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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 “ Ror C. We say that X has finite dimension n is there is a system of n linearly independentvectors tx1, . . . , xnu 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 toindicate 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 riseto the concept of convergence and continuity. A sequence txnu
8n“1 Ă X is said to converge to
x in the norm-metric, or strongly, if xn ´ x Ñ 0 as nÑ 8. A function f : X Ñ Y , whereY is a topological space, is continuous if fpxnq Ñ fpxq whenever xn Ñ x, or equivalently, iff´1pUq is open for any open U Ă Y .
Exercise 1.2. Prove that the norm ¨ : X Ñ R is a continuous function on X.
A seriesř8
n“1 xn is said to converge to its sum x P X if the sequence of partial sumsřNn“1 xn
tends to x. A series converges absolutely if the numerical seriesř8
n“1 xn converges. Thenatural fact that an absolutely convergent series itself converges does not necessarily hold ingeneral normed spaces. To ensure this property one has to impose an additional completenessassumption. A metric space pM,dq is complete if its every Cauchy sequence has a limit, i.e.if dpxn, xmq Ñ 0 implies Dx PM such that dpxn, xq Ñ 0. A complete normed space pX, ¨ qis called a Banach space.
Exercise 1.3. Prove that a normed space pX, ¨ q is Banach if and only if every absolutelyconvergent series in X is convergent.
1
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,
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
Brpx0q “ tx P X : x´ x0 ď ru “ x0 ` rBpXq
Drpx0q “ tx P X : x´ x0 ă ru “ x0 ` rDpXq.
(1)
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, ¨ 2q with the norm given by
x2 “
˜
nÿ
i“1
|xi|2
¸12
.
The Euclidean norm is generated by the inner product xx, yy “ř
xiyi via x2 “ xx, xy12.
The triangle inequality in this case is a consequence of the Cauchy-Schwatz inequality:
x` y2 “ x2 ` 2 Rexx, yy ` y2 ď x2 ` 2xy ` y2 ď px ` yq2.
There is a range of other natural norms on Kn, of which ¨ 2 is a part, and for which thetriangle inequality is not so straightforward. We introduce them next.
Let 1 ď p ă 8. We define `p as the space of sequences x “ px1, x2, . . .q such thatř
i |xi|p ă 8 and endow it with the norm
xp “
˜
8ÿ
i“1
|xi|p
¸1p
.
For p “ 8, `8 is the space of bounded sequences endowed with the supremum-norm
x8 “ supi|xi|.
The corresponding n-dimensional analogue is denoted `np . At this point, other than in casesp “ 1,8, it not clear whether `p is a linear space and ¨ p defines a norm on it. We willshow it next and establish several very important inequalities as we go along the proof.
Lemma 1.4. p`p, ¨ pq 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 `8is complete let xn “ txnpjqu
8j“1 be a Cauchy sequence. Hence, every numerical sequence
txnpjqun is Cauchy. This implies that xnpjq Ñ xpjq as n Ñ 8. Since txnu is Cauchy, wehave xn ´ xm8 ă ε for all n,m ą N . Thus, |xnpjq ´ xmpjq| ă ε for all j P N as well. Letus fix n and j and let m Ñ 8 in the last inequality. We obtain |xnpjq ´ 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
(2) ab ďap
p`bq
q, (Young’s Inequality)
LECTURES ON FUNCTIONAL ANALYSIS 3
whenever 1p` 1
q“ 1, p ě 1. Next, consider finite sequences x “ txiu
ni“1, y “ tyiu
ni“1 and
observe by (2)ˇ
ˇ
ˇ
ˇ
ˇ
nÿ
i“1
xiyi
ˇ
ˇ
ˇ
ˇ
ˇ
ď1
p
nÿ
i“1
|xi|p`
1
q
nÿ
i“1
|yi|q“
1
pxpp `
1
qyqq.
Thus, if xp “ yq “ 1, then |ř
i xiyi| ď 1. For general x and y, after normalization weobtain
(3)
ˇ
ˇ
ˇ
ˇ
ˇ
nÿ
i“1
xiyi
ˇ
ˇ
ˇ
ˇ
ˇ
ď xpyq, (Holder Inequality).
Finally,nÿ
i“1
|xi ` yi|pď
nÿ
i“1
|xi ` yi|p´1|xi| `
nÿ
i“1
|xi ` yi|p´1|yi|
ď
˜
nÿ
i“1
|xi ` yi|pp´1qq
¸1q
rxp ` yps “ x` ypqp rxp ` yps.
Thus,x` ypp ď x` y
pqp rxp ` yps,
and this impliesx` yp ď xp ` yp, (Minkowski’s inequality)
which is what we need only for finite sequences. It remains to notice that if x, y P `p arearbitrary, then the above inequality shows that the partial sums of the p-series of x` y areuniformly 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 “ txnpjqu8j“1 be Cauchy. Then, as before we can pass
to the limit in every coordinate xnpjq Ñ xpjq. For a fixed J P N, ε ą 0 and n,m largeenough, we have
Jÿ
j“1
|xnpjq ´ xmpjq|pă ε.
Letting mÑ 8, we obtainJÿ
j“1
|xnpjq ´ xpjq|pă ε.
In particular, this implies that all partial sums of the seriesř
j |xpjq|p are bounded, and
hence x “ txpjquj P `p. Now, let us let J Ñ 8 in the estimate above. We obtain thenxn ´ xp ď ε1p, thus xn Ñ x.
A normed space X is called separable if it contains a countable dense subset, i.e. if there isS Ă X, cardS “ 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 vectorstxAuAĂN, where xA is the characteristic function of A.
Our next classical example is c0. This is the space of sequences txju8j“1 such that limjÑ8 xj “
0 endowed with the uniform ¨ 8 norm.
4 R. SHVYDKOY
1.3. Constructing new spaces from old ones. There are many ways to construct newspaces from existing examples. A direct product of two linear spaces X and Y , denotedXˆY is the set of pairs tpx, yq : x P X, y P Y u endowed with the coordinate-wise operation ofsummation and multiplication by a scalar. This makes XˆY into a linear space. Identifyingelements of the product px, 0q with x , and p0, yq with y arranges a natural embedding of Xand Y into X ˆ Y . We thus can write x ` y “ px, yq. Let now pX, ¨ Xq and pY, ¨ Y q benormed and let 1 ď p ă 8. We can define a new norm on the product X ˆ Y by
x` yp “ pxpX ` y
pY q
1p.
The verification that this rule defined a norm is immediate from Minkowski’s inequalityestablished above. The obtained normed space is called the `p-sum of the X and Y anddenoted X‘pY . For p “ 8 we naturally define X‘8Y equipped with the norm x`y8 “maxtxX , yY u. Similarly, we define `p-sums of any number of spaces and even countablymany spaces by requiring a member of X1‘pX2‘p . . . to be a sequence of vectors x “ tx1, ...u
such that xp “´
ř8
j“1 xjpXj
¯1p
ă 8, or bounded in the case p “ 8.
Exercise 1.7. Verify that all of the newly introduced spaces are Banach if the original spacesare Banach.
Let us notice that for any pair of vectors x P SpXq and y P SpY q, the span rx, ys will beidentical to `2
p in the `p-product of spaces. So, for example, the unit ball of the X ‘1 R willlook like a symmetric tent with BpXq being the base and p0, 1q the top point. The ball ofX ‘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 isevery closed subspace. Thus, any closed subspace of a Banach space is Banach. Let us nowfix 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 iscalled the factor-space of X by Y , denoted XY , with the natural linear operations inheritedfrom X. We can endow XY with a norm too, called the factor-norm:
(4) rxs “ inftx` y : y P Y u “ disttx, Y u.
Exercise 1.8. Show that the above defines a norm on XY . Show that if X is complete, thenXY 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 aclosed linear subspace of X, and moreover, x` y “ x for all x P X, y P X0. It is easy toshow that (4) defines a norm on XX0, i.e. axiom (i) holds.
1.4. Norm comparison and equivalence. Let pX, ¨ q be a normed space and Y Ă Xis a subspace with another norm |||¨|||. We say that the norm |||¨||| is stronger than ¨ if thereexists 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 embeddingin both sides, cB¨pY q Ă B|||¨|||pY q Ă CB¨pY q. The stronger norm, therefore, defines a finertopology on Y , while equivalent norms define the same topology.
LECTURES ON FUNCTIONAL ANALYSIS 5
Example 1.9. We have `p Ă `q, for all 1 ď p ď q ď 8, and
(7) xq ď xp.
Indeed, assuming that x “ px1, ...q P Sp`pq implies that all |xi| ď 1. Hence, |xi|q ď |xi|
p, andthus, x P `q. Moreover, xq 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 oppositeembeddings for the Lebesgue spaces: Lqpdµq Ă Lppdµq, for all 1 ď p ď q ď 8, and
(8) fp ď µpΩq1p´ 1
q fq, for all f P Lqpdµq.
It readily follows from the Holder inequality,ż
Ω
|f |pdµ ď
ˆż
Ω
|f |qdµ
˙pq
µpΩq1´pq.
Thus, fpp ď µpΩq1´pqfpq , from which (8) follows.
Exercise 1.11. Show that in all of the examples above the norms are not equivalent on thesubspace in question.
Exercise 1.12. Verify the inequality
xq ď xp ď n1p´ 1
q xq,
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`n1 . So, let ¨ be a norm on Rn, and let teiu
ni“1 be the vectors of the standard unit basis.
Then for x “ř
xiei,
x ďÿ
|xi|ei ď maxteiuÿ
|xi| “Mx1.
To establish an inequality from below, let us consider the norm-function Npxq “ x. Bycompactness of Sp`n1 q and continuity of N , N attains its minimum on Sp`n1 q at x0. ThenNpx0q “ 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 cx1, by homogeneity.
1.5. Linear bounded operators. A map T between two linear spaces X Ñ Y is called alinear operator if T pαx`βyq “ αT pxq`βT pyq. We usually drop the parentheses, T pxq “ Tx,when a linear operator is in question. Suppose pX, ¨ Xq and pY, ¨ Y q are normed. A linearoperator T : X Ñ Y is called bounded or continuous if there exists a constant C ą 0 suchthat
(9) TxY ď CxX ,
holds for all x P X. We denote the set of all linear bounded operators between X and Y byLpX, 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;
6 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 boundedon the unit ball of X, i.e. there exists a C ą 0 such that, TxY ď 1, for all x P BpXq.If x P X is arbitrary, then x “ xx P BpXq, and hence T xY ď C. So, by linearity, weobtain (9).
The implication piq ñ pivq is also clear directly from (9). If pivq holds, and x0 P X isarbitrary, then for y Ñ 0, by linearity and continuity at the origin, we have
T px0 ` yq “ Tx0 ` Ty Ñ Tx0,
showing that T is continuous at x0. Thus, piiiq holds. Finally, if pivq holds, then there isa δ ą 0 such that xX ă δ implies TxY ă 1. So, if x is arbitrary, consider x1 “ δxx.Then Tx1 ď 1 implies Tx ď xδ 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, Tx ď Cx holds for all C ą 0 for which (9) holds. Thisshows that the infimum is in fact attained,
TxY ď T xX , @x P X.
Exercise 1.16. Show that
T “ supxPBpXq
TxY “ supxPSpXq
TxY .
These identities say that the norm of an operator is the measure of deformation of theunit 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 linearspace, and the above exercise shows that the operator norm endows it with a norm.
Theorem 1.18. If pX, ¨ Xq is normed and pY, ¨ Y q is Banach, then the space LpX, Y q isBanach in its operator norm.
Proof. Suppose tTnu Ă LpX, Y q is Cauchy. Then, in particular, for any fixed x P X,
Tnx´ TmxY ď Tn ´ TmxX Ñ 0,
as n,mÑ 8. Thus, tTnxu is Cauchy in Y . We, therefore can define the limit
T pxq “ limnÑ8
Tnx,
for each x P X. By linearity of Tn’s the limit is a linear operator as well. To show thatit is bounded, observe that the original sequence of operators is bounded, thus Tn ď Mfor some M and all n. So, TxY ď MxX 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 allx P BpXq we have
Tnx´ TmxY ă ε.
LECTURES ON FUNCTIONAL ANALYSIS 7
Let us keep n fixed and let mÑ 8. We already know that TmxÑ Tx, thus,
Tnx´ TxY ď ε
holds for all x P BpXq and all n large. This gives Tn ´ T ď ε for all n large, whichcompletes the proof.
Exercise 1.19. Suppose that T : X Ñ Y and S : Y Ñ Z are bounded. Prove that
S ˝ T ď ST .
Let us introduce some terminology associated with operators. Let T P LpX, Y q. Thekernel is T is defined by KerT “ T´1p0q “ tx P X : Tx “ 0u, image by ImT “ ty P Y : Dx PX,Tx “ yu. Note that for a bounded operator the kernel is always closed, while the imagemay not be. We say that T P LpX, Y q is an isomorphism if T is bijective and the inverseT´1 is bounded. This is equivalent to requiring that T is surjective and there are constantsc, C ą 0 such that
(10) cx ď Tx ď Cx, for all x P X.
Note that the sharp constants in (10) are in fact c “ T´1 and C “ T . If an isomorphismexists between two spaces we call the spaces isomorphic, and denote X « Y . T is calledan isometry, or isometric isomorphism, if Tx “ x for all x P X, and we denote X – Yfor isometrically isomorphic spaces. We generally don’t distinguish such spaces, and simplyrefer 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 – `2
8, 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 newterms 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 subjectivityassumption. In this case, the image of T is closed, and ImT « X.
1.6. The factor-map and criterion of compactness of BpXq. Let Y be a closed propersubspace of a normed space X. Let us consider the factor-map J : X Ñ XY defined by therule
Jx “ rxs.
From the definition of the factor-norm, it is clear that Jx ď x, thus making J a contrac-tion map. To show that in fact J “ 1, let us fix one x P X not in Y . Then rxs ą 0. Fora 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` yqx` y. Then
Jpx` yq “rxs
x` yě 1´
ε
rxs.
Since, x is fixed and ε is arbitrary, we obtain J “ 1. This observation shows that for anyclosed proper subspace Y we can find a vector on the unit sphere SpXq which is almost adistance 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, ¨Xq is compact if and only if dimX ă
8.
8 R. SHVYDKOY
Proof. We have already seen in Section 1.4 that the unit ball of a finite dimensional spacesis compact. Let us assume now that dimX “ 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 andxn ´ xm ě 12. 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 “ rx1s, andfind x2 P SpXq such that disttx2, Y1u ą 12. Then consider Y2 “ rx1, x2s and find x3 P SpXqwith disttx3, Y2u ą 12, and so on. The process will never terminate since X is not a spanof 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 “ kerT , we construct a new operator T :XX0 Ñ Y by the rule T ˝J “ T . One can easily check that this definition is not ambiguous.Moreover, one has T ď T J “ T . And on the other hand, if T rxs ě T ´ ε andrxs ă 1, then for some x0 P X0, x` x0 ă 1, and yet T rx` x0s “ T px` x0q. This showsthe opposite inequality T ď T . Thus, T “ T . Notice that the new operator hastrivial kernel.
1.7. Direct sums. Suppose Z is a linear space, X, Y Ă Z are subspaces such that Z “ X`Yand X X Y . This is equivalent to the statement that for every z P Z there exists a uniquecouple of vectors x P X, y P Y such that z “ x` y. We thus have two linear maps Pz “ x,Qz “ y, so that P ` Q “ Id, called projections. Suppose now, that Z is normed. It is notgiven that the operators P,Q are bounded. We say that Z is a direct sum of X and Y , andwrite Z “ X ‘ Y , if P , or equivalently, Q, is bounded. The summands of the direct sumare necessarily closed subspaces since X “ kerQ and Y “ kerP . A closed subspace X Ă Zis called complemented if there is a Y Ă Z such that Z “ X ‘ Y . In this case Y is called acomplement of X, and a complement is never a unique space. What defines Y uniquely isthe 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. Thenx´ y ě P ´1P px´ yq “ P ´1x “ 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 thatthere exists a sequence xn ` yn P SpZq such that xn Ñ 8. By the triangle inequality,
xn ď yn ď xn ` 1. Thus, ynxn
Ñ 1. We have›
›
›
xnxn
`ynxn
›
›
݄ 0. On the other hand,
›
›
›
›
xnxn
`ynxn
›
›
›
›
“
›
›
›
›
xnxn
`ynyn
`ynyn
ˆ
yn
xn´ 1
˙›
›
›
›
ď
›
›
›
›
xnxn
`ynyn
›
›
›
›
`
ˇ
ˇ
ˇ
ˇ
yn
xn´ 1
ˇ
ˇ
ˇ
ˇ
In view of all of the above,›
›
›
xnxn
`ynyn
›
›
݄ 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 withdimY ă 8.
Corollary 1.23. Every finite co-dimensional closed subspace is complemented.
Indeed, let Z “ X ` Y , X X Y , X closed, and dimY ă 8. If the spheres of X and Y arenot 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, whichby 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 ifand only if it is idempotent, P 2 “ P .
Exercise 1.26. Show that if Z “ X ‘ Y , then ZX « Y . Hint: consider the projectorP : Z Ñ Y along X, and its factor by the kernel P : ZX 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 operatoris called a linear functional. The space of linear functionals on X is denoted X 1, while thespace of all linear bounded functionals by X˚, called the dual space. It is often possible toidentify 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˚ “ X˚1 ‘q . . . ‘q X
˚n , 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 1zt0u, and let x0 P X be a vector
such that fpx0q ‰ 0. Then let x P X an arbitrary vector. Define y “ x ´ fpxqfpx0q
x0. Then
clearly, fpyq “ 0. It shows that for any x there exist unique y P kerpfq 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 thedistinction between bounded and unbounded functionals comes in the condition of closenessof the kernel, or even less restrictively, its density.
Lemma 1.29. Let f P X 1zt0u. The following are equivalent:
(a) f P X˚;(b) Kerpfq is closed;(c) Kerpfq is not dense in X.
Proof. The implications paq ñ pbq ñ pcq are trivial. Suppose that Kerpfq is not dense. Thenfor some ball Brpx0q XKerpfq “ H. Let y P SpXq. Then gptq “ fpx0 ` tyq “ fpx0q ` tfpyqis a continuous non-vanishing function on p´r, rq. This implies that the sign of it has toagree with that of fpx0q. Assuming fpx0q ą 0 we then have fpx0q ` tfpyq ą 0, and so,|fpyq| ď r´1fpx0q. This shows that f is bounded on the unit sphere and completes theproof.
Geometrically linear bounded functionals can be identified with affine hyperplanes. Thus,if f P X˚, then Hpfq “ tx P X : fpxq “ 1u, defines f uniquely. If f P SpX˚q, thenthe hyperplane is in some sense tangent to the unit sphere of X, namely, it does not deepinside the interior of the unit ball and it approaches arbitrarily close to SpXq. Note that afunctional may not necessarily attain its highest values on the sphere, i.e. the norm. Forexample, let X “ `1, and f “ p12, 23, 34, . . .q P Sp`8q. There is no sequence x P Sp`1q forwhich fpxq “ 1. If x P SpXq and f P SpX˚q with fpxq “ 1, then f is called a supportingfunctional of x. Existence of supporting functionals is not immediately obvious, and it bringsus to an even more fundamental question – does there exist at least one non-zero boundedlinear 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 extensiontheorem is to show that a given bounded functional defined on a linear subspace of X canbe extended boundedly to the whole space X retaining not only its boundedness but alsoits norm. The boundedness can be expressed as the condition of domination by the norm-function, i.e. if Y Ă X and f P Y 1 then f P Y ˚ if and only if
fpyq ď Cy,
for some C ą 0. We will in fact need a more general extension result that will be usefulwhen establishing separation theorems later in Section 2.4. We thus consider a positivelyhomogeneous convex functional p : X Ñ R Y t8u, which means that pptxq “ tppxq for allx 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 iffpyq ď ppyq for all y P Y .
Theorem 2.1 (Hahn-Banach extension theorem). Suppose Y Ă X, and p is a positivelyhomogeneous convex functional defined on X. Then every linear functional f P Y 1 dominatedby p on Y can be extended to a linear functional f P X 1 denominated by the same p on allof X.
In the core of the proof lies Zorn’s Lemma, which we recall briefly. Let P be a partiallyordered 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 suchthat b ě a, for all a P A. A maximal element m is an element with the property that ifa 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 amaximal 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 orderedby pf1, Y1q ď pf2, Y2q iff Y1 Ă Y2 and f2|Y1 “ f1. Let C Ă P be a chain. Define Y “ Ypf,Y qPCY ,
and let fpyq “ fpyq if y P Y . This defines an upper bound of C. By Zorn’s Lemma thereexists a maximal element m “ pf0, Y0q P P . Let us show that Y0 “ X. Indeed, if not, thenthere is a vector x0 P XzY0. Thus, for every element x P Z “ rx0, Y s there exist unique λ P Rand y P Y0 such that x “ λx0` y. We construct an extension f of f0 to Z so that pf, Zq P Pand run into contradiction with the maximality of m. In order to do that it suffices to finda value of f only on x0. Let c “ fpx0q, then to ensure that f is still dominated by p we needto satisfy
λc` f0pyq ď ppλx0 ` yq.
For λ ą 0 this is equivalent to
(11) c ď ppx0 ` y1q ´ f0py
1q,
and for λ ă 0 to
(12) c ě f0py2q ´ pp´x0 ` y
2q.
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 sideof (11) is no less than any number on the right hand side of (12), i.e.
ppx0 ` y1q ´ f0py
1q ě f0py
2q ´ pp´x0 ` y
2q,
for all y1, y2 P Y . This is true indeed, since in view of the convexity and dominance, we have
ppx0 ` y1q ` pp´x0 ` y
2q ě ppy1 ` y2q ě fpy1 ` y2q “ fpy1q ` fpy2q.
Let us discuss some immediate consequences of the Hanh-Banach theorem. First, everyvector on a normed space pX, ¨ q has a supporting functional. Indeed, let x P X, definef P rxs˚ by fpλxq “ λx. Then f “ 1, which means f is dominated by the norm. Theextension 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. Inthis 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˚xX , thus ipxqX˚˚ ď xX . On the other hand, letx˚ be a supporting functional. Then x˚X˚ “ 1, and pipxq, x˚q “ x˚pxq “ xX . Wewill think of X is a sunspace of X˚˚ with the natural identification of elements describedabove. If the embedding X ãÑ X˚˚ exhaust all elements of X˚˚, i.e. X “ X˚˚, then X iscalled reflexive. We will return to a discussion of reflexive spaces later as they possess veryimportant compactness properties.
Let T : X Ñ Y be a bounded operator. We can define the adjoint or dual operatorT ˚ : Y ˚ Ñ X˚ by the rule pT ˚y˚, xq “ py˚, Txq. Again, using the Hahn-Banach theorem weshow that
T “ T ˚.
First, |pT ˚y˚, xq| ď y˚Tx ď y˚T x. This shows T ˚ ď T . Let now ε ą 0 begiven. Find x P SpXq such that Tx ě T ´ ε. Then let y˚ P SpY ˚q be a supportingfunctional for Tx. We have pT ˚y˚, xq “ py˚, Txq “ Tx ě T ´ ε. This shows the oppositeinequality.
Exercise 2.3. Let T ˚˚ : X˚˚ Ñ Y ˚˚ be the second adjoint operator, i.e. T ˚˚ “ pT ˚q˚. Showthat T ˚˚|X “ T .
Exercise 2.4. Show that X˚ is always complemented in X˚˚˚. Hint: consider the adjointi˚ : 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 PX˚ : f |Y “ 0u. This is a closed subspace of X˚, called the annihilator of Y . Show that
Y ˚ – X˚Y K , and pXY 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 Afor 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 allconvex combinations of elements from A:
convA “
#
nÿ
i“1
λiai : ai P A,nÿ
i“1
λi “ 1, λi ě 0, n P N
+
.
It is the smallest convex subset of X containing A, or equivalently, the intersection
convA “č
AĂC,C convex
C.
The topological closure of the convex hull convA is the same as the smallest closed convexset containing A, or the intersection of such sets.
Theorem 2.8 (Caratheodori). Let A Ă Rn, then every point a P convA can be representedas a convex combination of at most n` 1 elements in A.
Proof. Suppose x “řNi“1 λiai, all λi ą 0,
ř
λi “ 1, and N ą n ` 1. We will find a wayto introduce a correction into the convex combination above as to reduce the number ofelements 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.So, we can find constants ti P R, not all of which are zero, such that
řNi“2 tipai ´ a1q “ 0.
Denoting t1 “ ´ř
ti, we can write
Nÿ
i“1
tiai “ 0.
By reversing the sign of all the ti’s if necessary, we can assume that at least one of them ispositive. We will now adjust the original convex combination by a constant multiple of thezero-sum above, thus not changing the x:
x “Nÿ
i“1
λiai ´ εNÿ
i“1
tiai “Nÿ
i“1
pλi ´ εtiqai.
Letting ε “ mintią0tλitiu ensures that µi “ λi ´ εti ě 0 for all i, and that for some i0,µi0 “ 0. Yet, clearly,
ř
µi “ 1. Thus, the new representation
x “Nÿ
i“1
µiai,
is at least one term shorter.
Corollary 2.9. If A Ă Rn is closed, then convA is closed too.
Indeed, simply use the previous theorem and pass to nested subsequences in all n ` 1terms by compactness. In the infinite dimensions closeness or even compactness of A isnot sufficient to conclude that convA is automatically closed. Let us consider the followingexample. Let X “ `2, and A “ t 1
n~enun Y t0u. It is easy to see that A is compact. Any
element of convA has only finitely many non-zero entries, yet x “ř8
n“11
2nn~en P convA.
LECTURES ON FUNCTIONAL ANALYSIS 13
2.3. Minkowski’s functionals. Let us recall from Section 2.4 that a function p : X Ñ
RY t8u is called convex if for any x, y P X one has ppλx` p1´ λqyq ď λppxq ` p1´ λqppyqfor 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 onehas qpλx ` p1 ´ λqyq ě λqpxq ` p1 ´ λqqpyq for all 0 ă λ ă 1. Is it easy to see that if p isconvex then the sub-level sets tp ď p0u are convex, and if q is concave then the super-levelsets tq ě q0u are convex.
Suppose A Ă X is convex and 0 P A. We associate to A a convex function, calledMinkowski’s functional, pA so that A is ”almost” given as a sub level set of pA. We definepApxq as follows. Suppose there is no t ě 0 for which x P tA, then pApxq “ 8. If x P tA forsome t ě 0, we set
pApxq “ 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 forpositively homogeneous functionals convexity is equivalent to triangle inequality, pApx`yq ďpApxq ` pApyq. So, let x, y P X. If any of pApxq or pApyq equal 8, the inequality becomestrivial. If both are finite, then for every ε ą 0 we can find t1 ă pApxq ` ε and t2 ă pApyq ` εsuch that x P t1A and y P t2A. Then
x` y P t1A` t2A “ pt1 ` t2q
ˆ
t1t1 ` t2
A`t2
t1 ` t2A
˙
Ă pt1 ` t2qA.
This shows that pApx ` yq ď pApxq ` pApyq ` ε, for all ε ą 0. Finally, pcq follows directlyfrom 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 concavefunctional to B as follows. If x P tB, for no t ě 0, then qBpxq “ ´8. Otherwise, we define
qBpxq “ suptt ě 0 : x P tBu.
Condition δBpXq X B “ H warrants that the supremum is finite for any x P X. Thefollowing 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 assumptionsabove, and let pA and qB be the corresponding Minkowski’s functionals. If pApxq ă 8, lett ě 0 be such that x P tA. Since 0 P A, the whole interval r0, xs is in tA and therefore notin tB. This in turn implies than x R sB for any s ě t, for if such s existed, then t
sx P tB
contradicting the previous. As a consequence, qBpxq ď t. We have shown that
(13) qBpxq ď pApxq, for all x P X.
2.4. Separation theorems.
Theorem 2.10 (Generalized Hahn-Banach theorem). Let p be convex, and q concave func-tionals defined on X. Let Y Ă X, f P Y 1 such that
(14) fpyq 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
(15) qpxq ď fpxq ď ppxq, for all x P X.
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 “ fpx0q, then we need
λc` fpyq 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 ` y1q ´ qpx1q ´ fpy1q,
while for λ ă 0,
c ě fpy2q ´ ppx2 ´ x0 ` y2q ` qpx2q.
The existence of c is ensured if
ppx1 ` x0 ` y1q ´ qpx1q ´ fpy1q ě fpy2q ´ ppx2 ´ x0 ` y
2q ` qpx2q,
which is true since
ppx1`x0`y1q`ppx2´x0`y
2q´ qpx1q´ qpx2q ě ppx1`x2`y1`y2q´ qpx1`x2q ě fpy1`y2q.
Theorem 2.11 (Separation Theorems). Let A,B be two disjoint convex subsets of a normedspace X.
(i) If A Ă B, then there exists f P X 1zt0u such that
(16) sup fpAq ď inf fpBq.
(ii) If A has a non-empty interior, then there exists f P X˚zt0u such that (16) holds.(iii) If A “ tx0u and B is closed then there exists f P X˚zt0u such that
(17) fpx0q ă inf fpBq.
Proof. Suppose A Ă B, then there exists x0 P A and δ ą 0 such that Bδpx0q X B “ H. Bymoving x0 to the origin we satisfy all the conditions on A and B as above, which allows usto define Minkowski’s functions, qB ď pA. Thus, we have (14) for f “ 0, and Y “ t0u. ByTheorem 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 andlet f be the functional constructed above. Since fpxq ď pApxq, we conclude that wheneverx ď ε, then fpxq ď 1. This shows f ď ε´1.
Finally if A “ tx0u and B is closed, we apply case piiq to A “ Bδpx0q for small δ ą 0. Sincef ‰ 0, there is y P SpXq for which fpyq ą 0. Thus, fpx0q ă fpx0q ` εfpyq ď inf fpBq.
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 strictlyseparated.
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~enun, and B “ 12A. These are convex disjoint sets.
Notice that for any b P B, let b “ 12
ř
i λi~ei, we have
b22 “1
4
ÿ
|λi|2ď
1
4
ÿ
λi “1
4.
Thus, B Ă 12BpXq. Since any ~en R
12BpXq, the conditions of Theorem 2.11 (i) are satisfied.
Next observe that 0 P A X B, by considering the sequences xn “ p~e1 ` ¨ ¨ ¨ ` ~enqn andyn “
12xn. Suppose f P `2t0u separates the two sets, i.e. sup fpAq ď c ď inf fpBq. Since 0
is in the closure of both sets, we conclude that c “ 0. Then f as a sequence is positive onthe one hand and negative on the other hand. Thus, f “ 0, a contradiction. It is easy toconstruct 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
convS “č
fPX˚
tx : fpxq ď sup fpSqu.
Indeed, the inclusion Ă is obvious. If however x0 R convS, then Theorem 2.11 (iii) providesa functional such that fpx0q ă inf fpconvSq ď inf fpSq. Reversing the sign of f shows thatx0 is not in one of the sets in the intersection.
2.5. Baire Category Theorem. Let us consider for a moment a general complete metricspace X, without necessarily a given linear structure. Let us ask ourselves how ”big” sucha space can be. One may say, a singleton X “ tx0u 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 centeredaround itself. It therefore is rather ”big”, in a sense. To make this discussion more preciselet us agree on what we mean by a ”small” subset of X. We say that F is nowhere densein X is for any open set U there exists an open subset V Ă U with no intersection withF . In other words, F has empty interior. A subset F Ă X is called of 1st Baire category ifF “
Ť8
i“1 Fi, where all Fi’s are nowhere dense. A subset is of 2nd Baire category if it is notof the 1st category. Thus, in the example above X itself is clearly of the 2nd category. Thisin fact holds in general.
Theorem 2.14 (Baire Category Theorem). Any complete metric space is a 2nd Baire cat-egory set.
Proof. Let us suppose, on the contrary, that X “Ť
Fi, and all Fi’s are nowhere dense. Then,there is a closed ball Bε1px1q with ε1 ă 1 disjoint from F1. Since F2 is however dense, thereis a ball Bε2px2q Ă Dε1px1q, with ε2 ă 12, disjoint from F2, continuing in the same mannerwe find a sequence of nested closed balls Bεnpxnq, with εn ă 1n. Clearly, dpxn, xmq ă 1m,for all n ą m. Thus, the sequence txnu 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 inthe exercises below.
Theorem 2.15 (Banach-Steinhauss uniform boundedness principle). Let F Ă LpX, Y qbe a family of bounded operators, and X is a Banach space. Suppose for any x P X,supTPF Tx ă 8. Then, the family is uniformly bounded, i.e. supTPF T ă 8.
16 R. SHVYDKOY
Proof. Let Fn “ tx P X : supTPF Tx ď nu. Not that each set Fn is closed, and byassumption X “
Ť
n Fn. Hence, by the Baire Category Theorem, one of Fn’s contains aball Brpx0q. This implies that for all x P BpXq, T px0 ` rxq ď n, for all T P F . Thus,Tx ď r´1pn` Tx0q ď r´1pn` supTPF Tx0q, 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. ThenS 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 thatDpY q Ă T pDpXqq, then 1
2DpY 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 D12pY q and let us fix a small ε ą 0 to be specified later. By(18) we can find x1 P 34DpXq such that y´y1 ă ε, where Tx1 “ y1. Since y´y1 P εDpY q,one finds x2 P εDpY q such that y´y1´y2 ă ε2, where y2 “ Tx2. Continuing this way, weconstruct a sequence txnu with xn ď ε2n. Let x “
ř
n xn. Then by construction Tx “ y,and x ď 34` 2ε ă 1, provided ε is small.
Theorem 2.18 (Open mapping theorem). Suppose T P LpX, Y q, and both spaces are Ba-nach. If, in addition, T is surjective, then T is an open mapping, i.e. T pUq is open for everyopen U .
Proof. Suppose, U is open. Let x0 P U , and let Dεpx0q Ă U . We prove the theorem if we showthat T pDεpx0qq contains an open neighborhood of Tx0. Since, T pDεpx0qq “ Tx0`εT pDpXqq,it amounts to showing that T pDpXqq contains a ball centered at the origin. Since T issurjective, we have Y “
Ť
n nT pDpXqq. By the Baire Category Theorem, one of the setsnT pDpXqq is dense in some ball Dδpy0q, and hence, by linearity, so is T pDpXqq. SinceT pDpXqq is convex and symmetric with respect to the origin,
δDpY q Ă convtDδpy0q, Dδp´y0qu Ă T pDpXqq
Applying Lemma 2.17 to the operator 1δT , we conclude δ
2DpY 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, ¨ 1q Ñ pX, ¨ 2q 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 thatif xn Ñ x and Txn Ñ y, then Tx “ y. Then T is bounded.
LECTURES ON FUNCTIONAL ANALYSIS 17
Proof. Let Γ “ tpx, Txq : x P Xu denote the graph of T . Thus, by the assumption, Γ isclosed in the product topology of the product X ˆ Y . Since it is also a linear subset, Γ asa closed subspace of the `1-sum, X b1 Y is a Banach spaces. Let us consider the projectiononto X restricted to Γ, P : Γ Ñ X, given by P px, Txq “ x. The operator is clearly boundedand easy to show that P is bijective. Hence, P´1 is bound, which implies
x ` Tx ď Cx,
and thus T is bounded.
3. Weak topologies
3.1. Topology of not necessarily metrizable spaces. Let pX, τq be a topological spacewith topology of open sets τ . A subset txαuαPA Ă X is called a net, if the index set A ispartially 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 tyβuβPB 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 txαuαPA is said to be convergentto x P X if for very open neighborhood U of x, there is α0 P A such that xα P U for allα ě α0. A function f : X Ñ Y , where Y is another topological space, is cooled continuousif f´1pGq 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 netlimαPA xα “ x, limαPA fpxαq “ fpxq.
Proof. Suppose f is continuous, and let limαPA xα “ x. For any open G containing fpxq,f´1pGq is open and contains x. Since eventually all xα are in f´1pGq, then all fpxαq will bein G.
Conversely, suppose there is open G Ă Y such that f´1pGq is not open. Thus, there is apoint x0 P f
´1pGq such that any open neighborhood U of x contains a point outside f´1pGq.Let us fix one such point xU for every U . Let A “ tU P τ : U open, x P Uu. It is a netordered by inclusion. Clearly, xU Ñ x, since for every U containing x, all elements of thenet, namely starting from xU , will fall inside U . Yet, fpxUq R G, and thus fpxUq Ñ fpxq.
Exercise 3.2. Show that a subset F Ă X is closed if and only if the limit of every convergentnet 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 txαuαPA Ă X be a net. First we let us establishexistence of a cluster point. A point y P X is a cluster point of a net if for every U P τcontaining y and every α0, there is α ě α0 such that xα P U . Suppose that our net doesnot have cluster points. Thus, for every y P X there is Uy and αy P A such that xα R Uyfor all α ě αy. Consider the open cover tUyuyPX . By compactness there is a finite sub coverUy1 , . . . , Uyn . Since A is a net, there is a α ě αyi for all i “ 1, ..., n. Then xα is in none ofthe 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 Uu be ordered by reverseinclusion on the first component, and by the order of A on the second. For β “ pU, αq, letyβ “ xα, and let npβq “ α. It is routine to show that tyβuβPB is a subnet converging to y.
Conversely, suppose every net has a converging subnet, and yet on the contrary, X is notcompact. This implies that there is an open cover U which has no finite subcover. Let us
18 R. SHVYDKOY
define A “ tα “ pU1, . . . , Unq : Ui P U , n P Nu ordered by α ě β if β Ă α. Clearly A is alsodirected. By assumption, for any α “ pU1, . . . , Unq there is xα R
Ť
i Ui. The net txαuαPAhas a converging subnet tyβuβPB, and y “ lim yβ. Since U is a cover, there is U P U withy P U . Let α “ pUq. By the definition of a subnet, there is β1 P B such that npβ1q ě αand yβ1 “ xnpβ1q, and there is another β2 ě β1 such that yβ2 P U . By monotonicity of n,npβ2q ě α, and yet xnpβ2q 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 iffor any point x P X any open neighborhood of x in τ2 contains an open neighborhood of xin 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. Showthat 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 ;(2) if F1, ¨ ¨ ¨ , Fn are elements of F , then
Şnj“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 Psatisfies 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 filterscontaining the given one satisfies Zorn’s Lemma as well. Ultrafilters can be characterized byadding one more condition to the three above: U is an ultrafilter if and only if it is a filterand
(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 inU , for otherwise, XzA is in U , and then H “ A X pXzAq P U 1. Conversely, if A Ă X issuch that XzA R U , then by p3q every F P U must intersect with A. Define a new familyU 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 isa filter. By maximality of U , U “ U 1, and hence A P U . An alternative to p4q is a formallystronger, 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 Anq. This is incompatible with p1q.
The compactness of a topological space can be restated in terms of convergence of ultra-filters. So, let pX, τq be a topological space and F be a filter on it. We say that limF “ xif every neighborhood of x has a non-empty intersection with any element of the filter. IfF is an ultrafilter, which will be our standard assumption, we showed above that every setthat intersects every element of F must lie in F . Thus, in this case limF “ x iff every openneighborhood of x is contained in F . If every two distinct points in X can be separated bydisjoint open neighborhoods, and such a space is called Hausdorff, then clearly, the limit isunique. 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 anypoint in X, then any point x P X is contained in Ux P τ with Ux R U . Thus, tUxuxPX forman open cover of X, which must contain a finite sub cover, U1, . . . , Un. But YjUj “ X Ă U ,so by p41q one of the sets must be in U , a contradiction.
LECTURES ON FUNCTIONAL ANALYSIS 19
Conversely, let C “ tUu be an open cover of X. Suppose that it contains no finite subcover.Thus, any finite intersection pXzU1qX. . .XpXzUnq in non-empty. This shows that the familyF “ tF Ă X : XzU Ă F, for some U P Cu is a filter, and let U be an ultrafilter containingF . By assumption, let x “ limU . 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 proofof Tihonov’s compacness theorem, which will be used later to establish the main result ofthis section, the Alaoglu Theorem.
Let tpXγ, τγquγPΓ be a collection of topological spaces. The product space, X “ś
γPΓXγ
is the set of functions x : Γ ÑŤ
γPΓXγ such that xpγq P Xγ. We usually denote xpγq “ xγand write x “ txγuγPΓ. Let πγ : X Ñ Xγ be the usual projection map. We define theproduct topology on X to be the topology generated by the sets π´1pUγq, 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 πγpxαq Ñ πγpxq forevery γ P Γ.
Theorem 3.7 (Tihonov’s compactness theorem). If all Xγ, γ P Γ, are compact, then theproduct X “
ś
γPΓXγ is compact in the product topology.
Proof. Let U be an ultrafilter on X. For every γ P Γ, consider Uγ “ πγpUq. Then Uγ isan ultrafilter. Since Xγ is compact, there exists a limit xγ “ limUγ. Let us show thatx “ txγuγPΓ “ limU . Let U be open neighborhood of x in the product topology. Then withx, U contains a finite intersection of basis sets Xni“1π
´1γipUiq. We have xγi P Ui, and hence
Ui P Uγi , which means there exist Ai P U such that πγipAiq “ Ui. Then Ai Ă π´1γipUiq, which
implies that π´1γipUiq P U , for each i, and hence Xni“1π
´1γipUiq 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 asthe topology with the following base of neighborhoods: for x P X, ε1, . . . , εn ą 0 andf1, . . . , fn P X
˚, let
(19) U f1,...,fnε1,...,εn
pxq “ ty P X : |fipyq ´ fipxq| ă εi, @i “ 1, nu.
Thus, a neighborhood is an intersection of open slabs of finite widths. We say that a sequence,or a net txαuαPA converges weakly to x, and denote xα
wÑ x, if it converges in the sense of
the weak topology.
Exercise 3.8. Show that xαwÑ x if and only if fpxαq Ñ fpxq for every functional f P X˚.
Exercise 3.9. Show that if a sequence xn converges weakly, then it is bounded in the norm-topology.
Exercise 3.10. Suppose dimX “ 8. Construct a net txαuαPA 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;Nq : 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, withxα 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 norm-tology, as opposed to ”weak” that refers to anything related to the weak topology. Forexample, ”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 anyneighborhood (19) is unbounded. Indeed, let H “ Xi Ker fi. This is a non-empty space, forotherwise X would have been a span of xi’s, with xi R Ker fi. Then U f1,...,fn
ε1,...,εnpxq contains all
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 1nu. Each contains a weak neighborhood of theorigin. We have shown every weak neighborhood is unbounded. Thus, we can find a xnwithin the nth ball with xn ą n. So, on the one hand, xn Ñ 0 weakly, and yet txnu isunbounded, in contradiction with Exercise 3.2.
The weak topology is still ”fine” enough to separate points, and even larger sets, by theSeparation Theorems.
Exercise 3.12. Show that if A is closed and B is strongly compact convex sets, then thereare two disjoint weakly open neighborhoods of A and B.
Lemma 3.13. Let txnu8n“1 Ă X be a sequence in any of the spaces X “ c0, `p, for 1 ă p ă
8. Then xnwÑ x if and only if txnu is norm bounded and converges to x pointwise, i.e.
xnpjq 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
xnwÑ 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 is bounded, let M “
supn xn, and let f P `1 “ c˚0 . Given ε ą 0, let N P N be such thatř
jąN |fpjq| ă εp2Mq.Then, for large n we have
ˇ
ˇ
ˇ
ˇ
ˇ
ÿ
jďN
xnpjqfpjq ´ÿ
jďN
xpjqfpjq
ˇ
ˇ
ˇ
ˇ
ˇ
ă ε2.
Thus,ˇ
ˇ
ˇ
ˇ
ˇ
ÿ
j
xnpjqfpjq ´ÿ
j
xpjqfpjq
ˇ
ˇ
ˇ
ˇ
ˇ
ď
ˇ
ˇ
ˇ
ˇ
ˇ
ÿ
jďN
xnpjqfpjq ´ÿ
jďN
xpjqfpjq
ˇ
ˇ
ˇ
ˇ
ˇ
`
ˇ
ˇ
ˇ
ˇ
ˇ
ÿ
jąN
xnpjqfpjq ´ÿ
jąN
xpjqfpjq
ˇ
ˇ
ˇ
ˇ
ˇ
ă ε.
Lemma 3.14. In `1, a sequence txnu8n“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 normand 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 becausethe 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 sometimescalled Kadets-Klee spaces.
As a weaker topology, the weak topology provides a smaller family of open sets than thenorm topology. Thus any weakly closed set is strongly closed as well 1. As a consequence ofthe Separation Theorem 2.11, it turns out that among the class of convex sets the propertyof 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.
Proof. Clearly, if C is strongly closed, and yet there is a point x0 P Cw
not in C. Then byTheorem 2.11, there is a functional f P X˚ so that fpx0q ą c ą sup fpCq. Thus x0 belongsto 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 Ccoincide. This has a rather interesting consequence for relationship between weak and strongconvergence of sequences.
Exercise 3.16. Show that if xn Ñ x weakly, then there is a sequence of convex combinationsmade 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
lim infnÑ8 xn ě x. More generally, if a net xαwÑ 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 ownweak topology determined by the functionals from ”upstairs”, i.e. X˚˚. However, one candefine 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˚-openneighborhood of f to be
(20) Ux1,...,xnε1,...,εn
pxq “ tg P X˚ : |gpxiq ´ fpxiq| ă εi, @i “ 1, nu.
Identifying element of X as vectors in X˚˚ we see it is just a special subclass of neighborhoodsdefined 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 necessarilybounded by the Banach-Steinhauss Theorem.
Exercise 3.18. Show that a net fαw˚Ñ f if and only if fαpxq Ñ fpxq for every x P X.
Exercise 3.19. Show that any weakly˚ convergent sequence in X˚ is strongly bounded.
Exercise 3.20 (Weak˚ lower-semi-continuity of norm). Show that if fnw˚Ñ f , then
lim infnÑ8
fn ě f.
More generally, if a net tfαuαPA converges weakly˚ to f , then for any ε ą 0 there is α0 P Aso that for all α ě α0, fα ě f ´ ε.
How much the weak˚ topology may be weaker than the weak topology is illustrated bythe 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 beclosed it becomes clear. If a set ”survives” weak limits from within itself, then it should definitely survivestrong limits.
22 R. SHVYDKOY
Exercise 3.21. Show that in `1, xnw˚Ñ x if and only if txnu 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, fpxq P r´x, xs. This naturally suggeststo consider BpX˚q as a subset of the product space T “
ś
xPXr´x, xs. By Tihonov’stheorem, this product space is compact in the product topology. It suffices to show thatBpX˚q is closed in T , because convergence of nets in the product topology is equivalent topointwise convergence, which for elements of BpX˚q amounts to weak˚ convergence.
To this end, let tfαuαPA be a net in BpX˚q with lim fα “ f P T . By linearity of fα’s andthe ”pointwise” sense of the limit above, we conclude that
fpλx` µyq Ð fαpλx` µyq “ λfαpxq ` µfαpyq Ñ λfpxq ` µfpyq.
Thus, f is linear, and since |fαpxq| ď x, we also have |fpxq| ď x for all x P X, whichidentifies f as an element of BpX˚q.
As an immediate consequence we see that for a reflexive Banach space X, the unit ball isweakly compact. This the property actually characterizes reflexiveness.
Theorem 3.23 (Kakutani). A Banach space X is reflexive if and only if its unit ball isweakly compact.
The theorem will follow from the next lemma.
Lemma 3.24. The unit ball BpXq is weakly˚ dense in BpX˚˚q. In particular, BpXqw˚
“
BpX˚˚q.
Indeed, let BpXq be weakly compact. In view of Lemma 3.24, for any x˚˚ P BpX˚˚q
we find a net xαw˚Ñ x˚˚, xα P BpXq. By compactness, there is a subnet yβ
wÑ x for some
x P BpXq, and yet that same subnet converges to x˚˚ in the weak˚ sense of X˚˚. Thus, forevery x˚ we have px˚˚, x˚q Ð pyβ, x
˚q “ px˚, yβq Ñ px˚, xq, which identifies x˚˚ as x, andthus BpXq “ BpX˚˚q, implying X “ X˚˚.
In order to prove Lemma 3.24, we have to go back to the separation theorem and makeone adjustment to it in the context of weak˚ topology.
Lemma 3.25. Suppose that f, f1, . . . , fn P X1, and Xnj“1 Ker fj Ă Ker f . Then f P rf1, . . . , fns.
Proof. We will prove the lemma by induction. Suppose n “ 1, and f1 ‰ 0 (otherwise thestatement is trivial). By the structure of the linear functionals discussed in Section 1.10,there is x1 P X with f1px1q “ 1, such that for every x P X we have x “ λx1 ` y, wherey P Ker f1. Since fpyq “ 0 we have fpxq “ λfpx1q “ fpx1qf1pxq, as desired.
Suppose the statement is true for n. Let us assume Xn`1j“1 Ker fj Ă Ker f . Consider
the space Y “ Ker fn`1. Then Xnj“1 Ker fj|Y Ă Ker f |Y . By the induction hypothesis,f |Y “
řnj“1 ajfj|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`1pxn`1q “ 1. Thus,
fpxq “ λfpxn`1q `
nÿ
j“1
ajfjpyq “ fpxn`1qfn`1pxq `nÿ
j“1
ajfjpyq
“ fpxn`1qfn`1pxq `nÿ
j“1
ajfjpxq ´ λnÿ
j“1
ajfjpxn`1q
“
˜
fpxn`1q ´
nÿ
j“1
ajfjpxn`1q
¸
fn`1pxq `nÿ
j“1
ajfjpxq
“
n`1ÿ
j“1
ajfjpxq.
(21)
Lemma 3.26. If x˚˚ P X˚˚ is continuous in the weak˚ topology, then x˚˚ P X.
Proof. By the assumption px˚˚q´1p´1, 1q contains a weak˚ neighborhood of the origin, say,Ux1,...,xnε1,...,εn
p0q. In particular, x˚˚ P p´1, 1q on Xj Kerxj. Since the latter is a linear space, x˚˚
must in fact vanish on it. Thus, by Lemma 3.25, x˚˚ P rx1, . . . , xns Ă X.
Theorem 3.27 (Separation Theorem for weak˚ topology). Suppose B is a weakly˚-closedconvex subset of X˚, and f R B. Then, there is x P X such that supBpxq ă 1 ă fpxq.
Proof. Let us follow the proof of Theorem 2.11. Let us assume f “ 0, and let A “ Ux1,...,xnε1,...,εn
p0qbe a weak˚ neighborhood of 0 disjoint from B. We then associate Minkowski’s functionals pAand qB to A and B respectfully. As a result of Hahn-Banach Theorem, we find a separatingfunctional F so that qB ď F ď pA. Since f P A implies pApfq ď 1, we see that F is boundedfrom above on A. Like in the proof of Lemma 3.26 we conclude that F vanishes on theintersection of the kernels of the x1, . . . , xn, and hence, F P X. By rescaling F if necessarywe can arrange the constant of separation to be 1.
Proof of Lemma 3.24. Let B “ BpXqw˚
. By Exercise 3.20, B Ă BpX˚˚q. Suppose there isF P BpX˚˚qzB. By Theorem 3.27 , we find a x˚ P X˚ such that Bpx˚q ă 1 ă F px˚q. Thefirst inequality holds, in particular, on BpXq, which shows that x˚ ď 1. This runs intocontradiction with the second inequality.
Exercise 3.28. Recall that pc0q˚˚ “ p`1q
˚ “ `8. Show that Bpc0q is weakly˚ sequentiallydense in Bp`8q, i.e. for every F P Bp`8q there is a sequence xn P Bpc0q converging weakly˚
to F .
Corollary 3.29. Let Y Ă X be a closed subspace of a reflexive space X. Then Y and XYare reflexive.
Proof. Indeed, BpY q “ Y XBpXq. Since the this set if convex and closed, it is weakly closedin BpXq and hence compact in the weak topology of X. However, by the Hahn-Banachextension theorem the topology induced on Y by the weak topology of X is exactly theweak topology of Y . Thus, BpY q is weakly compact in Y . Now, by Exercise 2.7 and bythe previous, pXY q˚ is a subspace of a reflexive space, which makes it reflexive. Then byExercise 2.6 pXY q itself is reflexive.
