THE OPEN MAPPING THEOREM AND RELATED THEOREMS

ANTON.R SCHEP

We start with a lemma,whose proof contains the most ingenious part of Banach's open

mapping theorem.Given a norm k k

i

we denote by B

i

(x;r) the open ball fy 2 X:

ky xk

i

< rg.

Lemma 1.Let X be a vector space with two norms k k

1

,k k

2

such that (X;k k

1

) is a

Banach space and assume that the identity map I:(X;k k

1

)!(X;k k

2

) is continuous.

If B

2

(0;1)

B

1

(0;r)

kk

2

,then B

2

(0;1) B

1

(0;2r) and the two norms are equivalent.

Proof.From the hypothesis we get B

2

(0;1) B

1

(0;r) +B

2

(0;

1

2

),so by scaling we get that

B

2

(0;

1

2

n

) B

1

(0;

r

2

n

) +B

2

(0;

1

2

n+1

) for all n 1.let now kyk

2

< 1.Then we can write

y = x

1

+y

1

,where kx

1

k

1

< r and ky

1

k

2

<

1

2

.Assume we have ky

n

k

2

<

1

2

n

we can write

y

n

= x

n+1

+y

n+1

,where kx

n+1

k

1

<

r

2

n

and ky

n+1

k

2

<

1

2

n+1

.By completeness of (X;k k

1

)

there exists x 2 X such that x =

P

1

n=1

x

n

,where the series converges with respect to the

norm k k

1

.By continuity of the identity map I:(X;k k

1

)!(X;k k

2

) it follows that

the same series also converges to x with respect to k k

2

.On the other hand the equation

y =

P

n+1

k=1

x

k

+y

n+1

shows that the series

P

1

n=1

x

n

converges to y with respect to k k

2

.

Hence y = x and thus kyk

1

= kxk

1

P

1

n=1

kx

n

k

1

< 2r.It follows thatB

2

(0;1) B

1

(0;2r)

and thus kyk

1

2rkyk

2

for all y 2 X.As the continuity of I gives that there exists C such

that kyk

2

Ckyk

1

for all y 2 X,we get that the two norms are equivalent.

Theorem 2.Let X be a vector space with two norms k k

1

,k k

2

such that (X;k k

1

) and

(X;k k

2

) are Banach spaces.Assume that the identity map I:(X;k k

1

)!(X;k k

2

) is

continuous.Then the norms k k

1

and k k

2

are equivalent.

Proof.Applying the Baire Category theorem in (X;k k

2

) to X = [

1

n=1

B

1

(0;n) we can

nd n

0

,x

0

and r

0

> 0 such that B

2

(x

0

;r

0

)

B

1

(0;n

0

)

kk

2

.Translating over x

0

we get

that B

2

(0;r

0

)

B

1

(x

0

;n

0

)

kk

2

.Now by the triangle inequality we get that B

2

(0;r

0

)

B

1

(0;n

0

+kx

0

k

1

)

kk

2

.By the above lemma the two norms are equivalent.

Theorem3 (Bounded Inverse Theorem).Let X,Y be Banach spaces and assume T:X!

Y is an one-to-one,onto continuous linear operator.Then T

1

:Y!X is continuous.

Proof.Dene kyk

T

= kT

1

yk.Then k k

T

is a norm on Y and kyk kTkkyk

T

for

y 2 Y,so I:(Y;k k

T

)!(Y;k k) is continuous.Moreover,if

P

1

n=1

ky

n

k

T

< 1,then

x

0

=

P

1

n=1

T

1

y

n

exists in X and kTx

0

P

N

n=1

y

n

k

T

= kx

0

P

N

n=1

T

1

y

n

k!0 as

N!1.Hence (Y;k k

T

) is also a Banach space.By the above Theorem the two norms

1

2 ANTON.R SCHEP

on Y are equivalent,so there exists C such that kT

1

(y)k Ckyk for all y 2 Y,i.e.T

1

is continuous.

We recall now that a linear map T:X!Y is called open if T(O) is open for all open

O X.It is easy to see that an open linear map is surjective.The Open Mapping theorem

gives a converse to that statement.Before stating and proving that theorem,we recall a few

basic facts about quotient maps.Let X be a Banach space and M X a closed subspace.

Then X=M is a Banach space with respect to the quotient norm k[x]k = inffkyk;y 2 [x]g.

Denote by Q the quotient map Q(x) = [x].Then Q is open.In fact,it is easy to see from

the denition of the quotient norm that Q(fx:kxk < 1g) = f[x]:k[x]k < 1g.

Theorem 4 (Open Mapping Theorem).Let X,Y be Banach spaces and assume T:X!

Y is an onto continuous linear operator.Then T is an open map.

Proof.Let Q:X!X=ker(T) be the quotient map.Then by the above remarks Q is an

open mapping.Let

^

T:X=ker(T)!Y be the induced map such that T =

^

T Q.Then

^

T

is one to one and onto,so by the above Theorem

^

T

1

is continuous,so

^

T is open and thus

T is open.

Let now A:D(A)!Y be a linear operator,where D(A) is a (not necessarily closed)

linear subspace of the Banach space X.The subspace D(A) is called the domain of A.

Given a linear operator A:D(A)!Y we dene the graph

(A) = (x;Ax):x 2 D(A);

It is clear that (A) is linear subspace of X Y.We can equip X Y with the product

normk(x;y)k = kxk+kyk.Then we say that A has a closed graph (or is a closed operator),

if (A) is a closed subspace of X Y.

Example 5.Let X = Y = C[0;1] with the supremum norm.Let D(A) = C

0

[0;1] the

subspace of X consisting of continuously dierentiable functions and dene A:D(A)!Y

by Af = f

0

.One can can see that A is not bounded,by taking f

n

(t) = t

n

,and noting

that kf

n

k = 1 and kAf

n

k = n.On the other hand A has a closed graph.To see that A

has a closed graph,let (f

n

;f

0

n

)!(f;g) in X Y.Then by the Fundamental Theorem of

Calculus f

n

(t) f

n

(0) =

R

t

0

f

0

n

(s) ds!

R

s

0

g(s) ds.It follows that f(t) = f(0) +

R

t

0

g(s) ds.

Hence f 2 D(A) and f

0

= g,i.e.,(f;g) 2 (A).

The following proposition is immediate from the denition.

Proposition 6.Let X and Y be Banach spaces and assume A:D(A)!Y is a linear

operator,where D(A) is a subspace of X.Then the following are equivalent.

(1) A has a closed graph.

(2) If x

n

2 D(A),x

n

!x 2 X,and Ax

n

!y 2 Y,then x 2 D(A) and Ax = y.

(3) D(A) is a Banach space with respect to the graph norm kxk

A

= kxk +kAxk.

Theorem 7 (Closed Graph Theorem).et X and Y be Banach spaces and assume A:

X!Y is a closed linear operator.Then A is bounded.

THE OPEN MAPPING THEOREM AND RELATED THEOREMS 3

Proof.Dene P:(A)!X by P(x;Ax) = x.Then P is clearly a bounded,one-to-one,

onto linear operator,so by the Bounded Inverse Theorem the inverse operator P

1

:X!

(A) is bounded.Hence there exists a constant C such that kxk + kAxk Ckxk,i.e.,

kAxk (C 1)kxk for all x 2 X.

We now present a proof of the Uniform Boundedness Principle,based on the Closed

Graph Theorem.

Theorem 8 (Banach-Steinhaus).Let X and Y be Banach spaces and assume A

2

L(X;Y ) ( 2 F) is a pointwise bounded family of bounded operators,i.e.,for all x 2 X

there exists a constant C

x

such that kA

xk C

x

for all 2 F.Then there exists a

constant C such that kA

jj C for all 2 F.

Proof.Dene the space

Y = f(y

):y

2 Y;sup

ky

k < 1g with norm k(y

)k =

sup

ky

k.It is straightforward to verify that

Y is also a Banach space.Now dene

T:X!

Y by Tx = (A

x).Note Tx 2

Y,since the collection A

is pointwise

bounded.Clearly T is linear and we claim that T is closed.To see this,let x

n

!0

and Tx

n

!(y

).Then A

x

n

!y

for all 2 F,but also A

x

n

!0 for all .Hence

(y

) = (0) and T is closed.By the Closed Graph Theorem T is bounded,i.e.,there exists

a constant C such that for all kxk 1 we have sup

kA

xk C.Hence kA

k C for all

2 F.

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