THE OPEN MAPPING THEOREM AND RELATED THEOREMS We ...

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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)
kk
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
)
kk
2
.Translating over x
0
we get
that B
2
(0;r
0
) 
B
1
(x
0
;n
0
)
kk
2
.Now by the triangle inequality we get that B
2
(0;r
0
) 
B
1
(0;n
0
+kx
0
k
1
)
kk
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.Dene 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 denition 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 dene 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 dierentiable functions and dene 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 denition.
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.Dene 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.Dene 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 dene
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.