Download LECTURES ON FUNCTIONAL ANALYSIS 1. Normed Spaces 1.1

LECTURES ON FUNCTIONAL ANALYSIS
R. SHVYDKOY
1. Normed Spaces
1.1. Basic concepts and notation. Let us consider a vector space X over the field K “ R
or C. We say that X has finite dimension n is there is a system of n linearly independent
vectors tx1 , . . . , xn u in X which spans X. We denote the linear span of a set S Ă X by rSs.
If X has no finite dimension, X is called infinite dimensional. A function
} ¨ } Ñ R`
is said to define a norm on X if the following axioms hold:
(i) }x} “ 0 iff x “ 0,
(ii) }αx} “ |α|}x} for all x P X, α P K,
(iii) }x ` y} ď }x} ` }y} (triangle inequality).
A function } ¨ } satisfying only (ii) and (iii) is called a pseudo-norm. We write pX, } ¨ }q to
indicate that X is equipped with the norm } ¨ } if it is not clear from the context.
Exercise 1.1. Show that (iii) is equivalent to
|}x} ´ }y}| ď }x ˘ y}.
Let us notice that a norm generates a metric, called norm-metric, on the space X via
dpx, yq “ }x ´ y}.
The corresponding topology is called norm-topology. The norm-topology naturally gives rise
to the concept of convergence and continuity. A sequence txn u8
n“1 Ă X is said to converge to
x in the norm-metric, or strongly, if }xn ´ x} Ñ 0 as n Ñ 8. A function f : X Ñ Y , where
Y is a topological space, is continuous if f pxn q Ñ f pxq whenever xn Ñ x, or equivalently, if
f ´1 pU q is open for any open U Ă Y .
Exercise 1.2. Prove that the norm } ¨ } : X Ñ R is a continuous function on X.
ř
řN
A series 8
x
is
said
to
converge
to
its
sum
x
P
X
if
the
sequence
of
partial
sums
n
n“1
n“1 xn
ř
tends to x. A series converges absolutely if the numerical series 8
}x
}
converges.
The
n
n“1
natural fact that an absolutely convergent series itself converges does not necessarily hold in
general normed spaces. To ensure this property one has to impose an additional completeness
assumption. A metric space pM, dq is complete if its every Cauchy sequence has a limit, i.e.
if dpxn , xm q Ñ 0 implies Dx P M such that dpxn , xq Ñ 0. A complete normed space pX, } ¨ }q
is called a Banach space.
Exercise 1.3. Prove that a normed space pX, } ¨ }q is Banach if and only if every absolutely
convergent series in X is convergent.
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2
R. SHVYDKOY
For two sets A, B Ă X we denote by A ` B their algebraic sum tx ` y : x P A, y P Bu,
and constant multiple by αA “ tαx : x P Au. Next,
(1)
BpXq “ tx P X : }x} ď 1u,
the closed unit ball
DpXq “ tx P X : }x} ă 1u,
the open unit ball
SpXq “ tx P X : }x} “ 1u,
the unit sphere
Br px0 q “ tx P X : }x ´ x0 } ď ru “ x0 ` rBpXq
Dr px0 q “ tx P X : }x ´ x0 } ă ru “ x0 ` rDpXq.
The family of open ball forms a basis for the norm-topology.
1.2. Classical examples. The simplest example of a normed space is the Euclidean space
`n2 “ pKn , } ¨ }2 q with the norm given by
˜
¸1{2
n
ÿ
}x}2 “
|xi |2
.
i“1
ř
The Euclidean norm is generated by the inner product xx, yy “ xi ȳi via }x}2 “ xx, xy1{2 .
The triangle inequality in this case is a consequence of the Cauchy-Schwatz inequality:
}x ` y}2 “ }x}2 ` 2 Rexx, yy ` }y}2 ď }x}2 ` 2}x}}y} ` }y}2 ď p}x} ` }y}q2 .
There is a range of other natural norms on Kn , of which } ¨ }2 is a part, and for which the
triangle inequality is not so straightforward. We introduce them next.
ř Let p1 ď p ă 8. We define `p as the space of sequences x “ px1 , x2 , . . .q such that
i |xi | ă 8 and endow it with the norm
˜
¸1{p
8
ÿ
}x}p “
|xi |p
.
i“1
For p “ 8, `8 is the space of bounded sequences endowed with the supremum-norm
}x}8 “ sup |xi |.
i
The corresponding n-dimensional analogue is denoted `np . At this point, other than in cases
p “ 1, 8, it not clear whether `p is a linear space and } ¨ }p defines a norm on it. We will
show it next and establish several very important inequalities as we go along the proof.
Lemma 1.4. p`p , } ¨ }p q is a Banach space for all 1 ď p ď 8.
Proof. First, let B “ 8. The axioms of norm in this case are trivial. To show that `8
is complete let xn “ txn pjqu8
j“1 be a Cauchy sequence. Hence, every numerical sequence
txn pjqun is Cauchy. This implies that xn pjq Ñ xpjq as n Ñ 8. Since txn u is Cauchy, we
have }xn ´ xm }8 ă ε for all n, m ą N . Thus, |xn pjq ´ xm pjq| ă ε for all j P N as well. Let
us fix n and j and let m Ñ 8 in the last inequality. We obtain |xn pjq ´ xpjq| ď ε for all j,
and hence }xn ´ x} ď ε, for all n ą N . We have shown that xn Ñ x.
Now let p ă 8. Let us prove the triangle inequality first. By concavity of lnpxq, we have
lnpλa ` µbq ě λ lnpaq ` µ lnpbq,
for all λ ` µ “ 1, λ, µ ě 0, and a, b ą 0. Exponentiating the above inequality we obtain
ap b q
(2)
ab ď
` , (Young’s Inequality)
p
q
LECTURES ON FUNCTIONAL ANALYSIS
3
whenever p1 ` 1q “ 1, p ě 1. Next, consider finite sequences x “ txi uni“1 , y “ tyi uni“1 and
observe by (2)
ˇ
ˇ
n
n
n
ˇÿ
ˇ 1ÿ
1ÿ q 1 p 1 q
ˇ
ˇ
|xi |p `
|yi | “ }x}p ` }y}q .
ˇ xi y i ˇ ď
ˇi“1
ˇ p i“1
q i“1
p
q
ř
Thus, if }x}p “ }y}q “ 1, then | i xi yi | ď 1. For general x and y, after normalization we
obtain
ˇ
ˇ
n
ˇÿ
ˇ
ˇ
ˇ
(3)
x
y
ˇ
i i ˇ ď }x}p }y}q , (Hölder Inequality).
ˇi“1
ˇ
Finally,
n
ÿ
p
|xi ` yi | ď
i“1
˜
n
ÿ
ď
n
ÿ
|xi ` yi |
p´1
n
ÿ
|xi | `
i“1
|xi ` yi |p´1 |yi |
i“1
¸1{q
r}x}p ` }y}p s “ }x ` y}p{q
p r}x}p ` }y}p s.
|xi ` yi |pp´1qq
i“1
Thus,
}x ` y}pp ď }x ` y}pp{q r}x}p ` }y}p s,
and this implies
}x ` y}p ď }x}p ` }y}p , (Minkowski’s inequality)
which is what we need only for finite sequences. It remains to notice that if x, y P `p are
arbitrary, then the above inequality shows that the partial sums of the p-series of x ` y are
uniformly bounded, which in turn implies that x ` y P `p and the triangle inequality (iii)
holds as desired.
To prove that `p is complete, let xn “ txn pjqu8
j“1 be Cauchy. Then, as before we can pass
to the limit in every coordinate xn pjq Ñ xpjq. For a fixed J P N, ε ą 0 and n, m large
enough, we have
J
ÿ
|xn pjq ´ xm pjq|p ă ε.
j“1
Letting m Ñ 8, we obtain
J
ÿ
|xn pjq ´ xpjq|p ă ε.
j“1
ř
In particular, this implies that all partial sums of the series j |xpjq|p are bounded, and
hence x “ txpjquj P `p . Now, let us let J Ñ 8 in the estimate above. We obtain then
}xn ´ x}p ď ε1{p , thus xn Ñ x.
A normed space X is called separable if it contains a countable dense subset, i.e. if there is
S Ă X, card S “ ω0 such that for every x P X and any ε ą 0 there is y P S with }x ´ y} ă ε.
Exercise 1.5. Show that `p is a separable space for all 1 ď p ă 8.
Exercise 1.6. Show that `8 is not a separable space. Hint: consider the set of vectors
txA uAĂN , where xA is the characteristic function of A.
Our next classical example is c0 . This is the space of sequences txj u8
j“1 such that limjÑ8 xj “
0 endowed with the uniform } ¨ }8 norm.
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R. SHVYDKOY
1.3. Constructing new spaces from old ones. There are many ways to construct new
spaces from existing examples. A direct product of two linear spaces X and Y , denoted
X ˆY is the set of pairs tpx, yq : x P X, y P Y u endowed with the coordinate-wise operation of
summation and multiplication by a scalar. This makes X ˆY into a linear space. Identifying
elements of the product px, 0q with x , and p0, yq with y arranges a natural embedding of X
and Y into X ˆ Y . We thus can write x ` y “ px, yq. Let now pX, } ¨ }X q and pY, } ¨ }Y q be
normed and let 1 ď p ă 8. We can define a new norm on the product X ˆ Y by
}x ` y}p “ p}x}pX ` }y}pY q1{p .
The verification that this rule defined a norm is immediate from Minkowski’s inequality
established above. The obtained normed space is called the `p -sum of the X and Y and
denoted X ‘p Y . For p “ 8 we naturally define X ‘8 Y equipped with the norm }x ` y}8 “
maxt}x}X , }y}Y u. Similarly, we define `p -sums of any number of spaces and even countably
many spaces by requiring a member of X1 ‘p X2 ‘p . . . to be a sequence of vectors x “ tx1 , ...u
¯1{p
´ř
8
p
ă 8, or bounded in the case p “ 8.
such that }x}p “
j“1 }xj }Xj
Exercise 1.7. Verify that all of the newly introduced spaces are Banach if the original spaces
are Banach.
Let us notice that for any pair of vectors x P SpXq and y P SpY q, the span rx, ys will be
identical to `2p in the `p -product of spaces. So, for example, the unit ball of the X ‘1 R will
look like a symmetric tent with BpXq being the base and p0, 1q the top point. The ball of
X ‘8 R would be the cylinder with base BpXq and hight 1.
A subspace Y of a linear space X is a subset which is closed under the linear operations.
Y is closed if it is closed in the norm-topology of X. If in addition X is complete, then so is
every closed subspace. Thus, any closed subspace of a Banach space is Banach. Let us now
fix a closed subspace Y Ă X and consider the equivalence relation x1 „ x2 off x1 ´ x2 P Y .
This defines a conjugacy class rxs, for every x P X. The space of all conjugacy classes is
called the factor-space of X by Y , denoted X{Y , with the natural linear operations inherited
from X. We can endow X{Y with a norm too, called the factor-norm:
(4)
}rxs} “ inft}x ` y} : y P Y u “ disttx, Y u.
Exercise 1.8. Show that the above defines a norm on X{Y . Show that if X is complete, then
X{Y is complete as well in the factor-norm.
If X is endowed with a pseudo-norm, } ¨ }, consider X0 “ tx P X : }x} “ 0u. This is a
closed linear subspace of X, and moreover, }x ` y} “ }x} for all x P X, y P X0 . It is easy to
show that (4) defines a norm on X{X0 , i.e. axiom (i) holds.
1.4. Norm comparison and equivalence. Let pX, } ¨ }q be a normed space and Y Ă X
is a subspace with another norm |||¨|||. We say that the norm |||¨||| is stronger than } ¨ } if there
exists a constant C ą 0 such that
(5)
}y} ď C |||y||| , for all y P Y.
The two norms are equivalent if there are c, C ą 0 for which
(6)
c |||y||| ď }y} ď C |||y||| , for all y P Y.
Geometrically, (5) means that B |||¨||| pY q Ă CB }¨} pY q, while (6) means that there is embedding
in both sides, cB }¨} pY q Ă B |||¨||| pY q Ă CB }¨} pY q. The stronger norm, therefore, defines a finer
topology on Y , while equivalent norms define the same topology.
LECTURES ON FUNCTIONAL ANALYSIS
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Example 1.9. We have `p Ă `q , for all 1 ď p ď q ď 8, and
(7)
}x}q ď }x}p .
Indeed, assuming that x “ px1 , ...q P Sp`p q implies that all |xi | ď 1. Hence, |xi |q ď |xi |p , and
thus, x P `q . Moreover, }x}q ď 1. The general inequality (7) follows by homogeneity.
Example 1.10. Let pΩ, Σ, µq be a finite measure space, µpΩq ă 8. We have the opposite
embeddings for the Lebesgue spaces: Lq pdµq Ă Lp pdµq, for all 1 ď p ď q ď 8, and
1
1
}f }p ď µpΩq p ´ q }f }q , for all f P Lq pdµq.
(8)
It readily follows from the Hölder inequality,
ˆż
˙p{q
ż
p
q
|f | dµ ď
|f | dµ
µpΩq1´p{q .
Ω
Thus,
}f }pp
ď µpΩq
1´p{q
}f }pq ,
Ω
from which (8) follows.
Exercise 1.11. Show that in all of the examples above the norms are not equivalent on the
subspace in question.
Exercise 1.12. Verify the inequality
1
1
}x}q ď }x}p ď n p ´ q }x}q ,
for vectors x P Rn . Can you interpret the upper bound as a particular case of (8)?
Exercise 1.13. Show that if µpΩq “ 8, then the Lp -norms are not comparable on Lp X Lq ,
i.e. neither is stronger than the other.
Theorem 1.14. On a finite dimensional linear space X all norms are equivalent.
Proof. By transitivity, it suffices to show that all norms on Rn are equivalent to the norm of
} be a norm on Rn , and let tei uni“1 be the vectors of the standard unit basis.
`n1 . So, let } ¨ ř
Then for x “ xi ei ,
ÿ
ÿ
}x} ď
|xi |}ei } ď maxt}ei }u |xi | “ M }x}1 .
To establish an inequality from below, let us consider the norm-function N pxq “ }x}. By
compactness of Sp`n1 q and continuity of N , N attains its minimum on Sp`n1 q at x0 . Then
N px0 q “ c ą 0, since N never vanishes on a non-zero vector. So, }x} ě c, for all x P Sp`n1 q,
and hence }x} ě c}x}1 , by homogeneity.
1.5. Linear bounded operators. A map T between two linear spaces X Ñ Y is called a
linear operator if T pαx`βyq “ αT pxq`βT pyq. We usually drop the parentheses, T pxq “ T x,
when a linear operator is in question. Suppose pX, } ¨ }X q and pY, } ¨ }Y q are normed. A linear
operator T : X Ñ Y is called bounded or continuous if there exists a constant C ą 0 such
that
(9)
}T x}Y ď C}x}X ,
holds for all x P X. We denote the set of all linear bounded operators between X and Y by
LpX, Y q. The following theorem justifies the terminology.
Theorem 1.15. Let T : X Ñ Y be a linear operator. The following are equivalent:
(i) T P LpX, Y q;
(ii) T maps bounded sets into bounded sets;
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R. SHVYDKOY
(iii) T is continuous as a map between X and Y endowed with their norm topologies;
(iv) T is continuous at the origin.
Proof. The implication piq ñ piiq is clear from (9). Conversely, T , in particular, is bounded
on the unit ball of X, i.e. there exists a C ą 0 such that, }T x}Y ď 1, for all x P BpXq.
If x P X is arbitrary, then x̄ “ x{}x} P BpXq, and hence }T x̄}Y ď C. So, by linearity, we
obtain (9).
The implication piq ñ pivq is also clear directly from (9). If pivq holds, and x0 P X is
arbitrary, then for y Ñ 0, by linearity and continuity at the origin, we have
T px0 ` yq “ T x0 ` T y Ñ T x0 ,
showing that T is continuous at x0 . Thus, piiiq holds. Finally, if pivq holds, then there is
a δ ą 0 such that }x}X ă δ implies }T x}Y ă 1. So, if x is arbitrary, consider x1 “ δx{}x}.
Then }T x1 } ď 1 implies }T x} ď }x}{δ giving us (9).
If T P LpX, Y q we define the norm T as follows:
}T } “ inftC ą 0 : (9) holdsu.
In particular, for any x P X, }T x} ď C}x} holds for all C ą 0 for which (9) holds. This
shows that the infimum is in fact attained,
}T x}Y ď }T }}x}X ,
@x P X.
Exercise 1.16. Show that
}T } “ sup }T x}Y “ sup }T x}Y .
xPBpXq
xPSpXq
These identities say that the norm of an operator is the measure of deformation of the
unit ball of X under T . In particular, if }T } ď 1, then T is called a contraction.
Exercise 1.17. Suppose that T, S P LpX, Y q. Then
}T ` S} ď }T } ` }S}.
The set LpX, Y q of all bounded linear operators between X and Y clearly forms a linear
space, and the above exercise shows that the operator norm endows it with a norm.
Theorem 1.18. If pX, } ¨ }X q is normed and pY, } ¨ }Y q is Banach, then the space LpX, Y q is
Banach in its operator norm.
Proof. Suppose tTn u Ă LpX, Y q is Cauchy. Then, in particular, for any fixed x P X,
}Tn x ´ Tm x}Y ď }Tn ´ Tm }}x}X Ñ 0,
as n, m Ñ 8. Thus, tTn xu is Cauchy in Y . We, therefore can define the limit
T pxq “ lim Tn x,
nÑ8
for each x P X. By linearity of Tn ’s the limit is a linear operator as well. To show that
it is bounded, observe that the original sequence of operators is bounded, thus }Tn } ď M
for some M and all n. So, }T x}Y ď M }x}X for all X, which proves boundedness. Finally,
to show that Tn Ñ T in operator norm, let us fix ε ą 0 , then for all n, m large and all
x P BpXq we have
}Tn x ´ Tm x}Y ă ε.
LECTURES ON FUNCTIONAL ANALYSIS
7
Let us keep n fixed and let m Ñ 8. We already know that Tm x Ñ T x, thus,
}Tn x ´ T x}Y ď ε
holds for all x P BpXq and all n large. This gives }Tn ´ T } ď ε for all n large, which
completes the proof.
Exercise 1.19. Suppose that T : X Ñ Y and S : Y Ñ Z are bounded. Prove that
}S ˝ T } ď }S}}T }.
Let us introduce some terminology associated with operators. Let T P LpX, Y q. The
kernel is T is defined by Ker T “ T ´1 p0q “ tx P X : T x “ 0u, image by Im T “ ty P Y : Dx P
X, T x “ yu. Note that for a bounded operator the kernel is always closed, while the image
may not be. We say that T P LpX, Y q is an isomorphism if T is bijective and the inverse
T ´1 is bounded. This is equivalent to requiring that T is surjective and there are constants
c, C ą 0 such that
(10)
c}x} ď }T x} ď C}x}, for all x P X.
Note that the sharp constants in (10) are in fact c “ }T ´1 } and C “ }T }. If an isomorphism
exists between two spaces we call the spaces isomorphic, and denote X « Y . T is called
an isometry, or isometric isomorphism, if }T x} “ }x} for all x P X, and we denote X – Y
for isometrically isomorphic spaces. We generally don’t distinguish such spaces, and simply
refer to them as equal, although sometimes specify the identification rule, i.e. T : X Ñ Y ,
between their elements.
Exercise 1.20. Show that `21 – `28 , but `n1 fl `n8 for all n ě 3.
Let us observe that equivalence between two norms introduced in Section 1.4 is in the new
terms equivalent to the identity i : pX, } ¨ }q Ñ pX, |||¨|||q being an isomorphism.
We say that T : X Ñ Y is an isomorphic embedding if (10) holds without subjectivity
assumption. In this case, the image of T is closed, and Im T « X.
1.6. The factor-map and criterion of compactness of BpXq. Let Y be a closed proper
subspace of a normed space X. Let us consider the factor-map J : X Ñ X{Y defined by the
rule
Jx “ rxs.
From the definition of the factor-norm, it is clear that }Jx} ď }x}, thus making J a contraction map. To show that in fact }J} “ 1, let us fix one x P X not in Y . Then }rxs} ą 0. For
a fixed ε ą 0, let us find y P Y such that
}rxs} ď }x ` y} ď }rxs} ` ε.
Consider the normalized unit vector x ` y “ px ` yq{}x ` y}. Then
}Jpx ` yq} “
}rxs}
ε
ě1´
.
}x ` y}
}rxs}
Since, x is fixed and ε is arbitrary, we obtain }J} “ 1. This observation shows that for any
closed proper subspace Y we can find a vector on the unit sphere SpXq which is almost a
distance 1 away from Y . This is more than enough to prove the following theorem.
Theorem 1.21. The unit ball of a normed space pX, }¨}X q is compact if and only if dim X ă
8.
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R. SHVYDKOY
Proof. We have already seen in Section 1.4 that the unit ball of a finite dimensional spaces
is compact. Let us assume now that dim X “ 8 and show that the ball is not compact.
It suffices to construct a separated sequence of vectors x1 , x2 , ... so that all }xn } “ 1 and
}xn ´ xm } ě 1{2. Indeed, any such sequence would not contain a convergent subsequence.
To this end, let us fix an arbitrary first vector x1 P SpXq. Consider the space Y1 “ rx1 s, and
find x2 P SpXq such that disttx2 , Y1 u ą 1{2. Then consider Y2 “ rx1 , x2 s and find x3 P SpXq
with disttx3 , Y2 u ą 1{2, and so on. The process will never terminate since X is not a span
of finitely many vectors. It is easy to see that the obtained sequence is as desired.
If T : X Ñ Y is a bounded operator and X0 “ ker T , we construct a new operator T̃ :
X{X0 Ñ Y by the rule T̃ ˝J “ T . One can easily check that this definition is not ambiguous.
Moreover, one has }T } ď }T̃ }}J} “ }T̃ }. And on the other hand, if }T̃ rxs} ě }T̃ } ´ ε and
}rxs} ă 1, then for some x0 P X0 , }x ` x0 } ă 1, and yet T̃ rx ` x0 s “ T px ` x0 q. This shows
the opposite inequality }T̃ } ď }T }. Thus, }T̃ } “ }T }. Notice that the new operator has
trivial kernel.
1.7. Direct sums. Suppose Z is a linear space, X, Y Ă Z are subspaces such that Z “ X`Y
and X X Y . This is equivalent to the statement that for every z P Z there exists a unique
couple of vectors x P X, y P Y such that z “ x ` y. We thus have two linear maps P z “ x,
Qz “ y, so that P ` Q “ Id, called projections. Suppose now, that Z is normed. It is not
given that the operators P, Q are bounded. We say that Z is a direct sum of X and Y , and
write Z “ X ‘ Y , if P , or equivalently, Q, is bounded. The summands of the direct sum
are necessarily closed subspaces since X “ ker Q and Y “ ker P . A closed subspace X Ă Z
is called complemented if there is a Y Ă Z such that Z “ X ‘ Y . In this case Y is called a
complement of X, and a complement is never a unique space. What defines Y uniquely is
the projection P : Z Ñ X. We say that P is the projection onto X along Y .
Theorem 1.22. An algebraic sum Z “ X ` Y is direct if and only if disttSpXq, SpY qu ą 0.
Proof. Suppose he projection P : Z Ñ X is bounded. Let x P SpXq and y P SpY q. Then
}x ´ y} ě }P }´1 }P px ´ yq} “ }P }´1 }x} “ }P }´1 . Thus, disttSpXq, SpY qu ě }P }´1 .
Suppose now that disttSpXq, SpY qu ą 0, and yet P is not bounded. It implies that
there exists a sequence xn ` yn P SpZq such that› }xn } Ñ 8.
› By the triangle inequality,
› xn
›
}yn }
}xn } ď }yn } ď }xn } ` 1. Thus, }xn } Ñ 1. We have › }xn } ` }xynn } › Ñ 0. On the other hand,
› ›
›
ˆ
˙›
› › xn
›
› xn
y
y
y
}y
}
n
n
n
n
›“›
›
›
`
`
`
´
1
›
› }xn } }xn } › › }xn } }yn } }yn } }xn }
›
› ˇ
ˇ
› xn
ˇ
yn ›› ˇˇ }yn }
ˇ
ď ››
`
`
´
1
ˇ
}xn } }yn } › ˇ }xn }
›
›
›
›
In view of all of the above, › }xxnn } ` }yynn } › Ñ 0, in contradiction with our assumption.
A subspace X Ă Z is called finite co-dimensional if Z “ X ` Y for some Y Ă Z with
dim Y ă 8.
Corollary 1.23. Every finite co-dimensional closed subspace is complemented.
Indeed, let Z “ X ` Y , X X Y , X closed, and dim Y ă 8. If the spheres of X and Y are
not separated, then there exist sequences yn P SpY q and xn P SpXq such that }xn ´ yn } Ñ 0.
By compactness we can choose a subsequence ynk Ñ y P SpY q. Thus, xn Ñ y as well, which
by the closedness of SpXq implies y P X, a contradiction.
LECTURES ON FUNCTIONAL ANALYSIS
9
Exercise 1.24. Show that the norm of any projection operator is at least 1.
Exercise 1.25. Prove that a bounded operator P : X Ñ X is a projection onto a subspace if
and only if it is idempotent, P 2 “ P .
Exercise 1.26. Show that if Z “ X ‘ Y , then Z{X « Y . Hint: consider the projector
P : Z Ñ Y along X, and its factor by the kernel P̃ : Z{X Ñ Y .
1.8. Completion and extension by continuity.
1.9. The dual space. If the target space of a linear operator is the scalar field , the operator
is called a linear functional. The space of linear functionals on X is denoted X 1 , while the
space of all linear bounded functionals by X ˚ , called the dual space. It is often possible to
identify the dual of a Banach space up to an isometry.
Example 1.27. sdfsdfdsf
Exercise 1.28. Show that if X “ X1 ‘p . . . ‘p Xn , then X ˚ “ X1˚ ‘q . . . ‘q Xn˚ , where p, q
are conjugates. Show that the same is true for infinite `p -sums if 1 ď p ă 8.
1.10. Structure of linear functionals. Suppose f P X 1 zt0u, and let x0 P X be a vector
pxq
such that f px0 q ‰ 0. Then let x P X an arbitrary vector. Define y “ x ´ ffpx
x0 . Then
0q
clearly, f pyq “ 0. It shows that for any x there exist unique y P kerpf q and λ P R such that
x “ λx0 ` y.
In particular it shows that the kernel of f is one co-dimensional. We will show that the
distinction between bounded and unbounded functionals comes in the condition of closeness
of the kernel, or even less restrictively, its density.
Lemma 1.29. Let f P X 1 zt0u. The following are equivalent:
(a) f P X ˚ ;
(b) Kerpf q is closed;
(c) Kerpf q is not dense in X.
Proof. The implications paq ñ pbq ñ pcq are trivial. Suppose that Kerpf q is not dense. Then
for some ball Br px0 q X Kerpf q “ H. Let y P SpXq. Then gptq “ f px0 ` tyq “ f px0 q ` tf pyq
is a continuous non-vanishing function on p´r, rq. This implies that the sign of it has to
agree with that of f px0 q. Assuming f px0 q ą 0 we then have f px0 q ` tf pyq ą 0, and so,
|f pyq| ď r´1 f px0 q. This shows that f is bounded on the unit sphere and completes the
proof.
Geometrically linear bounded functionals can be identified with affine hyperplanes. Thus,
if f P X ˚ , then Hpf q “ tx P X : f pxq “ 1u, defines f uniquely. If f P SpX ˚ q, then
the hyperplane is in some sense tangent to the unit sphere of X, namely, it does not deep
inside the interior of the unit ball and it approaches arbitrarily close to SpXq. Note that a
functional may not necessarily attain its highest values on the sphere, i.e. the norm. For
example, let X “ `1 , and f “ p1{2, 2{3, 3{4, . . .q P Sp`8 q. There is no sequence x P Sp`1 q for
which f pxq “ 1. If x P SpXq and f P SpX ˚ q with f pxq “ 1, then f is called a supporting
functional of x. Existence of supporting functionals is not immediately obvious, and it brings
us to an even more fundamental question – does there exist at least one non-zero bounded
linear functional on a given normed space?
10
R. SHVYDKOY
2. Fundamental Principles
2.1. The Hahn-Banach extension theorem. The essence of the Hahn-Banach extension
theorem is to show that a given bounded functional defined on a linear subspace of X can
be extended boundedly to the whole space X retaining not only its boundedness but also
its norm. The boundedness can be expressed as the condition of domination by the normfunction, i.e. if Y Ă X and f P Y 1 then f P Y ˚ if and only if
f pyq ď C}y},
for some C ą 0. We will in fact need a more general extension result that will be useful
when establishing separation theorems later in Section 2.4. We thus consider a positively
homogeneous convex functional p : X Ñ R Y t8u, which means that pptxq “ tppxq for all
x P X and t ě 0, and ppλx ` p1 ´ λqyq ď λppxq ` p1 ´ λqppyq, for all 0 ă λ ă 1, x, y P X.
The latter is equivalent to the triangle inequality, ppx ` yq ď ppxq ` ppyq. Note that a norm,
or a quasi-norm, is an example of such a functional. We say that p dominates f on Y if
f pyq ď ppyq for all y P Y .
Theorem 2.1 (Hahn-Banach extension theorem). Suppose Y Ă X, and p is a positively
homogeneous convex functional defined on X. Then every linear functional f P Y 1 dominated
by p on Y can be extended to a linear functional f P X 1 denominated by the same p on all
of X.
In the core of the proof lies Zorn’s Lemma, which we recall briefly. Let P be a partially
ordered set. A subset C of P is called a chain if its every two elements are comparable, i.e.
@a, b P C either a ď b or b ď a. An upper bound for a set A Ă P is an element b P P such
that b ě a, for all a P A. A maximal element m is an element with the property that if
a ě m, then a “ m. Generally, it may not be unique.
Lemma 2.2 (Zorn’s Lemma). If every chain of P has an upper bound, then P contains a
maximal element.
Zorn’s lemma is equivalent to the Axiom of Choice.
Proof of Theorem 2.1. Let us introduce the set of pairs P “ tpf, Y q : f P Y 1 , f ď pu ordered
by pf1 , Y1 q ď pf2 , Y2 q iff Y1 Ă Y2 and f2 |Y1 “ f1 . Let C Ă P be a chain. Define Ỹ “ Ypf,Y qPC Y ,
and let f˜pyq “ f pyq if y P Y . This defines an upper bound of C. By Zorn’s Lemma there
exists a maximal element m “ pf0 , Y0 q P P . Let us show that Y0 “ X. Indeed, if not, then
there is a vector x0 P XzY0 . Thus, for every element x P Z “ rx0 , Y s there exist unique λ P R
and y P Y0 such that x “ λx0 ` y. We construct an extension f of f0 to Z so that pf, Zq P P
and run into contradiction with the maximality of m. In order to do that it suffices to find
a value of f only on x0 . Let c “ f px0 q, then to ensure that f is still dominated by p we need
to satisfy
λc ` f0 pyq ď ppλx0 ` yq.
For λ ą 0 this is equivalent to
(11)
c ď ppx0 ` y 1 q ´ f0 py 1 q,
and for λ ă 0 to
(12)
c ě f0 py 2 q ´ pp´x0 ` y 2 q.
LECTURES ON FUNCTIONAL ANALYSIS
11
In order for such a c to exist one has to make sure that any number on the right hand side
of (11) is no less than any number on the right hand side of (12), i.e.
ppx0 ` y 1 q ´ f0 py 1 q ě f0 py 2 q ´ pp´x0 ` y 2 q,
for all y 1 , y 2 P Y . This is true indeed, since in view of the convexity and dominance, we have
ppx0 ` y 1 q ` pp´x0 ` y 2 q ě ppy 1 ` y 2 q ě f py 1 ` y 2 q “ f py 1 q ` f py 2 q.
Let us discuss some immediate consequences of the Hanh-Banach theorem. First, every
vector on a normed space pX, } ¨ }q has a supporting functional. Indeed, let x P X, define
f P rxs˚ by f pλxq “ λ}x}. Then }f } “ 1, which means f is dominated by the norm. The
extension then has the same norm 1 and supports x.
For a normed space X, one can consider the dual of the dual space, X ˚˚ , called second dual.
There is a canonical isometric embedding i : X ãÑ X ˚˚ defined as follows: ipxqpx˚ q “ x˚ pxq.
It is convenient to use parentheses to indicate action of a functional: x˚ pxq “ px˚ , xq. In
this notation pipxq, x˚ q “ px˚ , xq or simply, px, x˚ q “ px˚ , xq. To show that i is an isometry,
notice that |pipxq, x˚ q| ď }x˚ }X ˚ }x}X , thus }ipxq}X ˚˚ ď }x}X . On the other hand, let
x˚ be a supporting functional. Then }x˚ }X ˚ “ 1, and pipxq, x˚ q “ x˚ pxq “ }x}X . We
will think of X is a sunspace of X ˚˚ with the natural identification of elements described
above. If the embedding X ãÑ X ˚˚ exhaust all elements of X ˚˚ , i.e. X “ X ˚˚ , then X is
called reflexive. We will return to a discussion of reflexive spaces later as they possess very
important compactness properties.
Let T : X Ñ Y be a bounded operator. We can define the adjoint or dual operator
˚
T : Y ˚ Ñ X ˚ by the rule pT ˚ y ˚ , xq “ py ˚ , T xq. Again, using the Hahn-Banach theorem we
show that
}T } “ }T ˚ }.
First, |pT ˚ y ˚ , xq| ď }y ˚ }}T x} ď }y ˚ }}T }}x}. This shows }T ˚ } ď }T }. Let now ε ą 0 be
given. Find x P SpXq such that }T x} ě }T } ´ ε. Then let y ˚ P SpY ˚ q be a supporting
functional for T x. We have pT ˚ y ˚ , xq “ py ˚ , T xq “ }T x} ě }T } ´ ε. This shows the opposite
inequality.
Exercise 2.3. Let T ˚˚ : X ˚˚ Ñ Y ˚˚ be the second adjoint operator, i.e. T ˚˚ “ pT ˚ q˚ . Show
that T ˚˚ |X “ T .
Exercise 2.4. Show that X ˚ is always complemented in X ˚˚˚ . Hint: consider the adjoint
i˚ : X ˚˚˚ Ñ X ˚ .
Exercise 2.5. Prove that T : X Ñ Y is an isomorphism if and only if T ˚ : Y ˚ Ñ X ˚ is.
Exercise 2.6. Show that X is reflexive if and only if X ˚ is reflexive.
Exercise 2.7. Let Y Ă X be a closed subspace, and X is Banach. Define Y K “ tf P
X ˚ : f |Y “ 0u. This is a closed subspace of X ˚ , called the annihilator of Y . Show that
Y ˚ – X ˚ {Y K , and pX{Y q˚ – Y K .
2.2. Convex sets. We say that a set A Ă X is convex if x, y P A implies λx ` p1 ´ λqy P A
for all 0 ă λ ă 1, i.e. with every pair of points A contains the interval connecting them.
A direct consequence of homogeneity and triangle inequality of the norm is that any ball is
12
R. SHVYDKOY
a convex set. For an arbitrary set A Ă X we define the convex hull of A as the set of all
convex combinations of elements from A:
#
+
n
n
ÿ
ÿ
conv A “
λi ai : ai P A, λi “ 1, λi ě 0, n P N .
i“1
i“1
It is the smallest convex subset of X containing A, or equivalently, the intersection
č
conv A “
C.
AĂC, C convex
The topological closure of the convex hull conv A is the same as the smallest closed convex
set containing A, or the intersection of such sets.
Theorem 2.8 (Caratheodori). Let A Ă Rn , then every point a P conv A can be represented
as a convex combination of at most n ` 1 elements in A.
ř
ř
λi “ 1, and N ą n ` 1. We will find a way
Proof. Suppose x “ N
i“1 λi ai , all λi ą 0,
to introduce a correction into the convex combination above as to reduce the number of
elements in the sum by 1. Then the proof follows by iteration.
First, let us observe that since N ą n`1, the number of elements in the family a2 ´a1 , a3 ´
a1 , . . . , aN ´ a1 is larger then the dimension and hence they are not linearly
independent.
řN
t
pa
So, we can find constants
t
P
R,
not
all
of
which
are
zero,
such
that
i
i“2 i i ´ a1 q “ 0.
ř
Denoting t1 “ ´ ti , we can write
N
ÿ
ti ai “ 0.
i“1
By reversing the sign of all the ti ’s if necessary, we can assume that at least one of them is
positive. We will now adjust the original convex combination by a constant multiple of the
zero-sum above, thus not changing the x:
x“
N
ÿ
i“1
λ i ai ´ ε
N
ÿ
ti ai “
i“1
N
ÿ
pλi ´ εti qai .
i“1
Letting ε “ minti ą0 tλ
ři {ti u ensures that µi “ λi ´ εti ě 0 for all i, and that for some i0 ,
µi0 “ 0. Yet, clearly, µi “ 1. Thus, the new representation
x“
N
ÿ
µ i ai ,
i“1
is at least one term shorter.
Corollary 2.9. If A Ă Rn is closed, then conv A is closed too.
Indeed, simply use the previous theorem and pass to nested subsequences in all n ` 1
terms by compactness. In the infinite dimensions closeness or even compactness of A is
not sufficient to conclude that conv A is automatically closed. Let us consider the following
example. Let X “ `2 , and A “ t n1 ~en un Y t0u. It is easy to see that A is compact. Any
ř
1
element of conv A has only finitely many non-zero entries, yet x “ 8
en P conv A.
n“1 2n n ~
LECTURES ON FUNCTIONAL ANALYSIS
13
2.3. Minkowski’s functionals. Let us recall from Section 2.4 that a function p : X Ñ
R Y t8u is called convex if for any x, y P X one has ppλx ` p1 ´ λqyq ď λppxq ` p1 ´ λqppyq
for all 0 ă λ ă 1. A function q : X Ñ R Y t´8u is called concave if for any x, y P X one
has qpλx ` p1 ´ λqyq ě λqpxq ` p1 ´ λqqpyq for all 0 ă λ ă 1. Is it easy to see that if p is
convex then the sub-level sets tp ď p0 u are convex, and if q is concave then the super-level
sets tq ě q0 u are convex.
Suppose A Ă X is convex and 0 P A. We associate to A a convex function, called
Minkowski’s functional, pA so that A is ”almost” given as a sub level set of pA . We define
pA pxq as follows. Suppose there is no t ě 0 for which x P tA, then pA pxq “ 8. If x P tA for
some t ě 0, we set
pA pxq “ inftt ě 0 : x P tAu.
We list the basic properties of the Minkowski’s functional.
(a) pA is positively homogeneous and convex;
(b) tpA ă 1u Ă A Ă tpA ď 1u.
For α ą 0, x P tA if and only if αx P αtA. This readily implies (a). Notice that for
positively homogeneous functionals convexity is equivalent to triangle inequality, pA px`yq ď
pA pxq ` pA pyq. So, let x, y P X. If any of pA pxq or pA pyq equal 8, the inequality becomes
trivial. If both are finite, then for every ε ą 0 we can find t1 ă pA pxq ` ε and t2 ă pA pyq ` ε
such that x P t1 A and y P t2 A. Then
ˆ
˙
t1
t2
x ` y P t1 A ` t2 A “ pt1 ` t2 q
A`
A Ă pt1 ` t2 qA.
t1 ` t2
t1 ` t2
This shows that pA px ` yq ď pA pxq ` pA pyq ` ε, for all ε ą 0. Finally, pcq follows directly
from the definition and that 0 P A.
Suppose now B is another convex set not containing a small ball around the origin, i.e.
there is δ ą 0 such that δBpXq X B “ H. We can associate a similar, but now concave
functional to B as follows. If x P tB, for no t ě 0, then qB pxq “ ´8. Otherwise, we define
qB pxq “ suptt ě 0 : x P tBu.
Condition δBpXq X B “ H warrants that the supremum is finite for any x P X. The
following list of properties can be established in a similar fashion:
(a) qB is positively homogeneous and concave;
(b) tqB ą 1u Ă B Ă tqB ě 1u.
Suppose now that we have two disjoint convex sets A and B satisfying all the assumptions
above, and let pA and qB be the corresponding Minkowski’s functionals. If pA pxq ă 8, let
t ě 0 be such that x P tA. Since 0 P A, the whole interval r0, xs is in tA and therefore not
in tB. This in turn implies than x R sB for any s ě t, for if such s existed, then st x P tB
contradicting the previous. As a consequence, qB pxq ď t. We have shown that
(13)
qB pxq ď pA pxq, for all x P X.
2.4. Separation theorems.
Theorem 2.10 (Generalized Hahn-Banach theorem). Let p be convex, and q concave functionals defined on X. Let Y Ă X, f P Y 1 such that
(14)
f pyq ď ppx ` yq ´ qpxq, for all y P Y, x P X.
14
R. SHVYDKOY
Then f can be extended to all of X, f˜ P X 1 , satisfying
qpxq ď f˜pxq ď ppxq, for all x P X.
(15)
Proof. The proof goes exactly the same way as before. We only need to check that if Y Ă X,
and x0 P XzY , then we can extend Y to Z “ rx0 , Y s preserving the domination property
(14). If c “ f px0 q, then we need
λc ` f pyq ď ppx ` λx0 ` yq ´ qpxq,
for all x P X and y P Y and λ P R. Again, for λ ą 0 this is equivalent to
c ď ppx1 ` x0 ` y 1 q ´ qpx1 q ´ f py 1 q,
while for λ ă 0,
c ě f py 2 q ´ ppx2 ´ x0 ` y 2 q ` qpx2 q.
The existence of c is ensured if
ppx1 ` x0 ` y 1 q ´ qpx1 q ´ f py 1 q ě f py 2 q ´ ppx2 ´ x0 ` y 2 q ` qpx2 q,
which is true since
ppx1 ` x0 ` y 1 q ` ppx2 ´ x0 ` y 2 q ´ qpx1 q ´ qpx2 q ě ppx1 ` x2 ` y 1 ` y 2 q ´ qpx1 ` x2 q ě f py 1 ` y 2 q.
Theorem 2.11 (Separation Theorems). Let A, B be two disjoint convex subsets of a normed
space X.
(i) If A ­Ă B, then there exists f P X 1 zt0u such that
(16)
sup f pAq ď inf f pBq.
(ii) If A has a non-empty interior, then there exists f P X ˚ zt0u such that (16) holds.
(iii) If A “ tx0 u and B is closed then there exists f P X ˚ zt0u such that
(17)
f px0 q ă inf f pBq.
Proof. Suppose A ­Ă B, then there exists x0 P A and δ ą 0 such that Bδ px0 q X B “ H. By
moving x0 to the origin we satisfy all the conditions on A and B as above, which allows us
to define Minkowski’s functions, qB ď pA . Thus, we have (14) for f “ 0, and Y “ t0u. By
Theorem 2.10, there exists and extension f for which (15) holds, and thus, in view of (13),
f separates A and B.
If A has a non-empty interior then clearly piq holds. Let us assume that εBpXq Ă A and
let f be the functional constructed above. Since f pxq ď pA pxq, we conclude that whenever
}x} ď ε, then f pxq ď 1. This shows }f } ď ε´1 .
Finally if A “ tx0 u and B is closed, we apply case piiq to A “ Bδ px0 q for small δ ą 0. Since
f ‰ 0, there is y P SpXq for which f pyq ą 0. Thus, f px0 q ă f px0 q ` εf pyq ď inf f pBq.
Exercise 2.12. If a strict inequality holds in (16), then A and B are called strictly separated.
Show that if A is compact and B closed convex disjoint sets, then they can be strictly
separated.
The condition of (i) is not sufficient to guarantee that a bounded separator would exist.
Let us consider the following example. Let X “ `2,0 be the linear space of finite sequences
LECTURES ON FUNCTIONAL ANALYSIS
15
endowed with the `2 -norm, A “ convt~en un , and B “ 21 A. These are convex disjoint sets.
ř
Notice that for any b P B, let b “ 12 i λi~ei , we have
1ÿ
1ÿ
1
}b}22 “
|λi |2 ď
λi “ .
4
4
4
Thus, B Ă 12 BpXq. Since any ~en R 21 BpXq, the conditions of Theorem 2.11 (i) are satisfied.
Next observe that 0 P A X B, by considering the sequences xn “ p~e1 ` ¨ ¨ ¨ ` ~en q{n and
yn “ 21 xn . Suppose f P `2 {t0u separates the two sets, i.e. sup f pAq ď c ď inf f pBq. Since 0
is in the closure of both sets, we conclude that c “ 0. Then f as a sequence is positive on
the one hand and negative on the other hand. Thus, f “ 0, a contradiction. It is easy to
construct an unbounded separator though by taking f “ p1, 1, ....q.
Corollary 2.13. Let S Ă X be a subset of a normed space. Then
č
conv S “
tx : f pxq ď sup f pSqu.
f PX ˚
Indeed, the inclusion Ă is obvious. If however x0 R conv S, then Theorem 2.11 (iii) provides
a functional such that f px0 q ă inf f pconv Sq ď inf f pSq. Reversing the sign of f shows that
x0 is not in one of the sets in the intersection.
2.5. Baire Category Theorem. Let us consider for a moment a general complete metric
space X, without necessarily a given linear structure. Let us ask ourselves how ”big” such
a space can be. One may say, a singleton X “ tx0 u is an obvious example of a ”small”
complete metric space. Well, it is not actually that small compared to its own standards.
After all that one point is closed and open, and it is in fact a ball of any radius centered
around itself. It therefore is rather ”big”, in a sense. To make this discussion more precise
let us agree on what we mean by a ”small” subset of X. We say that F is nowhere dense
in X is for any open set U there exists an open subset V Ă U with no intersection with
F . InŤother words, F has empty interior. A subset F Ă X is called of 1st Baire category if
F “ 8
i“1 Fi , where all Fi ’s are nowhere dense. A subset is of 2nd Baire category if it is not
of the 1st category. Thus, in the example above X itself is clearly of the 2nd category. This
in fact holds in general.
Theorem 2.14 (Baire Category Theorem). Any complete metric space is a 2nd Baire category set.
Ť
Proof. Let us suppose, on the contrary, that X “ Fi , and all Fi ’s are nowhere dense. Then,
there is a closed ball Bε1 px1 q with ε1 ă 1 disjoint from F1 . Since F2 is however dense, there
is a ball Bε2 px2 q Ă Dε1 px1 q, with ε2 ă 1{2, disjoint from F2 , continuing in the same manner
we find a sequence of nested closed balls Bεn pxn q, with εn ă 1{n. Clearly, dpxn , xm q ă 1{m,
for all n ą m. Thus, the sequence txn u is Cauchy. By completeness, there exists a limit x,
which belongs to all the balls, and hence not in any Fj ’s, a contradiction.
There are many consequences of the Baire category theorem, some of which are given in
the exercises below.
Theorem 2.15 (Banach-Steinhauss uniform boundedness principle). Let F Ă LpX, Y q
be a family of bounded operators, and X is a Banach space. Suppose for any x P X,
supT PF }T x} ă 8. Then, the family is uniformly bounded, i.e. supT PF }T } ă 8.
16
R. SHVYDKOY
Proof. Let Fn “ tx
Ť P X : supT PF }T x} ď nu. Not that each set Fn is closed, and by
assumption X “ n Fn . Hence, by the Baire Category Theorem, one of Fn ’s contains a
ball Br px0 q. This implies that for all x P BpXq, }T px0 ` rxq} ď n, for all T P F. Thus,
}T x} ď r´1 pn ` }T x0 }q ď r´1 pn ` supT PF }T x0 }q, implying that desired result.
Corollary 2.16. Let S Ă X be a subset such that for every x˚ P X ˚ , sup |x˚ pSq| ă 8. Then
S is bounded.
Indeed, if viewed as a subset of X ˚˚ , S is a pointwise bounded family of operators. Hence,
it is norm-bounded by the Banach-Steinhauss theorem.
2.6. Open mapping theorem.
Lemma 2.17. Suppose T : X Ñ Y is bounded, and X is a Banach space. Suppose that
DpY q Ă T pDpXqq, then 12 DpY q Ă T pDpXqq.
Proof. Let us note, by linearity of T , that
(18)
rDpY q Ă T prDpXqq
for any r ą 0. Let us fix y P D1{2 pY q and let us fix a small ε ą 0 to be specified later. By
(18) we can find x1 P 3{4DpXq such that }y ´y1 } ă ε, where T x1 “ y1 . Since y ´y1 P εDpY q,
one finds x2 P εDpY q such that }y ´ y1 ´ y2 } ă ε{2, where
ř y2 “ T x2 . Continuing this way, we
n
construct a sequence txn u with }xn } ď ε{2 . Let x “ n xn . Then by construction T x “ y,
and }x} ď 3{4 ` 2ε ă 1, provided ε is small.
Theorem 2.18 (Open mapping theorem). Suppose T P LpX, Y q, and both spaces are Banach. If, in addition, T is surjective, then T is an open mapping, i.e. T pU q is open for every
open U .
Proof. Suppose, U is open. Let x0 P U , and let Dε px0 q Ă U . We prove the theorem if we show
that T pDε px0 qq contains an open neighborhood of T x0 . Since, T pDε px0 qq “ T x0 `εT pDpXqq,
it amounts to showing that
Ť T pDpXqq contains a ball centered at the origin. Since T is
surjective, we have Y “ n nT pDpXqq. By the Baire Category Theorem, one of the sets
nT pDpXqq is dense in some ball Dδ py0 q, and hence, by linearity, so is T pDpXqq. Since
T pDpXqq is convex and symmetric with respect to the origin,
δDpY q Ă convtDδ py0 q, Dδ p´y0 qu Ă T pDpXqq
Applying Lemma 2.17 to the operator 1δ T , we conclude 2δ DpY q Ă T pDpXqq and the proof is
finished.
As a direct consequence of the open mapping theorem we obtain the following.
Corollary 2.19. Suppose T P LpX, Y q is bijective. Then T ´1 is automatically bounded.
Corollary 2.20. Two norms on a Banach space are either equivalent or incomparable.
Indeed, if C} ¨ }1 ě } ¨ }2 , then the identity operator i : pX, } ¨ }1 q Ñ pX, } ¨ }2 q is bounded.
Hence, the inverse is bounded, which establishes the equivalence.
Theorem 2.21 (Closed graph theorem). Suppose T : X Ñ Y is a linear operator such that
if xn Ñ x and T xn Ñ y, then T x “ y. Then T is bounded.
LECTURES ON FUNCTIONAL ANALYSIS
17
Proof. Let Γ “ tpx, T xq : x P Xu denote the graph of T . Thus, by the assumption, Γ is
closed in the product topology of the product X ˆ Y . Since it is also a linear subset, Γ as
a closed subspace of the `1 -sum, X b1 Y is a Banach spaces. Let us consider the projection
onto X restricted to Γ, P : Γ Ñ X, given by P px, T xq “ x. The operator is clearly bounded
and easy to show that P is bijective. Hence, P ´1 is bound, which implies
}x} ` }T x} ď C}x},
and thus T is bounded.
3. Weak topologies
3.1. Topology of not necessarily metrizable spaces. Let pX, τ q be a topological space
with topology of open sets τ . A subset txα uαPA Ă X is called a net, if the index set A is
partially ordered and directed, i.e. for every pair α, β P A there is γ P A with γ ě α, γ ě β.
A subnet is a net tyβ uβPB with a map n : A Ñ B such that yβ “ xnpβq , n is monotone,
and for every α P A there is β P B with npβq ě α. A net txα uαPA is said to be convergent
to x P X if for very open neighborhood U of x, there is α0 P A such that xα P U for all
α ě α0 . A function f : X Ñ Y , where Y is another topological space, is cooled continuous
if f ´1 pGq P τ for any open G Ă Y .
Lemma 3.1. A function f : X Ñ Y , is continuous if and only if for any convergent net
limαPA xα “ x, limαPA f pxα q “ f pxq.
Proof. Suppose f is continuous, and let limαPA xα “ x. For any open G containing f pxq,
f ´1 pGq is open and contains x. Since eventually all xα are in f ´1 pGq, then all f pxα q will be
in G.
Conversely, suppose there is open G Ă Y such that f ´1 pGq is not open. Thus, there is a
point x0 P f ´1 pGq such that any open neighborhood U of x contains a point outside f ´1 pGq.
Let us fix one such point xU for every U . Let A “ tU P τ : U open, x P U u. It is a net
ordered by inclusion. Clearly, xU Ñ x, since for every U containing x, all elements of the
net, namely starting from xU , will fall inside U . Yet, f pxU q R G, and thus f pxU q ­Ñ f pxq. Exercise 3.2. Show that a subset F Ă X is closed if and only if the limit of every convergent
net inside F is contained in F .
We say that X is compact if every open cover of X contains a finite sub cover.
Lemma 3.3. X is compact if and only if every net contains a convergent subnet.
Proof. Suppose X is compact, and let txα uαPA Ă X be a net. First we let us establish
existence of a cluster point. A point y P X is a cluster point of a net if for every U P τ
containing y and every α0 , there is α ě α0 such that xα P U . Suppose that our net does
not have cluster points. Thus, for every y P X there is Uy and αy P A such that xα R Uy
for all α ě αy . Consider the open cover tUy uyPX . By compactness there is a finite sub cover
Uy1 , . . . , Uyn . Since A is a net, there is a α ě αyi for all i “ 1, ..., n. Then xα is in none of
the open sets above, which shows that they don’t form a cover.
So, let y be a cluster point. Let B “ tpU, αq : y P U, U P τ, xα P U u be ordered by reverse
inclusion on the first component, and by the order of A on the second. For β “ pU, αq, let
yβ “ xα , and let npβq “ α. It is routine to show that tyβ uβPB is a subnet converging to y.
Conversely, suppose every net has a converging subnet, and yet on the contrary, X is not
compact. This implies that there is an open cover U which has no finite subcover. Let us
18
R. SHVYDKOY
define A “ tα “ pU1 , . . . , Un q : Ui P U, n P Nu ordered by α ě β if βŤĂ α. Clearly A is also
directed. By assumption, for any α “ pU1 , . . . , Un q there is xα R i Ui . The net txα uαPA
has a converging subnet tyβ uβPB , and y “ lim yβ . Since U is a cover, there is U P U with
y P U . Let α “ pU q. By the definition of a subnet, there is β 1 P B such that npβ 1 q ě α
and yβ 1 “ xnpβ 1 q , and there is another β 2 ě β 1 such that yβ 2 P U . By monotonicity of n,
npβ 2 q ě α, and yet xnpβ 2 q P U , in contradiction with the construction.
Exercise 3.4. A topology τ1 on X is said to be stronger than another topology τ2 on X if
for any point x P X any open neighborhood of x in τ2 contains an open neighborhood of x
in τ1 . We denote it τ1 ě τ2 . If τ1 ě τ2 and τ2 ě τ1 , then the topologies are called equivalent.
For example, equivalent norms on a normed space X define equivalent norm-topolgies. Show
that in general, τ1 ě τ2 if and only if a net converging in τ1 also converges in τ2 .
Let X be a set. A family of subsets F Ă 2X is called a filter if
(1) H R F;
Ş
(2) if F1 , ¨ ¨ ¨ , Fn are elements of F, then nj“1 Fj P F;
(3) if F P F and F Ă S, then S P F.
Let P be the set of all filters in X ordered by inclusion. A routine verification shows that P
satisfies the conditions of Zorn’s Lemma. Every maximal element of P is called an ultrafilter.
In fact, for any filter F there is an ultrafilter containing F, for the subset of P of filters
containing the given one satisfies Zorn’s Lemma as well. Ultrafilters can be characterized by
adding one more condition to the three above: U is an ultrafilter if and only if it is a filter
and
(4) for any subset A Ă X either A P U or XzA P U.
Indeed, if p4q holds and U 1 is another filter containing U, then any set A P U 1 should be in
U, for otherwise, XzA is in U, and then H “ A X pXzAq P U 1 . Conversely, if A Ă X is
such that XzA R U, then by p3q every F P U must intersect with A. Define a new family
U 1 “ tS : F X A Ă S, F P Uu. Clearly, U Ă U 1 , A P U 1 , and one can easily check that U 1 is
a filter. By maximality of U, U “ U 1 , and hence A P U. An alternative to p4q is a formally
stronger, but equivalent condition:
(41 ) if A1 Y . . . Y An P U, then some Ai P U.
Indeed, if non of Ai ’s belongs to U, then all the complements do, and hence their intersection,
which is XzpA1 Y . . . Y An q. This is incompatible with p1q.
The compactness of a topological space can be restated in terms of convergence of ultrafilters. So, let pX, τ q be a topological space and F be a filter on it. We say that lim F “ x
if every neighborhood of x has a non-empty intersection with any element of the filter. If
F is an ultrafilter, which will be our standard assumption, we showed above that every set
that intersects every element of F must lie in F. Thus, in this case lim F “ x iff every open
neighborhood of x is contained in F. If every two distinct points in X can be separated by
disjoint open neighborhoods, and such a space is called Hausdorff, then clearly, the limit is
unique. What follows, however, does not require this assumption.
Lemma 3.5. X is compact if and only if every ultrafilter in X converges to a point in X.
Proof. Suppose X is compact, and let U be an ultrafilter on X. If U does not converge to any
point in X, then any point x P X is contained in Ux P τ with Ux R U. Thus, tUx uxPX form
an open cover of X, which must contain a finite sub cover, U1 , . . . , Un . But Yj Uj “ X Ă U,
so by p41 q one of the sets must be in U, a contradiction.
LECTURES ON FUNCTIONAL ANALYSIS
19
Conversely, let C “ tU u be an open cover of X. Suppose that it contains no finite subcover.
Thus, any finite intersection pXzU1 qX. . .XpXzUn q in non-empty. This shows that the family
F “ tF Ă X : XzU Ă F, for some U P Cu is a filter, and let U be an ultrafilter containing
F. By assumption, let x “ lim U. Since C is a cover, there is U P C with x P U . But,
XzU P U, so U cannot be in U, a contradiction.
Ultrafilters are useful for many purposes. In particular, they provide an economical proof
of Tihonov’s compacness theorem, which will be used later to establish the main result of
this section, the Alaoglu Theorem.
ś
Let tpXγ , τγ quγPΓ be a collection of topological spaces. The product space, X “ γPΓ Xγ
Ť
is the set of functions x : Γ Ñ γPΓ Xγ such that xpγq P Xγ . We usually denote xpγq “ xγ
and write x “ txγ uγPΓ . Let πγ : X Ñ Xγ be the usual projection map. We define the
product topology on X to be the topology generated by the sets π ´1 pUγ q, where Uγ P τγ .
This is also the minimal topology on X in which all the projection maps are continuous.
Exercise 3.6. Show that xα Ñ x in the product topology if and only if πγ pxα q Ñ πγ pxq for
every γ P Γ.
Theorem 3.7
ś (Tihonov’s compactness theorem). If all Xγ , γ P Γ, are compact, then the
product X “ γPΓ Xγ is compact in the product topology.
Proof. Let U be an ultrafilter on X. For every γ P Γ, consider Uγ “ πγ pUq. Then Uγ is
an ultrafilter. Since Xγ is compact, there exists a limit xγ “ lim Uγ . Let us show that
x “ txγ uγPΓ “ lim U. Let U be open neighborhood of x in the product topology. Then with
pUi q. We have xγi P Ui , and hence
x, U contains a finite intersection of basis sets Xni“1 πγ´1
i
pUi q, which
Ui P Uγi , which means there exist Ai P U such that πγi pAi q “ Ui . Then Ai Ă πγ´1
i
n
´1
´1
implies that πγi pUi q P U, for each i, and hence Xi“1 πγi pUi q P U. Since U contains that
intersection, it must itself be in the ultrafilter U.
3.2. Weak topology. Let X be a Banach space. We define the weak topology on X as
the topology with the following base of neighborhoods: for x P X, ε1 , . . . , εn ą 0 and
f1 , . . . , fn P X ˚ , let
(19)
,...,fn
Uεf11,...,ε
pxq “ ty P X : |fi pyq ´ fi pxq| ă εi , @i “ 1, nu.
n
Thus, a neighborhood is an intersection of open slabs of finite widths. We say that a sequence,
w
or a net txα uαPA converges weakly to x, and denote xα Ñ x, if it converges in the sense of
the weak topology.
w
Exercise 3.8. Show that xα Ñ x if and only if f pxα q Ñ f pxq for every functional f P X ˚ .
Exercise 3.9. Show that if a sequence xn converges weakly, then it is bounded in the normtopology.
Exercise 3.10. Suppose dim X “ 8. Construct a net txα uαPA in X such that xα Ñ 0 weakly,
yet for every α0 P A and N ą 0, there is α ě α0 such that }xα } ě N . Hint: let
A “ tpf1 , . . . , fn ; N q : fj P X ˚ , n P N, N ą 0u.
Define a directed partial order on A, and for every α P A pick an xα P Xj Ker fj , with
}xα } ą N . Show that xα Ñ 0 weakly and is frequently unbounded.
20
R. SHVYDKOY
From now on we will use the term ”strong” with respect to anything related to the normtology, as opposed to ”weak” that refers to anything related to the weak topology. For
example, ”strongly compact set” v.s. ”weakly compact set”, or ”strong convergence” v.s.
”weak convergence”.
It is clear that the weak topology is weaker than the norm-topology on any normed space.
In fact, on an infinite dimensional space it is strictly weaker. To see this, we show that any
neighborhood (19) is unbounded. Indeed, let H “ Xi Ker fi . This is a non-empty space, for
,...,fn
otherwise X would have been a span of xi ’s, with xi R Ker fi . Then Uεf11,...,ε
pxq contains all
n
of x ` H.
Lemma 3.11. The weak topology is not metrizable on an infinite dimensional space.
Proof. Suppose, on the contrary that there is a metric dp¨, ¨q that defines the weak topology.
Consider the sequence of balls tdpx, 0q ă 1{nu. Each contains a weak neighborhood of the
origin. We have shown every weak neighborhood is unbounded. Thus, we can find a xn
within the nth ball with }xn } ą n. So, on the one hand, xn Ñ 0 weakly, and yet txn u is
unbounded, in contradiction with Exercise 3.2.
The weak topology is still ”fine” enough to separate points, and even larger sets, by the
Separation Theorems.
Exercise 3.12. Show that if A is closed and B is strongly compact convex sets, then there
are two disjoint weakly open neighborhoods of A and B.
Lemma 3.13. Let txn u8
n“1 Ă X be a sequence in any of the spaces X “ c0 , `p , for 1 ă p ă
w
8. Then xn Ñ x if and only if txn u is norm bounded and converges to x pointwise, i.e.
xn pjq Ñ xpjq, for all j P N.
Proof. We present the proof for X “ c0 and leave the `p -case as an exercise. So, suppose
w
xn Ñ x. Then the sequence is bounded by Exercise 3.2. Moreover, taking f “ ~ej , we
obtain the pointwise convergence. Conversely, if xn Ñ x pointwise, and
ř is bounded, let M “
˚
supn }xn }, and let f P `1 “ c0 . Given ε ą 0, let N P N be such that jąN |f pjq| ă ε{p2M q.
Then, for large n we have
ˇ
ˇ
ˇ
ˇÿ
ÿ
ˇ
ˇ
xpjqf pjqˇ ă ε{2.
xn pjqf pjq ´
ˇ
ˇ
ˇjďN
jďN
Thus,
ˇ
ˇ ˇ
ˇ ˇ
ˇ
ˇÿ
ˇ ˇÿ
ˇ ˇÿ
ˇ
ÿ
ÿ
ÿ
ˇ
ˇ ˇ
ˇ ˇ
ˇ
xn pjqf pjq ´
xpjqf pjqˇ ` ˇ
xn pjqf pjq ´
xpjqf pjqˇ
ˇ xn pjqf pjq ´ xpjqf pjqˇ ď ˇ
ˇj
ˇ
ˇ
ˇ
ˇ
ˇ
j
jďN
jďN
jąN
jąN
ă ε.
Lemma 3.14. In `1 , a sequence txn u8
n“1 converges weakly to x if and only if it converges to
x strongly.
Let us note that in spite of Exercise 3.4, the lemma above does not imply that the norm
and weak topologies are equivalent, because it deals only with sequences. Loosely speaking,
the reason why on `1 weak and strong convergences for sequences are equivalent is because
the dual `8 is ”very large”, so large that weak convergence is just as hard to arrange as
LECTURES ON FUNCTIONAL ANALYSIS
21
strong convergence. Banach spaces with the property stated in Lemma 3.14 are sometimes
called Kadets-Klee spaces.
As a weaker topology, the weak topology provides a smaller family of open sets than the
norm topology. Thus any weakly closed set is strongly closed as well 1. As a consequence of
the Separation Theorem 2.11, it turns out that among the class of convex sets the property
of being closed is the same in weak and strong topologies.
Lemma 3.15. A convex set C Ă X is strongly closed if and only if it is weakly closed.
w
Proof. Clearly, if C is strongly closed, and yet there is a point x0 P C not in C. Then by
Theorem 2.11, there is a functional f P X ˚ so that f px0 q ą c ą sup f pCq. Thus x0 belongs
to the open set U “ tf ą cu disjoint from C, a contradiction.
The exact same argument shows that the weak and strong closures of a convex set C
coincide. This has a rather interesting consequence for relationship between weak and strong
convergence of sequences.
Exercise 3.16. Show that if xn Ñ x weakly, then there is a sequence of convex combinations
made of xn ’s that converge to x strongly.
Exercise 3.17 (Weak lower-semi-continuity of norm). Show that if xn Ñ x weakly, then
w
lim inf nÑ8 }xn } ě }x}. More generally, if a net xα Ñ x, then for any ε ą 0 there is α0 so
that for all α ě α0 , }xα } ě }x} ´ ε.
3.3. Weak˚ topology. Consider now the dual space X ˚ . As any Banach space it has its own
weak topology determined by the functionals from ”upstairs”, i.e. X ˚˚ . However, one can
define a weaker Hausdorff topology on X ˚ determined by the functionals from ”downstairs”,
i.e. X. Let x1 , . . . , xn P X and ε1 , . . . , εn ą 0, and f P X ˚ . We define a weak˚ -open
neighborhood of f to be
(20)
,...,xn
Uεx11,...,ε
pxq “ tg P X ˚ : |gpxi q ´ f pxi q| ă εi , @i “ 1, nu.
n
Identifying element of X as vectors in X ˚˚ we see it is just a special subclass of neighborhoods
defined earlier in (19). It is still a Hausdorff topology as pairs of distinct functionals in X ˚
can be separated by elements of X. A sequence xn converging weak˚ to x is necessarily
bounded by the Banach-Steinhauss Theorem.
w˚
Exercise 3.18. Show that a net fα Ñ f if and only if fα pxq Ñ f pxq for every x P X.
Exercise 3.19. Show that any weakly˚ convergent sequence in X ˚ is strongly bounded.
w˚
Exercise 3.20 (Weak˚ lower-semi-continuity of norm). Show that if fn Ñ f , then
lim inf }fn } ě }f }.
nÑ8
More generally, if a net tfα uαPA converges weakly˚ to f , then for any ε ą 0 there is α0 P A
so that for all α ě α0 , }fα } ě }f } ´ ε.
How much the weak˚ topology may be weaker than the weak topology is illustrated by
the following example (c.f. Lemma 3.14).
1It
sounds a bit counterintuitive from the linguistic point. But if you think what it takes for a set to be
closed it becomes clear. If a set ”survives” weak limits from within itself, then it should definitely survive
strong limits.
22
R. SHVYDKOY
w˚
Exercise 3.21. Show that in `1 , xn Ñ x if and only if txn u is bounded and xn Ñ x pointwise.
Theorem 3.22 (Alaoglu). The unit ball of a dual space is compact in the weak˚ -topology.
Proof. Notice that for any f P BpX ˚ q, and x P X, f pxq P r´}x},
ś }x}s. This naturally suggests
˚
to consider BpX q as a subset of the product space T “ xPX r´}x}, }x}s. By Tihonov’s
theorem, this product space is compact in the product topology. It suffices to show that
BpX ˚ q is closed in T , because convergence of nets in the product topology is equivalent to
pointwise convergence, which for elements of BpX ˚ q amounts to weak˚ convergence.
To this end, let tfα uαPA be a net in BpX ˚ q with lim fα “ f P T . By linearity of fα ’s and
the ”pointwise” sense of the limit above, we conclude that
f pλx ` µyq Ð fα pλx ` µyq “ λfα pxq ` µfα pyq Ñ λf pxq ` µf pyq.
Thus, f is linear, and since |fα pxq| ď }x}, we also have |f pxq| ď }x} for all x P X, which
identifies f as an element of BpX ˚ q.
As an immediate consequence we see that for a reflexive Banach space X, the unit ball is
weakly compact. This the property actually characterizes reflexiveness.
Theorem 3.23 (Kakutani). A Banach space X is reflexive if and only if its unit ball is
weakly compact.
The theorem will follow from the next lemma.
w˚
Lemma 3.24. The unit ball BpXq is weakly˚ dense in BpX ˚˚ q. In particular, BpXq
BpX ˚˚ q.
“
Indeed, let BpXq be weakly compact. In view of Lemma 3.24, for any x˚˚ P BpX ˚˚ q
w˚
w
we find a net xα Ñ x˚˚ , xα P BpXq. By compactness, there is a subnet yβ Ñ x for some
x P BpXq, and yet that same subnet converges to x˚˚ in the weak˚ sense of X ˚˚ . Thus, for
every x˚ we have px˚˚ , x˚ q Ð pyβ , x˚ q “ px˚ , yβ q Ñ px˚ , xq, which identifies x˚˚ as x, and
thus BpXq “ BpX ˚˚ q, implying X “ X ˚˚ .
In order to prove Lemma 3.24, we have to go back to the separation theorem and make
one adjustment to it in the context of weak˚ topology.
Lemma 3.25. Suppose that f, f1 , . . . , fn P X 1 , and Xnj“1 Ker fj Ă Ker f . Then f P rf1 , . . . , fn s.
Proof. We will prove the lemma by induction. Suppose n “ 1, and f1 ‰ 0 (otherwise the
statement is trivial). By the structure of the linear functionals discussed in Section 1.10,
there is x1 P X with f1 px1 q “ 1, such that for every x P X we have x “ λx1 ` y, where
y P Ker f1 . Since f pyq “ 0 we have f pxq “ λf px1 q “ f px1 qf1 pxq, as desired.
n`1
Suppose the statement is true for n. Let us assume Xj“1
Ker fj Ă Ker f . Consider
n
the space
řn Y “ Ker fn`1 . Then Xj“1 Ker fj |Y Ă Ker f |Y . By the induction hypothesis,
f |Y “ j“1 aj fj |Y . By the structure of fn`1 we have for any x P X, x “ λxn`1 ` y for some
LECTURES ON FUNCTIONAL ANALYSIS
23
y P Y , and where fn`1 pxn`1 q “ 1. Thus,
n
n
ÿ
ÿ
f pxq “ λf pxn`1 q `
aj fj pyq “ f pxn`1 qfn`1 pxq `
aj fj pyq
j“1
j“1
“ f pxn`1 qfn`1 pxq `
n
ÿ
aj fj pxq ´ λ
j“1
(21)
˜
“
f pxn`1 q ´
n
ÿ
n`1
ÿ
aj fj pxn`1 q
j“1
¸
aj fj pxn`1 q fn`1 pxq `
j“1
“
n
ÿ
n
ÿ
aj fj pxq
j“1
aj fj pxq.
j“1
Lemma 3.26. If x˚˚ P X ˚˚ is continuous in the weak˚ topology, then x˚˚ P X.
Proof. By the assumption px˚˚ q´1 p´1, 1q contains a weak˚ neighborhood of the origin, say,
,...,xn
p0q. In particular, x˚˚ P p´1, 1q on Xj Ker xj . Since the latter is a linear space, x˚˚
Uεx11,...,ε
n
must in fact vanish on it. Thus, by Lemma 3.25, x˚˚ P rx1 , . . . , xn s Ă X.
Theorem 3.27 (Separation Theorem for weak˚ topology). Suppose B is a weakly˚ -closed
convex subset of X ˚ , and f R B. Then, there is x P X such that sup Bpxq ă 1 ă f pxq.
,...,xn
Proof. Let us follow the proof of Theorem 2.11. Let us assume f “ 0, and let A “ Uεx11,...,ε
p0q
n
˚
be a weak neighborhood of 0 disjoint from B. We then associate Minkowski’s functionals pA
and qB to A and B respectfully. As a result of Hahn-Banach Theorem, we find a separating
functional F so that qB ď F ď pA . Since f P A implies pA pf q ď 1, we see that F is bounded
from above on A. Like in the proof of Lemma 3.26 we conclude that F vanishes on the
intersection of the kernels of the x1 , . . . , xn , and hence, F P X. By rescaling F if necessary
we can arrange the constant of separation to be 1.
w˚
Proof of Lemma 3.24. Let B “ BpXq . By Exercise 3.20, B Ă BpX ˚˚ q. Suppose there is
F P BpX ˚˚ qzB. By Theorem 3.27 , we find a x˚ P X ˚ such that Bpx˚ q ă 1 ă F px˚ q. The
first inequality holds, in particular, on BpXq, which shows that }x˚ } ď 1. This runs into
contradiction with the second inequality.
Exercise 3.28. Recall that pc0 q˚˚ “ p`1 q˚ “ `8 . Show that Bpc0 q is weakly˚ sequentially
dense in Bp`8 q, i.e. for every F P Bp`8 q there is a sequence xn P Bpc0 q converging weakly˚
to F .
Corollary 3.29. Let Y Ă X be a closed subspace of a reflexive space X. Then Y and X{Y
are reflexive.
Proof. Indeed, BpY q “ Y X BpXq. Since the this set if convex and closed, it is weakly closed
in BpXq and hence compact in the weak topology of X. However, by the Hahn-Banach
extension theorem the topology induced on Y by the weak topology of X is exactly the
weak topology of Y . Thus, BpY q is weakly compact in Y . Now, by Exercise 2.7 and by
the previous, pX{Y q˚ is a subspace of a reflexive space, which makes it reflexive. Then by
Exercise 2.6 pX{Y q itself is reflexive.