)

converges and it is sufficient to show

P

∞

i=1

|x

ⁱ

− x

ⁱ⁻¹

|

^converges.

Since

f

is contractive, we have

|x

ⁱ

− x

ⁱ⁻¹

| = |f(x

ⁱ⁻¹

) − f(x

ⁱ⁻²

)|

≤ λ|x

ⁱ⁻¹

− x

ⁱ⁻²

|

≤ λ

²

|x

ⁱ⁻²

− x

ⁱ⁻³

|

.. .

≤ λ

ⁱ⁻¹

|x

¹

− x

⁰

|.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

³⁹

Proof: We first show that

lim

k→∞

x

_k exists. Since

x

_k

= x

₀

+(x

₁

−x

⁰

)+(x

₂

−x

¹

)+· · ·+(x

^k

−x

^k−1

) = x

₀

+

k

X

i=1

(x

_i

−x

ⁱ⁻¹

), {x

^k

}

^k≥0 converges if and only if

P

∞

i=1

(x

_i

− x

ⁱ⁻¹

)

converges and it is sufficient to show

P

∞

i=1

|x

ⁱ

− x

ⁱ⁻¹

|

^converges.

Since

f

is contractive, we have

|x

ⁱ

− x

ⁱ⁻¹

| = |f(x

ⁱ⁻¹

) − f(x

ⁱ⁻²

)|

≤ λ|x

ⁱ⁻¹

− x

ⁱ⁻²

|

≤ λ

²

|x

ⁱ⁻²

− x

ⁱ⁻³

|

.. .

≤ λ

ⁱ⁻¹

|x

¹

− x

⁰

|.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

⁴⁰

Then we have

∞

X

i=1

|x

ⁱ

− x

ⁱ⁻¹

| ≤

∞

X

i=1

λ

ⁱ⁻¹

|x

¹

− x

⁰

|

= |x

¹

− x

⁰

|

∞

X

i=1

λ

ⁱ⁻¹

= 1

1 − λ |x

¹

− x

⁰

|

since

0 ≤ λ < 1

. This show that

P

∞

i=1

|x

ⁱ

− x

ⁱ⁻¹

|

is bounded, hence it converges. Let

lim

k→∞

x

_k

= x

^∗.

f

is a contractive mapping

⇒ f

is continuous

⇒ x

^∗

= lim

k→∞

x

_k

= lim

k→∞

f (x

_k−1

) = f ( lim

k→∞

x

_k−1

) = f (x

^∗

)

⇒ x

^∗ is a fixed point of

f

^.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

⁴⁰

Then we have

∞

X

i=1

|x

ⁱ

− x

ⁱ⁻¹

| ≤

∞

X

i=1

λ

ⁱ⁻¹

|x

¹

− x

⁰

|

= |x

¹

− x

⁰

|

∞

X

i=1

λ

ⁱ⁻¹

= 1

1 − λ |x

¹

− x

⁰

|

since

0 ≤ λ < 1

. This show that

P

∞

i=1

|x

ⁱ

− x

ⁱ⁻¹

|

is bounded, hence it converges. Let

lim

k→∞

x

_k

= x

^∗.

f

is a contractive mapping

⇒ f

is continuous

⇒ x

^∗

= lim

k→∞

x

_k

= lim

k→∞

f (x

_k−1

) = f ( lim

k→∞

x

_k−1

) = f (x

^∗

)

⇒ x

^∗ is a fixed point of

f

^.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

⁴⁰

Then we have

∞

X

i=1

|x

ⁱ

− x

ⁱ⁻¹

| ≤

∞

X

i=1

λ

ⁱ⁻¹

|x

¹

− x

⁰

|

= |x

¹

− x

⁰

|

∞

X

i=1

λ

ⁱ⁻¹

= 1

1 − λ |x

¹

− x

⁰

|

since

0 ≤ λ < 1

^.

This show that

P

∞

i=1

|x

ⁱ

− x

ⁱ⁻¹

|

is bounded, hence it converges. Let

lim

k→∞

x

_k

= x

^∗.

f

is a contractive mapping

⇒ f

is continuous

⇒ x

^∗

= lim

k→∞

x

_k

= lim

k→∞

f (x

_k−1

) = f ( lim

k→∞

x

_k−1

) = f (x

^∗

)

⇒ x

^∗ is a fixed point of

f

^.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

⁴⁰

Then we have

∞

X

i=1

|x

ⁱ

− x

ⁱ⁻¹

| ≤

∞

X

i=1

λ

ⁱ⁻¹

|x

¹

− x

⁰

|

= |x

¹

− x

⁰

|

∞

X

i=1

λ

ⁱ⁻¹

= 1

1 − λ |x

¹

− x

⁰

|

since

0 ≤ λ < 1

^.

This show that

P

∞

i=1

|x

ⁱ

− x

ⁱ⁻¹

|

is bounded, hence it converges. Let

lim

k→∞

x

_k

= x

^∗.

f

is a contractive mapping

⇒ f

is continuous

⇒ x

^∗

= lim

k→∞

x

_k

= lim

k→∞

f (x

_k−1

) = f ( lim

k→∞

x

_k−1

) = f (x

^∗

)

⇒ x

^∗ is a fixed point of

f

^.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

⁴⁰

Then we have

∞

X

i=1

|x

ⁱ

− x

ⁱ⁻¹

| ≤

∞

X

i=1

λ

ⁱ⁻¹

|x

¹

− x

⁰

|

= |x

¹

− x

⁰

|

∞

X

i=1

λ

ⁱ⁻¹

= 1

1 − λ |x

¹

− x

⁰

|

since

0 ≤ λ < 1

. This show that

P

∞

i=1

|x

ⁱ

− x

ⁱ⁻¹

|

is bounded,

hence it converges. Let

lim

k→∞

x

_k

= x

^∗.

f

is a contractive mapping

⇒ f

is continuous

⇒ x

^∗

= lim

k→∞

x

_k

= lim

k→∞

f (x

_k−1

) = f ( lim

k→∞

x

_k−1

) = f (x

^∗

)

⇒ x

^∗ is a fixed point of

f

^.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

⁴⁰

Then we have

∞

X

i=1

|x

ⁱ

− x

ⁱ⁻¹

| ≤

∞

X

i=1

λ

ⁱ⁻¹

|x

¹

− x

⁰

|

= |x

¹

− x

⁰

|

∞

X

i=1

λ

ⁱ⁻¹

= 1

1 − λ |x

¹

− x

⁰

|

since

0 ≤ λ < 1

. This show that

P

∞

i=1

|x

ⁱ

− x

ⁱ⁻¹

|

is bounded, hence it converges.

Let

lim

k→∞

x

_k

= x

^∗.

f

is a contractive mapping

⇒ f

is continuous

⇒ x

^∗

= lim

k→∞

x

_k

= lim

k→∞

f (x

_k−1

) = f ( lim

k→∞

x

_k−1

) = f (x

^∗

)

⇒ x

^∗ is a fixed point of

f

^.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

⁴⁰

Then we have

∞

X

i=1

|x

ⁱ

− x

ⁱ⁻¹

| ≤

∞

X

i=1

λ

ⁱ⁻¹

|x

¹

− x

⁰

|

= |x

¹

− x

⁰

|

∞

X

i=1

λ

ⁱ⁻¹

= 1

1 − λ |x

¹

− x

⁰

|

since

0 ≤ λ < 1

. This show that

P

∞

i=1

|x

ⁱ

− x

ⁱ⁻¹

|

is bounded, hence it converges. Let

lim

k→∞

x

_k

= x

^∗.

f

is a contractive mapping

⇒ f

is continuous

⇒ x

^∗

= lim

k→∞

x

_k

= lim

k→∞

f (x

_k−1

) = f ( lim

k→∞

x

_k−1

) = f (x

^∗

)

⇒ x

^∗ is a fixed point of

f

^.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

⁴⁰

Then we have

∞

X

i=1

|x

ⁱ

− x

ⁱ⁻¹

| ≤

∞

X

i=1

λ

ⁱ⁻¹

|x

¹

− x

⁰

|

= |x

¹

− x

⁰

|

∞

X

i=1

λ

ⁱ⁻¹

= 1

1 − λ |x

¹

− x

⁰

|

since

0 ≤ λ < 1

. This show that

P

∞

i=1

|x

ⁱ

− x

ⁱ⁻¹

|

is bounded, hence it converges. Let

lim

k→∞

x

_k

= x

^∗.

f

is a contractive mapping

⇒ f

is continuous

⇒ x

^∗

= lim

k→∞

x

_k

= lim

k→∞

f (x

_k−1

) = f ( lim

k→∞

x

_k−1

) = f (x

^∗

)

⇒ x

^∗ is a fixed point of

f

^.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

⁴⁰

Then we have

∞

X

i=1

|x

ⁱ

− x

ⁱ⁻¹

| ≤

∞

X

i=1

λ

ⁱ⁻¹

|x

¹

− x

⁰

|

= |x

¹

− x

⁰

|

∞

X

i=1

λ

ⁱ⁻¹

= 1

1 − λ |x

¹

− x

⁰

|

since

0 ≤ λ < 1

. This show that

P

∞

i=1

|x

ⁱ

− x

ⁱ⁻¹

|

is bounded, hence it converges. Let

lim

k→∞

x

_k

= x

^∗.

f

is a contractive mapping

⇒ f

is continuous

⇒ x

^∗

= lim

k→∞

x

_k

= lim

k→∞

f (x

_k−1

) = f ( lim

k→∞

x

_k−1

) = f (x

^∗

)

⇒ x

^∗ is a fixed point of

f

^.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

⁴⁰

Then we have

∞

X

i=1

|x

ⁱ

− x

ⁱ⁻¹

| ≤

∞

X

i=1

λ

ⁱ⁻¹

|x

¹

− x

⁰

|

= |x

¹

− x

⁰

|

∞

X

i=1

λ

ⁱ⁻¹

= 1

1 − λ |x

¹

− x

⁰

|

since

0 ≤ λ < 1

. This show that

P

∞

i=1

|x

ⁱ

− x

ⁱ⁻¹

|

is bounded, hence it converges. Let

lim

k→∞

x

_k

= x

^∗.

f

is a contractive mapping

⇒ f

is continuous

⇒ x

^∗

= lim

k→∞

x

_k

= lim

k→∞

f (x

_k−1

) = f ( lim

k→∞

x

_k−1

) = f (x

^∗

)

⇒ x

^∗ is a fixed point of

f

^.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

⁴⁰

Then we have

∞

X

i=1

|x

ⁱ

− x

ⁱ⁻¹

| ≤

∞

X

i=1

λ

ⁱ⁻¹

|x

¹

− x

⁰

|

= |x

¹

− x

⁰

|

∞

X

i=1

λ

ⁱ⁻¹

= 1

1 − λ |x

¹

− x

⁰

|

since

0 ≤ λ < 1

. This show that

P

∞

i=1

|x

ⁱ

− x

ⁱ⁻¹

|

is bounded, hence it converges. Let

lim

k→∞

x

_k

= x

^∗.

f

is a contractive mapping

⇒ f

is continuous

⇒ x

^∗

= lim

k→∞

x

_k

= lim

k→∞

f (x

_k−1

) = f ( lim

k→∞

x

_k−1

) = f (x

^∗

)

⇒ x

^∗ is a fixed point of

f

^.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

⁴¹

To prove the uniqueness, let

x

^and

y

both be fixed points of

f

^.

Then

|x − y| = |f(x) − f(y)| ≤ λ|x − y|.

Since

λ < 1

, this forces

|x − y| = 0

^{. That is,}

x = y

^.

Theorem 5 If

f ∈ C[a, b]

^{such that}

a ≤ f(x) ≤ b

^{for all}

x ∈ [a, b]

^{, then}

f

^has a fixed point in

[a, b]

. Suppose, in addition, that

f

⁰

(x)

^{exists in}

(a, b)

and there exists a positive constant

M < 1

^{such that}

|f

⁰

(x)| ≤ M < 1

^for

all

x ∈ (a, b)

. Then the fixed point is unique.

Proof: If

f (a) = a

^or

f (b) = b

^{, then}

a

^or

b

is a fixed point of

f

^{and we are}

done. Otherwise, it must be

g(a) > a

^and

g(b) < b

^.

Let

h(x) = f (x) − x

^.

⇒ h ∈ C[a, b]

^since

f ∈ C[a, b]

^{, and}

h(a) = f (a) − a > 0

^,

h(b) = f (b) − b < 0

^.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

⁴¹

To prove the uniqueness, let

x

^and

y

both be fixed points of

f

^{. Then}

|x − y| = |f(x) − f(y)| ≤ λ|x − y|.

Since

λ < 1

, this forces

|x − y| = 0

^{. That is,}

x = y

^.

Theorem 5 If

f ∈ C[a, b]

^{such that}

a ≤ f(x) ≤ b

^{for all}

x ∈ [a, b]

^{, then}

f

^has a fixed point in

[a, b]

. Suppose, in addition, that

f

⁰

(x)

^{exists in}

(a, b)

and there exists a positive constant

M < 1

^{such that}

|f

⁰

(x)| ≤ M < 1

^for

all

x ∈ (a, b)

. Then the fixed point is unique.

Proof: If

f (a) = a

^or

f (b) = b

^{, then}

a

^or

b

is a fixed point of

f

^{and we are}

done. Otherwise, it must be

g(a) > a

^and

g(b) < b

^.

Let

h(x) = f (x) − x

^.

⇒ h ∈ C[a, b]

^since

f ∈ C[a, b]

^{, and}

h(a) = f (a) − a > 0

^,

h(b) = f (b) − b < 0

^.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

⁴¹

To prove the uniqueness, let

x

^and

y

both be fixed points of

f

^{. Then}

|x − y| = |f(x) − f(y)| ≤ λ|x − y|.

Since

λ < 1

, this forces

|x − y| = 0

^{. That is,}

x = y

^.

Theorem 5 If

f ∈ C[a, b]

^{such that}

a ≤ f(x) ≤ b

^{for all}

x ∈ [a, b]

^{, then}

f

^has a fixed point in

[a, b]

. Suppose, in addition, that

f

⁰

(x)

^{exists in}

(a, b)

and there exists a positive constant

M < 1

^{such that}

|f

⁰

(x)| ≤ M < 1

^for

all

x ∈ (a, b)

. Then the fixed point is unique.

Proof: If

f (a) = a

^or

f (b) = b

^{, then}

a

^or

b

is a fixed point of

f

^{and we are}

done. Otherwise, it must be

g(a) > a

^and

g(b) < b

^.

Let

h(x) = f (x) − x

^.

⇒ h ∈ C[a, b]

^since

f ∈ C[a, b]

^{, and}

h(a) = f (a) − a > 0

^,

h(b) = f (b) − b < 0

^.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

⁴¹

To prove the uniqueness, let

x

^and

y

both be fixed points of

f

^{. Then}

|x − y| = |f(x) − f(y)| ≤ λ|x − y|.

Since

λ < 1

, this forces

|x − y| = 0

^{. That is,}

x = y

^.

Theorem 5 If

f ∈ C[a, b]

^{such that}

a ≤ f(x) ≤ b

^{for all}

x ∈ [a, b]

^{, then}

f

^has a fixed point in

[a, b]

. Suppose, in addition, that

f

⁰

(x)

^{exists in}

(a, b)

and there exists a positive constant

M < 1

^{such that}

|f

⁰

(x)| ≤ M < 1

^for

all

x ∈ (a, b)

. Then the fixed point is unique.

Proof: If

f (a) = a

^or

f (b) = b

^{, then}

a

^or

b

is a fixed point of

f

^{and we are}

done. Otherwise, it must be

g(a) > a

^and

g(b) < b

^.

Let

h(x) = f (x) − x

^.

⇒ h ∈ C[a, b]

^since

f ∈ C[a, b]

^{, and}

h(a) = f (a) − a > 0

^,

h(b) = f (b) − b < 0

^.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

⁴¹

To prove the uniqueness, let

x

^and

y

both be fixed points of

f

^{. Then}

|x − y| = |f(x) − f(y)| ≤ λ|x − y|.

Since

λ < 1

, this forces

|x − y| = 0

^{. That is,}

x = y

^.

Theorem 5 If

f ∈ C[a, b]

^{such that}

a ≤ f(x) ≤ b

^{for all}

x ∈ [a, b]

^{, then}

f

^has a fixed point in

[a, b]

. Suppose, in addition, that

f

⁰

(x)

^{exists in}

(a, b)

and there exists a positive constant

M < 1

^{such that}

|f

⁰

(x)| ≤ M < 1

^for

all

x ∈ (a, b)

. Then the fixed point is unique.

Proof: If

f (a) = a

^or

f (b) = b

^{, then}

a

^or

b

is a fixed point of

f

^{and we are}

done.

Otherwise, it must be

g(a) > a

^and

g(b) < b

^.

Let

h(x) = f (x) − x

^.

⇒ h ∈ C[a, b]

^since

f ∈ C[a, b]

^{, and}

h(a) = f (a) − a > 0

^,

h(b) = f (b) − b < 0

^.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

⁴¹

To prove the uniqueness, let

x

^and

y

both be fixed points of

f

^{. Then}

|x − y| = |f(x) − f(y)| ≤ λ|x − y|.

Since

λ < 1

, this forces

|x − y| = 0

^{. That is,}

x = y

^.

Theorem 5 If

f ∈ C[a, b]

^{such that}

a ≤ f(x) ≤ b

^{for all}

x ∈ [a, b]

^{, then}

f

^has a fixed point in

[a, b]

. Suppose, in addition, that

f

⁰

(x)

^{exists in}

(a, b)

and there exists a positive constant

M < 1

^{such that}

|f

⁰

(x)| ≤ M < 1

^for

all

x ∈ (a, b)

. Then the fixed point is unique.

Proof: If

f (a) = a

^or

f (b) = b

^{, then}

a

^or

b

is a fixed point of

f

^{and we are}

done. Otherwise, it must be

g(a) > a

^and

g(b) < b

^.

Let

h(x) = f (x) − x

^.

⇒ h ∈ C[a, b]

^since

f ∈ C[a, b]

^{, and}

h(a) = f (a) − a > 0

^,

h(b) = f (b) − b < 0

^.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

⁴¹

To prove the uniqueness, let

x

^and

y

both be fixed points of

f

^{. Then}

|x − y| = |f(x) − f(y)| ≤ λ|x − y|.

Since

λ < 1

, this forces

|x − y| = 0

^{. That is,}

x = y

^.

Theorem 5 If

f ∈ C[a, b]

^{such that}

a ≤ f(x) ≤ b

^{for all}

x ∈ [a, b]

^{, then}

f

^has a fixed point in

[a, b]

. Suppose, in addition, that

f

⁰

(x)

^{exists in}

(a, b)

and there exists a positive constant

M < 1

^{such that}

|f

⁰

(x)| ≤ M < 1

^for

all

x ∈ (a, b)

. Then the fixed point is unique.

Proof: If

f (a) = a

^or

f (b) = b

^{, then}

a

^or

b

is a fixed point of

f

^{and we are}

done. Otherwise, it must be

g(a) > a

^and

g(b) < b

^.

Let

h(x) = f (x) − x

^.

⇒ h ∈ C[a, b]

^since

f ∈ C[a, b]

^{, and}

h(a) = f (a) − a > 0

^,

h(b) = f (b) − b < 0

^.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

⁴¹

To prove the uniqueness, let

x

^and

y

both be fixed points of

f

^{. Then}

|x − y| = |f(x) − f(y)| ≤ λ|x − y|.

Since

λ < 1

, this forces

|x − y| = 0

^{. That is,}

x = y

^.

Theorem 5 If

f ∈ C[a, b]

^{such that}

a ≤ f(x) ≤ b

^{for all}

x ∈ [a, b]

^{, then}

f

^has a fixed point in

[a, b]

. Suppose, in addition, that

f

⁰

(x)

^{exists in}

(a, b)

and there exists a positive constant

M < 1

^{such that}

|f

⁰

(x)| ≤ M < 1

^for

all

x ∈ (a, b)

. Then the fixed point is unique.

Proof: If

f (a) = a

^or

f (b) = b

^{, then}

a

^or

b

is a fixed point of

f

^{and we are}

done. Otherwise, it must be

g(a) > a

^and

g(b) < b

^.

Let

h(x) = f (x) − x

^.

⇒ h ∈ C[a, b]

^since

f ∈ C[a, b]

^{, and}

h(a) = f (a) − a > 0

^,

h(b) = f (b) − b < 0

^.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

⁴²

By the intermediate value theorem,

∃ x

^∗

∈ [a, b]

^{such that}

h(x

^∗

) = 0

^.

⇒ f(x

^∗

) − x

^∗

= 0

^and

f (x

^∗

) = x

^∗.

⇒ f

has a fixed point

x

^{∗ in}

[a, b]

^.

Suppose that

p 6= q

are both fixed points of

f

ⁱⁿ

[a, b]

. By the Mean-Value theorem, there exists

ξ

^between

p

^and

q

^{such that}

f

⁰

(ξ) = f (p) − f(q)

p − q = p − q

p − q = 1.

However, this contradicts to the assumption that

f

⁰

(x) ≤ M < 1

^{for all}

x

ⁱⁿ

[a, b]

. Therefore the fixed point of

f

^{is unique.}

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

⁴²

By the intermediate value theorem,

∃ x

^∗

∈ [a, b]

^{such that}

h(x

^∗

) = 0

^.

⇒ f(x

^∗

) − x

^∗

= 0

^and

f (x

^∗

) = x

^∗.

⇒ f

has a fixed point

x

^{∗ in}

[a, b]

^.

Suppose that

p 6= q

are both fixed points of

f

ⁱⁿ

[a, b]

. By the Mean-Value theorem, there exists

ξ

^between

p

^and

q

^{such that}

f

⁰

(ξ) = f (p) − f(q)

p − q = p − q

p − q = 1.

However, this contradicts to the assumption that

f

⁰

(x) ≤ M < 1

^{for all}

x

ⁱⁿ

[a, b]

. Therefore the fixed point of

f

^{is unique.}

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

⁴²

By the intermediate value theorem,

∃ x

^∗

∈ [a, b]

^{such that}

h(x

^∗

) = 0

^.

⇒ f(x

^∗

) − x

^∗

= 0

^and

f (x

^∗

) = x

^∗.

⇒ f

has a fixed point

x

^{∗ in}

[a, b]

^.

Suppose that

p 6= q

are both fixed points of

f

ⁱⁿ

[a, b]

. By the Mean-Value theorem, there exists

ξ

^between

p

^and

q

^{such that}

f

⁰

(ξ) = f (p) − f(q)

p − q = p − q

p − q = 1.

However, this contradicts to the assumption that

f

⁰

(x) ≤ M < 1

^{for all}

x

ⁱⁿ

[a, b]

. Therefore the fixed point of

f

^{is unique.}

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

⁴²

By the intermediate value theorem,

∃ x

^∗

∈ [a, b]

^{such that}

h(x

^∗

) = 0

^.

⇒ f(x

^∗

) − x

^∗

= 0

^and

f (x

^∗

) = x

^∗.

⇒ f

has a fixed point

x

^{∗ in}

[a, b]

^.

Suppose that

p 6= q

are both fixed points of

f

ⁱⁿ

[a, b]

^.

By the Mean-Value theorem, there exists

ξ

^between

p

^and

q

^{such that}

f

⁰

(ξ) = f (p) − f(q)

p − q = p − q

p − q = 1.

However, this contradicts to the assumption that

f

⁰

(x) ≤ M < 1

^{for all}

x

ⁱⁿ

[a, b]

. Therefore the fixed point of

f

^{is unique.}

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

⁴²

By the intermediate value theorem,

∃ x

^∗

∈ [a, b]

^{such that}

h(x

^∗

) = 0

^.

⇒ f(x

^∗

) − x

^∗

= 0

^and

f (x

^∗

) = x

^∗.

⇒ f

has a fixed point

x

^{∗ in}

[a, b]

^.

Suppose that

p 6= q

are both fixed points of

f

ⁱⁿ

[a, b]

. By the Mean-Value theorem, there exists

ξ

^between

p

^and

q

^{such that}

f

⁰

(ξ) = f (p) − f(q)

p − q = p − q

p − q = 1.

However, this contradicts to the assumption that

f

⁰

(x) ≤ M < 1

^{for all}

x

ⁱⁿ

[a, b]

. Therefore the fixed point of

f

^{is unique.}

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

⁴²

By the intermediate value theorem,

∃ x

^∗

∈ [a, b]

^{such that}

h(x

^∗

) = 0

^.

⇒ f(x

^∗

) − x

^∗

= 0

^and

f (x

^∗

) = x

^∗.

⇒ f

has a fixed point

x

^{∗ in}

[a, b]

^.

Suppose that

p 6= q

are both fixed points of

f

ⁱⁿ

[a, b]

. By the Mean-Value theorem, there exists

ξ

^between

p

^and

q

^{such that}

f

⁰

(ξ) = f (p) − f(q)

p − q = p − q

p − q = 1.

However, this contradicts to the assumption that

f

⁰

(x) ≤ M < 1

^{for all}

x

ⁱⁿ

[a, b]

^.

Therefore the fixed point of

f

^{is unique.}

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

⁴²

By the intermediate value theorem,

∃ x

^∗

∈ [a, b]

^{such that}

h(x

^∗

) = 0

^.

⇒ f(x

^∗

) − x

^∗

= 0

^and

f (x

^∗

) = x

^∗.

⇒ f

has a fixed point

x

^{∗ in}

[a, b]

^.

Suppose that

p 6= q

are both fixed points of

f

ⁱⁿ

[a, b]

. By the Mean-Value theorem, there exists

ξ

^between

p

^and

q

^{such that}

f

⁰

(ξ) = f (p) − f(q)

p − q = p − q

p − q = 1.

However, this contradicts to the assumption that

f

⁰

(x) ≤ M < 1

^{for all}

x

ⁱⁿ

[a, b]

. Therefore the fixed point of

f

^{is unique.}

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

Sol. Non-linear Fun.

⁴³

在文檔中 Sol. Non-linear Fun. (頁 149-188)