Contentsifarah/teaching/6461W19/6461w19notes.pdf · An alternative proof of the Riesz Representation Theorem 65 16. Some dual spaces 71 Date: The version of 12:10 on Thursday 7th

FUNCTIONAL ANALYSIS I, MATH 6461, WINTER2019 CLASS NOTES.

I. FARAH

Contents

0.1. Partial Prerequisites 11. Topological vector spaces 11.1. Normed spaces 62. Properties of Topological Vector Spaces 72.1. The unit ball 93. Infinite-dimensional normed spaces. Banach spaces 103.1. Constructions of spaces: Direct products and direct sums,

quotients, the completion 133.2. Examples of function spaces 164. Baire Category Theorem and its Consequences 185. The Open Mapping Theorem 206. The dual space 257. Weak topologies, Hahn–Banach separation 318. The weak∗ topology. Banach–Alaoglu theorem 339. Inner product, Hilbert space 379.1. Polarization, parallelogram and Pythagoras 389.2. 1

2+ 1

2= 1: Hilbert space 39

9.3. Separability 4110. Lp spaces 4311. Dual spaces of Lp spaces 4912. Dual spaces of Lp spaces 5012.1. An isometric embedding of Lq into (Lp)∗ 5012.2. Hölderfest—more on Lp spaces 5413. The dual of C0(X) 5614. Riesz Representation Theorem 5714.1. An outline of the proof of Riesz Representation Theorem 59Positive functionals 6014.2. Stone–Čech compactification 6115. An alternative proof of the Riesz Representation Theorem 6516. Some dual spaces 71

Date: The version of 12:10 on Thursday 7th February, 2019.1

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17. Convexity. The Krein–Milman theorem 7118. Applications of the Krein–Millman theorem 7419. Fixed points: the Kakutani–Markov Theorem 7920. The Stone–Weierstraß Theorem 8221. Operator theory on a Hilbert space 8522. Spectral theory of self-adjoint compact operators 89Appendix A. Topology 93A.1. Nets 95References 96

0.1. Partial Prerequisites. Set theory. Countability. Axiom of Choice.Topology (see Appendix A) Metric spaces. Completeness in metric

spaces. Topological spaces. Hausdorff topological spaces. Compact-ness. Convergence via nets.

Measure theory. Basics of the Lebesgue measure.Suggested references: [4, Chapters 1–3], [8], or subsets of [5], [1], [6]

1. Topological vector spaces

All vector spaces are assumed to be over R or C. For quite a while(until we start spectral theory) the choice of the field will not matter,and K shall always stand for ‘R or C.’ The elements of K are scalars.We consider K with its standard topology. Recall that this topologyis locally compact, separable and completely metrizable, where themetric is given by d(λ, η) = |λ− η|. If I write r < ∞ or r > 0 then itis understood that r ∈ R (and that R ⊆ K).

Definition 1.1. A vector space X over K is a topological vector space(TVS) if it is equipped with a Hausdorff topology such that each of thefollowing two maps:

X2 3 (x, y) 7→ x+ y ∈ Xand

K×X 3 (λ, x) 7→ λx ∈ Xis continuous.

‘Let X be a topological vector space’ stands for ‘Let X be a topo-logical vector space over K’ and ‘Let X and Y be topological vectorspaces’ stands for ‘Let X and Y be topological vector spaces over (thesame) K.’

Example 1.2. Here are some examples of TVSs.

FUNCTIONAL ANALYSIS I, MATH 6461, WINTER 2019 CLASS NOTES. 3

(1) if n ≥ 1 then Rn, with the standard Euclidean topology andthe standard vector space structure, is a TVS over R.

(2) if n ≥ 1 then Cn, with the standard Euclidean topology andthe standard vector space structure, is a TVS over C. It is alsoa TVS over R of dimension 2n.

(3) The space of all continuous real-valued functions on [0, 1]. Vec-tor space operations are defined pointwise. This space carriesmore than one important topology. Here are some of them.(a) The uniform topology, associated with the metric

d(f, g) := supt∈[0,1]

|f(t)− g(t)|.

(b) The topology of pointwise convergence. Open neighbour-hoods of g are given by F b [0, 1]1 and ε > 0:

Ug,F,ε := {f ∈ C([0, 1]) : maxt∈F|f(t)− g(t)| < ε}.

(c) With µ denoting the Lebesgue measure, the topology isassociated with the metric d1(f, g) :=

∫|f − g| dµ. Since

we are considering only the continuous functions, this isindeed a metric (exercise!).

(4) The space of all continuous complex-valued functions on [0, 1]with topologies defined analogously to those in the previousexample. It can be considered as a TVS over R or over C.

Lemma 1.3. Assume X is a topological vector space.

(1) For every a ∈ X map X 3 x 7→ x+a ∈ X is a homeomorphismof X onto X. In particular, if U0 is a neighbourhood basis at 0then {a+ U : U ∈ U0} is a neighbourhood basis at a.

(2) For every λ ∈ K \ {0} the map x 7→ λx is a homeomorphism ofX onto X.

(3) Suppose X is a topological vector space and let U be an openneighbourhood of 0. For every x ∈ X there exists r > 0 in Rsuch that sx ∈ U for all −r < s < r.

Proof. (1) The map x 7→ x + a is a continuous bijection of X onto X,and so is its inverse x 7→ x− a.

(2) Both this map and its inverse, x 7→ λ−1x, are continuous bijec-tions of X onto itself.

(3) The set {s ∈ R : sx ∈ U} is by (2) an open neighbourhood of 0in R. �

1A b B will always stand for ‘A is a finite subset of B’.

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For λ ∈ K and A ⊆ X writeλA := {λx : x ∈ A}.

The following is an immediate consequence of (3) in the lemma above.

Corollary 1.4. If U is an open neighbourhood of 0 then⋃∞n=1 nU = X.

While we are at it, for subsets A and B of X define

A+B := {x+ y : x ∈ A, y ∈ B}.(Exercise: If A is open then A+B is open for every B.)

Lemma 1.5. A linear map between topological vector spaces is contin-uous if and only if it is continuous at 0.

Proof. Only the converse direction requires a proof. Suppose T : X →Y is a linear map between topological vector spaces and let U ⊆ Ybe an open set such that T−1(U) is not open. Since the latter set isclearly nonempty, pick a ∈ X such that a = T (x) is in U . If a ∈ U thenU −a = {b−a : b ∈ U} is an open neighbourhood of 0 whose preimageT−1(U − a) = T−1(U) − x contains 0 and it is not open. Hence T isdiscontinuous at 0. �

A neighbourhood W of 0 is balanced (some authors use circled) iffor every x ∈ W the line segment {rx : 0 ≤ r ≤ 1} is included in W .This is equivalent to stating that λW = W whenever |λ| ≤ 1. 2

Lemma 1.6. Every open neighbourhood U of 0 includes a balancedopen neighbourhood of 0.

Proof. Since multiplication by scalars g : K×X → X is continuous at(0,0), g−1(U) is an open neighbourhood of (0, 0) in K×X. Thus thereexists ε > 0 and an open V ⊆ X such that 0 ∈ V and

{λ ∈ K : |λ| < ε} × V ⊆ g−1(U).Let

W :=⋃δ


A proof of the following lemma is a bit more work than I made itlook in the class.3

Lemma 1.7. For every x ∈ X \ {0} the map λ 7→ λx is a homeomor-phism of K onto the 1-dimensional subspace spanned by x.

Proof. The map f(λ) = λx is continuous since the multiplication byscalars is continuous. It is also an injection, since λx 6= 0 wheneverλ 6= 0.

It therefore remains to prove that the image of an open subset of Kis a relatively open subset of span(x).

Let us first prove that the image of an open neighbourhood U of 0such that U is compact is relatively open. Since f is continuous, f [U ]is a compact subset of span(x). By Proposition A.6, the restriction off to U is a homeomorphism. Therefore f [U ] is relatively open in f [U ].Since 0 ∈ U and f is linear, W is an open neighbourhood of 0. ByLemma 1.6 there exists a balanced open neighbourhood of 0, call it V ,included in W . Therefore V ∩ span(x) is relatively open in span(x).

We claim that V ∩ span(x) ⊆ f [U ]. Assume otherwise, and let µ besuch that µx /∈ f [U ]. Let r = inf{s ≥ 0 : sµx ∈ f [U ]}. Then 0 < r < 1(since µx /∈ f [U ]). Also, rµx /∈ f [U ] because f [U ] is a relatively opensubset of f [U ]. But this implies rµx /∈ W , which is a contradiction sinceW is balanced. This implies that W ∩ span(x) ⊆ f [U ], as claimed.

We have now proved that the image of every open neighbourhood of0 whose closure is compact is relatively open. By linearity, the imageof every open set whose closure is compact is relatively open, and bylocal compactness of K, the image of every open set is relatively open.

We have proved that f is a bijection that is both continuous andopen. It is therefore a homeomorphism. �

Exercise 1.8. Assume X is a TVS and U is a nonempty open neigh-bourhood of 0. Prove that there exists a nonempty open neighbourhoodV of 0 such that V + V ⊆ U .

The right notion of an isomorphism in the category of topologicalvector spaces is that of a linear homeomorphism. The natural mor-phisms are continuous linear maps. In general a linear map need notbe continuous. However this is the case with the finite-dimensionalspaces.

3Thanks to Gates Wang for pointing out to an issue with the proof. As you cansee, the proof is not too different from the proof of Proposition 1.10 given later on.Both results will follow easily once we develop more theory (and, even better, wewill not need them until then).

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Lemma 1.9. If X is a TVS then for every n ≥ 1, every linear mapf : Kn → X is continuous.

Proof. The proof is by induction on n. The case n = 1 is (an easy por-tion of) Lemma 1.7. Suppose that the assertion holds for n. Supposethat f : Kn+1 → X. Let f0 : Kn → X be defined by f0(ā) = f(ā, 0), forall ā ∈ Kn and let f1 : K→ X be defined by f1(a) = f(0, 0, . . . , 0, a).

Both f0 and f1 are continuous by the inductive assumption, andf(ā, b) = f0(ā) + f1(b) is continuous because the addition is continuouson X. �

As pointed out in the paragraph preceding Lemma 1.7, there is noreason to read the proof of the following proposition now. This isbecause it will be an immediate consequence of a fundamental resultabout Banach spaces.

Proposition 1.10. Every linear isomorphism between two finite di-mensional topological vector spaces is a homeomorphism.

Proof. It suffices to prove that if X is an n-dimensional TVS andf : Kn → X is a vector space isomorphism, then f is a homeomor-phism. It is continuous by Lemma 1.9 and it is a bijection by linearitysince n is finite.

For every r < ∞ the set Br := {ā ∈ Zn : ‖ā‖ ≤ r} is a compactsubset of Kn (Heine–Borel). Then f [Br] is compact, and thereforeclosed in X. The restriction of f to Br is a homeomorphism betweenBr and f [Br] (a compact Hausdorff topology is the weakest Hausdorfftopology).

It remains to prove that f−1 is continuous, or equivalently, that ifU ⊆ X is open then f [U ] is open in Kn.

This will follow from a little geometric argument. Let

A := {ā ∈ Kn : ‖x‖ = 1}(the unit sphere in Kn). Then for every nonzero vector b̄ ∈ Kn thereexists a unique positive scalar λ = |b̄‖ such that λb̄ ∈ A. ThereforeK\A is the union of two disjoint open sets,

I = {rx̄ : 0 ≤ r < 1, x̄ ∈ A}, J = {rx̄ : 1 < r


W ∩ f [A] = ∅. Therefore for every nonzero x̄ ∈ W there exists λ > 1such that λx̄ ∈ f [A]. This implies that W ⊆ {λf [A] : λ < 1}, andf−1[W ] ⊆ Z1. Since f is a homeomorphism between Z1 and f [Z1],f−1(U) is open.

Since f−1 is continuous at 0, it is continuous and this concludes theproof. �

Theorem 1.11. If X and Y are two finite-dimensional TVS (over thesame field K) then they are linearly homeomorphic if and only if theyhave the same dimension.

Proof. By Proposition 1.10, if ker(f) = {0} then f is a linear homeo-morphism between Y and X and therefore continuous. If it is nontriv-ial, then it is clearly a linear subspace of Y . Let g : Kn → Y be a linearisomorphism. Then f ◦ g : Kn → X is continuous by the first para-graph of this proof. Also, g is a linear homeomorphism, and thereforef = f ◦ g ◦ g−1 is continuous. �

The following can be extracted from the above proof (‘linear sub-space’ means ‘vector subspace’).

Exercise 1.12. Prove that every linear subspace of a finite-dimensionalTVS is closed.

1.1. Normed spaces. The first examples of topological vector spaceswill be equipped with a natural metric.

A seminorm on a vector space X is a function ‖ · ‖ : X → [0,∞)with the following properties.

(1) ‖x‖ ≥ 0 for all x ∈ X.(2) ‖x+ y‖ ≤ ‖x‖+ ‖y‖.(3) ‖sx‖ = |s|‖x‖ for every scalar s.

A seminorm is a norm if in addition we have

(4) ‖x‖ = 0 if and only if x = 0.Example 1.13. For n ≥ 1 each of the following is a norm on Rn.

(1) ‖x̄‖1 =∑n

i=1 |xi|(2) ‖x̄‖∞ = max1≤i≤n |xi|.(3) ‖x̄‖2 = (

∑ni=1 |xi|2)1/2. The triangle inequality is not com-

pletely trivial in this case. Define the inner product on Rnby

(x̄|ȳ) =∑

1≤j≤n

xjyj.

Then ‖x̄‖2 = (x̄|x̄)1/2, and the Cauchy–Schwarz inequality holds:|(x̄|ȳ)|2 ≤ (x̄|x̄)(ȳ|ȳ).

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The triangle inequality for ‖·‖2 is a straightforward consequenceof the Cauchy–Schwarz inequality.

The analogous formulas define norms in Cn. (Later on we shall define‖ · ‖p for other values of p ≥ 1.)

Exercise 1.14. Suppose that (X, ‖ · ‖) is a normed space. Prove thatd(x, y) = ‖x− y‖

defines a metric on X and that X is a topological vector space withrespect to the topology induced by d.

It is not difficult to find a metric space (X, d) and x ∈ X such thatthe closure of the open ball of diameter 1 does not include the closedball of diameter 1. This cannot happen in a normed space.

Lemma 1.15. Suppose that (X, ‖ · ‖) is a normed space. For all r > 0the closure of Y = {x ∈ X : ‖x‖ < r} is equal to Z = {x ∈ X : ‖x‖ ≤}.

Proof. Clearly Y ⊆ Z. We first prove that Z is closed. If x̄ /∈ Zthen ‖x̄‖ > r. Let δ := ‖x̄‖ − r. The open δ/2-ball centered at x,{y : ‖x − y‖ < δ/2} is by the triangle inequality disjoint from Z.Therefore an arbitrary point in X \ Z has an open neighbourhooddisjoint from Z, and Z is closed. This implies Y ⊆ Z. Conversely, ifx ∈ Z then let ȳn := nn+1 x̄. We have ‖ȳn‖ ≤

nn+1

r < r, therefore ȳn ∈ Yfor all n. Also, ‖x̄ − ȳn‖ = ‖ 1n x̄‖ ≤

rn, and therefore limn ȳn = x̄. We

conclude that Z ⊆ Y . �

2. Properties of Topological Vector Spaces

A subset A of a TVS X is convex if for all a and b in A the linesegment

{ra+ (1− r)b : 0 ≤ r ≤ 1}is included in A. A subset A of a TVS X is bounded if for every openneighbourhood U of 0 there is n such that nU ⊇ A.

Definition 2.1. A TVS X is

(1) Normable if it can be equipped with a norm compatible withits topology.4

(2) Locally convex if 0 has a basis consisting of convex open sets.(3) Metrizable if its topology is metrizable.(4) Locally bounded if 0 has a basis consisting of bounded sets.

Exercise 2.2. Prove the following implications hold for TVS.

4Warning: This is a nonstandard terminology.


(1) Normable implies locally convex.(2) Normable implies metrizable.(3) Normable implies locally bounded.(4) Metrizable implies 1st countable (i.e., 0 has a countable local

basis).

Try to guess which (if any) of the implications are reversible, andwhether there are TVS which are not metrizable and/or not locallyconvex.

2.0.1. Examples of finite-dimensional spaces. Each of the following ex-amples of topological vector spaces (or rather, metrizable vector spaces)has both real and complex versions.

(1) For n ≥ 1 consider Kn with respect to the Euclidean metric,

d(x̄, ȳ) := (∑j

|xj − yj|2)1/2.

(The convention that I am using here and elsewhere: x̄ standsfor the tuple (xj : 1 ≤ j ≤ n) where n is clear from the context.I shall also write

∑j in place of

∑j≤n when n is clear from the

context.)(2) For n ≥ 1 consider Kn with respect to the `1-metric,

d(x̄, ȳ) :=∑j

|xj − yj|.

(3) For n ≥ 1 consider Kn with respect to the `∞-metric

d(x̄, ȳ) := maxj|xj − yj|.

Exercise 2.3. Suppose X is a (Hausdorff) TVS.

(1) Prove that if A and B are subsets of X and one of them is openthen

A+B := {a+ b : a ∈ A, b ∈ B}is open.

(2) Prove that if A is open then λA is open for all λ 6= 0.(3) Prove that if A is closed then λA is closed for all λ.(4) Prove that if A is a compact subset of X and B is a closed

subset of X then A+B is a closed subset of X.(5) Is the following always true? If A is a closed subset of X and

B is closed subset of X then A + B is a closed subset of X.Justify your answer (i.e., prove the statement or find a coun-terexample.)

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2.1. The unit ball. A bit of notation and terminology is in order.The closed unit ball in a normed space (X, ‖ · ‖) is

BX‖·‖ := {x ∈ X : ‖x‖ ≤ 1}.We shall omit X or ‖ · ‖ (but never both) whenever they are clear fromthe context, e.g. in the following exercise.

Exercise 2.4. Let (X, ‖ · ‖) be a normed space. Prove that BX has thefollowing properties.

(1) BX is convex.(2) BX is symmetric: −BX = BX (where −A := (−1)A).(3) BX is balanced: λBX = BX for all λ ∈ K satisfying |λ| ≤ 1.

Lemma 2.5. Assume X is a vector space and that ‖ · ‖ and ‖| · ‖| arenorms on X. Then the following are equivalent for r 0 then B‖‖‖·‖‖ = 1rB‖|·‖|. Byputting this together, we have that ‖x‖ ≤ r‖|x‖| if and only if (with‖‖ · ‖‖ as defined above) B‖·‖ ⊇ BB‖‖·‖‖ =

1rB‖|·‖|, as required. �

A subset A of a topological vector space X is bounded if for everyopen neighbourhood U of 0 there exists r 0 such that

f−1(εBY ) ⊇ δBX .


By the linearity, this is equivalent to εδ−1f−1(BY ) ⊇ BX and thereforeto εδ−1BY ⊇ f(BX). �

Exercise 2.8. Assume X and Y are normed spaces. Let B(X, Y )denote the vector space of all bounded linear operators from X to Y .For f ∈ B(X, Y ) define the operator norm

‖f‖ := inf{r ≥ 0 : f(BX) ⊆ rBY }.Check that (B(X, Y ), ‖ · ‖) is a normed space.

Exercise 2.9. Let X be a normed space. Prove that X is finite-dimensional if and only if its unit ball is compact.

(Hint: If X is infinite-dimensional prove then the unit ball of Xcontains an infinite set xn, for n ∈ N, such that ‖xm − xn‖ ≥ 1/2 form 6= n.)

3. Infinite-dimensional normed spaces. Banach spaces

Two norms ‖·‖ and ‖| ·‖| on X are equivalent if they induce the samevector space topology. If X is a TVS then a norm on X is compatibleif it induces the original topology on X.

Lemma 3.1. Norms ‖ · ‖ and ‖| · ‖| on X are equivalent if and only ifthere are 0 < r ≤ R

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3.0.1. Examples of sequence spaces.

(1) Let KN be the space of all sequences of elements of K. We shalldescribe the neighbourhood basis of x̄ ∈ RN. For a finite F ⊆ Nand ε > 0 let

UF,ε(x̄) := {ȳ ∈ KN : |xj − yj| < ε for all j ∈ F}.

Sets UE,ε(x̄) form a basis a topology that turns KN into a topo-logical vector space.

(2) Let `2 := {x̄ ∈ KN :∑

j |xj|2


Exercise 3.5. Prove that every n-dimensional subspace of `2 is iso-metric to `2(n).

(Hint: Gram–Schmidt.)Show that for some n not every n-dimensional subspace of c0 is iso-

metric with `∞(n).

Definition 3.6. Banach space is a normed vector space (X, ‖·‖) whichis complete with respect to metric d associated with the norm ‖ · ‖.

Lemma 3.7. `∞, c and c0 are Banach spaces.

Proof. Assume (x̄(k)) is a Cauchy sequence in `∞. Since ‖x̄ − ȳ‖ ≥‖xn− yn‖ for all n, for every n ∈ N the sequence x(k)n is Cauchy in Kand therefore convergent. Let yn := limk x(k)n. Then ‖ȳ − x̄(k)‖ → 0as k →∞.

To check that c is a Banach space we need to check it is a closedsubspace of `∞ (using Exercise 3.12). Let ȳ ∈ `∞ \c. Then limk yk doesnot exist, so r := lim supk yk − lim infk yk > 0. Then ‖x̄ − ȳ‖ < r/3implies x̄ /∈ c and therefore ȳ has an open neighbourhood disjoint fromc.

The proof that c0 is a closed subspace of c is similar. �

The following proposition states that X is a Banach space iff everyabsolutely convergent series is convergent.

Proposition 3.8. For a normed space X the following are equivalent.

(1) X is a Banach space.(2) Whenever xn, for n ∈ N, is a sequence in X such that

∑n ‖xn‖ <

∞ the sequence yn :=∑

j≤n xj, for n ∈ N is convergent.

Proof. (1) implies (2): Assume (1). If xn is as in (2) then by the Cauchycriterion for convergence the sequence yn is Cauchy and therefore con-vergent by (1).

(2) implies (1): Assume (2). Let zn be a Cauchy sequence in X.It suffices to find a subsequence of zn that converges. Let z

′n be a

subsequence such that ‖z′m − z′n‖ < 2−m for m < n. Let x1 := z1 andxn+1 := z

′n+1 − z′n. Then

∑j≤n xj = z

′j and by (2) there exists y ∈ X

such that ‖z′j − y‖ → 0 as j →∞, as required. �

Lemma 3.9. The space `2 is a Banach space.

Note that the proof of Lemma 3.7 does not work here. Why?

Proof. By Proposition 3.8 it suffices to check that if a(n) ∈ `2 are suchthat

∑n ‖a(n)‖2

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series∑

n |a(n)j| is absolutely convergent, and the sum bj =∑

n a(n)jis well-defined.

This defines b̄ ∈ KN. To prove that b̄ ∈ `2 and that limm ‖b̄ −∑n≤m ā(n)‖2 = 0, fix ε > 0. Let m be such that

∑∞n=m+1 ‖ā(n)‖2 < ε

and fix k ≥ m+ 1. The triangle inequality and our assumption imply

‖∑k

n=m+1 ā(n)‖2 ≤∑m

n=m+1 ‖ā(n)‖2 < ε.

Therefore

‖b̄−∑

n≤m ā(n)‖2 = ‖∑∞

n=m+1 ā(n)‖2 ≤ supk∑k

n=m+1 ‖ā(n)‖2 ≤ ε.

Since ε > 0 was arbitrary, b̄ is the limit (in ‖ · ‖2) of the partial sums,and that it belongs to `2. �

The proof of the following is omitted, since it is very similar to theproof of Lemma 3.9.

Lemma 3.10. The space `1 is a Banach space. �

If X is a TVS and Y is its linear subspace then we can consider Ywith the subspace topology. A linear subspace of a TVS X is a vectorsubspace of X. More precisely, it is a set of vectors closed under theaddition and multiplication by scalars.

Exercise 3.11. Prove that a subspace of a TVS is a TVS.

Exercise 3.12. Prove that a closed subspace of a Banach space is aBanach space with respect to the induced norm.

Exercise 3.13. Prove that a finite-dimensional subspace of a TVS isalways closed.

3.1. Constructions of spaces: Direct products and direct sums,quotients, the completion. I shall now describe some constructionsof normed spaces. The details of the proofs are straightforward andwatching someone write down the details is about as illuminating andhelpful as watching someone ride a bike. Therefore almost every sen-tence is an exercise left for you to work out. It is important that youfill in sufficient amount of detail to convince yourself that you couldprovide the remaining (sometimes onerous) details if your life dependedon that. The last thing that you should do is look up the details in theliterature without making an attempt to work them out.

3.1.1. Products and sums. Suppose X is a normed space. Let

`∞(X) := {x̄ ∈ XN : supj|xj|


the space of all bounded sequences in X. On `∞(X) define

‖x̄‖∞ := supn‖xn‖.

Prove that this is a norm on X, and that `∞(X) is a Banach space ifX is a Banach space.

3.1.2. Direct sums and products. Given a family of spaces Xj for j ∈ Jdefine (here x̄ = (xj : j ∈ J); the most common case is when J = N)∏

j

Xj = {x̄ : supj‖xj‖

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3.1.3. Quotient. SupposeX is a normed space over K and Y is its linearsubspace. From basic algebra we know that the relation ∼ defined byletting

x ∼ y if and only if x− y ∈ Yis an equivalence relation and the quotient space X/Y is a vector spaceover K. Its elements are cosets x+Y , for x ∈ X. On X/Y the formula

N(x+ Y ) = infy∈Y‖x+ y‖

defines a function N : X/Y → K.

Lemma 3.18. The function N is a seminorm on X/Y . It is a normif and only if Y is a closed subspace.

Proof. If x ∈ X and λ ∈ K we need to check that N(λx + Y ) =|λ|N(x + Y ). This is trivial if λ = 0, so we may assume λ 6= 0. ThenN(λx + Y ) = infy∈Y ‖‖λx + y‖ = infy∈Y |λ|‖x + λ−1y‖ = N(x + Y )because y ∈ Y if and only if λ−1y ∈ Y .

The proof that N is additive is also straightforward (but worth work-ing out if you haven’t seen it before).

For x ∈ X we have N(x + Y ) = 0 if and only infy∈Y ‖x + y‖ = 0.This is equivalent to having x ∈ Y (the closure of Y ), and therefore weconclude that N(x+ Y ) = 0 if and only if x ∈ Y holds for all x ∈ X ifand only if Y is a closed subspace. �

Lemma 3.19. If X is a Banach space and Y is a closed subspace ofX then X/Y is a Banach space.

Proof. Lemma 3.18 implies that X/Y is a normed space if X is anormed space and Y is a closed subspace of X. It remains to checkthat the completeness of X implies the completeness of X/Y . The ‘keytriviality’ in this proof the following.

Claim 3.20. If x and y are in X and such that ‖Q(x− y)‖ < r, thenthere is y′ ∈ X such that Q(y′) = Q(y) and ‖x− y′‖ < r.

Proof. Let z ∈ Y be such that ‖x− y − z‖ < r. Then y′ := y + z is asrequired. �

Suppose zn is a Cauchy sequence in X/Y . By going to a subsequence,we can suppose ‖zn+1 − zn‖ < 2−n for all n. It suffices to prove thatthis subsequence is convergent. Pick x0 ∈ X such that x0 + Y = z0.By using Claim, we can recursively find xn ∈ X for n ≥ 1 such thatQ(xn) = zn and ‖xn−1−xn‖ < 2−n for all n. The sequence xn is Cauchy,and therefore convergent, in X. Its limit y satisfies Q(zn)→ Q(y). �


3.1.4. The completion of a normed space. Suppose X is a normed spacethat is not a Banach space (e.g. X = c00 with respect to the ‖ ·‖∞-norm). Let Y be the subspace of `∞(X) consisting of all Cauchysequences.

Lemma 3.21. The space Y is a closed linear subspace of `∞(X)

Proof. The fact that Y is a linear subspace is an immediate consequencethat sums of Cauchy sequences are Cauchy and that scalar multiple ofa Cauchy sequence is Cauchy.

In order to prove that Y is closed, choose z̄ ∈ `∞(X) \ Y . As z̄is not Cauchy, there exists ε > 0 such that for every m there existsn > m such that ‖zm − zn‖ ≥ ε. Let U be an open ball containing z̄of diameter ε/3. Then for every x̄ ∈ U the following is true. For everym there exists n > m such that ‖zm − zn‖ ≥ ε/3 (this is proved bythe ‘ε/3-trick’ - work it out if you haven’t seen it yet). Therefore everypoint z̄ /∈ Y has an open neighbourhood disjoint from Y , and Y is aclosed subspace of `∞(X). �

Now consider Y0, the subspace of Y consisting of all Cauchy se-quences converging to 0. This is a closed subspace of Y and the proofis similar to the proof of the lemma above.

Let X̃ := Y/Y0. By §3.1.3 it is a Banach space. Map f : X → `∞(X)that sends x to the constant sequence x̄ = (x, x, x, . . . ) is a linearisometry, and its range is contained in Y . Moreover, f−1(Y0) = {0}.Therefore Q ◦ f maps X isometrically into X̃.

We shall identify X with its image (Q ◦ f)(X) in X̃.We say that X̃ is the completion of X. The following proposition

shows that the completion is uniquely defined.

Proposition 3.22. Let X and X̃ be as above.

(1) X is a dense subspace of X̃.(2) Suppose Z is a Banach space and g : X → Z is a bounded linear

map. Then g can be extended to a bounded linear map g̃ : X̃ →Z of the same norm. Also, g̃(X̃) is equal to the closure of g(X)and in particular if g(X) is dense in Z then g̃ is surjective.

3.2. Examples of function spaces.

(6) Suppose X is a compact Hausdorff space (think X = [0, 1] andyou lose little). Let

CR(X) := {f : X → R : f is continuous}.C(X) := {f : X → C : f is continuous}.

18 I. FARAH

Each of these spaces is equipped with the sup metric,

‖f‖∞ = supx∈X|f(x)|.

By compactness, the sup is attained and ‖f‖∞ = maxx∈X |f(x)|.(This also implies that the norm is finite and therefore well-defined.)

(7) Suppose X is a locally compact Hausdorff space (think X = Ror X = (0, 1) and you lose little). Since not all continuousreal-valued functions on X are bounded, we let

C00(X) := {f : X → C : supp(f) is relatively compact}.C0(X) := {f : X → C : f is continuous and vanishes at ∞}.

Here supp(f) = {x ∈ dom(f) : f(x) 6= 0} and a set is relativelycompact if its closure is compact (think bounded subsets of Rand Heine–Borel). f vanishes at ∞ means that the followingholds for every ε > 0:

There exists a compact K ⊆ X such that

supx∈X\K

|f(x)| ≤ ε.

We equip each of these spaces with the sup metric

‖f‖∞ := supx∈X|f(x)|.

(8) Let µ be a σ-finite, strictly positive Borel measure on a compactHausdorff space X (think X = R and µ=the Lebesgue measureand you lose little). On the space of all continuous functions fon X such that µ(supp(f)) 0. Hence ‖f‖2 ≥ εµ(U) > 0 (since U is strictly positive).

The completion of this space is L2(X,µ). (This is not theonly way to define L2 or other Lp spaces. One should keep inmind that either way the elements of L2(X,µ) are not functionsbut equivalence classes of functions.)


(9) With X and µ as in (8), on C0(X) define

‖f‖1 :=∫|f | dµ.

This is a norm. We prove that ‖f‖1 = 0 implies f = 0. If f 6= 0then U = {x : |f(x)| ≥ ε} is (by the continuity) a nonemptyopen set for some ε > 0. Hence ‖f‖1 ≥ εµ(U) > 0 (since U isstrictly positive).

The completion of this space is L1(X,µ).(10) An alternative way to define L2(X,µ) is to consider the vector

space of all step functions on X whose support has finite mea-sure and take the completion with respect to ‖ · ‖2. One canprove that this space is isometrically isomorphic to L2(X,µ).The analogous remark applies to L1(X,µ).

4. Baire Category Theorem and its Consequences

An intersection of finitely many dense open subsets of a metric spaceis dense and open (check). If the space is complete we can say more.

Theorem 4.1 (Baire Category Theorem). Assume (X, d) is a com-plete metric space. An intersection of a countable family of dense opensubsets of X is dense in X.

Proof. Let Un, for n ∈ N, be dense open subsets of X. We writeBε(x) = {y : d(x, y) < ε}.

Let V be a nonempty open subset of X. We shall prove that V ∩⋂n Un 6= ∅. Since U1 is dense we can choose x1 ∈ U1 ∩ V . Since U1

and V are open we can choose ε(1) > 0 so that B2ε(1)(x1) is includedin U1 ∩ V . Let V2 := Bε(1)(x1).

Now choose x2 ∈ U2 ∩ V2 and ε(2) > 0 so that B2ε(2)(x2) is includedin U2 ∩ V2. Let V3 := Bε(2)(x2). While we are at it, shrink ε(2) ifnecessary to assure ε(2) < 1/2.

Continuing in this manner, choose xn, Vn and ε(n) so that

(1) xn ∈ Un ∩ Vn,(2) ε(n) < 1/n.(3) B2ε(n)(xn) ⊆ Un ∩ Vn.(4) Vn+1 = Bε(n)(xn).

Note that V = V1 ⊇ V2 ⊇ V3 ⊇ . . . . This has two important conse-quences:

First, Vn ⊆ Uj for all j < n. Second, if m > n then xm ∈ Vn andd(xn, xm) < ε(n). As ε(n) < 1/n this means that (xn) is a Cauchysequence.

20 I. FARAH

Since X is complete x := limn xn exists. We have

d(xn, x) ≤ supm≥n

d(xn, xm) ≤ 1/n.

Therefore x ∈⋂n Vn, and by the above x ∈

⋂n Un. Also, x ∈ V 2 ⊆ V

and therefore V ∩⋂n Un 6= ∅.

As V was an arbitrary nonempty open subset of X, this completesthe proof. �

A subset F of a topological space X is nowhere dense if everynonempty open U ⊆ X has a nonempty open subset disjoint fromF . Equivalently, F is nowhere dense if and only if X \F is dense open(proof by reading the definitions). A subset of a topological space issaid to be of the first category (or meager) if it can be covered bycountably many nowhere dense subsets. As the closure of a nowheredense set is nowhere dense (check), a set is of the first category if andonly if it can be covered by countably many closed nowhere dense sets.We have a reformulation of the Baire Category Theorem.

Corollary 4.2. Suppose X is a complete metric space. Then X cannotbe covered by countably many nowhere dense sets. �

Sets which are not of the first category are said to be of secondcategory (or nonmeager). Corollary 4.2 is sometimes expressed as ‘Acomplete metric space is not meager in itself.’

A family J of subsets of X is a σ-ideal if the following hold.

(1) If A ∈ J and B ⊆ A then B ∈ J .(2) If An ∈ J for n ∈ N, then

⋃nAn ∈ J .

A σ-ideal is proper if X /∈ J . An example of a proper σ-ideal is theideal of subsets of R of Lebesgue measure zero. For any topologicalspace X its subsets of first category clearly form a σ-ideal and BaireCategory Theorem states that this ideal is proper if X is a completemetric space.

Exercise 4.3. Prove that the Baire Category Theorem is also true forcompact Hausdorff spaces, as well as locally compact Hausdorff spaces.That is, a (locally) compact Hausdorff space cannot be covered by count-ably many nowhere dense subsets.

Exercise 4.4. Suppose X is a complete metric space.

(1) Prove that if Vn, for n ∈ N, is a sequence of nonempty opensubsets of a complete metric space X such that (i) Vn+1 ⊆ Vnfor all n and (ii) limn diam(Vn) = 0 then the set

⋂n VN =

⋂n Vn

has exactly one element.


(2) Use (1) to give a (very slightly) different proof of Theorem 4.1.(3) Show that both (i) and (ii) are necessary conditions in (1).

Back to functional analysis.

Exercise 4.5. Let X be a TVS and let Y be a closed subspace of X.Prove that Y is nowhere dense if and only if Y 6= X.Exercise 4.6. Prove that there is a linear bijection f : `2 → `2 suchthat neither f nor its inverse is continuous.

We shall denote the open unit ball by

B̊X := {x ∈ X : ‖x‖ < 1}.As Å usually denotes the interior of a set A, the notation is justifiedby the following exercise.

Exercise 4.7. Let X be a normed space. Prove that the interior ofthe closed unit ball BX is equal to the open unit ball B̊X and that theclosure of the open unit ball is the closed unit ball.

(You may also want to prove that this is false in general metricspaces.)

5. The Open Mapping Theorem

The following simple lemma is (together with the much deeper BaireCategory Theorem) key to the proof of the Open Mapping Theorem.

Lemma 5.1. Assume X and Y are normed spaces, X is a Banachspace, and f : X → Y is linear, bounded and such that f(B̊X) is densein rB̊Y for some r > 0. Then f(BX) includes rB̊Y .

Proof. Choose b ∈ rB̊Y =⋃s 0.

We record a simple consequence of the linearity of f :

f(B̊X) is dense in rB̊Y if and only if

f(sB̊X) is dense in srB̊Y

for some (equivalently, all) s > 0.

Using this, choose a1 ∈ X such that ‖a1‖ < 1−δ and ‖b−f(a1)‖ < δ2r.(We can make ‖b− f(a1)‖ as small as we want, but δ2r will do.) Nowchoose a2 ∈ X such that ‖a2‖ < δ2 and

‖b− f(a1)− f(a2)‖ <δ

4r.

Continuing in this manner find an ∈ X for n ≥ 2 such that ‖an‖ < δ2n−1and ‖b −

∑nj=1 f(aj)‖ <

δ2n

. Since X is a Banach space, a :=∑

j aj

22 I. FARAH

is well-defined and satisfies ‖a‖ ≤ (1 − δ) +∑∞

j=1 2−jδ ≤ 1. Also, the

sequence cn := f(∑n

j=1 aj) in Y converges to b and therefore f(a) = b,as required.

Since b was arbitrary, this proves the lemma. �

Recall that a map between topological spaces is open if the image ofevery open set is open.

Exercise 5.2. Convince yourself that the property of being open isindependent from being continuous by finding examples (if you haven’talready done so in a topology course).

Exercise 5.3. Assume f is a linear map between topological vectorspaces. Prove that the following are equivalent.

(1) f is open.(2) f(U) includes an open neighbourhood of 0 (in Y ) for every open

neighbourhood of 0 (in X).

If in addition spaces X and Y are normed then this is equivalent to

(3) f(BX) includes y + rBY for some y ∈ Y and some r > 0.

Exercise 5.4. Suppose U is a nonempty open subset of a TVS X.Then U − U = {a− b : a ∈ U, b ∈ U} includes an open neighbourhoodof the identity.

We shall prove the Open Mapping Theorem and some of its impor-tant consequences.

The following lemma is not needed in the proof below.

Lemma 5.5. Suppose f : X → Y is linear, both X and Y are normedspaces, and the closure of f(BX) has a nonempty interior. Then f(BX)includes an open neighbourhood of 0 in Y .

Proof. Let b ∈ Y and r > 0 be such that b + rBY is included in theinterior of f(BX). Then the closure of 2f(BX) includes (b + rBY ) −(b + rBY ) which in turn includes 2rBY ; hence the closure of f(BX)includes the interior of rBY . �

Exercise 5.6. State and prove a TVS version of Lemma 5.5

Theorem 5.7 (Open Mapping Theorem). Suppose X and Y are Ba-nach spaces and f : X → Y is a bounded linear map such that f(X) isof second category in Y . Then f is open.

Proof. By Exercise 5.3 it suffices to show that f(BX) includes an openneighbourhood of 0. As

⋃n nB

X = X and f(X) is of second categoryin Y ,

⋃n nf(B

X) is of second category in Y .


As Y is a Banach space, there exists n such that the closure ofnf(BX) has a nonempty interior. By Lemma 5.5 the closure of nf(BX)includes an open neighbourhood of 0; in other words, it includes rBY

for some r > 0. By Lemma 5.1 nf(BX) includes r2BY . Therefore the

image of any neighbourhood of 0 includes a neighbourhood of 0 andthe map is open.

Also note that by the linearity,⋃n f(B

X) includes Y . �

One measure of the importance of a theorem is the variety of itscorollaries. Here we go.

Corollary 5.8. Every bounded bijection between Banach spaces has abounded inverse.

Proof. If f : X → Y is a bounded bijection between Banach spacesthen the image of X is of the second category in Y and therefore f isopen. This means that f−1 is continuous. �

Corollary 5.9. If f : X → Y is a bounded linear map between Banachspaces then f(X) is either equal to Y or it is of the first category in Y .

Proof. If f(X) is of the second category in Y then f is open by the

Open Mapping Theorem. In particular, U := f(B̊X) is a nonemptyopen subset of f(X) containing 0. Clearly f(X) ⊇

⋃n nU and by

Corollary 1.4,⋃n nU = Y . �

Corollary 5.10. Suppose

(1) X is a vector space,(2) ‖ · ‖ and ‖| · ‖| are two Banach space norms on X, and(3) there exists r 0 such that ‖a‖ ≥ s‖|a‖| for all a.Proof. The identity map from (X, ‖| ·‖|) into (X, ‖·‖) is by our assump-tion bounded (its norm is ≤ r). It is clearly surjective, and thereforeby Corollary 5.8 it has bounded inverse. Then s := ‖f−1‖−1 is asrequired. �

Exercise 5.11. Can you improve the conclusion of Corollary 5.9 toprove that if f : X → Y is a bounded linear map between normed spacesthen f(X) is either nowhere dense or equal to Y ?

Exercise 5.12. Does every bounded bijection between TVS have abounded inverse? Does every bounded bijection between normed spaceshave a bounded inverse?

Exercise 5.13. Find examples showing that each of the assumptions(1), (2) and (3) is necessary in Corollary 5.10.

24 I. FARAH

Exercise 5.14. If X and Y are TVS, then X × Y with respect to theproduct topology is a TVS. If X and Y are normed then this topologyon X × Y is compatible with the norm ‖(a, b)‖ := max{‖a‖, ‖b‖).

If f : X → Y is a linear map between vector spaces then its graphΓf := {(x, f(x)) : x ∈ X}

is a vector subspace of X × Y .

Theorem 5.15 (Closed Graph Theorem). If X and Y are Banachspaces and f : X → Y is a linear map. then f is continuous if andonly if Γf is a closed subspace of X × Y (see Exercise 5.14).

Proof. Assume Γf is a closed subspace of X × Y . Let πX and πY bethe projections of X × Y onto X and Y , respectively. (More precisely,πX(x, y) = x and πY (x, y) = y.)

As ‖(x, y)‖ = max(‖x‖, ‖y‖), both πX and πY are bounded. By Ex-ercise 3.12 Γf is a Banach space, being a closed subspace of a Banachspace. Note that πX(Γf ) = X since X = dom(f) and that the restric-tion of πX to Γf is an injection since f is a function (the ‘vertical linetest’). Hence the restriction of πX to Γf has an inverse, g : X → Γf .

By Corollary 5.8 applied to the restriction of πX to Γf we concludethat g is bounded. As f = πY ◦ g and composition of continuous mapsis continuous, we conclude that f is continuous.

The converse direction is both less interesting and easier. Suppose‖f‖ ≤ r. In order to prove that Γf is a closed subspace of X×Y choose(x, y) ∈ X × Y \ Γf . Let z := f(x). Then the open ball around (x, y)with diameter

δ :=1

2‖z − y‖min(‖f‖−1, 1)

is disjoint from Γf . If (x′, y′) belongs to this ball then

‖x− x′‖ < 12‖z − y‖‖f‖−1

and therefore ‖f(x)− f(x′)‖ = ‖z − f(x′)‖ < 12‖z − y‖. �

If X and Y are normed spaces then

B(X, Y ) := {f : X → Y : f is linear and bounded}.

Exercise 5.16. Prove that B(X, Y ) is a normed vector space with re-spect to the operator norm,

‖f‖ = supx∈X,‖x‖≤1

‖f(x)‖.

Prove that B(X, Y ) is a Banach space if Y is a Banach space.


The following result is also known as the Banach–Steinhaus Theo-rem.

Theorem 5.17 (The Uniform Boundedness Principle). Suppose X andY are Banach spaces and F ⊆ B(X, Y ) is such that for every x ∈ Xthe set

Ox := {f(x) : f ∈ F}is bounded. Then F is a bounded subset of B(X, Y ) (i.e., sup{‖f‖ :f ∈ F}

26 I. FARAH

6. The dual space

Linear functional on a TVS X is a linear map from X into K.We shall almost always consider only the continuous (equivalently,bounded) linear functionals. If X is a TVS then X∗ denotes the spaceof all continuous linear functionals (i.e. B(X,K)).Lemma 6.1. Suppose X is finite-dimensional. Then X and X∗ arelinearly homeomorphic.

Proof. A functional ϕ on X is uniquely determined by its restrictionto the basis. As X is finite-dimensional, every linear functional isbounded.

Let n := dim(X) and let xj, for 1 ≤ j ≤ n, denote the basis for X.Define x∗j by (using Kronecker’s delta, δij = 1 if i = j and δij = 0 ifi 6= j)

x∗j(xk) = δjk.

Linearly extension of x∗j to X is a bounded linear functional (as ev-ery linear map between finite-dimensional TVS is continuous, Theo-rem 1.11). Also, every ϕ ∈ X∗ can be uniquely written as ϕ(

∑j≤n αjxj) =∑

j≤n αjx∗j . Therefore x

∗j , for 1 ≤ j ≤ n, form a basis for X∗. Hence

dimx∗ = dimX and the conclusion follows by Theorem 1.11. �

Lemma 6.1 is rather specific for the finite-dimensional spaces, al-though its conclusion is true for the Hilbert space.

Proposition 6.2. Suppose X is a normed space. Then X∗ is a Banachspace with respect to the norm

‖ϕ‖ = sup‖x‖≤1

|ϕ(x)|.

Proof. This is the operator norm on X∗ = B(X,K) (Exercise 2.8).Suppose ϕn, for n ∈ N, is a Cauchy sequence in X∗. Since

|ϕm(x)− ϕn(x)| ≤ ‖ϕm − ϕn‖,the sequence ϕn(x), for n ∈ N, is Cauchy for every x ∈ X. It istherefore also bounded, and by Corollary 5.18 there exists ϕ ∈ X∗such that limn ϕn(x) = ϕ(x) for all x.

But we are not done—pointwise convergence is in general muchweaker than convergence in norm. (We will be returning to this point.)For every ε > 0 there exists m such that ‖ϕm−ϕn‖ ≤ ε for all n ≥ m.Then ‖ϕm − ϕ‖ ≤ ε, and we conclude that limm ϕm = ϕ in norm. �

Our next objective is to prove the Hahn–Banach theorem, a resultto the effect that every normed space (as well as every locally convexTVS) has a rich dual space.


Let X be a Banach space over field K.A Minkowski functional is a function µ : X → R which satisfies the

following two conditions. It is subadditive, µ(x+y) ≤ µ(x)+µ(y) for allx, y and positive homogeneous, µ(ax) = aµ(x) for every a ∈ [0,∞). Wenote that X may be a vector space over R or C, that every seminormis a Minkowski functional (but not necessarily vice versa), and that aMinkowski functional is rarely a functional as defined above.

Every seminorm is a Minkowski functional.

Exercise 6.3. Suppose that U is a convex open neighbourhood of 0 ina TVS X. Prove that µU(x) := inf{r ∈ R+ : r−1x ∈ U} is a Minkowskifunctional and that U = {x : |µU(x)| ≤ 1}.

A functional ϕ is dominated by µ if ϕ(x) ≤ µ(x) for all x in thedomain of ϕ. (We are implicitly assuming that dom(ϕ) ⊆ dom(µ).)

Lemma 6.4. Suppose X is a real vector space and µ is a Minkowskifunctional on X. Furthermore suppose Y is a linear subspace of X andϕ is a linear functional on Y dominated by µ (i.e. ϕ(x) ≤ µ(x) forall x ∈ Y ). Them ϕ can be extended to a linear functional ψ on Xdominated by µ.

Corollary 6.5. Assume X is a normed space and x ∈ X. Then thereexists ϕ ∈ X such that |ϕ(x)| = ‖x‖ and ‖ϕ‖ = 1.

Proof. Let Y be the linear span of x and let ϕ(ax) = a‖x‖. Defineµ : X → R by letting µ(y) := ‖y‖ for all y. As µ is Minkowski func-tional, we can apply the Hahn–Banach extension theorem and obtainϕ as required. �

The functional ϕ as in Corollary 6.5 is the norming functional for x.

Corollary 6.6. Assume X is a normed space and Y is its proper closedsubspace. Then for every vector x ∈ X \ Y there exists a functionalϕ ∈ X∗ such that ‖ϕ‖ = 1, Y ⊆ ker(ϕ), and ϕ(x) = dist(x, Y ).

Proof. Apply Corollary 6.5 to the quotient space X/Y and x + Y toobtain ψ ∈ (X/Y )∗ such that ψ(x + Y ) = ‖x + Y ‖. Then ϕ is thecomposition of the quotient map π : X → X/Y with ψ. �

The proof of Lemma 6.4 will make use of some set theory (and thisis, arguably, the more straightforward component of the proof). Amaximal element in a partially ordered set P is a such that no otherelement of P is strictly greater than a. A subset of a partially orderedset is totally ordered (or linearly ordered) if every two of its elementsare comparable.

28 I. FARAH

Zorn’s Lemma. Suppose P is a partially ordered set such that everytotally ordered subset of P has an upper bound. Then P has a maximalelement.

This is technically not a ‘Lemma’ because one cannot prove it fromthe ‘obvious’ axioms of set theory alone. Zorn’s Lemma is equivalent tothe Axiom of Choice (see [4, Theorem 1.1.6]). A typical use of Zorn’sLemma in functional analysis is given in the proof of Lemma 6.4.

Proof of Lemma 6.4. The proof has two components. First, we let Pbe the set of all pairs (Z, θ) where Z is a linear subspace of X includingY and θ is a linear functional on Z dominated by µ and extending ϕ.We order P by letting (Z, θ) ≤ (Z ′, θ′) if Z ⊆ Z ′ and θ′ extends θ.

Thus P is the set of all partial extensions of ϕ to subspaces of X, andwe need to find ψ such that (X,ψ) ∈ P. Let us check that P satisfies theassumption of Zorn’s Lemma. If C is a totally ordered subset of P, thenZ ′ :=

⋃(Z,θ)∈C Z is a linear subspace of X (check). For every z ∈ Z ′

there is the unique s ∈ R with the property that for every (Z, θ) ∈ Csatisfying z ∈ Z we have θ(z) = s. Let θ′(z) be this s. Then θ′ is alinear functional on Z ′ that extends ϕ and is dominated by µ.

By Zorn’s Lemma P has a maximal element (Z, ψ).5 If Z = X thenψ is as required.

We shall therefore assume that Z is a proper subset of X and provethat this leads to contradiction. Fix x ∈ X \ Z and let Z ′ be the spanof Z and {x}. We shall find ψ′ such that (Z ′, ψ′) is an element of P thatmajorizes (Z, ψ). In order to define ψ′ we only (!) need to find the rightvalue for ψ′(x). As every element of Z ′ can be uniquely written as z+txfor unique z ∈ Z and t ∈ R, if ψ′(x) = α then ψ′(z + tx) = ψ(z) + tα.In order that ψ′ be majorized by µ we need to have

ψ(z) + tα = ψ′(z + tx) ≤ µ(z + tx).By the positive homogeneity of µ only two values of t matter, 1 and−1. Hence we only need to find α that satisfies the following twoinequalities.

ψ(z) + α = ψ′(z + x) ≤ µ(z + x)ψ(z)− α = ψ′(z − x) ≤ µ(z − x).

In other words, we need

α ≤ µ(z + x)− ψ(z)α ≥ ψ(z)− µ(z − x).

5This maximal element is by no means unique; if it were we would not needZorn’s Lemma to help us find it.


for all z ∈ Z. In order to prove the existence of a good α, we need toknow that

supz∈Z

ψ(z)− µ(z − x) ≤ infz∈Z

µ(z + x)− ψ(z)

or equivalently ψ(y)− µ(y− x) ≤ µ(z + x)−ψ(z) for all y and z in Z.Fix y and z in Z. Then y + z ∈ Z and therefore (using the linearity

of ψ, the fact that ψ is dominated by µ, and the subadditivity of µ,respectively) we have

ψ(y)+ψ(z) = ψ(y+z) ≤ µ(y+z) = µ((y−x)+(z+x)) ≤ µ(y−x)+µ(z+x).This is equivalent to the required inequality. Since y and z were arbi-trary, we can choose α as required.

Hence ψ can be extended to the span of Z ∪ {x}. This contradictsthe maximality of the pair (Z, ψ) and completes the proof. �

The proof of the Hahn–Banach theorem provides the first situationin which the cases when the field of scalars is equal to R or to C differ.

Lemma 6.7. Suppose X is a complex normed space and ϕ ∈ X∗.If we consider X as a real normed space, then 0 such that ε < r andlet x ∈ BX be such that ‖ϕ(x)‖ > ‖ϕ‖ − ε. With λ := ‖ϕ(x)‖/ϕ(x)we have x′ := λx satisfies x′ ∈ BX and r − ε. Sinceε > 0 was arbitrarily small, we have ‖

30 I. FARAH

Let X be a normed space. For a subspace Y of X the annihilatorof Y is

Y ⊥ := {ϕ ∈ X∗|Y ⊆ ker(ϕ)}.The annihilator of a subspace Z of X∗ is defined to be

Z⊥ := {x ∈ X|ϕ(x) = 0 for all ϕ ∈ Z}.

Proposition 6.9. If X is a normed space and Y is a subspace of Xthen (Y ⊥)⊥ is the norm-closure of Y . In particular, if Y is a closedsubspace then (Y ⊥)⊥ = Y .

Proof. As kerϕ is a closed subspace, the annihilator of a subspace isequal to the annihilator of its closure. Clearly Y ⊆ (Y ⊥)⊥. If x is notin the closure of Y , then by Corollary 6.6 there exists ϕ ∈ Y ⊥ such thatϕ(x) 6= 0, and therefore x /∈ (Y ⊥)⊥. This proves the reverse inclusion,(Y ⊥)⊥ ⊆ Y , and completes the proof. �

For Z ⊆ X∗ we have (Z⊥)⊥ ⊇ Z, but the equality does not holdin general. However, the operation Z 7→ (Z⊥)⊥ behaves like a closureoperator in topology. This topology is very important, and we willstudy it in some detail.

Corollary 6.10. If X is a normed space, then the completion of X isisometrically isomorphic to a subspace of the second dual X∗∗ of X.

Proof. For x ∈ X define a functional x∗∗ on X∗ byx∗∗(ϕ) := ϕ(x).

This functional is linear and by Corollary 6.5 its norm is equal to‖x‖. Therefore x 7→ x∗∗ is a linear isometry of X into X∗∗. Since thelatter is a Banach space (by Proposition 6.2), by the universality of thecompletion X̃ of X this isometry can be extended to an isometry of X̃into X∗∗. �

Definition 6.11. Normed spaces X and Y are in algebraic duality ifthere exists a bilinear form

〈·, ·〉 : X × Y → Ksuch that

(1) 〈·, y〉 ∈ X∗ for all y ∈ Y and 〈x, ·〉 ∈ Y ∗ for all x ∈ X.(2) 〈·, Y 〉 separates points of X,(3) 〈X, ·〉 separates points of Y .

One example of spaces in duality is X,X∗, where the bilinear formis given by the functional evaluation

〈x, ϕ〉 := ϕ(x).


Some other examples are X∗, X (well, the notion is symmetric) andX∗, X∗∗, also with the natural functional evaluation.

Example 6.12. The spaces c0 and `1 are in duality, via

〈x̄, ȳ〉 =∑n

xnyn.

It is not difficult to check (but do check) that this is a bilinear formand that its norm is 1.

Example 6.13. The spaces `∞ and `1 are in duality, via

〈x̄, ȳ〉 =∑n

xnyn.

It is not difficult to check (but do check) that this is a bilinear formand that its norm is 1.

We start with a leftover from the last class. Suppose X and Y arenormed spaces and T ∈ B(X, Y ). Define T ∗ : Y ∗ → X∗ by the equation

〈x, T ∗ϕ〉 = 〈Tx, ϕ〉

for all x ∈ X and ϕ ∈ Y ∗. Then T ∗ is well-defined (to see this read theabove formula as T ∗ϕ(x) := 〈Tx, ϕ〉). Also, T ∗ is linear (check).

Proposition 6.14. With the notation as above, the map T 7→ T ∗ fromB(X, Y ) into B(Y ∗, X∗) is a linear isometry.

Proof. For ϕ ∈ Y ∗ we have

‖T ∗ϕ‖ = sup{|〈Tx, ϕ| : x ∈ BX} ≤ sup{‖T‖‖x‖‖ϕ‖ : x ∈ BX} = ‖T‖‖ϕ‖.

Therefore ‖T ∗‖ ≤ ‖T‖ and T ∗ ∈ B(Y ∗, X∗).In order to prove that ‖T ∗‖ ≥ ‖T‖ fix ε > 0. Let x ∈ BX be such

that ‖Tx‖ ≥ ‖T‖ − ε. If ϕ ∈ Y ∗ is a norming functional for Tx as inCorollary 6.5 then

‖T ∗ϕ‖ ≥ ‖T ∗ϕ(x)‖ = |〈Tx, ϕ〉| = ‖Tx‖ ≥ ‖T‖ − ε.

As ε > 0 was arbitrary, we conclude that ‖T ∗‖ ≥ ‖T‖ and therefore‖T ∗‖ = ‖T‖.

It is straightforward to check that

(αS + βT )∗ = αS∗ + βT ∗

for all S and T and scalars α and β, and this completes the proof. �

32 I. FARAH

7. Weak topologies, Hahn–Banach separation

Definition 7.1. Let X be a vector space and let F be a separatingfamily of seminorms on X. (That is, for all distinct x and y in X thereis m ∈ F such that m(x − y) 6= 0.) The weak topology induced by Fis the weakest topological vector space topology on X with respect towhich all m ∈ F are continuous.

If ϕ is a linear functional on X, then |ϕ| is a seminorm. A weaktopology induced by a set of functionals is defined to be the weaktopology induced by the associated set of seminorms.

One situation in which this definition applies is when the spaces Xand Y are in algebraic duality (Definition 6.11); we can consider theweak topology on X induced by Y and the weak topology on Y inducedby X (see Corollary 8.2).

We should note that Definition 7.1 differs from the usual topolog-ical definition, as the weakest topology that makes all functions in Fcontinuous is often not a TVS topology. One way to define the weaktopology is to first define the neighbourhood basis of zero and thentranslate it to other points.

Lemma 7.2. Suppose F is a separating family of seminorms on X.Then the weak topology induced by X has the sets

U(m, ε) := {x : m(x) < ε},for m ∈ F and ε > 0, as its subbasis.

Suppose X is a TVS and U ⊆ X is an open neighbourhood of 0. LetmU : X → [0,∞) be defined by mU(0) = 0 and

mU(x) := inf{r : r−1x ∈ U}for x 6= 0. Since U is open, for every x ∈ X we have εx ∈ U for a smallenough ε > 0 and therefore mU(x) 0. We need tocheck that mU(x+ y) ≤ mU(x) +mU(y). Choose s > 0 and t > 0 such


that s > mU(x) and t > mU(y), so that s−1x ∈ U and t−1y ∈ U . Then

the convex combination

s

s+ ts−1x+

t

s+ tt−1y =

1

s+ t(x+ y)

belongs to U , and mU(x + y) ≤ s + t. As s and t were arbitrary, wehave mU(x+ y) ≤ mU(x) +mU(y), as required.

(3) As in (2) only the direct implication requires a proof. By (2)mU isa Minkowski functional and as U is balanced we have mU(x) = mU(αx)for every scalar α satisfying |α| = 1, mU(αx) = |α|mU(x) for all x andall α. �

Exercise 7.4. Suppose X is a TVS.

(1) Prove that X is locally convex if and only if its topology is in-duced by a family of seminorms.

(2) Prove that X is normable if and only if its topology is inducedby a finite family of separating seminorms. (Hint: Lemma 1.6— but be careful!)

(3) Prove that the topology on RN (as in Example 3.0.1) is inducedby a separating family of seminorms.

We are now ready (and very pleased) to present the Hahn–BanachTheorem, pt. II. Note that the space X is not assumed to be normed,or even locally convex.

Theorem 7.5 (Hahn–Banach Separation Theorem). Suppose X is areal TVS and W and A are its disjoint convex subsets. Suppose more-over that W is open. Then there exist a continuous linear functionalϕ ∈ X∗ and r ∈ R such that

ϕ(x) < r ≤ ϕ(a)

for all x ∈ W and all a ∈ A.

We shall prove this theorem after stating the complex version; theproofs are very similar.

Theorem 7.6 (Hahn–Banach Separation Theorem, complex version).Suppose X is a complex TVS and W and A are its disjoint convex sub-sets. Suppose moreover that W is open. Then there exist a continuouslinear functional ϕ ∈ X∗ and r ∈ R such that

34 I. FARAH

Proof of the Hahn–Banach Separation Theorem, both versions. We mayassume both W and A are nonempty. Fix b ∈ W and c ∈ A and let

U := −b+W + c− A = {−b+ x+ c− a : x ∈ U, a ∈ A}.Then we have the following.

(i) 0 ∈ b−W and 0 ∈ −c+ A hence 0 ∈ U .(ii) As W is open, so is U .(iii) As both b−W and −c+ A are convex, so is U .

Therefore U satisfies the assumptions of Lemma 7.3 and

mU(x) := inf{s : s−1x ∈ U}is a Minkowski functional such that U = {x : mU(x) < 1}.

As W ∩ A = ∅, a − b /∈ U and therefore mu(a − b) ≥ 1. On thesubspace K(a− b) spanned by a− b let ψ be the functional defined by

ψ(α(a− b)) := α.As a− b /∈ U , we have ψ(x) ≤ mU(x) on K(a− b) ∩ U .

Let us now consider the case when K = R. By Lemma 6.4 we canextend ψ to functional ϕ on X dominated by mU . Fix x ∈ W andy ∈ A. Then

ϕ(x)− ϕ(y) = ϕ(−b+ x+ a− y)− ϕ(aa− b) > 0and therefore r := supx∈W ϕ(x) satisfies r ≤ ϕ(y) for all y ∈ A andϕ(x) < r for all x ∈ W .

If K = C then as in the proof of the Hahn–Banach Extension The-orem consider X as a real space and find ϕ0 satisfying the conclusion.Then

ϕ(x) := ϕ(x)− iϕ(ix)is continuous and it satisfies


Lemma 8.1. Suppose X is a vector space, n ≥ 1, and ψ and ϕj, forj ≤ n, are linear functionals on X. The following are equivalent.

(1) ψ is a linear combination of ϕj, for J ≤ n.(2) There exists a constant K 0 such that

{x : |ψ(x)| < ε} ⊇⋂j≤n

{x||ϕj(x)| < 1}.

Therefore |ψ| ≤ ε−1 maxj≤n |ϕ|, and Lemma 8.1 implies that ψ ∈span{ϕj : j ≤ n}. Since Y is a vector space, this proves X∗ ⊆ Y . �

A local way to state the previous corollary is worth noting.

Corollary 8.3. If X and Y are in algebraic duality, Y separates thepoints of X, and ψ is a linear functional on X. Then ψ is continuouswith respect to the weak topology induced by Y if if and only if ψ ∈ Y .Corollary 8.4. Let X be a normed space. The dual space of X∗, whenX∗ is considered as a topological vector space with respect to its weak∗-topology, is naturally isomorphic to X. �

We shall use two standard results from general topology to proveTheorem 8.7, the main result of today’s lecture.

Theorem 8.5. If K is a compact Hausdorff space and L is a closedsubspace of K, then L is compact (and certainly Hausdorff) in thesubspace topology. �

36 I. FARAH

If Kγ, for γ ∈ J , is a nonempty family of compact spaces thenon the Cartesian product

∏γKγ one defines the product topology as

follows. The basic open sets are defined by a finite F ⊆ J and opensets Uλ ⊆ Kλ for λ ∈ F . Given such F and Ū , let

W (F, Ū) = {x ∈∏γ

Kγ : x(λ) ∈ Uλ for all λ ∈ F}.

It is straightforward to check that if all Kγ are Hasudorff so is theirproduct. The following is much deeper (and actually equivalent to theAxiom of Choice).

Theorem 8.6 (Tychonoff’s Theorem). Suppose Kγ, for γ ∈ J , is anonempty family of compact topological spaces. Then the product space∏

γKγ is also compact. �

Theorem 8.7 (Banach–Alaoglu). Suppose X is a TVS and U is aneighbourhood of 0 in X. Then the polar of U ,

Z := {ϕ ∈ X∗ : supx∈U|ϕ(x)| ≤ 1}

is weak*-compact.

Before starting the proof of this theorem we state its most importantconsequence:

Corollary 8.8. If X is a normed space then the unit ball of X∗ isweak*-compact. �

Proof of Theorem 8.7. For x ∈ U let Kx := {z ∈ K : |z| ≤ 1} with theusual topology. The space K :=

∏x∈U Kx is by Tychonoff’s theorem

compact (and it is Hausdorff).Let F : Z → K be defined by

F (ϕ)(x) = ϕ(x)

for x ∈ U . Then F is an injection since two linear functionals differ ifftheir restrictions to an open neighbourhood of 0 differ. Since every ϕis continuous on Z by the definition, and since K is taken with respectto the product topology, F is continuous. For every functional ϕ ∈ X∗the set {x ∈ X : |ϕ(x)| < 1} is nonempty (it contains 0) and open.This implies that F is an open map from Z into F [Z].

Being continuous, open, and an injection, F is a homeomorphismbetween Z and F [Z].

By Theorem 8.5 it suffices to prove that F [Z] is a closed subset of K.Choose ψ ∈ K \ F (Z). It is a map ψ : U → K, and as it is not in

the range of F (Z) it fails to be the restriction of a linear map to U .Therefore (at least) one of the following two possibilities applies.


(i) There are x and y in U such that x+ y ∈ U but ψ(x) +ψ(y) 6=ψ(x+ y).

(i) There are x ∈ U and α ∈ K such that αx ∈ K but ψ(αx) 6=αψ(x).

Suppose (i) holds and let ε := |ψ(x) + ψ(y)− psi(x+ y)|/3. The set

W := {ϕ ∈ K : maxz∈{x,y,x+y}

|ϕ(z)− ψ(z)| < ε}

is an open subset of K, and any ϕ ∈ W satisfies ϕ(x+y) 6= ϕ(x)+ϕ(y).Now suppose (ii) holds. Let ε := |αψ(x)− ψ(αx)|/2 and let

W := {ϕ ∈ K : max{|α||ϕ(x)− ψ(x)|, |ϕ(αx)− ψ(αx)} < ε}.

As in (i), any ϕ ∈ W satisfies ϕ(αx) 6= αϕ(x)In either case, ψ has an open neighbourhood disjoint from F (Z).

Since ψ was arbitrary, this proves that F (Z) is closed and the theoremfollows. �

Remark 8.9. The ‘right’ way to complete the proof of Theorem 8.7 isto observe that conditions in (i) and (ii) are both ‘open.’ More precisely,if x and y are fixed then the set {η ∈ K : η(x) + η(y) = η(x + y)} isa closed subset of K (this is really what the proof of case (i) shows).Similarly, for fixed x and α the set {η ∈ K : η(αx) = αη(x)} is closed,as the proof of (ii) shows.)

Exercise 8.10. Suppose X is an infinite-dimensional normed space.Prove that X∗ is infinite-dimensional.

Corollary 8.11. Suppose X is a normed space such that its dual X∗

is infinite-dimensional. Then the weak*-topology on X∗ is weaker thanits norm topology.

Proof. By Theorem 8.7 the unit ball BX∗

is weak*-compact but by ahomework question it is not compact in norm. �

Proposition 8.12. Suppose X is a normed space. and Z is a weak*-closed subspace of X∗. Then for every ϕ ∈ X∗ \Z (if any) there existsx ∈ Z⊥ such that ϕ(x) 6= 0.

Proof. As Z is weak*-closed, there exists a weak*-open neighbourhoodU of ϕ disjoint from Z. As the weak*-topology is locally convex (everyweak topology is locally convex), we can choose U to be convex. Bythe Hahn–Banach separation theorem there exists a weak*-continuousfunctional ζ on X∗ that annihilates Z and is such that ζ(ϕ) 6= 0. Asζ is weak*-continuous, there is x ∈ X such that ζ(ψ) = ψ(x) for allψ ∈ X∗, and in particular x is as required. �

38 I. FARAH

Corollary 8.13. Let X be a normed space. A subset Z of X∗ is equalto Y ∗ for some Y ⊆ X if and only if Z is a weak*-closed subspace.

Proof. Every set of the form Y ⊥ is a subspace and weak*-closed (check!).What we need to check is that Z is a weak*-closed subspace of X∗ ifand only if Z = (Z⊥)⊥, because then Y := Z⊥ is as required.

Suppose Z is a weak*-closed subspace of X∗ and ψ /∈ Z. By Proposi-tion 8.12 there exists x ∈ Z⊥ such that ψ(x) 6= 0. Therefore, ψ /∈ (Z⊥)⊥and the conclusion follows. �

Proposition 8.14. Suppose X is a normed space and Z is a subspaceof X∗. Suppose Z is weak*-closed and separates the points of X. ThenZ = X∗.

9. Inner product, Hilbert space

Definition 9.1. If H is a vector space a sesquilinear form on H is a(·|·) : H2 → K such that

(1) It is sesquilinear : Linear in the first variable and conjugate-linear in the second.

(2) (ξ|η) = (η|ξ),(3) (ξ|ξ) ≥ 0.

If in addition it satisfies

(4) (ξ|ξ) = 0 if and only if ξ = 0then (·|·) is a pre-inner product.

Lemma 9.2. If (·|·) is a pre-inner product on H, then the Cauchy–Schwarz inequality holds:

|(x|y)|2 ≤ (x|x)(y|y)for all x and y in H. Moreover, the equality holds if and only if x andy are linearly dependent. �

If (·|·) is a pre-inner product on H, define‖x‖2 = (x|x)1/2.

Lemma 9.3. The function ‖cdot‖2 is a norm on H.

Proof. We have ‖x‖2 = 0 if and only if x = 0 and ‖ax‖2 = |a|‖x‖2for all x ∈ H and all scalars a. The triangle inequality for ‖ · ‖2 is astraightforward consequence of the Cauchy–Schwarz inequality:

(x+y|x+y) = (x|x)+2


Example 9.4. Here are two examples of pre-inner products.

(1) With H = `2: If ξ and η belong to `2 then let

(ξ|η) :=∑j

ξj η̄j.

(2) With H = C([0, 1]), let

(f |g) =∫ 1

0

f(x)g(x) dx

9.1. Polarization, parallelogram and Pythagoras. Notably, in aHilbert space the inner product can be defined directly from the normby the polarization identities :

(ξ|η) = 14

(‖ξ + η‖22 − ‖ξ − η‖22)

if K = R and

(ξ|η) = 14

3∑j=0

ij‖ξ + ijη‖2.

Hilbert space also satisfies the parallelogram identity

2(‖ξ‖22 + ‖η‖22) = ‖ξ + η‖22 + ‖ξ − η‖22.

Finally, if (ξ|η) = 0 we say that ξ and η are orthogonal. In this casewe have the Pythagorean equality

‖ξ + η‖22 = ‖ξ‖2 + ‖η‖22.

If H is a vector space equipped with function (·|·) satisfying the aboveproperties then we say that H is a pre-Hilbert space. One can think ofH as a normed space (with respect to ‖ξ‖2 := (ξ|ξ)1/2) in which a pre-inner product is defined by the polarization identity. The completionof H (Proposition 3.22) is a Banach space H̃. The pre-inner productis uniformly continuous with respect to (·|·) and it can be continuouslyextended to H̃. This completion is a Hilbert space.

Lemma 9.5. Suppose H is a Hilbert space.

(1) If C is a closed convex subset of H then there exists ξ ∈ C suchthat ‖ξ‖ = infη∈C ‖η‖.

(2) If C is a closed convex subset of H and ζ ∈ H then there existsξ ∈ C such that ‖ζ − ξ‖ = infη∈C ‖ζ − η‖.

(3) The vector ξ as in (1) and (2) is unique.

40 I. FARAH

Proof. (1) Let r = infη∈C ‖η|. Choose a sequence ξn, for n ∈ N, suchthat ‖ξn‖ < r − 1/n. For m and n we have

‖ξm − ξn‖2 = 2 = 2(‖ξm‖22 + ‖ξn‖22)− ‖21

2(ξm + ξn)‖2

≤ 2((r + 1/m)2 + (r + 1/n)2)− 4r2

which can be made arbitrarily small. Therefore (ξm) is a Cauchysequence. Since C is closed, vector ξ := limm ξm belongs to C and‖ξ‖ = limm ‖ξm‖ = r.

(2) Apply (1) to C − ζ.(3) It suffices to prove uniqueness in (1). Suppose ξ and η are both

in C and ‖ξ‖2 = ‖η‖2 = r (r is as in (1)). Then ζ := 12(ξ + η) ∈ C and

4r2 = ‖2ζ‖22 = 2(‖ξ‖22 + ‖η‖22)− ‖ξ − η‖22 = 4r2 − ‖ξ − η‖22and therefore ‖ξ − η‖2 = 0. �

Proposition 9.6. Suppose H is a Hilbert space and K is a closedsubspace of H.

(1) Then K⊥ = {x ∈ H : (x|y) = 0 for all y ∈ K} is a closedsubspace of H.

(2) For every x ∈ H there are the unique x0 ∈ K and x1 ∈ K⊥such that x = x0 + x1.

(3) We have ‖x‖22 = ‖x0‖22 + ‖x+ 1‖2.(4) The function pK : H → K defined by pK(x) = y where y ∈ K

is such that ‖x− y‖2 = infy∈K ‖x− y‖ is a linear and bounded,with ‖pK‖ = 1.

Proof. For (1), check that K⊥ is a linear subspace. It is norm-closedby the continuity of the inner product.

(2) Fix x. For y ∈ K, we have ‖x− y‖ = infy∈K ‖x− y‖ if and onlyif x− y ∈ K⊥ (draw a picture!).

By Lemma 9.5, there is a unique x0 ∈ K such that ‖x − x0‖ =infy∈K ‖x − y‖. Let x1 = x − x0. By the previous line, x1 ∈ K⊥ andthe uniqueness follows by the uniqueness of x0.

(3) Computation.(4) By (3), for x ∈ H we have ‖pK(x)‖ ≤ ‖x‖. Also pK(x) = x if

and only if x ∈ K. The linearity is clear. �

9.2. 12

+ 12

= 1: Hilbert space. We say that ξ is orthogonal to η if(ξ|η) = 0.

Exercise 9.7. If K is 1-dimensional and ξ ∈ K is a unit vector, thenpK(η) = (η|ξ)ξ.


Exercise 9.8. If H1 and H2 are Hilbert spaces equip the vector spaceH1 ⊕H2 (H1 ⊕2 H2 is a more precise notation) with the norm

(ξ, η) := (‖ξ‖22 + ‖η‖22)1/2.(1) Check that this is a Hilbert space.(2) Check that H1 is isometric to {(ξ, 0) : ξ ∈ H1} and H2 is iso-

metric to {(0, ξ) : ξ ∈ H2}.(3) Generalize this to infinite sums of Hilbert spaces (cf. §3.1.2).

Exercise 9.9. Suppose H is a Hilbert space and K is its closed sub-space.

(1) K⊥ := ker(pK) is a closed subspace such that the map ξ 7→(pK(ξ), ξ − pK(ξ)) is an isometric isomorphism between H andK ⊕K⊥.

(2) If L is a closed subspace of K then L = K⊥ if and only ifpL = 1− pK.

(3) (K⊥)⊥ = K.

Proof. The kernel of a bounded linear operator is always a closed sub-space, and the fact that this is an isometry follows from the Pythagoreanequality. �

Exercise 9.10. If K is a closed subspace of H, prove that H/K (see§3.1.3) is Hilbert space isometrically isomorphic to K⊥ via

ξ +K 7→ ξ − pK(ξ).

Definition 9.11. An indexed set {ej : j ∈ J} in H is orthonormal if(ej|ek) = δjk

for all j, k.

Exercise 9.12. If {ej : j ∈ J} is an orthonormal set and K is theclosure of its linear span, then

pK(η) =∑j∈J

(η|ej)ej

Proposition 9.13. (1) Every orthonormal set in a Hilbert spacecan be extended to a maximal orthonormal set.

(2) If {ej : j ∈ J} is a maximal orthonormal set, then for everyξ ∈ H the set {j ∈ J : (ξ|ej) 6= 0} is countable6 and

ξ =∑j

(ξ|ej)ej, ‖ξ‖2 =∑j

|(ξ|ej)|2,

6This does not exclude the possibility of the set being finite.

42 I. FARAH

Proof. In a separable Hilbert space (1) follows by using the Gram-Schmidt process. In general we use Zorn’s Lemma. Let P be the setof all orthonormal sets in H extending the given set, ordered by theextension. That is, (ej : j ∈ J) extends (fj : j ∈ I) if J ⊇ I andej = fj for all j ∈ I. If C is a totally ordered subset of P let

JC :=⋃{J : (ej : j ∈ J) ∈ C}.

For j ∈ J , the vector ej is uniquely defined. For all i and j in J thereexists a single element of C such that both ei and ej are in it, andtherefore (ei|ej) = δij. We claim that (ej : j ∈ JC) is maximal. Let Kbe the closed linear span of (ej : j ∈ J). This set is not maximal iffK⊥ 6= 0. If K⊥ 6= 0 then we can choose a unit vector ξ ∈ K⊥. Then(ej|ξ) = 0 for all j.

(2) This follows easily by Exercise 9.12. �

9.3. Separability. A TVS is separable if it has a countable dense sub-space. If D ⊆ X then span(D) is the set of all finite linear combinationsof elements of D. Equivalently, it is the smallest linear subspace of Xthat includes D.

Exercise 9.14. Prove that the following are equivalent for a real TVS X.

(1) X is separable(2) there are vectors xn, for n ∈ N, such that span{xn : n ∈ N} is

dense in X(3) There exists a countable dense Z ⊆ X that is a vector subspace

over Q.

Spaces with a countable basis are said to be second countable. Thisproperty implies separability.

Exercise 9.15. Prove that a metrizable space is separable if and onlyif it is second countable.

Exercise 9.16. Assume X is locally compact and non-compact Haus-dorff space and let X ∪ {∞} be its one-point compactification.

(1) Prove that C0(X) is isometric to a subspace of C(X ∪ {∞}).(2) Give an example showing that C00(X) is a proper subspace of

C0(X) for some X.(3) Prove that c00 and c0 are special cases of C00(X) and C0(X) for

certain locally compact space X.(4) Prove that Lp(X,µ) is separable if X is locally compact metriz-

able and µ is a σ-finite Borel measure on X.(Hint: By Lemma 11.2 step functions with finite measure

support and range in Q + iQ are dense. The regularity of µ


implies that simple functions that are constant on the elementsof the basis of X are dense in Lp. The assumption on X impliesthat X is second countable, i.e. has a countable basis.)

(5) Prove that L∞(X,µ) is separable if and only if there exists anull set Y ⊆ X such that X \ Y is finite.

Exercise 9.17. (1) If 1 < p, q < ∞ and 1p

+ 1q

= 1 then f ∈ Lp iff|f |q/p ∈ Lq.

(2) If f : X → K is measurable then there exists a measurable func-tion sgn(f) : X → K such that f sgn(f) = |f | and | sgn(f)| = 1 a.e.

Theorem 9.18. All separable, infinite-dimensional Hilbert spaces overK are isometrically isomorphic.

Proof. Suppose H1 and H2 are Hilbert spaces over K with orthonormalbases E1 and E2 of the same cardinality. Fix a bijection f : E1 → E2.Extend f to a linear map f̃ : span(E1)→ span(E2). Since E1 and E2 areorthonormal, f̃ preserves the inner product. It is therefore an isometry,and it can be continuously extended to a linear map g : H1 → H2. Thecontinuous linear extension of f̃−1 is easily checked to be an isometricinverse of g, and therefore g is as required.

Since every Hilbert space H has an orthonormal basis E , it remainsto check that H has a countably infinite orthonormal basis if and onlyif it is separable. Since K has a countable dense sub-field K0 (Q in thereal case and Q + iQ in the complex case), if E is countable then theset of all K0-linear combinations of the elements of E is a countable set.If its closure is a proper subset, L, of H, then L⊥ 6= {0} contradictingthe maximality of E .

Conversely, suppose H is separable and let D be a countable densesubset of H. For every x ∈ D the set supp(x) = {y ∈ E : (x|y) 6= 0}is countable. (Otherwise, there is n ≥ 1 and an uncountable set ofy ∈ E such that |(x|y)| ≥ 1/n. But for every x ∈ H we have ‖x‖2 =supy∈E |(x|y)|2, and therefore only finitely many of the coefficients (x|y)can be nonzero.)

Therefore E0 =⋃x∈D supp(x) is a countable subset of E . It suffices

to prove that E0 = E . If y ∈ E \ E0, then (y|x) = 0 for all x ∈ D; since‖y‖ = 1, this contradicts the assumption that D was dense in H. �

A maximal orthonormal set in a Hilbert space is orthonormal ba-sis (or just basis). Many Banach spaces do not have a basis. Thedimension of a Banach space is the cardinality of its basis.

44 I. FARAH

Theorem 9.19. For every ϕ ∈ H∗ there exists a unique ξ = ξ(ϕ) ∈ Hsuch that

ϕ(η) = (η|ξ)for all η. The map ϕ 7→ ξ(ϕ) is a conjugate-linear isometry from H∗onto H.

Proof. If ϕ = 0 then ξ = 0 clearly works. We may therefore assumeϕ 6= 0. Let L = ker(ϕ). This is a closed subspace of H and L⊥ isone-dimensional.

Fix a unit vector ξ0 ∈ L⊥ and letξ := ϕ(ξ0)ξ0.

We claim that ξ is as required.For η ∈ H we have

ϕ(η) = ϕ(η − pL(η) + pL(η)) = ϕ(η − pL(η)) = ϕ(pL⊥(η)).

Fix η ∈ L⊥. Since L⊥ = span(ξ0), η = (η|ξ)ξ. Thenϕ(η) = ϕ((η|ξ)ξ) = (η|ξ)ϕ(ξ) = (η|ξ).

This proves that ϕ(η) = (η|ξ(ϕ)) for every η ∈ H. A computationproves that the map ϕ 7→ ξ(ϕ) is a conjugate linear map from H∗ intoH with a trivial kernel. Since every ξ ∈ H defines a linear functionalη 7→ (η|ξ), this map is onto.

It remains to prove that it is an isometry, that is ‖ϕ‖ = ‖ξϕ‖2. Wemay assume ϕ 6= 0; then we have

‖ϕ‖ = sup‖η‖2=1

|ϕ(η)| = sup‖η‖2=1

|(η|ξϕ)| ≤ ‖ξϕ‖2.

Since ϕ 6= 0 we have ξϕ 6= 0 and η := ‖ξϕ‖−1ξϕ is a unit vector. Then‖ϕ‖ ≥ |ϕ(η)| = |(η|ξϕ)| = ‖ξϕ‖. This completes the proof. �

10. Lp spaces

The following discussion belongs to the grey area between measuretheory and functional analysis. Although these notes cover the topic inthe full generality of Lp spaces associated to arbitrary measure spaces,in class (and the exam) we will be covering only the sequence space `p.It corresponds to the counting measure on N.

The material is taken mostly from [7, §1.3]. An another great refer-ence for measure theory is [3]. Fix 1 ≤ p < ∞ and (if p > 1) q suchthat

1

p+

1

q= 1.

In particular 1 < p and 1 < q.


Lemma 10.1 (A silly reference lemma). Suppose 1 ≤ p ≤ ∞ and 1 ≤q ≤ ∞. Then the following are equivalent (using the usual conventionssuch as 1/∞ = 0).

(1) 1p

+ 1q

= 1

(2) q = pp−1

(3) q = 1 + qp

(4) 1− 1q

= 1p

Now let X be a locally compact7 Hausdorff space.8 Let Σ be a σ-algebra of subsets of X containing all open subsets of X. The field ofBorel subsets of X is the smallest such σ-algebra by definition.

Fix a function µ : Σ→ [0,∞] satisfying(1) µ(K)

46 I. FARAH

There are other important examples of measures but these will sufficefor our purposes.

Recall that Σ is a σ-algebra of subsets of X. A function f : X → Kis Σ-measurable (or just measurable if Σ is clear from the context) iff−1(U) ∈ Σ for every open U ⊆ K. A function f is Borel if it is Borel-measurable. (Conveniently enough, this is equivalent to the graph off being Borel.)

By basic results from measure theory, every measurable set is equalto a Borel—actually Gδ—set modulo a null set, and every measurablefunction is continuous off a set of arbitrarily small measure (Luzin’stheorem). Either of these results implies that for every measurablefunction f : X → K there exists a Borel-measurable function equal tof almost everywhere (abbreviated as a.e.).

For measurable functions f and g on X we write

f ≤ g

if f(t) ≤ g(t) for almost all t.

Definition 10.3. Let Lp(X,µ) (or just Lp if X,µ is fixed; and we shallfix it throughout) be defined as follows.

Lp(X,µ) := {f : X → K : f is measurable and∫X

|f |p dµ


Lemma 10.6. If a > 0, b > 0, 1 < p, q 0, and ‖g‖q > 0.By renormalizing (i.e. replacing f with ‖f‖−1p f and g with ‖g‖−1q g)we may assume ‖f‖p = 1 = ‖g‖q. For all x such that f(x)g(x) 6= 0Lemma 10.6 implies

|f(x)g(x)| ≤ 1p|f(x)|p + 1

q|g(x)|q.

By integrating we get ‖fg‖1 =∫|fg| dµ ≤ 1

p‖f‖p + 1q‖g‖q = 1. �

Note that the case when p = q = 2 of Hölder’s inequality gives theCauchy inequality.

Lemma 10.8 (Minkowski’s inequality). Suppose 1 ≤ p < ∞ and fand g belong to Lp(X,µ). Then f + g ∈ Lp(X,µ) and

‖f + g‖p ≤ ‖f‖p + ‖g‖p.

Proof. Since |f(x) + g(x)| ≤ 2 max(|f(x)|, |g(x)|) for all x, we have|f + g| ∈ Lp(X,µ).

The inequality is evident p = 1, so we may assume 1 < p 1, we can write (using the triangle inequality)

‖f + g‖pp =∫|f + g|p dµ =

∫|f + g| |f + g|p−1 dµ

≤∫|f | |f + g|p−1 dµ+

∫|g| |f + g|p−1 dµ.

By the Hölder’s inequality, and using (p− 1)q = p,∫|f | |f + g|p−1 dµ ≤ (

∫|f |p)1/p(

∫|f + g|(p−1)q)1/q = ‖f‖p ‖f + g‖p/qp .

48 I. FARAH

Similarly,∫|g||f + g|p−1 dµ ≤ ‖g‖p ‖f + g‖p/qp , and we have

‖f + g‖pp ≤ (‖f‖p + ‖g‖p)‖f + g‖p/qp .

Dividing both sides with ‖f + g‖p/qp and using p− p/q = 1, we obtainthe desired inequality. �

Proposition 10.9. For all 1 ≤ p


(Since f is compactly supported we have that ‖f‖p < ∞.) (We havealready seen the special cases when p = 1 and p = 2, defined as normson C0(X).)

A function θ from a linear space into R is convex if for all x and yand 0 < t < 1 we have

θ(tx+ (1− t)y) ≥ tθ(x) + (1− t)θ(y).

Fix r > 0. Then the function R 3 x 7→ rx is convex (draw a picture!).Hence each Lp is a seminormed space. Note that ‖f‖p = 0 does

not necessarily imply f = 0; it only implies f = 0 a.e. We can taketwo routes to define the Lp spaces, analogous to Example 10.10 andExample 10.11.

A special case, when X = N and µ is the counting measure, are thesequence spaces `p for 1 ≤ p

50 I. FARAH

11. Dual spaces of Lp spaces

As in the previous lecture, let µ be a regular Radon measure on alocally compact space X. For a measurable f : X → R define

ess sup f = sup{r : µ(f−1(r,∞)) > 0}.For a continuous f we have ess sup(f) = sup(f) but this is not truefor measurable functions. For example, the characteristic function ofthe rational numbers 1Q satisfies ess sup 1Q = 0 < sup(1Q) = 1.

Let

L∞(X,µ) := {f : X → K : f is measurable and ess sup(|f |) 0 partition the range of gmn intofinitely many sets of diameter < ε/µ(supp(gmn).

9The sequence space version of this lemma states that c00 is dense in `p for all1 ≤ p


Enumerate these sets as J1, . . . , Jl. For j ≤ l let Aj := g−1(Jj) andchoose λj ∈ Jj. Then h :=

∑lj=1 λjχAj is a simple function with finite

measure support, and since |h− gmn| ≤ ε on Xn, we have

‖h− gmn‖p ≤ µ(Xn)1/pε.

The above proof shows that every positive function is a limit of positivesimple functions with finite measure support. Since every f ∈ Lpcan be written as a linear combination of four positive functions, theconclusion follows. �

Exercise 11.3. Prove that simple functions are dense in L∞.Prove that simple functions with finite measure support are dense in

L∞ if and only if µ is a finite measure.

12. Dual spaces of Lp spaces

12.1. An isometric embedding of Lq into (Lp)∗. Fix finite p and qsuch that 1

p+ 1

q= 1. If f ∈ Lq then for every g ∈ Lp Hölder’s inequality

implies that ‖fg‖1 ≤ ‖f‖q‖g‖p

52 I. FARAH

Exercise 12.1. Suppose that f : X → K is a measurable function andr < ∞ is such that µ{x : |f(x)| ≥ r} > 0. Prove that there existsg ∈ L1 such that ‖g‖1 = 1 and

∫fg dµ ≥ r.

(Hint: For every measurable set A of positive measure the functiong = µ(A)−11A is in the unit sphere of L1.)

Because of the following Theorem (and its general case, Theorem 12.4below) we routinely identify `∗p with `q (and L

∗p with Lq).

Theorem 12.2. For 1 ≤ p < ∞ and q such that 1p

+ 1q

= 1 we

have that the isometric embedding of `q into `∗p given by the algebraic

duality between `p and `q is a surjection. Therefore (`p)∗ is isometrically

isomorphic to `q.

Proof. For ā ∈ `p and b̄ ∈ `q let

(ā|b̄) :=∑n

anbn.

By Hölder’s inequality, the sum on the right-hand side is not greaterthan ‖ā‖p‖b̄‖q and therefore finite. This is clearly a bilinear map. Forb̄ ∈ `q

ϕb̄(ā) := (ā|b̄)is an element of `∗p.

Claim 12.3. For b̄ ∈ `q we have ‖ϕb̄‖ = ‖b̄‖q.

Proof. Fix b̄. We have ≤ by Hölder’s inequality.We prove ≥ in case when K = R first. Let an := bq/pn if bn ≥ 0 and

an := −bq/pn if bn < 0. Then∑

n |an|p =∑

n |bn|q = ‖b̄‖qq, in particularā ∈ `p and ‖ā‖p = ‖b̄‖q/pq . Let c̄ = ‖b̄‖−q/pq ā. Then c̄ ∈ `p and ‖c̄‖p = 1.We have (since 1 + q

p= q and q − q

p= 1

p)

ϕb̄(c̄) = ‖b̄‖−q/pq∑n

(±|bn|q/p)(∓|bn|) = ‖b̄‖−q/pq∑n

|bn|q = 1

For the complex case, define an := 0 if bn = 0 and an :=bnbn|bn|q/p

otherwise. A calculation analogous to the above shows that ā ∈ `p,ϕb̄(ā) =

∑n |bn|q and ‖ϕb̄‖ = ‖b̄‖q. �

Now fix ϕ ∈ `∗p. In order to find b̄ ∈ `q such that ϕb̄ = ϕ, letbn := ϕ(en). We need to prove that b̄ ∈ `q. To this end, for n ≥ 1consider c(n) ∈ c00 such that c(n)j = bj for j ≤ n and c(n)j = 0if j > n. Then ϕ(ā) = (ā|c̄(n)) if supp(ā) ⊆ {1, . . . , n}. Therefore‖c̄(n)‖q ≤ ‖ϕ‖ for all n. But this implies that b̄ ∈ `q and ‖barb‖q ≤ supn ‖c̄(n)‖q ≤ ‖ϕ‖. �


Theorem 12.4. For 1 ≤ p < ∞ and q such that 1p

+ 1q

= 1 we have

that (Lp)∗ is isometrically isomorphic to Lq.

Proof. The paragraph before the theorem shows that the map Lq 3f 7→ ϕf ∈ (Lp)∗ is an isometry (and this map is clearly linear). Ittherefore remains to prove that this map is surjective.

We shall firs

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