Implicit Function Theorems

III )

Section 3.4 Implicit Function Theorems

144

Chapter 3 Calculus in Banach Spaces

X =

^L�I

^GlX;

^{and if}

f

is a mapping from an open set of

X

into a normed space Y, then the partial derivatives

D;/(x),

if they exist, are continuous linear maps

(D;/)(x)

^£(X;,

Y) such t.hat

The connection between partial derivatives and a "total" derivative is as one expects from multivariable calculus. That relationship is formalized next.

( 1 )

Theorem 10. Let

f

be defined on an open set n in the direct-sum space

X = L�=I ^GlXj

and take values in a normed space _Y.Assume that all the partial derivatives

D;/

^{exist in}ⁿand are continuous at a point

x

^{in n .}^Then

f

is Fnkhet differentiable at

x,

and its Frechet derivative is given by

J'(x)h = ^L

_;=1

n D;/(x)h; (h

^{E X)}

Proof. Equat.ion ( 1 ) defines a linear transformation from

X

to Y, and

n n

1\J'(x)hll � L

_;=1

IID;f(x)hdl � L

_.=1

IID;/(x)II Ilhill

� I';;J';;n

^max

IIDjf(x)1I ^L

^•

.

⁼¹

n Ilhdl

= l�J�n

^max

IIDjf(x)II IIhil

Thus Equation

(1)

defines a bounded linear transformation. Let

G(h) = ^f(x ⁺ h) - f(x) - ^L

_;=1

n D;/(x)hi

We want to prove that

IIG(h)11

⁼

o(llhll).

For sufficiently small

h, x ⁺ h

^{is in}^n,

and the partial derivatives of

G

^{exist at}

x ⁺ h.

^{They are}

D;G(h)

⁼

D;/(x ^{+ II)} - D;/(x)

c

is a given positive number, we use the assumed continuity of

D;/

^at

x

^to

find a positive 0 such that for

Ilhil

^<⁰^{we have}

IIDiG(h)1I

c,

^for

^{1 �} ⁱ ^�

^n.

Then, by the mean value theorem,

IIG(h)11 ^� IIG(hh h2, · · · , hn) - G(O, h2,

^{• • •}

, hn)11 n

+ IIG(O, h2,. · . , hn) - G(O, 0, h3,· · · , hn)11

+ . . .

+ IIG(O,O, .. . ;hn) - G(O,O, .. . ,0)1\

� L

_;=1

cllhdl = cllhll

Section 3.5 Extremum Problems and Lagrange Multipliers 145 Since ^cwas arbitrary, this shows that

II

^C(h)

¹ I

⁼^o(

lI

lD

Problems 3.4

•

1. Let U be a closed ball in a Banach space. Let _F: U+ ^--tU, where U+ is an open set containing U. Prove that if sup{IIF'(x)1I : x E U} < ^{1 ,}then _Fhas a unique fixed point in U.

2. Let ^Xbe a Banach space and 0 an open set in ^Xcontaining a, b, Xo, and the line segment

S joining a to b. Prove that

IIf(b) - f(a) - !,(xo)(b ^-a)1

I ^�

^{IIb -}^a

ll

sup 1I!'(x) - !'(xo)1I .

:rES

[Suggestion: Use the function g(x) ⁼f(x) - f'(xo)x.J Determine whether the same inequality is true when f'(xo) is replaced by an arbitrary linear operator. In this problem, f : ^X^--t^Y,where ^Yis any normed space.

3. Suppose F(xo, YO) ⁼O. If Xl is close to Xo, there should be a Yl such that F(XI , Yl ) = O.

Show how Newton's method can be used to obtain Yl. (Here F : ^X^x^Y^--t^Z,and X, Y, Z are Banach spaces.)

4. Prove Theorem 2.

5. Let f : n --t Y be a continuously differentiable map, where n is an open set in a Banach space, and ^Yis a normed linear space. Assume that f'(x) is invertible for each X E ii,

and prove that f(n) is open.

6. Let 0 be the point in [0, 1 J where cos 0 ⁼o. Define X to be the vector space of all continuously differentiable functions on [O, IJ that vanish at the point o. Define a norm on X by writing "xII = sUPO:<::.:<:: 1 Ix'(t)l. Prove that there exists a positive number /j

such that if Y E X and lIyiI <

�

hen there exists an x E X satisfying sin ox + x 0 cos = Y

7. Let f be a continuous map from an open set n in a Banach space X into a Banach space ^Y.Suppose that for some Xo in n, f'(xo) exists and is invertible. Prove that f is one-to-one in some neighborhood of Xo.

8. In Example 2, with the nonlinear integral equation, show that the mapping x H f'(x) is continuous; indeed, it satisfies a Lipschitz condition.

9. Rework Example 2 when the term 2x(0) is replaced by ox(O), for an arbitrary constant o. In particular, treat the case when 0 = O.

3.5 Extremum Problems and Lagrange Multipliers

A minimum

point of a real-valued function f defined on a set Af is a point Xo such that f(xo)

�

f(x) for all x E

M. M

has a topology, then the concept of

relative

minimum point is defined as a point Xo E

M

such that for some neighborhood N of Xo we have f(xo)

�

f(x), for all x in N.

Theorem ^1. Necessary Condition for Extremum. Let n be an open set in a normed linear space, and let f : n -t IR. If Xo is a minimum point of f and if f'(xo ) exists, then f'(xo) =

o.

146 Chapter 3 Calculus in Banach Spaces

Proof.

Let X be the Banach space, and assume

f'(xo)

#- 0. Then there exists

v

^E^Xsuch that

f'(xo)v

⁼

^{- 1.}

By the definition of

f'(xo)

we can take

A ^>

⁰

and so small that

Xo ⁺ AV

is in n and

If(xo ⁺ AV) - f(xo) - AJ'(xo)vl / A ^ll v ^ll

(2I1vID- 1

This means that

H f(xo ⁺ AV) - f(xo)]

is within distance

4

from

-1,

and so is negative. This implies

f(xo ⁺ AV)

f(xo).

^•

In this section we will be concerned mostly with

constrained

extremum problems. A simple illustrative case is the following. We have two nice functions,

f

^and

g,

^on

JR2

^to

JR.

^{We put}

^M

⁼

{(x, y) : g(x, y)

⁼

^O}

and seek an extremum of

fiM.

(That means

f

restricted to

M.)

If the equation

g(x, y)

= 0 defines

y

as a function of

x,

^say

y

⁼

y(x),

then we can look for an

unrestricted

extremum of

<J>(x)

⁼

f(x, y(x)).

Hence we try to solve the equation

<J>'(x)

⁼

O.

This leads to

0 =

h(x,y(x)) ⁺ f2(x,y(x))y'(x)

h (x, y(x)) - f2(x, Y(X))gl (x, Y(X))/g2(X, y(x))

Thus we must solve simultaneously

( 1 ) g(x,y)

^{= 0} ^and

The method of Lagrange multipliers introduces the function

H(x, y, A)

⁼

f(x, y) ⁺ Ag(X, y)

and solves simultaneously

HI

⁼

H2

⁼

H3

⁼

O.

^Thus

fI(x,y) ⁺ Agl(X,y)

⁼

f2(x,y) ⁺ Ag2(X,y)

⁼

g(x,y)

⁼⁰

g2(X, y)

f 0, then

A

⁼ ^-

f2(x, y)/g2(X,

^V),and we recover system

( 1).

^The

method of Lagrange multipliers treats

x

^and

y

symmetrically, and includes both cases of the implicit function theorem. Thus

y

can be a differentiable function of

x,

^or

x

can be a differentiable function of

y.

Example 1. Let

f

^and⁹be functions from

JR2

^to

JR

defined by

f(x, y)

⁼

x2 ⁺ y2, g(x, y)

⁼

x - y ⁺

1 . The set

M

{(x, y) : g(x, y)

⁼

^O}

is the straight line shown in Figure 3.3. Also shown are some level sets of

f,

i.e., sets of the type

((x,y) : f(x,y)

⁼

^c}.

At the solution, the gradient of

f

is parallel to the gradient of

g.

The function

H

^is

H(x, y, A)

⁼

x2 ⁺ y2 ⁺ A(X - y ^{+ 1),}

^{and the}

three equations to be solved are

2 x ⁺ A

⁼

² y - A

⁼

x - y ^{+ 1}

⁼

^O.

The solution

(

^{4 , 4)·}

^•

Dalam dokumen Analysis for Applied Mathematics (Halaman 151-155)

III )

Section 3.4 Implicit Function Theorems

144

X =

GlX;

f

X

D;/(x),

(D;/)(x)

£(X;,

f

X = L�=I GlXj

D;/

x

f

x,

J'(x)h = L

n D;/(x)h; (h

X

n n

1\J'(x)hll � L

IID;f(x)hdl � L

IID;/(x)II Ilhill

� I';;J';;n

IIDjf(x)1I L

.

n Ilhdl

= l�J�n

IIDjf(x)II IIhil

(1)

G(h) = f(x + h) - f(x) - L

n D;/(x)hi

IIG(h)11

o(llhll).

h, x + h

G

x + h.

D;G(h)

D;/(x + II) - D;/(x)

c

D;/

x

Ilhil

IIDiG(h)1I

c,

1 � i �

IIG(h)11 � IIG(hh h2, · · · , hn) - G(O, h2,

, hn)11 n

+ IIG(O, h2,. · . , hn) - G(O, 0, h3,· · · , hn)11

+ IIG(O,O, .. . ;hn) - G(O,O, .. . ,0)1\

� L

cllhdl = cllhll

II

1 I

lI

lD

I �

ll

�

�

A minimum

�

M.

M

relative

M

�

o.

146 Chapter 3 Calculus in Banach Spaces

Proof.

f'(xo)

v

f'(xo)v

- 1.

f'(xo)

A >

Xo + AV

If(xo + AV) - f(xo) - AJ'(xo)vl / A ll v ll

(2I1vID- 1

^GlX;

^£(X;,

X = L�=I ^GlXj

J'(x)h = ^L

IIDjf(x)1I ^L

G(h) = ^f(x ⁺ h) - f(x) - ^L

h, x ⁺ h

x ⁺ h.

D;/(x ^{+ II)} - D;/(x)

^{1 �} ⁱ ^�

IIG(h)11 ^� IIG(hh h2, · · · , hn) - G(O, h2,

¹ I

I ^�

^{- 1.}

A ^>

Xo ⁺ AV

If(xo ⁺ AV) - f(xo) - AJ'(xo)vl / A ^ll v ^ll

H f(xo ⁺ AV) - f(xo)]

f(xo ⁺ AV)

^M

^O}

h(x,y(x)) ⁺ f2(x,y(x))y'(x)

f(x, y) ⁺ Ag(X, y)

fI(x,y) ⁺ Agl(X,y)

f2(x,y) ⁺ Ag2(X,y)

x2 ⁺ y2, g(x, y)

x - y ⁺

^O}

^c}.

x2 ⁺ y2 ⁺ A(X - y ^{+ 1),}

2 x ⁺ A

² y - A

x - y ^{+ 1}

^O.