Chapter 4 Fundamental Theorems For Normed And Banach Spaces
4.1 Zorn’s Lemma
4.1-1 Definition. A partially ordered set is a set M with a partial order relation
 satisfying the following:
(PO1) a  a for all aM.
(PO2) If a  b and b  a, then a = b.
(PO3) If a  b and b  c, then a  c.
4.1-2 Definition. An upper bound of a subset W of a partially ordered set M is
uM such that x  u for all xM. A maximal element of M is mM such that
if m  x, then m = x. A subset C of M is called a chain or a totally ordered set
if for any a, bC, a  b or b  a ( or both ).
4.1-3 Zorn’s Lemma. Let M be a non empty partially ordered set. If every
chain C  M has an upper bound, then M has at least one maximal element.
4.1-4 Theorem. ( Total Orthonormal Set ) In every Hilbert space H  {0}
there exists a total orthonormal set.
Proof. Let M be the set of all orthonormal subsets of H. Since H  {0}, then
there is xH and x  0. Hence {
x
|| x||
} is an orthonormal subset of H. Hence
M  . Also note that (M, ) defines a partially ordered set ( set inclusion). Let
C  M be any chain, then X = A is an upper bound of C. By Zorn’s Lemma
AC
M has a maximal element F. Suppose that F is not total in H. Then by Theorem
z
3.6-2, F  {0} and so there is zF and z  0. Then F1 = F  { || z || } is
orthonormal and F is a proper subset of F1 which is a contradiction of
maximality of F. Therefore, the orthonormal se F is total in H
4.2 Hahn-Banach Theorem
4.2-1 Definition. A sublinear functional is a real valued function p on a vector
space X which is a) subadditive, p(x + y)  p(x) + p(y) for all x, yX.
b) positive homogeneous, p(x ) = p(x) for all xX and all   0.
4.2-2 Hahn-Banach Theorem. ( Extension of Linear Functionals )
Let X be a real vector space and p is a sublinear functional on X. Furthermore,
let f be a linear functional which is defined on a subspace Z of X and satisfies
~
f(x)  p(x) for all xZ. Then f has a linear extension f from Z to X satisfying
~
f (x)  p(x) for all xX.
Proof. (a) Let E be the set of all linear extensions g of f that satisfy g(x)  p(x)
for all xD(g). Since fE, then E  . Define on E a partial ordering by g  h


if and only if h is an extension of g. For any chain C  E, define g by g (x) =


g(x) if xD(g) and gC. Then g is a linear functional and D( g ) =
1
gC
D( g )

which is a vector space since C is a chain (why?). It is clear that g  g for all

gC. Then g is an upper bound of C. Hence by Zorn’s Lemma E has a
~
maximal element say, f . By the definition of E this is a linear extension of f
~
~
which satisfies f (x)  p(x) for all xD( f ).
~
~
~
b) We show that D( f ) = X. Suppose that D( f )  X. Choose y1X\D( f ) and
~
~
let Y1 be the subspace spanned by D( f ) and y1. Since 0D( f ), then y1  0.
~
Any xY1 can be written uniquely as x = y + y1, yD( f ). To see this
~
~
suppose that x = y + y1 and x = y + y1. Then ( - ) y1 = y - y. However,
~
~
~
~
y1D( f ) and y - yD( f ), then  -  = 0, so that  =  and y = y. This
~
proves the uniqueness. Define g1 : Y1  R by g1(y + y1) = f (y) + c, where
c is any real constant. It is easy to see that g1 is linear ( How?). If  = 0 then
~
~
~
g1(y) = f (y) for all yD( f ). Hence g1 is a proper extension of f . However,
g1(x)  p(x) for all x = y + y1D(g1), [see its proof below] ..………..……(1)
~
~
Hence g1E which is contradicts the maximality of f . Therefore, D( f ) = X.
~
~
~
(c) Finally we prove (1). Let y, z D( f ). For any fixed y1X, f (y) - f (z) =
~
f (y - z)  p(y - z) = p(y + y1 - y1- z)  p(y + y1) + p(- y1- z). Then
~
~
~
~
-p(-y1- z) - f (z)  p(y + y1) - f (y). Hence u = sup{-p(-y1- z)- f (z): zD( f )}
~
~
 inf{ p(y + y1) - f (y): yD( f )} = w. Then there is a real number c such that
~
~
u  c  w. So that -p(-y1- z) - f (z)  c for all zD( f ) .………………..(2),
~
~
and c  p(y + y1) - f (y) for all yD( f ) ..……………..………………..(3)
To prove (1) we have three steps. First step, when  < 0. Put z =  -1y in (2),
~
then multiply both sides by - to get, p(-y1-  -1y) + f (y)  -c. Then, for
~
x = y + y1 g1(x) = f (y) + c  -p(-y1-  -1y) = p(y + y1) = p(x) since p
is sublinear. Thus (1) is true when  < 0. Second step, when  = 0. Then x =
~
~
yD( f ) and g1(x) = f (y)  p(y) = p(x). Thus (1) is true when  = 0. Third
step, when  > 0. By (3), but replace y by  -1y, and then multiply both sides
~
~
by  to get c  p( -1y + y1) - f ( -1y) = p(x) - f (y). Hence g1(x) =
~
f (y) + c  p(x). Thus (1) is true when  > 0. Therefore, (1) is true for all ,
that is g1(x)  p(x) for all x = y + y1D(g1). This completes the proof
H. W. 5-10. H.W.* 8, 10.
2
4.3 Hahn-Banach TheoremFor Complex Vector Spaces And Normed Spaces
4.3-1 Hahn-Banach Theorem. ( Generalized ) Let X be a real or a complex
vector space and p is a subadditive real-valued functional on X, and for any
scalar , p(x) = ||p(x) ……………………………………………………...(1)
Furthermore, let f be a linear functional which is defined on a subspace Z of X
and satisfies |f(x)|  p(x) for all xZ…………………………………………(2)
~
~
Then f has a linear extension f from Z to X satisfying | f (x)|p(x) for all xX.
Proof. Left to the reader.
4.3-2 Hahn-Banach Theorem. ( Normed Space ) Let f be a bounded linear
functional on a subspace Z of a normed space X. Then there exists a bounded
~
linear functional f on X which is an extension of f to X and has the same
~
~
~
norm, || f ||X = || f || Z, where || f ||X = sup { | f (x)| : xX, || x || = 1 },
|| f || Z = sup{ | f(x) | : xZ, || x || = 1}, and || f || Z = 0 in the trivial case Z = {0}.
~
Proof. If Z = {0}, then f = 0 and the extension is f = 0. Let Z  {0}. Then for
any xZ we have | f(x) |  || f ||Z || x ||. Define p : X  R by p(x) = || f ||Z || x ||.
Then | f(x) |  p(x) for all xZ. Furthermore, for any x, yX and any R we
have, (1) p( x + y ) = || f ||Z || x + y ||  || f ||Z ( || x || + || y || ) = p( x ) + p( y )
and (2) p(x ) = || f ||Z || x || = |  | p( x ). Hence by Theorem 4.3-1, there
~
exists a linear functional f on X which is an extension of f and satisfies
~
~
| f (x)|  p(x) = || f ||Z || x || for all xX. Then || f ||X  || f ||Z. However, || f ||Z
~
~
~
= || f ||Z  || f ||X. Hence || f ||X = || f || Z
4.3-3 Theorem. ( Bounded Linear Functional ) Let X be a normed space and
let x0  0 be any element of X. Then there exists a bounded linear functional
~
~
~
f on X such that || f || = 1 and f ( x0 ) = || x0 ||.
Proof. Consider the subspace Z = { x0 :  is a scalar }. Define f : Z  k by
f(x) = f(x0) =  || x0 ||, where k is the scalar field ( real or complex ). It is easy
to see that f is linear (how?). Also f is bounded, where |f(x)| = |f(x0)| = ||x0|| =
|| x || for all xX. Hence || f || = 1. By Theorem 4.3-2, f has a linear extension
~
~
~
f from Z to X of norm || f || = || f || = 1. Moreover, f ( x0 ) = f( x0 ) = || x0 ||
4.3-4 Corolllary. For any x in a normed space X we have, || x || = sup{
| f ( x )|
|| f ||
: fX / , f  0 }. Hence if x0 is such that f( x0 ) = 0 for all fX /, then x0 = 0.
~
~
~
Proof. By Theorem 4.3-3, there exists f X / such that || f || = 1 and f ( x ) =
|| x || where x is a nonzero element in X. So that sup{
3
| f ( x )|
|| f ||
: fX / , f  0 } ≥
~
| f ( x )|
~
|| f ||
= || x ||. However, | f(x) | ≤ || f || || x ||, then sup{
|| x ||. Therefore, sup{
| f ( x )|
|| f ||
| f ( x )|
|| f ||
: fX / , f  0 } ≤
: fX / , f  0 } = || x ||. If x0 is such that f( x0 ) = 0
for all fX /, then || x0 || = sup{
| f ( x )|
|| f ||
: fX / , f  0 } = . Hence x0 = 0
H. W. 1, 2, 4, 5, 8, 11, 15. H.W.* 4.
4.5 Adjoint operator
4.5-1 Definition. Let T : X  Y be a bounded linear operator, where X and
Y are normed spaces. Then the adjoint operator T ×: Y /  X / of T is defined by
(T ×g)(x) = g(Tx) = f(x), where X / and Y / are the dual spaces of X and Y,
respectively, and xX, fX / and gY /.
4.5-2 Theorem. The adjoint operator of T in the above definition is linear and
bounded with || T × || = || T ||.
Proof. For any xX, g1, g2Y / and any scalar  we have, (T × (g1 + g2))(x) =
(g1 + g2)(Tx) = g1(Tx) + g2(Tx) = (T ×g1)(x) + (T ×g2)(x). Hence
T ×(g1 + g2) = T ×g1 + T ×g2. Therefore, T is linear. Since gY / and T are
bounded then, | f(x) | = | g(Tx) | ≤ || g || || Tx || ≤ || g || || T || || x || and so || f || ≤
|| g || || T ||. By the definition of T ×, T ×g = f, and so || T ×g || = || f || ≤ || g || || T ||.
Hence || T × || = sup{|| T ×g || : gY /, || g || = 1} ≤ || T ||……………………(1)
This means that T × is bounded. By Theorem 4.3-3 for any x0  0 in X there
exists g0Y / such that || g0 || = 1 and g0( Tx0 ) = || Tx0 ||. Let f0 = T ×g0.
Then || Tx0 || = g0( Tx0 ) = f0( x0 ) ≤ || f0 || || x0 || = || T ×g0 || || x0 || ≤ || T ×|| || g0 ||
|| x0|| = || T ×|| || x0||. This implies that || T || ≤ || T × ||. By this and (1) we have,
|| T ×|| = || T ||
4.5-3 Theorem. If S, T : X  Y and W : Y  Z are bounded linear operators,
where X, Y and Z are normed spaces. Then
(1) ( S + T ) × = S× + T × and (T)× = T× for any scalar .
(2) ( WT ) × = T×W×.
(3) If T -1BL(Y, X), then (T ×) -1BL(X/, Y/ ) and (T ×) -1 = (T -1) ×.
Proof. Left to the reader.
4.5-4 Theorem. Let H1 and H2 be two Hilbert spaces, T BL(H1, H2). Then
there exist A1 : H/1  H1 and A2 : H/2  H2 such that T* = A1T×A2-1 where T*
and T× are the Hilbert adjoint and the adjoint operators of T, respectively, and
both A1 and A2 are bijective, isometric and conjugate linear.
Proof. Let T : H1  H2 be a bounded linear operator and T× : H/2  H/1 be its
adjoint. Then T×g = f where g(Tx) = f(x), fH/1, gH/2 and xH1. Then by
4
Theorem 3.8-1 there exist a unique x0H1 and a unique y0H2 such that f(x) =
< x, x0>, || f || = || x0 ||, g(y) = < y, y0> and || g || = || y0 ||. By noting that x0 and
y0 are uniquely determined by f and g, respectively we can define A1: H/1  H1
and A2 : H/2  H2 by A1f = x0 and A2g = y0. Then || A1f || = || x0 || = || f ||. Hence
A1 is isometric and so it is one to one. Let hH1. Then f : H1  k defined by
f(x) = < x, h > for all xH1 is a bounded linear operator ( k is the scalar field).
Then A1f = h, hence A1 is onto. Similarly, for A2. To show that A1 is conjugate
linear, let f1, f2 H/1 and  any scalar. Then there exist x1, x2H1 such that
f1(x) = < x, x1 > and f2(x) = < x, x2 > for all xH1. Then (f1 + f2)(x) = f1(x) +
__
__
f2(x) = < x, x1 > + < x, x2 > = < x,  x1+ x2 >. Hence A1(f1 + f2) =  x1+ x2
__
=  A1f1 + A1f2. Therefore, A1 is conjugate linear. Similarly for A2. For any
y0H2 there exists gH/2 such that A2g = y0, so A2-1y0 = g. Then (A1T×A2-1)(y0)
= (A1T×)(g) = A1f = x0. Also, < Tx, y0 > = g(Tx) = f(x) = < x, x0 > =
< x, (A1T×A2-1)(y0) >. However, < Tx, y0 > = < x, T*y0 > and T* is unique,
hence T* = A1T×A2-1
H. W. 1-5, 8,9 H.W.* 8.
4.6 Reflexive Spaces
4.6-1 Lemma. For every fixed x in a normed space X, the functional gx
defined on X/ by gx(f) = f(x) ( for all fX/ ) is a bounded linear functional so
that gxX// and has the norm || gx || = || x ||.
Proof. Left to the reader.
4.6-2 Lemma. Let X be a normed space. Then the canonical mapping C:XX//
defined by C(x) = gx, where gx(f) = f(x) ( for all fX/ ) is an isomorphism from
X onto R(C), the range of C.
Proof. For any fX/, x, yX and any scalar , C(x + y)(f) = g x + y(f) =
f(x + y) = f(x) + f(y) = gx(f) + gy(f) = C(x)(f) + C(y)(f).
Hence C(x + y) = C(x) + C(y), and so C is linear. Note that for all fX/,
g x – y (f) = f(x–y) = f(x) – f(y) = gx(f) – gy(f) = (gx–gy)(f). Hence g x – y = gx–gy.
By Lemma 4.6-1, || C(x) – C(y) || = || gx– gy || = || g x – y || = || x – y ||. This
means that C is isometric and so it is one to one. Therefore, C is an
isomorphism from X onto R(C)
4.6-3 Definition. A normed space X is said to be reflexive if R(C) = X //,
where C as in Lemma 4.6-2.
Notes. (1) By Lemma 4.6-2, the normed space X is isomorphic to a subspace
of X//. Hence X is embeddable in X//, and in this case C is called the canonical
embedding of X into X//.
(2) If the normed space X is reflexive, then it is isomorphic with X//.
5
4.6-4 Theorem. If a normed space X is reflexive, then it is complete.
Proof. Since X is reflexive, then X is isomorphic to X//. However, X// is a
Banach space, then X is a Banach space
4.6-5 Theorem. Every finite dimensional normed space is reflexive.
Proof. Since dimX is finite, then X/ = X*. By Theorem 2.9-3 dimX* = dimX,
then X// = X**. However, C : X  X** is an isomorphism, so that C : X  X//
is an isomorphism. Hence X is reflexive
Note. Since (
p
) // =
p
( 1 < p <  ), then
p
is reflexive.
4.6-6 Theorem. Every Hilbert space is reflexive.
Proof. Left to the reader.
4.6-7 Lemma. Let Y be a proper closed subspace of a normed space X. Let
~
~
x0X-Y be arbitrary and δ = inf { || y - x0 || : y Y}, the distance from x0 to Y.
~
~
~
Then there exists f X/ such that || f || = 1, f (y) = 0 for all yY and
~
f ( x0) = δ.
Proof. Consider the subspace Z = { y + x0 : yY,  is a scalar } of X, and
define the functional f on Z by f(z) = f(y + x0) = δ. Then f is linear (how?)
Since Y is closed, then δ > 0 and so f ≠ 0. Note that for all yY, f(y) = f(y +
0x0) = 0 and f(x0)=δ (let y = 0 and  =1). Now we show that f is bounded. If 
= 0, then f(z) = 0. If  ≠ 0, then -  yY for all yY and | f(z)| = | f(y + x0) |
1
~
~
= ||δ = || inf{|| y - x0 || : y Y}  || || -  y - x0 || = || y + x0 || = || z ||.
Hence f is bounded and || f ||  1 ……………………………………………..(1)
By the definition of infimum there is a sequence (yn) of elements in Y such that
1
|f ( y n  x 0 )|
|| y n  x 0 ||
|| yn - x0 ||  δ as n . Then || f || 
=

|| y n  x 0 ||
 1 as n .
By this an (1) we have || f || = 1. By Hahn-Banach Theorem 4.3-2, there exists
~
~
~
~
f X/ such that || f || = 1, f (y) = 0 for all yY and f ( x0) = δ
4.6-8 Theorem. If the dual space X/ of a normed space X is separable, then X
itself is separable.
Remark. A separable normed space X with a nonseparable dual space X/ can’t
be reflexive.
Proof. Suppose that X/ is nonseparable but X is separable and reflexive. Then
X// isomorphic to X which implies the separability of X// . Then by Theorem
4.6-8, X/ is separable which is a contradiction. Therefore, X can’t be reflexive
1
Example.
is not reflexive.
1
Proof. Since
is separable but (
reflexive
H. W. 1, 5, 7-9 H.W.* 8.
1 /
) =
6

is not separable, then
1
is not
4.7 Category Theorem. Uniform Boundedness Theorem
4.7-1 Definition. A subset M of a metric space X is said to be
__
__
(a) rare ( or nowhere dense ) in X if M has no interior point ( ( M )0 = φ ),
(b) meager ( or of first category ) in X if M is the union of countably many sets
each of which is rare in X.
(c) nonmeager ( or of the second category ) in X if M is not meager in X.
4.7-2 Bair Category Theorem. If a metric space X ≠ φ is complete then it is

nonmeager in it self. Hence if X ≠ φ is complete and X = Ak , Ak is closed
k 1
for all k then at least one Ak contains a nonempty open subset.
Proof. Suppose that the complete metric space X ≠ φ is meager in itself. Then
__

X=
Mk
with each Mk is rare ( nowhere dense ) in X. Then ( M 1 )0 = φ, that
k 1
__
means M 1 does not contain a nonempty open set. But X does ( for example X
__
__
__
itself ), so that M 1 ≠ X. Hence (~ M 1 ) = X - M 1 is nonempty and open, so
__
we can choose p1(~ M 1 ) and an open ball about it, say B1 = B(p1; ε1 ) 
__
__
(~ M 1 ) and ε1 < ½ . Since M2 is nowhere dense in X, then M 2 does not
__
contain a nonempty open set. Hence B(p1; ½ε1 ) is not a subset of M 2 which
__
implies that (~ M 2 ) ∩ B(p1; ½ε1 ) is nonempty and open, so we can choose
__
p2(~ M 2 ) ∩ B(p1; ½ε1 ) and an open ball about it, say B2 = B(p2; ε2 ) 
__
(~ M 2 ) ∩ B(p1; ½ε1 ) and ε2 < ½ ε1. So by induction we obtain a sequence of
balls Bk = B(pk; εk ), εk < 2-k such that Bk+1  B(pk; ½εk )  Bk. Since εk < 2-k,

then
 d ( pk , pk 1 ) ≤
k 1

2
k
and so it converges. Hence (pk) is a Cauchy
k 0
sequence in the complete metric space X so it converges, say lim pk = pX.
Also for all m and all n > m we have, Bn  B(pm; ½εm ) so that d(pm, p) ≤
d(pm, pn ) + d(pn, p) < ½ εm + d(pn, p) → ½ εm as n → ∞ Hence pBm for all m.
__
Since Bm  (~ M m ), then p  Mm for all m, so that p 

Mk
= X. This
k 1
contradicts pX. Therefore, X must be nonmeager in it self
4.7-3 Uniform Boundedness Theorem. Let (Tn ) be a sequence of bounded
linear operators Tn : X  Y from a Banach space X into a normed space Y
such that ( || Tn || ) is bounded for every xX, say || Tnx || ≤ cx ..………….(1)
where cx is a real number. Then the sequence of the norms || Tn || is bounded,
that is, there is c such that || Tn || ≤ c …………………………………….(2)
7
Proof. For every kN, let Ak  X be the set of all x such that || Tnx || ≤ k for
__
all n. To show that Ak is closed, let x Ak , then there is a sequence (xj ) of
elements in Ak such that lim xj = x. Then for any fixed n, we have || Tnxj || ≤ k,
hence by the continuity of Tn and the continuity of the norm we have, || Tnx || ≤
k and so xAk. Therefore, Ak is closed. Since for every xX there exists cx
such that || Tnx || ≤ cx for all n. Then each xX belongs to some Ak. Hence X =

Ak , however, X is complete, then by Bairs Category Theorem there exists
k 1
__
Ak0 contains an open ball, say B0 = B(x0; r )  Ak0 = Ak0 . Now let xX be
r
2|| x||
arbitrary and nonzero and let z = x0 + γx, γ =
≠ 0 ………………(3)
r
Then || z - x0 || = || γx || = 2 < r. Hence zB0 and so z Ak0. Then by the
definition of Ak0 we have, || Tnz || ≤ k0 for all n. Since x0B0  Ak0, then
1
|| Tnx0|| ≤ k0 for all n. By (3), x =  (z - x0). Then for all n, || Tnx || =
1

1
1
|| Tn(z - x0) || ≤  ( || Tnz || + || Tnx0 || ) ≤  (2k0) =
4k0
r
|| Tn || ≤
. Therefore, || Tn || ≤ c for all n, where c =
2|| x||
r
4k0
r
(2k0).
Hence

4.7-4 Space of Polynomials. The normed space X of all polynomials is not
complete under the norm defined by || x || = max |  j | , where α0, α1, α2, … are
j
the coefficients of x.
Proof. We construct a sequence of bounded linear functionals on X which
satisfies (1) but not (2). Let x be a nonzero polynomial of degree Nx, then

x(t) =
 t
j 0
j
j
, where αj = 0 for j > Nx. Let ( fn ) be a sequence of functionals
that are defined on X by fn(0) = 0, fn(x) = α0 + α1 + α2 … + αn-1. It is left to
the reader to show that for any fixed n, fn is linear. Since |αj |≤ max |  j | = || x ||,
j
then | fn(x) | ≤ n|| x ||. Hence fn is bounded. Furthermore, for each fixed xX,
| fn(x) | ≤ (Nx + 1) max |  j | = cx, (since x is a polynomial of degree Nx that has
j
Nx + 1 coefficients ). Therefore, (| fn(x) |) satisfies (1). Finally, we show that
n
(fn) does not satisfy (2). Choose a polynomial x defined by x(t) =
t
j
. Then
j 0
| f n ( x )|
|| x || = 1 and fn(x) = 1 + 1 + … + 1 = n = n || x ||. Hence || fn || ≥ || x|| = n.
Hence the sequence (|| fn ||) is unbounded and so (2) is not satisfied. Therefore,
by Uniform Boundedness Theorem, X is not complete
H. W. 1, 3, 9, 5, 11, 13. H.W.*1 4.
8
4.8 Strong and Weak Convergence
4.8-1 Definition. A sequence ( xn ) in a normed space X is said to be strongly
convergent (or convergent in the norm) if there is xX such that || xn – x ||  0
as n  ∞. This is written as lim xn = x or xn  x. x is called the strong limit of
( xn ) and we say that ( xn ) converges strongly to x.
4.8-2 Definition. A sequence ( xn ) in a normed space X is said to be weakly
w
convergent (xn  x) if there is xX such that for every fX /, lim f(xn ) = f(x).
x is called the weak limit of ( xn ) and we say that ( xn ) converges weakly to x.
4.8-3 Lemma. Let ( xn ) be a weakly convergent sequence in a normed space
w
X, say xn  x. Then
(a) The weak limit x of ( xn ) is unique.
(b) Every subsequence of ( xn ) converges weakly to x.
(c) The sequence ( || xn || ) is bounded.
Proof. The proof of (a) and (b) are left to the reader.
w
(c) Since xn  x, then f(xn )  f(x) for all fX /, so that the sequence of
numbers ( f(xn ) ) is bounded. That is, there is cf a constant depending on f
such that | f (xn) | ≤ cf for all n. Using the canonical mapping C: X  X //, we
can define gnX // by gn(f) = f(xn ) for all f X /. Then for all n, | gn(f) | = | f(xn) |
≤ cf , that is the sequence (| gn(f) |) is bounded for all fX/. However, X/ is
complete, Then by uniform bounded Theorem 4.7-3, ( || gn || ) is bounded, and
by Lemma 4.6-1, || gn || = || xn ||. Hence ( || xn || ) is bounded
4.8-4 Theorem. Let ( xn ) be a sequence in a normed space X. Then
(a) Strong convergence implies weak convergence with the same limit.
(b) The converse of (a) is not generally true.
(c) If dim X < ∞, then weak convergence implies strong convergence.
Proof. The proof of (a) is left to the reader.
(b) Let (en ) be an orthonormal sequence in a Hilbert space H. By Riez’s
representation Theorem, any fH / has a representation f(x) = < x, z >. Hence

f(en) = <en, z >. By Bessel’s inequality  | en , z |2 ≤ || z || 2. Hence
n 1

 | e , z |
2
n 1
n
converges. So that |<en, z >| 0 as n∞. This implies that f(en) = <en, z>0 =
2
w
f(0) as n  ∞. Since fH/ was arbitrary, then en  0. However, (en) does not
converge strongly, because for all m ≠ n, || em - en ||2 = <em, em > + <en, en> = 2
w
(c) Suppose that xn  x and dim X = k. Let { e1, e2, …, ek } be any basis of X
k
and, say, xn =  (j n ) e j and x =
j 1
k
 e
j 1
j
j
. By assumption, f(xn )  f(x) for all
fX /, in particular fj(xn )  fj(x) for all j = 1, 2, …, k, where fj(ej) = 1 and
fj(em) = 0 ( m ≠ j ). Hence fj(xn ) = αj (n) and fj(x) = αj and so αj (n)  αj. Then
9
|| xn – x || = ||
k
 ( (jn)   j )e j || ≤
j 1
k
| 
j 1
(n)
j
  j | || ej ||  0 as n  ∞. Therefore,
(xn ) converges strongly to x
Examples
w
4.8-5 Hilbert Space. In a Hilbert space H, xn  x if and only if
<xn, z >  < x, z > for all z in H.
Proof. Left to the reader.
p
w
p
4.8-6 Space
. In
, where 1 < p < ∞, xn  x if and only if:
(A) The sequence ( || xn || ) is bounded.
(B) For every fixed j we have j (n) j as n  ∞; here, xn = (j (n)) and x = (j).
Proof. Left to the reader.
4.8-7 Lemma ( Weak Convergence ). In a normed space X we have,
w
xn  x if and only if:
(A) The sequence ( || xn || ) is bounded.
(B) For every element f of a total subset M of X/ we have, f(xn) f(x) as n∞.
w
Proof. If xn  x, then (A) follows from Lemma 4.8-3, and (B) follows from
the definition of weak convergence. Conversely, suppose that (A) and (B) hold.
By (A), || xn || ≤ c for all n and || x || ≤ c for sufficiently large c. Since M is total
in X/, then for every fX/ there is a sequence (fi) in span M such that fi  f
(since span M is dense in X / ).Then for every  > 0 there is j such that || fi - f || <

. Moreover, since fispan M, then by (B) there is N such that | fi(xn) - fi(x) |
3c
<

3
for all n > N. Then for all n > N we have, | f(xn) - f(x) | ≤ | f(xn) - fi(xn) |
+ | fi(xn) - fi(x) | + | f i(x) - f(x) | < || f - fi || || xn || +
+

3
+

3c

3
+ || fi - f || || x || <

3c
c
w
c = . Since fX/ was arbitrary, then xn  x
H. W. 1-10 H.W.* 4, 5.
4.9 Convergence of sequences of operators and functionals
4.9-1 Definition. Let X and Y be normed spaces. A sequence ( Tn ) of
operators in B(X, Y) is said to be
(1) Uniformly convergent if ( Tn ) converges in the norm of B(X, Y), that is
there is T :XY such that || Tn – T ||  0 as n  ∞.
(2) Strongly convergent if ( Tnx) converges strongly in Y for every xX, that is
there is T :XY such that || Tnx – Tx ||  0 as n  ∞ for every xX.
(3) Weakly convergent if ( Tnx) converges weakly in Y for every xX, that is
there is T :XY such that || f(Tnx) – f(Tx) ||  0 as n  ∞ for every xX and
for every fY /.
10
Remark. Uniform operator convergence implies strongly operator
convergence implies weakly operator convergence.
Proof. Left to the reader.
Examples
2
4.9-2 Space
. The strong operator convergence need not imply the uniform
operator convergence.
2
2
Proof. Let ( Tn ) be a sequence of operators, where Tn:
 is defined by
Tnx = ( 0, 0, ….., 0, n+1, n+2, ……. ), ( note: the number of zeros is n ) where x
2
= (j)
. It is easy to see that Tn is linear for any n. Also, for any n, since


||Tnx|| = (  |  j |2 ) ≤ ( |  j |2 ) = || x ||, then Tn is bounded and || Tn || ≤ 1.
1
2
j  n 1
1
2
j 1
Moreover, if we choose x0 = ( 0, 0, ….., 0, n+1, n+2, ……. ) we get, Tnx0 = x0
and so || Tn || ≥
||Tn x0 ||
|| x0 ||
= 1. Therefore, || Tn || = 1 and so || Tn – 0 || = || Tn || = 1,
which implies that ( Tn) is not uniformly convergent to the zero operator. But, it
2
is clear that for all x we have, Tnx( 0, 0, …..) as n  ∞. Hence ( Tn) is
strongly operator convergent to 0
2
4.9-3 Space
. The weak operator convergence need not imply the strong
operator convergence.
Proof. Let ( Tn ) be a sequence of operators, where Tn:
2

2
is defined by
2
Tnx = ( 0, 0,..., 0, 1, 2, ... ), where the number of zeros is n and x = (j)
It is easy to see that Tn is linear and bounded for any n (how?) Since
Hilbert space, then every linear functional f on
f(x) = < x, z > =


j 1
=


k 1

| 
k 1
|2
j
 j , where z = (j)
2
.
is a
has a Riesz representation
. But f(Tnx) = <Tnx, z > =


j  n 1
__
j n
j
__
k
 n  k . By Cauchy-Schwarz inequality, | f(Tnx) | 2 = | <Tnx, z > | 2 ≤

k
__
2
2
 |
m  n 1
m
|2 0 as n  ∞. Therefore, f(Tnx) 0 = f(0x) for all x
2
. Thus
(Tn) is weakly operator convergent to 0. But, (Tn) is not strongly operator
convergent to 0, because for x = (1, 0, 0,...), we have || Tmx–Tnx || = 2 for all
m ≠ n
Note. Since the real and the complex fields are finite dimensional, then by
Theorem 4.8-4 (c), the weak and the strong convergence of any sequence of
bounded linear operators are the same.
4.9-4 Definition. Let ( fn ) be a sequence of bounded linear functionals on a
normed space X, then
(1) strongly convergence of ( fn) means that there is fX/ such that || fn – f || 0
as n  ∞. This written as fn  f
11
(2) weak* convergence of ( fn) means that there is fX/ such that fn(x)  f(x)
w
as n  ∞ for every xX. This written as fn  f
Remark. Let X and Y be normed spaces, and ( Tn ) be a sequence of elements
in B(X, Y).
(1) If ( Tn ) converges uniformly to T, then TB(X, Y).
(2) Strongly or weakly convergent of ( Tn ) to some T :XY need not imply
that TB(X, Y).
Proof. The easy proof of (1) is left to the reader. To verify (2), consider the
2
2
subspace X = {x = (j) : x has finitely many nonzero} of
. Let Tn:XX
be defined by Tnx = (1, 22, 33, ..... , nn, n+1, n+2, ……. ) and T:XX be
defined by Tx = (jj). It is left to the reader to show that for any fixed n, Tn is
linear and bounded. Moreover, T is linear but is not bounded, then TB(X).
Finally, it is clear that ( Tn ) converges strongly to T, where for all xX,
|| Tnx – Tx ||  0 as n  ∞
4.9-5 Lemma. Let X be a Banach space,Y a normed space and ( Tn ) be a
sequence of elements in B(X, Y) that converges strongly to a limit T :XY,
then TB(X, Y).
Proof. It is left to the reader to prove that T is linear. Since Tnx  Tx for all
xX, then by Lemma 1.4-2(a), the sequence (Tnx) is bounded. However, X is
complete, then by U. B. Theorem 4.7-3, there is c > 0 such that || Tn || ≤ c for
all n. Hence, || Tnx || ≤ || Tn || || x || ≤ c || x ||. Then by using the continuity of the
norm we have, || Tx || = lim || Tnx || ≤ c || x ||. Hence T is bounded and then
TB(X, Y)
4.9-6 Theorem. Let X and Y be Banach spaces and ( Tn ) be a sequence of
elements in B(X, Y). Then, ( Tn ) converges strongly if and only if:
(A) The sequence ( || Tn || ) is bounded.
(B) The sequence (Tnx) is Cauchy in Y for every x in a total subset M of X.
Proof. If Tnx  Tx for all xX, then (Tnx) is bounded and so by U. B.
Theorem 4.7-3, ( || Tn|| ) is bounded. (B) follows from the definition of strongly
operator convergent and the fact that every convergent sequence is Cauchy.
Conversely, suppose that (A) and (B) hold, then there is c > 0 such that ||Tn|| ≤ c
for all n. Since M is total in X, then span M is dense in X. Hence for every
xX and every  > 0 there is yspan M such that || x – y || <

3c
. By (B), the

sequence (Tny) is Cauchy. Hence there is kN such that || Tny – Tmy || < 3 for
m, n > k. Therefore, for all m, n > k and any fixed xX we have,
|| Tnx – Tmx || ≤ || Tnx – Tny || + || Tny – Tmy || + || Tmy – Tmx || < || Tn|| || x – y ||




+ 3 + || Tm|| || y – x || < c 3c + 3 + c 3c = . Hence (Tnx) is Cauchy in Y for
all xX. However, Y is complete, then (Tnx) converges in Y; that is ( Tn )
converges strongly
12
4.9-7 Corollary. A sequence ( fn ) of bounded linear functional on a Banach
space X is weak* convergent, the limit being a bounded linear functional on X
if and only if:
(A) The sequence ( || fn || ) is bounded.
(B) The sequence (fnx) is Cauchy for every x in a total subset M of X.
H. W. 1-4 H.W.* 4.
4.12 Open Mapping Theorem
4.12-1 Definition. Let X and Y be metric spaces. Then T : D(T)  Y with
D(T)  X is called an open mapping if the image of any open set in D(T) is an
open set in Y.
4.12-2 Lemma. A bounded linear operator T from a Banach space X onto a
Banach space Y has the property that the image T(B) of the open unit ball
B = B(0; 1)  X contains an open ball a bout 0Y.
4.12-3 Open Mapping Theorem, Bounded Inverse Theorem. A bounded
linear operator from a Banach space X onto a Banach space Y is an open
mapping. Hence if T is bijective, then T -1 is continuous and thus is bounded.
Proof. Let A be an open set in X. Let y = Tx be an element in T(A). Since A is open,
then A contains an open ball with center x. Hence A-x contains an open ball with
1
center 0; let the radius of the ball be r, and set k = r . Then k(A-x) contains the open
unit ball B(0; 1). Then by Lemma 4.12.2, T(k(A-x)) = k[T(A) – T(x)] contains a ball a
bout 0Y, and so does T(A) – T(x). Hence T(A) cotains an open ball a bout Tx = y.
But y was arbitrary in T(A), then T(A) is open. Therefore, T is an open mapping.
Finally, if T is bijective, then T -1 exists and linear by Theorem 2.6-10. However, T is
open then by Theorem 1.3-4 T is continuous, and by Theorem 2.7-9, T is bounded
H. W. 1, 2, 5-7 H.W.* 6.
13
4.13 Closed Linear Operators. Closed Graph Theorem
4.13-1 Definition. Let X and Y be normed spaces and T : D(T)  Y be a
linear operator with D(T)  X. Then T is called a closed linear operator if its
graph G(T) = { (x, y) : xD(T), y = Tx } is closed in the normed space XY,
where the two algebraic operations of the vector space XY are defined as
(x1, y1) + (x2, y2) = (x1+ x2, y1+ y2);  (x, y) = (x, y) ( is scalar) and the
norm on XY is defined by || (x, y) || = || x || + || y ||.
4.13-2 Closed Graph Theorem. Let X and Y be Banach spaces and let
T : D(T)  Y be a closed linear operator with D(T)  X. If D(T) is closed in
X, then T is bounded.
Proof. Let (zn) be any Cauchy sequence in XY, where zn = (xn, yn). Then for
every  > 0 there is kN such that for all m > n > k, || xn – xm || + || yn – ym || =
|| ( xn – xm, yn – ym )|| = || zn – zm || <  ……………………………………...(1)
Hence (xn) and (yn) are Cauchy sequences in X and Y, respectively. However,
X and Y are complete, then there are xX and yY such that xnx and yn y.
Let m∞ in (1) to get, || zn – (x, y) || = || xn – x || + || yn – y || <  for all n > k.
This means that (zn ) converges to z = (x, y)XY. Therefore, XY is
complete. Since G(T) is closed in XY and D(T) is closed in X, then G(T) and
D(T) are complete. Now consider the mapping P : G(T)  D(T), that defined
by P((x, Tx)) = x. It is left to the reader to show that P is linear. Since, || P((x,
Tx)) || = || x || for all xX, then P is bounded. It is easy to see that P is bijective
(how?) Hence P-1 : D(T)  G(T) exists and is defined by P-1(x) = (x, Tx).
Since G(T) and D(T) are complete, then by open mapping Theorem 4.12-2, P -1
is bounded, say || (x, Tx) || = || P-1(x) || ≤ b || x || for some b > 0 and all xD(T).
Hence || Tx || ≤ || Tx || + || x || = || (x, Tx) || ≤ b || x ||. Therefore, T is bounded.
4.13-3 Theorem. Let T:D(T)Y be a linear operator with D(T)X and X and Y
be normed spaces. Then, T is closed if and only if it has the following property:
if xnx, where xnD(T) for all n and Txny, then xD(T) and Tx = y………(2)
Proof. Suppose that T is closed, then G(T) is closed. Let xnx, where xnD(T)
for all n and Txny, then (xn, Txn)G(T) and (xn, Txn)(x, y). Hence (x,
y)G(T). This means that xD(T) and Tx = y. Conversely, suppose that the
_______
property (2) holds. Let (x, y) G (T ) . Then there is ( (xn, Txn) ) a sequence of
elements in G(T) such that (xn, Txn)(x, y). Hence xnx, xnD(T) for all n
and Txny. Then by assumption [ the property (2) ], xD(T) and Tx = y; that
is (x, y)G(T). Therefore, G(T) is closed, and so T is closed.
4.13-4 Example Defferential operator. Let X = C[0, 1] and T:D(T)X be such
that Tx = x /, where x / is the derivative of x and D(T) is a subspace of functions
xX which have a continuous derivative. Then T is not bounded, but is closed.
14
Proof. By 2.7-5, T is linear but not bounded. Let xnD(T) for all n such that
xnx and Txn = xn /  y. By 1.5-5, the convergence in the normed space C[0,
t
1] is uniform convergence. Then
 y ( )d
0
t
t
=  lim x ( )d = lim  xn/ ( )d = lim
0
n 
/
n
0
t
xn(t) - xn(0) = x(t) - x(0); that is x(t) = x(0) +
 y ( )d . Then xD(T) and Tx
0
/
= x = y. Then by Theorem 4.13-3, T is closed.
Note. D(T) in Example 4.13-4, is not closed , because if it is closed then by
Theorem 4.13-2, T is bounded, because T is closed. Then we have a
contradiction with T is not bounded. Therefore, D(T) in Example 4.13-4, is not
closed.
Remark. (1) Closedness does not imply boundedness of linear operator. [ see
Example 4.13-4 above. ]
(2) Boundedness does not imply closedness of linear operator.
Proof. (2) Let T : D(T)  D(T) be the identity operator on D(T), where D(T)
is a proper dense subspace of a normed space X. It is clear that T is linear and
bounded. But T is not closed. This follows from Theorem 4.13-3 if we take an
x X – D(T) and a sequence (xn) in D(T) which converges to x.
4.13-5 Lemma. Let T : D(T)  Y be a bounded linear operator with D(T)  X,
where X and Y are normed spaces. Then,
(a) If D(T) is a closed subset of X, then T is closed.
(b) If T is closed and Y is complete, then D(T) is a closed subset of X.
Proof. Left to the reader.
H. W. 3-6, 11-13. H. W*. 6, 12.
15