MATH541 Functional Analysis, Spring 2021

Lectures delivered by Marius Junge

Notes by Xinran Yu

April 4, 2021

Warning: I’m typing the notes slowly. Given that lecture recordings are not uploaded

regularly, you can expect no updates for weeks.

The first several lectures contains a review on the materials from Real Analysis, which

I will omit in this notes.

Contents

1 Baire’s Categroy Theorem 20210125 3

2 Baire’s Categroy Theorem Cont. 20210127 3

3 Basic Banach Space Theory 20210129 4

4 Basic Banach Space Theory Cont. 20210201 6

5 Hahn-Banach Theorem 20210203 9

6 Hahn-Banach Theorem Cont. 20210205 10

7 Vector Space 20210208 13

1

8 Locally Convex Topological Vector Spaces 20210210 16

9 Hahn-Banach Separation Theorem 20210212 18

10 Weak Topology 20210215 21

11 Weak Topology Cont. 20210219 23

12 Alaoglu’s Theorem 20210222 25

13 Reflexive Spaces 20210224 27

14 Reflexive Spaces Cont. 20210226 30

15 Riesz-Thorin Theorem 20210301 33

16 Clarkson’s Inequality 20210303 36

17 Uniform convexity of Lp 20210305 39

18 Uniform Boundedness and Open Mapping 20210308 41

19 ‖Pn‖ is Unbounded 20210310 44

20 Krein–Milman Theorem 20210312 49

21 Krein–Milman Theorem Cont. 20210315 49

2

1 Baire’s Categroy Theorem 20210125

Ref: A Course in Functional Analysis, John B. Conway, 1985

1. Metric space

2. Chicago suburb distance Rb compact = closed and bounded no longer true

3. Cauchy sequence, completeness

4. Open, closed ball

5. Nowhere dense set, dense set, closure, interior.

Y is nowhere dense ⇐⇒ Y C is open and dense.

Theorem 1.1 (Baire’s theorem). In a complete metric space, the countable

union of nowhere dense sets is again nowhere dense.

Lemma 1.2. The intersection of open dense sets is again open dense.

Using the above lemma + induction to prove Baire’s theorem.

Dense, nowhere dense, somewhere dense. Stack Exchange Theorem in notes: count-

able intersection of open dense is dense, then countable union does not have interior

points. Need X complete metric space, so that the limit point is in X.

2 Baire’s Categroy Theorem Cont. 20210127

Last time: open set, closed sets, theorem: let (X, d) be a complete metric space, On

open dense, then ∩nOn is dense.

1. intuition dense set ∼=, taking away a countable set of points

2. proof idea completeness → geometric series.

3. Use Baire’s theorem to show no function f : [0, 1]→ R continuous exactly at Q4. proof hard works is to find complete metric space and makes the theorem work

5. Normed space. A normed space is complete if absolute convergent sequences

are convergent. Banach space.

3

6. isometry

7. ‖f(x)‖C(K) = supk∈K|f(k)|.

Question 2.1. Let Cb(R) be the set of continuous and bounded function. Is Cb(R) =

C(K) for some compact K? — Yes.

Want to do: Start with Banach space, create new ones.

Lemma 2.2. Let T : X → Y be a linear map between normed spaces. TFRE

1. T is continuous.

2. T is continuous at 0.

3. ‖T‖op = sup‖x‖≤1

‖Tx‖ is finite.

4. T is Lipschitz.

Homogeneity, duality

Lemma 2.3. Let X be a normed sapce and Y be Banach. Then the vector

space L(X, Y ) with the norm ‖ ‖op becomes a Banach space.

L(normed,Banach) is Banach.

Corollary 2.4. X∗ = L(X,C) is Banach.

3 Basic Banach Space Theory 20210129

proof of Lemma 2.3. Step 1. (Tn) Cauchy implies (Tn(xk)) Cauchy.

4

Step 2. Let f(x) := limTn(x). Prove lim sup ‖Tn(x)− f(x)‖ = 0.

‖Tn − T‖ = ‖Tn − limTm‖ = lim ‖Tn − Tm‖≤ lim sup

m,n≥N‖Tn − Tm‖ < ε

‖Tn − T‖ < ε implies ‖Tn(x)− T (x)‖ < ε, and so lim sup ‖Tn(x)− f(x)‖ = 0.

Step 3. f is bounded, and Tn → f .

Corollary 3.1. X Banach, then L(X,X) = L(X) is Banach algebra.

Definition 3.2. A Banach algebra is a Banach space (A), ‖ ‖ together with a

product · : A×A → A, with ‖ab‖ ≤ ‖a‖‖b‖.1. closed subset of Banach is Banach.

2. K(X, Y ) := T : X → Y | T (BX) compact is closed .

3. In finite dimension, linear bounded T is compact.

Definition 3.3 (Totally bounded).

∀ε,∃N s.t. Y ⊂N⋃j=1

B(xj, ε)

This is equivalent to relatively compact. Ref

Theorem 3.4.

K(H,H)∗∗ = B(H,H)

(We’ll this theorem later.)

5

Theorem 3.5.

∃ι : X → X∗∗; ι(x)(f) = f(x), with f : X → K1. ι is an isometry.

2. ι(x) is the completion of X.

Part 1 follows from Hahn-Banach.

Definition 3.6. (X, d) is a metric space. A completion (Y, d′) is given by

1. ι : X → Y is an isometry.

2. ι(X) is dense.

3. (Y, d′) is complete.

Completion is unique.

4 Basic Banach Space Theory Cont. 20210201

Completion problem: see Theorem 3.5

proof of Theorem 3.5.

Claim 4.1. ‖ι(x)‖X∗∗ ≤ ‖x‖X .

Note that

‖ι(x)‖X∗∗ = sup‖f(x)‖X∗≤1

|ι(x)(f)| (by definition)

= sup‖f(x)‖X∗≤1

|f(x)| (ι inclusion)

≤ sup‖f‖X∗≤1

‖x‖ ≤ ‖x‖.

By definition ‖f‖X∗ ≤ 1 ⇐⇒ |f(x)| ≤ ‖x‖.

For a normed space the completion achieves in X∗∗.

6

Banach space

Lemma 4.2. Cb(x, x0) is a Banach space.

Cb(x, x0) = f : X → R | continuous and ∃C, |f(x)| ≤ Cd(x, x0) .

Norm: ‖f‖ = supx|f(x)|d(x,x0)

.

An embedding isometry ι : X → Cb(X)∗; i(x)(f) = f(x). Hint: use evaluation map

sup‖f‖≤1

|f(x)− f(x0)| = d(x, x0).

Distance attaining function is f(x) = d(x, x0), where x 6= x0.

Theorem 4.3 (Hahn-Banach Extension). Given a vector space X, a sublinear

map q : X → R s.t.

q(x+ y) ≤ q(x) + q(y) (subadditive) and q(sx) = sq(x), s > 0.

Let Y ⊂ X and f : Y → R linear, with f ≤ q, then ∃F : X → R linear F ≤ q

and F |Y = f .

warning This theorem is completely algebraic. There is no topology.

Lemma 4.4. We can always add an extra dimension.

Proof. Step 1. Y ⊂ X = y + tx0 | t ∈ R . Candidates for F (extend 1-dim):

F (y + tx0) = F (y) + tF (x0) = f(y) + ta0 for some a0. What is a0? trick

F (y + tx0) ≤ q(y + tx0)

F (y − tx0) ≤ q(y − tx0)=⇒ f(y) + ta0 ≤ q(y + tx0)

f(y)− sa0 ≤ q(y − sx0)

7

=⇒ a0 ≤q(y + tx0)− f(y)

t, t > 0

a0 ≥f(y)− q(y − sx0)

s, s > 0

=⇒ a0 ≤ infq(y + tx0)− f(y)

t, t > 0

a0 ≥ supf(y)− q(y − sx0)

s, s > 0

Check the sup is less than inf:

f(y)− q(y − sx0)s

≤ q(z + tx0)− f(z)

t

⇐⇒ f(y)t− q(y − sx0)t ≤ q(z + tx0)s− f(z)s

f(y)t+ f(z)s ≤ q(z + tx0)s+ q(y − sx0)tf(yt+ sz) ≤ q(yt+ tsx0 − tsx0 + sz)

≤ q(yt− stx0) + q(tsx0 + sz)

≤ tq(y − sx0) + sq(tx0 + z)

This exactly fits the assumption, so we can pick a0 = sup f(y)−q(y−sx0)s

.

Step 2. Use Zorn’s lemma. Consider

L = (Z, F ) | Y ⊂ Z, F ≤ q on Z, F |Y = f .

Order on the set: (Z1, F1) ≤ (Z2, F2) if Z1 ⊂ Z2 and F2|Z1 = F1. Every chain

has an upper bound Z∞ = ∪Zi, F = ∪Fi. Hence there exists a maximal element

(Zmax, Fmax) ∈ L.

Claim 4.5. Zmax = X.

If not, ∃x0 6∈ Zmax apply lemma to Fmax, Zmax +Rx0 admits F ′max. Contradiction.

Remark 4.6. Hahn-Banach is also true for C.

8

5 Hahn-Banach Theorem 20210203

Lemma 5.1. Take C convex, 0 ∈ C. The Minkowski functional

qC(x) = infλ | x ∈ λC

is sublinear.

Proof. x, y ∈ V . Let ε > 0, choose λ, µ s.t. x ∈ λC, y ∈ µC.

qC(x) ≤ λ ≤ (1 + ε) qC(x)

qC(y) ≤ µ ≤ (1 + ε) qC(y).

Then z = λλ+µ

xλ

+ µλ+µ

yµ∈ C. Therefore x+ y = (λ+ µ)

(x

λ+µ+ y

λ+µ

). So

qC(x+ y) ≤ λ+ µ ≤ (1 + ε) (qC(x) + qC(y)).

Send ε→ 0.

Corollary 5.2. Let C,D be nonempty convex sets C ∩ D = ∅. There there exists

f : V → R s.t. f(x) ≤ f(y) for all x ∈ C, y ∈ D.

Proof. Take x0 ∈ C, y0 ∈ D. trick Shifting trick: let

B := C −D − (x0 − y0),

where C − D := x − y | x ∈ C, y ∈ D . Since x − y 6= 0, y0 − x0 6∈ B. Let

Y = R(y0 − x0).Claim 5.3. qB(x0 − y0) ≥ 1.

Define f(t(y0 − x0)) = t, then f ≤ qB. Hahn-Banach extension gives F : V → R,

with F ≤ q and F (y0 − x0) = 1. Note that qB(x− y − (x0 − y0)) ≤ 1 implies

F (x− y − (x0 − y0)) ≤ 1

=⇒ F (x− y)− F (x0 − y0) ≤ 1

F (x) ≤ F (y) + 1− F (y0 − x0) = F (y)

9

Theorem 5.4. For X a normed space and q(x) = ‖x‖, X subset of complex vector

space, ∀x with unit norm, ∃ a complex linear functional f ≤ ‖ · ‖ with |f(x)| = 1.

Proof. Consider X as a real normed space. Take x0 in X and let Y = Rx0 + iRx0,‖x0‖. Define f(zx0) = Re(z). Note that f ≤ q as

f(zx0) = Re(z) ≤ |z| = ‖zx0‖ ≤ (zx0).

Then ∃F : X → R with F (x) ≤ ‖x‖ real linear and F (x0) = 1.

Fabrication: want to define G(x) = F (x) − iF (ix). If G is complex linear and

F = ReG, G(x) = ReG(x) + ImG(x) = F (x)− Re(iG(x)).

Claim 5.5.

1. G(x) = F (x)− iF (ix) is complex linear

2. |G(x)| ≤ ‖x‖

6 Hahn-Banach Theorem Cont. 20210205

Theorem 6.1 (Complex version Hahn-Banach). Let X be a complex vector space.

If f : Y → C is a complex linear functional on a complex linear subspace Y ⊂ X,

and q : X → [0,∞] a sublinear function and q(zx) = q(x), |z| = 1 (semi-norm). If

|f | ≤ q, then there exists F : X → C, such that |F | ≤ q, F |Y = f

Proof. Apply the real Hahn-Banach to f = Re f . F : X → R. Define a new F by

F (x) = F (x)− iF (ix).

10

Check F is complex linear.

Hahn-Banach separation.

Lemma 6.2. Let C be a convex set and qC is a Minkowski functional

1. x ∈ C then qC(x) ≤ 1

2. x 6∈ C then qC(x) ≥ 1.

x | qC(x) < 1 ⊂ X ⊂ x | qC(x) ≤ 1 .

And the inclusions are strict.

Proof. qC(y) = infλ | yλ∈ C . For part 1, x ∈ C so qC(x) ≤ λ = 1.

For part 2, assume qC(x) < 1, then ∃λ < 1 such that xλ∈ C. This (together with

convexity) implies

x = (1− λ) · 0 + λ · xλ∈ C,

contradiction.

C may or may not contain the boundary.

1. Topology

2. filter

3. continuous

Definition 6.3. A filter on a set X is a subset F ⊂ 2X such that

1. If A,B ∈ F then A ∩B ∈ F2. If A ⊂ B and A ∈ F , then B ∈ F .

It is nontrivial if ∀A ∈ F , A 6= ∅.

Definition 6.4. A neighbourhood filter is a collection (FX)x∈X of filters.

11

Remark 6.5.

1. (Topology ⇒ Filter)

Given topology τ , FX is generated by the non-empty open sets.

FX = A ⊂ X | ∃O open , x ∈ O ⊂ A .

Neighbourhood filter.

2. (Filter ⇒ Topology)

Given a filter FX , define O is open iff ∀x ∈ O, O ∈ FX . intuition A topology

can equivalently be defined by open sets or neighbourhood filters.

Lemma 6.6. (τF)τ = τ .

Definition 6.7. f is continuous at x if ∀B ∈ Ff(x), f−1(B) ∈ FX .

Recall: If f : X → Y continuous and K ⊂ X compact, then f(K) compact

Definition 6.8. A space (X,+, ·, τ) is a topological vector spaces if

1. (X,+, ·) is a vector space

2. + : X ×X → X continuous

· : K×X → X continuous

Example 6.9. 1. R2 with the Chicago railway metric is not a topological vector

space. + not continuous.

2. Let (Ω,Σ, µ) be a measure space. Define

L0(Ω,Σ, µ) = f : Ω→ R | f measurable, µ(|f | > ε)→ 0 as ε→∞.

Define

d(f, 0) := inf ε | µ(|f | > ε) < ε , d(f, g) = d(f − g, 0).

This is a translation invariant metric. Hence a translation invariant topological

vector space.

12

7 Vector Space 20210208

1. Topological space

2. Topological vector space (X,+, ·, τ), in particular, the translation map Tx :

X → X; y 7→ Tx(y) = x+ y is a homeomorphism

3. Application to Hahn-Banach

4. Tychonoff’s theorem

Motivational lemma

Lemma 7.1. Let X be a topological vector space, f : X → R be a linear nonzero

continuous map, then the image of an open convex set is open.

Proof. If f is linear and O is convex then f(O) is convex. Convex sets of R is intervals.

Assume f(O) = (a, b] or [a, b]. That is there is a x ∈ O, f(x) = supy∈O

f(y), then

f(x) = b. Since f(x0) 6= 0 with f(x0) = 1, (f 6= 0), we consider x(t) = x+ tx0. Then

O open implies there is a t0, for all |t| < t0, x + tx0 ∈ O (translation is continuous).

But now

f(x+ tx0) = f(x) + tf(x0) = b+ t · 1 > b.

Contradiction.

later Extension is continuous.

Theorem 7.2 (Tychonoff). For each j ∈ J , let Xj be a topological space. If

each Xj is compact, then X =∏

j∈J Xj is compact in the product topology.

Clarification: x = (xi)i∈I , O is a neighborhood of x if there are ij, Oj such that

O = (yi) | yij ∈ Oij .

Example 7.3. Let Xi be a metric space, the index set I = N. Now the following

13

defines a distance of the product topology

d((xn), (yn)) =∑n≥0

2−nd(xn, yn)

1 + d(xn, yn).

(yn) | d((xn), (yn)) < ε ⊃ (yn) | dist(xj, yj) <ε

2, j = 1, · · ·n .

Proof. Assume d(xj, yj) ≤ ε2

for all j. Then

d((xn), (yn)) =∑n≥0

2−nd(xn, yn)

1 + d(xn, yn)

≤m∑n=1

2−nd(xn, yn)

1 + d(xn, yn)+∑n>m

2−m

≤ ε

2+ε

2≤ ε

(choose large m so that the second term is less than ε2). Any continuity condition

only depends on finitely many terms

1. non-trivial filter, filter converges to x, ultra filter

intuition Filter is the analogue of sequence converging to something. They want to

being small.

Definition 7.4. We say that a filter F converges to x if F ⊃ Nx.

Every neighbourhood is contained in the filter.

Definition 7.5. A maximal non-trivial filter is called a ultra filter.

Remark 7.6. Let U be an ultra filter then for every A ⊂ X, either A ∈ U or AC ∈ U .

Proof. Fix A ⊂ X.

Case 1. A ∈ U done.

14

Case 2. A 6∈ U then AC ∈ U . (Prove by contradiction, assume AC 6∈ U) Define Uto be the smallest filter which contains AC and elements in U . (Show U is again a

filter). Indeed this new filter U is closed by superset. Need to show if A, B ∈ Uimplies A ∩ B ∈ U .

• A, B ∈ U done.

• A, B ⊃ AC done.

• A ∈ U , B ⊃ AC . We know B ⊃ AC implies BC ⊂ A, and we know A 6= A, so

A ∩ B = ∅.Then U is a filter, contradicting to the fact U is an ultra filter.

Corollary 7.7. Every ultra filter on an interval converges.

Lemma 7.8. (X, τ) is compact iff every ultra filter converges.

Proof. Ref.

(⇒) Let (X, τ) be compact and U be an ultra filter. Assume U does not converge to

any point. Then ∀x ∈ X, Nx 6⊂ U . Then every point has a neighbourhood Ox which

is not in U .

Take the open cover ∪xOx of X, Ox as above. By compactness, there is a finite

subcover Ox1∪· · ·∪Oxn . Since U is an ultra filter, OCxi∈ U , and the finite intersection

of OCxi

’s is in U . But (n⋂i=1

OCxi

)C

=n⋃i=1

Oxi = X

implies ∩ni=1OCxi

= ∅ ∈ U , contradiction.

(⇐) Let X ⊂ ∪xOx, Ox open. Assume that X 6⊂ ∪ni=1Oxi for any finite subset of

15

indices. Then ∩ni=1OCxi6= ∅. Define

F =

A

∣∣∣∣∃ i1, · · · , in s.t.n⋂i=1

OCxi⊂ A

.

This is a filter, let U be the ultra filter contains F . Then U converges, say to some

x0 ∈ X, then Nx0 ⊂ U . Then there is a neighbourhood of x0 which is contained in

U , and then OCx ∈ F ⊂ U . But Ox ∩OC

x = ∅, contradiction.

proof of Theorem 7.2. Ref.

Let X = (∏

iXi, τi), F be an ultra filter. Let πi : X → Xi be the projection to the

i-th term. Note that πi(F) is also an ultra filter, so it converges to some xi ∈ Xi.

Then F converges to (xi)i∈I .

Claim 7.9. Let x = (xi)i∈I , if O ∈ Nx then O ∈ U .

This means O ⊃ Oi1 × · · · × Oin × Xj1 × Xj1 × · · · . Now π−1ik (Oik) = Wk open

and belongs to U , as Oik ∈ U . Hence, the finite intersection of Wk’s is in U . Then

O ∈ U .

8 Locally Convex Topological Vector Spaces 20210210

Recall

1. Topological vector spaces (X,+, ·)2. Tychonoff theorem

3. intuition An ultra filter is a generalisation of sequence converging to a point.

Definition 8.1. A topological vector space is called locally convex if ∀x, ∀O ∈ Nx,∃W convex such that x ∈ W ⊂ O.

Example 8.2.

1. Let X is a normed space, Nx = O | ∃x > 0, int(Br) + x ⊂ O .2. LetX = C∞(R), K a compact subset, with semi-norm ‖f‖K,n = sup

x∈Ksup1≤i≤n

|f (i)(x)|.

16

(This is a semi-norm because supp f can be in KC) The resulting topology is

locally convex.

Example 8.3 (Non-examples).

1. L0 = f : R→ R | f measurable , with

d(f, 0) = inf ε | µ(|f | > ε) < ε .

2. ‖f‖p = (E|f |p)l/p with 0 < p < 1. Bp = f | ‖f‖p < 1 . (Cannot put in a

convex set if it is infinite dimension). The first example is when p → 0. (E is

expectation?)

Theorem 8.4. Let (X, τ) be a topological space, the following are equivalent.

1. X is a Locally convex topological vector spaces (LCTVS)

2. ∃(qi)i∈I of semi-norms on X

3. O 3 N0 iff ∃ i, ∃ r s.t. x | qi(x) < r ⊂ O.

intuition The topology is determined by many different shaped balls. Open iff contain

one of the balls.

Proof of Theorem 8.4

(⇐) Take a point x ∈ X and O is an open neighbourhood of x. Define a translation

map T−x : X → X, by T−x(y) = y − x. Note that T−x is a homeomorphism, so

T−x(O) =: W is an open neighbourhood of 0. By (iii), ∃ i s.t. y | qi(y) < 1 ⊂ W .

Define V = x+ W = y | qi(y − x < 1) ⊂ O.

Definition 8.5. A set W 3 0 is called absolutely convex if

n∑j=1

|λj| ≤ 1 =⇒n∑j=1

λjxj ∈ W.

Definition 8.6. A set W is balanced if |z| = 1, zw = w for all w ∈ W .

17

Remark 8.7. W is absolutely convex if W is convex and balanced.

(⇒) Prove existance of seminorms. Take K = R let O be open and ∃W ⊂ O

containing 0 and convex. Since − : X → X; −x 7→ x is continuous, we know

(−)−1(W ) ⊃ V is convex, V ∈ Nx (Aside: W ∩ −W is convex and balanced).

Define qV (x) = infλ | xλ∈ V

Lemma 8.8. qV is a semi-norm.

That is, qV (λx) = |λ|qV (x) and subadditive qV (x+ y) ≤ qV (x) + qV (y).

Then1

4⊂ y | qV (y) <

1

2(ball of some semi-norm) ⊂ V.

For every neighbourhood of 0 can choose a semi-norm

For K = C. Want for any set O, find a W which is convex and contained in ∩|z|=1zO

(in a intersection of rotations). (∩|z|=1zO)C = ∪|z|=1(zO)C .

Question: Is B = ∪|z|=1(zO)C closed? – Yes. Let T = z | |z| = 1 . The map

T ×X → X; (z, x) 7→ zx is continuous and T is compact.

Lemma 8.9. B is closed. (A compact translation of a closed set is closed.)

Proof. Let A be an ordered index set, xα ∈ B, xα → x meaning for a neighbourhood

O of x, ∃α0, ∀α > α0, x ∈ O.

Then 0 6∈ B, and ∃W ⊂ ∩|z|=1zO)C convex and ∩|z|=1zw is balanced convex set.

9 Hahn-Banach Separation Theorem 20210212

Lemma 9.1. Let X, Y be locally convex topological vector spaces. A linear map

T : X → Y is continuous if and only if T is continuous at 0.

18

Propersition 9.2. Let X be a locally convex topological space and f : X → R be

a linear and continuous map. Let W be an open convex neighbourhood of 0. Then

either f(W ) = 0 or f(W ) is open.

Theorem 9.3 (Hahn-Banach Separation Theorem). Let C be a convex

nonempty subset in a topological space X and x 6∈ C, then

1. there exists a linear map f : X → R such that f(y) ≤ f(x), ∀y ∈ C,

2. if in addition X is a locally convex topological vector space and C is open,

then f is continuous, nontrivial and f(y) < f(x), ∀y ∈ C.

Proof. (1) Let x0 ∈ C, then C = C −x0 contains 0, by Lemma 5.1, the Minkowski

functional qC = infλ | y ∈ λC is sublinear. Let V = R(x − x0) and define

f(t(x − x0)) = t, which is linear. Then x − x0 6∈ C − x0. By Lemma 6.2, y ∈ Cimplies qC(y − x0) ≤ 1. Therefore

f(y − x0) ≤ f(x− x0) = 1 =⇒ f(y) ≤ f(x).

(2) Now if C is open then C = C − x0 is open (here we only require a topological

space, we don’t actually need locally convexity). Consider g : X ×X → X, g(x, y) =

x− y. This map is continuous, 0 ∈ C.

There exists V1, V2 neighbourhoods of 0, such that V1 − V2 ⊂ C. Define V = V1 ∩ V2(V is a neighbourhood of 0). Then 0 ∈ V − V ⊂ C. By previous part f |C ≤ 1.

check Hence

f(V − V ) ⊂ f(C) ⊂ y | f(y) ≤ r .

Then for all y = a − b ∈ V − V , f(y) ≤ 1 and −y = b − a ∈ V − V so f(−y) ≤ 1.

This means f is bounded. Hence f is continuous at 0. By previous Lemma, f is

continuous and f(C) is open (image of open convex set is open). Then f(y− x0) < 1

for all y ∈ C.

19

Theorem 9.4. Let C,D be nonempty convex sets. If C ∩D = ∅, then there

is a linear functional f on X such that f(x) < f(y), for all x ∈ C, y ∈ D.

Proof. trick Consider C = C −D = x − y | x ∈ C, y ∈ D . Note that C is open

if either C or D is open, and 0 6∈ C. Now shift the set, i.e. let D = C − (x0 − y0).Apply previous theorem 0 6∈ C, so there exists a f 6= 0 and continuous, f(z) < f(0),

for all z ∈ C = C −D. Say z = x− y, for x ∈ C and y ∈ D. Then f(x) < f(y).

Theorem 9.5. Let C be a closed convex set and D be a compact convex

set in a locally convex topological vector space. Then there exists a continuous

nontrivial f and r < s such that f(x) < r < s < f(y) for all x ∈ D and y ∈ C.

Proof. Assume C is closed and D is compact. CC is open, D∩C = ∅. For any x ∈ Dthere is a Wx convex such that (x+Wx) ∩ C = ∅.

Consider the open sets x+ Wx

2, their union ∪(x+ Wx

2) gives an open cover of D. Then

there is a finite subcover D ⊂ ∪i(xi +Wxi

2). Take W = ∩i

Wxi

2, and let y = d + w ∈

D +W . Then there exists an xj such that d = xj +Wxj

2. Therefore,

y = d+ w ∈ xj +Wxj

2+W ⊂ xj +

Wxj

2+Wxj

26⊂ C.

(Convexity impliesWxj

2+Wxj

2⊂ Wxj .) analogue Triangle inequality on metric spaces.

Hence we have a strict separation between D+W and C, and we can find a nontrivial

continuous f such that f(x) < f(d+w), where x ∈ C, d ∈ D and w ∈ W . Note that

f(D) is compact as D is compact, so f(D) is a closed interval [a, b]. Then

f(D +W ) = f(D) + f(W ) = [a, b] + (−α, β)

(f(W ) is a neighbourhood of 0 check ). So for all x ∈ C,

f(x) ≤ a− α < a ≤ inf f(y) | y ∈ D .

20

Example 9.6.

1. Let X be a normed space, and C = x | ‖x‖ ≤ 1 = BX . Take x0 such that

‖x0‖ > 1, then D = x0 compact. There exists f such that f(x) ≤ 1, ‖f‖ ≤ 1

and f(x0) > 1.

2. Take a ball BX and a triangle D.

Next, we want to make the separation line unique.

10 Weak Topology 20210215

Definition 10.1. Let X be a Banach space and Y ⊂ X∗ a subspace. Then σ(X,Y )-

topology is the coarsest topology making all the functional y ∈ Y continuous. This

means the semi-norms defining this topology are given by

qy1,··· ,yn(x) = maxi=1,··· ,n

|yi(x)|.

Every locally convex space is given by semi-norms. Semi-norms are indexed by finite

subsets of Y .

Theorem 10.2. The dual space of (X, σ(X, Y )) is Y (as a set). That is,

(X, σ(X, Y ))∗ = Y.

Note the two spaces only equal as a set, not necessarily as a topological space. Because

Y on the LHS can be taken as a algebraic dual without topological assumptions,

whereas Y on the RHS is a topological vector space (may with its own norm).

Remark 10.3. Let X be a locally convex topological vector space and Y a Banach

21

space or locally convex topological vector space, then L(X, Y ) is also a locally convex

topological vector space.

Proof. Step 1. Y ⊂ (X, σ(X, Y ))∗.

Claim 10.4. For every y ∈ Y , fy(x) = y(x) is continuous with respect to the new

topology.

It suffice to show f is continuous at 0: ∀ε, ∃O ∈ σ(X, Y ) containing 0, such that if

x ∈ O, then |f(x)| < ε. (f(0) = 0). In this new topology open neighbourhood means

there exists a semi-norm in system such that O ⊃ x | q(x) < δ , i.e there exists

some Bq(δ) ⊂ O. This is equivalent to say |f(x)| ≤ C · q(x), for some semi-norm

q. compare In Banach space we don’t have a choice of the norm, so we require

|f(x)| ≤ C · ‖x‖.

In our case, the semi-norm qy(x) = |y(x)| does the job, because |fy(x)| = |y(x)| =

qy(x). More generally, the semi-norm is given by qy(x) = maxj |yj(x)|.

Step 2. (X, σ(X, Y ))∗ ⊂ Y .

Let f : X → K be continuous. By definition there exists a q such that |f(x)| ≤ q(x)

and q(x) = maxj |yj(x)|. Fix y1, · · · , yn and define a map

φ : X −→ Kn

x 7−→ (y1(x), · · · , yn(x)).

Then φ(X) ⊂ Kn is a subspace. Denote Z = φ(X), then z = (y1(x), · · · , yn(x)).

Consider the map

ψ : Z −→ K

z 7−→ f(x).

This map is well-defined, linear, and |ψ(z)| ≤ maxj |zj| = ‖z‖∞. By Hahn-Banach,

there exists ψ : lm∞ → K, such that ψ|Z(z) = ψ(z) and ‖ψ‖ = ‖ψ‖ ≤ ‖z‖∞. Note that

ψ(z) ∈ (lm∞)∗ = lm1 . This means there exists α1, · · · , αn such that ψ(z) =∑

j αjzj.

22

This means

f(x) = ψ(φ(x)) = ψ(φ(x)) =∑j

αjφj(x) = fy(x),

where y =∑

j αjyj.

Example 10.5. Let X be a space and take Y = X∗. Then

• σ(X,X∗) is called the weak topology of X and (X, σ(X,X∗)) = X∗,

• σ(X∗, X) is called the weak* topology of X∗ and (X∗, σ(X∗, X)) = X.

11 Weak Topology Cont. 20210219

Theorem 11.1 (Goldstine). Let X be a Banach space, then the image of the

closed unit ball BX ⊂ X under the canonical embedding ι into the closed unit

ball BX∗∗ of the bidual space X∗∗ is weak*-dense.

BXσ(X∗∗,X∗)

= BX∗∗

intuition The unit ball with weak*-topology is compact. In finite dimension, close

+ bounded = compact. Generalisations of finite dimension.

Proof. Recall that X∗∗ is a locally convex topological vector space with respect to

σ(X∗∗, X∗)-topology. This topology is given by the semi-norm q(x∗∗) = maxj |x∗∗(x∗j)|,with x∗1, · · · , x∗j ∈ X∗.

The canonical embedding ι : X → X∗∗, is an isometry (Hahn-Banach Theorem) and

ι|BX : BX → BX∗∗ . We want to show the closure ι(BX) with respect to the σ(X∗∗, X∗)

topology satisfies ι(BX) = B∗∗X . Prove by contradiction.

Assume that x∗∗ 6∈ ι(BX), with ‖x∗∗‖X∗∗ ≤ 1. Note that ι(BX) is closed, compact

and convex. By Hahn-Banach separation (Theorem 9.5), there exists a nontrivial

continuous map f : X∗∗ → R so that |f(ι(x))| ≤ 1 < s < |f(x∗∗)| for all x ∈ BX . On

23

one hand we have

‖f‖X∗∗ = sup‖x‖≤1

|f(x∗∗)| = sup‖x‖≤1

|f(ι(x))| ≤ 1.

Then by definition,

|x∗∗(f)| ≤ ‖x∗∗‖X∗∗ · ‖f‖X∗∗ ≤ 1.

On the other hand we have |x∗∗(f)| = f(x∗∗) > 1. Contradiction.

Example 11.2. Let X = C0 = (xn) | limn xn = 0 , with ‖(xn)‖ = supn ‖xn‖.Then X∗ = l1 because

‖yn‖1 =∑n

yn = supk

k∑i=1

|yk|

= supk< y, ε1, · · · , εk, 0, · · · , 0 > .

where εi = sgn(yi) and < y, z >=∑ynzn. And X∗∗ = l∞ = (xn) | supn |xn| <∞.

What is σ(l∞, l1)-topology? The answer is pointwise convergence on bounded set.

Consider bounded sequences xα (‖xα‖ ≤ C). Then xα → x ∈ l∞ iff for all y ∈ l1,

xα(y)→ x(y).

For bounded sets ‖xα‖ ≤ 1, ∀α,

xα → x ⇐⇒ xαn → xn,∀n.

(⇒) Take yn = (0, · · · , 1, · · · , 0) ∈ l1.

(⇐) Let y ∈ l1 and ε > 0 then there exists n0 such that∑n>n0

|yn| < ε2. There exists α0

such that any α > α0, |xαn − xn| < ε2

for all n > n0. We need

|xα(y)− x(y)| ≤< ε

2+ε

2= ε.

Let yN = (y1, · · · , yN , 0, · · · , 0), yN → y in σ(l∞, l1) because we can use pointwise

convergence.

24

12 Alaoglu’s Theorem 20210222

Alaoglu’s Theorem says that the closed unit ball in X∗ is compact in the weak*-

topology.

Theorem 12.1 (Alaoglu). Given a topological vector space X, and let BX∗ =

x∗ ∈ X∗ | ‖x∗‖ ≤ 1 be the closed unit ball in X∗. Then BX∗ is compact in

X∗ with respect to the weak*- topology on X∗.

Proof. Ref. or see Conway p.134

Let the set Dx = z ∈ K | |z| ≤ 1 . Consider the product D :=∏

x∈Bx Dx. Since Dx

is compact in K, Tychonov’s theorem says that D compact in the product topology.

Elements in D are functionals, given by µ ∈ K, µ(x) = µx ∈ D ⊂ C, although they

need not to be linear.

The inclusion

ι : BX∗ ⊂∏x∈Bx

D =: K

is given by

ι(x∗)(x) = x∗(x).

Note that ι(BX∗) ⊂ K. Indeed, if ‖x‖ ≤ 1 and ‖x∗‖ ≤ 1, then |x∗(x)| ≤ 1 ∈ D.

Claim 12.2. ι(BX∗) is closed. Hence, ι(BX∗) is a compact subspace of K.

Proof of the claim. Take a net (x∗α) in BX∗ which converges to f ∈ D pointwisely. So

f(x) = limα→∞ x∗α(x). In particular |f(x)| ≤ 1 for all ‖x‖ ≤ 1. (Need to show f is in

the range. We can not take N as index set, instead replacing N by a partially ordered

set. Usually the index set is given by the neighbourhood basis of f . Let Oi ∈ Nf ,i = 1, 2, then O1 ∩O2 ∈ Nf and O1 ∩O2 ≥ Oi.)

For x ∈ X, define F (x) = β−1f(βx) for some β such that ‖βx‖ ≤ 1 (check this

is well defined). Then F agrees with f on BX . We claim that F is linear. Take

25

xi ∈ X, i = 1, 2. Consider y = x1+x2‖x1‖+‖x2‖ . If we take λ = ‖x1‖

‖x1‖+‖x2‖ , then by convexity

y = λ x1‖x1‖ + (1− λ) x2

‖x2‖ ∈ BX . Then

f(y) = limαx∗α(y) = lim

αx∗α

(x1

‖x1‖+ ‖x2‖

)+ x∗α

(x2

‖x1‖+ ‖x2‖

)= f

(x1

‖x1‖+ ‖x2‖

)+ f

(x2

‖x1‖+ ‖x2‖

).

So

F (x1 + x2) = f(y) · (‖x1‖+ ‖x2‖)

=

(f

(x1

‖x1‖+ ‖x2‖

)+ f

(x2

‖x1‖+ ‖x2‖

))· (‖x1‖+ ‖x2‖)

= F (x1) + F (x2).

We have a linear functional F ∈ X∗ satisfying |F (x)| ≤ 1 when ‖x‖ ≤ 1. This means

‖F‖X∗ ≤ 1. So F ∈ BX∗

Definition 12.3. A Banach space is reflexive if X∗∗ = X.

Goal: to show X is reflexive iff X∗ is reflexive.

Propersition 12.4. A closed subspace of a reflexive Banach space is reflexive.

Proof. The following diagram is commutative. (Check)

X X∗∗

Y Y ∗∗

ι

j

ιY

j∗∗

Step 1. Y ∗∗ = Y . Take an element y∗∗ ∈ Y ∗∗, note that

j∗∗(y∗∗)(x∗) = y∗∗ j∗(x∗) = y∗∗(x∗ j) = x∗|Y ∈ Y ∗.

So we can apply y∗∗ to this element, and define φ(x∗) = y∗∗(x∗|Y )

26

Lemma 12.5. If T : Y → X is isometric, then T ∗∗ : Y ∗∗ → X∗∗ is also isometric.

The above lemma says Y ∗∗ embeds isometrically into X∗∗ (we will prove this later).

If in addition, X∗∗ = X, we deduce that for every y∗∗ there exists an x ∈ X such that

y∗∗(x∗|Y ) = x∗(x).

We want to show x ∈ Y . We claim that y∗∗ ∈ Y , otherwise by Hahn-Banach sep-

aration there exists x∗ such that x∗(y∗∗) = 1 and x∗|Y = 0. The last equation says

x∗(x) = y∗∗(x∗|Y ) = x∗|Y = 0. A contradiction (as y∗∗ ∈ Y ∗∗ ⊂ X∗∗ = X).

Lemma 12.6. If T : X → Y is isometric then T ∗ : Y ∗ → X∗ sends closed unit ball

to closed unit ball.

Proof. Note that T ∗(BY ∗) ⊂ BX∗ . Indeed,

‖T ∗‖ = sup‖y∗‖≤1

‖T ∗(y∗)‖ = sup‖y∗‖≤1

‖y∗ T‖

= sup‖y∗‖≤1,|x|≤1

|y∗ T (x)| = sup‖y∗‖≤1,|x|≤1

|y∗(x)| ≤ 1.

So |T ∗(y∗)| ≤ ‖T ∗‖‖y∗‖ ≤ 1.

To show T ∗ is onto, take x∗ ∈ BX∗ . Can define f(Tx) = x∗(x), ‖f‖ ≤ 1. By

Hahn-Banach there exists y∗ such that y∗(Tx) = f(Tx) = x∗(x). T ∗(y) = x∗.

Lemma 12.7. If T : Y → X is a surjection, then T ∗ : X∗ → Y ∗ is an isometry.

Proof of the Lemma 12.5. The previous two lemma gives the result.

13 Reflexive Spaces 20210224

Theorem 13.1. X is reflexive ⇐⇒ X∗ is reflexive.

27

Proof. (⇒) Assume that X = X∗∗. Then BX∗ is closed in σ(X∗, X∗∗) = σ(X∗, X).

Take an element x∗∗∗ in BX∗∗∗ , there exists a sequence x∗α → x∗∗∗ in σ(X∗∗∗, X∗∗)

topology. Since Bx∗ is closed in σ(X∗, X), there is an x∗ such that x∗α → x∗. This

means x∗∗∗ = x∗.

(⇐) If X∗ is reflexive then X∗∗ is reflexive, but X ⊂ X∗∗ as a closed subspace.

Remark 13.2. X is reflexive iff BX∗ is σ(X∗, X∗∗) closed.

Definition 13.3. A Banach space is called uniformly convex, if ∀ε > 0, ∃δ > 0

such that ‖x‖ ≤ 1, ‖y‖ ≤ 1 and ‖x− y‖ > ε, then ‖x+y2‖ ≤ 1− δ.

Lemma 13.4. Take (xn) a sequence with

lim supn‖xn‖ ≤ 1 and lim

ninfm>n

∥∥∥∥ xn + xm2

∥∥∥∥ = 1.

Then (xn) is Cauchy.

Proof. Let ε > 0. Since lim supn ‖xn‖ ≤ 1, we can choose ε0 > 0, ∃n0 such that

‖xn‖ ≤ 1 + ε0, for all n > n0. So ‖ xn1+ε0‖ ≤ 1, for all n > n0. Then∥∥∥∥ xn + xm

2(1 + ε0)

∥∥∥∥ =

∥∥∥∥ xn + xm2

∥∥∥∥ · 1

1 + ε0≥ 1

(1 + ε0)2,

for all n > n0.

Taking 1(1+ε0)2

= 1− δ. Using uniform convexity (contrapositive), we have ∀n, ∃m∥∥∥∥ xn − xm2(1 + ε0)

∥∥∥∥ < ε.

Conclusion: Above shows ∀ε, ∃n0, ∀n > n0, ∃m, such that ‖xn − xm‖ < 2ε(1 + ε0).

We use this for ε = 2−k, then there exists a converging subsequence xnk such that

‖xnk − xnk+1‖ ≤ 2−k.

28

Theorem 13.5 (Milman-Pettis). Uniformly convex Banach spaces are reflex-

ive.

Proof. Ref.

Let x∗∗ ∈ BX∗∗ , ‖x∗∗‖ = 1. Then by definition of ‖x∗∗‖, for all n, there exists

x∗n ∈ BX∗ , such that x∗∗(x∗n) ≥ 1− 1n. Since BX ⊂ BX∗∗ is dense in σ(X∗∗, X∗). Let

qn(y) = |x∗n(y)|. There exists (xk) in BX such that

|qn(x∗∗ − xk)| = |x∗n(xk)− x∗∗(x∗n)| ≤ 1

2k, for n = 1, · · · , k.

In particular, apply the above to n = k, then

|x∗k(xk)− x∗∗(x∗k)| ≤1

2k=⇒ − 1

2k+ x∗∗(x∗k) ≤ x∗k(xk).

Recall x∗∗(x∗k) ≥ 1− 1k. So 1− 3

2k≤ x∗n(xk) ≤ 1 (RHS because x∗n is in the unit ball).

Then take m > k, we have

2− 6

2k≤ 1− 3

2k+ 1− 3

2m≤ x∗k(xk) + x∗m(xm) ≤ x∗k(xk + xm) ≤ ‖xk + xm‖ ≤ 2. (1)

Taking lim inf on both sides we get lim inf ‖xk+xm2‖ = 1, and lim sup ‖xk‖ ≤ 1. By

the above lemma (xn) is Cauchy.

Remark 13.6. Assume there are two sequences xn, xn satisfies the property (1),

then then limxn = lim xn.

Now if (y∗n) is another family using the above construction, then there exists (xn) in

BX such that

|y∗n(xk)− x∗∗(x∗n)| ≤ 1

2k.

Then x∗(xk) → x∗∗(x) and y∗(xk) → x∗∗(y) implies x∗∗ = limxn = lim xn in

σ(X∗∗, X∗).

29

14 Reflexive Spaces Cont. 20210226

Real analysis: Lp(Ω,Σ, µ) = [f ] | f : Ω → K, f measurable,∫|f |p dµ < ∞, where

Ω is a set, Σ is a σ-algebra and µ is a σ-additive measure. Recall

• Simple functions f =∑n

j=1 αj1Ej are dense.

• ‖f‖p = sup‖g‖p′≤1

|∫fg dµ|.

Use Holder inequality, say ‖f‖p = 1, then g = sgn(f) · |f |p/p′ .

Corollary 14.1. If 1 ≤ p ≤ ∞, then Lp′ embeds isometrically into L∗p,

ιp′ : Lp′ → L∗p

g 7→(ιp′(g) : f 7→ ιp′(g)(f) =

∫fg dµ

)and ‖f‖p = ‖ιp′(g) : Lp → K‖.

Theorem 14.2. Let 1 < p <∞ and assume Lp is reflexive. Then L∗p′ = Lp.

(Here we check isometric isomorphism, there are two type of isomorphisms for Banach

spaces, see more here)

Proof. Let ϕ : Lp′ → K with ‖ϕ‖L∗p′

= 1. Recall Lp′ → L∗p is an isometry. By

Hahn-Banach extension, there exists a ϕ : L∗p → K, with ϕ|Lp′ = ϕ.

Lp′ L∗p

K

ιp′

ϕ∃ϕ

To show ιp′ is surjective, take η ∈ L∗p. If we can find g ∈ Lp′ such that∫fg dµ = η(f),

30

then ιp′(g) = η and we are done. Such a g exists by commutativity and reflexivity

ϕ(g) = ϕ(ιp′(g)) = ιp′(g)(f) =

∫fg dµ =⇒ ιp′(g)(f) = η(f).

Example 14.3 (Discrete case). Let Ω = I, Σ = 2I , µ be the counting measure. If

I = N, then

Lp(N,Σ, µ) = `p = (xn) |∑n

|xn|p <∞.

What is the f defining the functional ϕ : `(N) → K? Well, f is given by a sequence

(yn) = ((0, 0, · · · , 1n, · · · , 0, 0)). One can show that the

‖yn‖p′ = supn

(n∑k=1

|yk|p′

) 1p′

= supϕ(xn) | ‖xn‖ ≤ 1 .

Prove using Holder.

Theorem 14.4. `∗p = `p′ for 1 < p <∞.

Remark 14.5. For I = N, let c0 = (xn) ∈ `∞ | limxn = 0 . Then c∗0 = `1,

c∗∗0 = `∗1 = `∞.

Corollary 14.6. B`1 ⊂ B`∗∞ is σ(`∗∞, `∞)-dense.

This means for any ϕ ∈ `∗∞, for any fi ∈ `∞, there exists g ∈ `1, with ‖g‖`1 ≤ ‖ϕ‖,such that

|ϕ(fi)− fj(g)| ≤ ε i.e. arbitrarily closed.

Or there exists a net (gα) ∈ `1 with ‖gα‖`1 ≤ ‖ϕ‖, such that

ϕ(f) = limαf(gα) = lim

α

∑n∈N

f(n)gα(n).

Remark 14.7. Let ϕ : `∞ → K, and assume ϕ(1) = 1. TFAE

31

• ‖ϕ‖ = 1

• ∀g ≥ 0, ϕ(g) ≥ 0.

We call this positive functionals.

Define the state space S(`∞) = ϕ | ϕ(1) = 1, ‖ϕ‖ = 1 . Then discrete probability

measures are dense in the state space. Indeed if ϕ(1) = 1 and ‖ϕ‖ = 1, then there is

gα ∈ `1 with gα(1) = 1, ‖gα‖ ≤ 1 and gα(f)→ ϕ(f). That is gα → ϕ in σ(`∗∞, `∞).

Lemma 14.8. ‖gα‖`1 = 1 and∑

n gα(n) = 1 implies gα ≥ 0.

This means gα are discrete probability measures because ϕ(f) = limα

∑n∈N f(n)gα(n)

exists.

Theorem 14.9. Let be ϕ : C(K) → C be such that ϕ(1) = 1 and ‖ϕ‖ = 1. Then

there exists a net (xj)n(α)j=1 (λαj )

n(α)j=1 , where

∑λαj = 1 such that

ϕ(f) = limα

n(α)∑j=1

f(xαj ) · λαj .

Proof. The Banach space C(K) embeds into the Banach space `∞(K) (view this as

a discrete index set, no topology) isometrically via ι(f)(k) = f(k).

C(K) `∞(K)

K

ι

ϕ∃ϕ

As previous seen, ϕ exists by Hahn-Banach extension. Also have ϕ(1) = 1, ‖ϕ‖ = 1

and then ϕ ∈ S(`∞(K)). By previous remark, and also the fact that every function

in `1 is support on a countable number of points

ϕ(F ) = limα

∑(tj)

F ((tαj )) · λαj

32

where∑

α λαj = 1. Can replace LHS of this equation by limα limM

∑Mj=1 λ

α,Mj · F (tαj )

with∑M

j=1 λα,Mj = 1 (technical detail skipped). But

F = ι(f) = limα′

M(α′)∑j=1

λα′

j · f(tα′

j ).

Consider C[0, 1]. It is separable (admits a countable dense subset), whereas `∞(N) is

non-separable.

Corollary 14.10. If ϕ : C[0, 1]→ C, with ϕ(1) = 1 and ‖ϕ‖ = 1. Then there exists

a sequence (tnj )(λnj ), where∑λnj = 1 such that

ϕ(f) = limα

M(n)∑j=1

f(xnj ) · λnj .

15 Riesz-Thorin Theorem 20210301

Theorem 15.1 (Riesz-Thorin). Let A be a linear operator and let 1 ≤ p0, p1, q0, q1 ≤∞ where p0 6= p1 and q0 6= q1. Suppose A : Lp0 → Lq0 is bounded and A : Lp1 → Lq1is bounded. Let

1

p=

1− θp0

+θ

p1and

1

q=

1− θq0

+θ

q1

where θ ∈ (0, 1). Then

‖A‖Lp→Lq ≤ ‖A‖1−θLp0→Lq0· ‖A‖θLp1→Lq1 .

If we call ‖A‖1−θLp0→Lq0= M0 and ‖A‖θLp1→Lq1 = M1, then ‖A‖Lp→Lq ≤M1−θ

0 ·M θ1 .

In class Lp is replaced with `p, but there is a more generalized version in literature.

I leave Lp in the Theorem to reminds myself this fact. For 1 < p < q < r < ∞,

Lp ∩ Lq ⊂ Lr ⊂ Lp + Lq. In our case (finite dimensional), the same matrix makes

sense and A : `p0 ∩ `p1 → `q0 + `q1 .

33

We will use the following lemma to prove Riesz-Thorin Theorem.

Lemma 15.2 (Hadamard’s Three-Line Theorem). Suppose f(z) is bounded and con-

tinuous function on 0 ≤ Re(z) ≤ 1 and analytic in the interior. Denote

Mθ = supy∈R|f(θ + iy)|.

Then Mθ ≤M1−θ0 M θ

1 for θ ∈ (0, 1).

If we control the function on boundary then we control the function in the interior.

Example 15.3. Map from a strip to a disk. Let f(z) =∑anz

n be an analytic

function, a0 = f(0) = 12πi

∫ f(z)z

dz. Then

|a0| ≤∫|f(z)| dz =

1

2πi

∫f(eiθ) dθ ≤ sup |f(eiθ)|.

Proof. Ref.

Recall `p → `∗p′ isometrically. So

‖A‖`p→`q = sup ∑

kj

yj · Ajk · xk∣∣∣ ∑ |xi|p ≤ 1,

∑|yj|q

′ ≤ 1.

Assume∑|xi|p = 1 and

∑|yj|q

′= 1. Define a function

xk(z) =xk|xk||xk|

p ·(

1−zp0

+ zp1

)and yj(z) =

yj|yj||yj|

q′ ·(

1−zq0′ +

zq1′

).

Then F (z) =∑

jk yj(z) · Ajk · xk(z) is also analytic. Take 0 ≤ Re(z) ≤ 1 and define

G(z) = M z−10 M−z

1 F (z).

Claim 15.4. |G(it)| ≤ 1 and |G(1 + it)| ≤ 1.

34

Take z = it, then

G(it) =∑jk

xk|xk||xk|

p ·(

1−itp0

+ itp1

)· Ajk ·

yj|yj||yj|

q′ ·(

1−itq0′ +

itq1′

)

=∑jk

αk|xk|pp0 · Ajk · βj|yj|

q′q0′

= ‖A‖`p0→`q0 ·∥∥∥∥∑

jk

αk|xk|pp0

∥∥∥∥p0p0

·∥∥∥∥∑

jk

βj|yj|q′q0′

∥∥∥∥q0′

q0′

≤ 1,

where |αk|, |βk| = 1 (???). Similarly for G(1 + it).

The Three-Line Lemma gives |G(θ)| ≤ 1. Note that

G(θ) = M θ−10 M−θ

1

∑jk

xk|xk||xk|

p ·(

1−θp0

+ θp1

)· Ajk ·

yj|yj||yj|

q′ ·(

1−θq0′ +

θq1′

)

= M θ−10 M−θ

1

∑jk

xk|xk||xk|p ·

1p · Ajk ·

yj|yj||yj|q

′ · 1q′

= M θ−10 M−θ

1

∑jk

xk · Ajk · yj.

This implies |∑

jk xk · Ajk · yj| ≤M1−θ0 M θ

1 .

Corollary 15.5. Assume x, y are complex numbers and r ≤ s ≤ r′ then

(|x+ y|r + |x− y|r)1r ≤ 21− 1

s · (|x|s + |y|s)1s .

Example 15.6. When r = 2, x, y ∈ R, then we get the parallelogram law

(|x+ y|2 + |x− y|2)12 = (x2 + 2xy + y2 + x2 − 2xy + y2)

12 =√

2 · (x2 + y2)12 .

Proof. Take the matrix A =

(1 1

1 −1

). Then

A

(x

y

)= (|x+ y|r + |x− y|r)

1r ≤ 21− 1

s · (|x|s + |y|s)1s

35

For the case s ≥ 2,

‖A‖`2∞→`2∞ = sup

max(|x+ y|, |x− y|)∣∣∣ |x| ≤ 1, |y| ≤ 1

≤ 2.

‖A‖`22→`22 ≤ (|x+ y|2 + |x− y|2)12 =√

2 · (x2 + y2)12 ≤√

2.

Using Riesz-Thorin Theorem we obtain

‖A‖`2s→`2s ≤ 21−θ ·√

2θ

= 21− θ2 = 21− 1

s ,

with the last step given by 1s

= 1−θ∞ + θ

2.

For 1 ≤ s ≤ 2, we note that r ≤ s ≤ r′ implies s′ ≤ r. It suffices to consider r = s′.

Again Riesz-Thorin Theorem gives

‖A‖`s→`s ≤ ‖A‖1−θ`1→`∞ · ‖A‖θ`2→`2 ≤ 11−θ ·

√2θ

= 21s′ = 21− 1

s ,

with 1s

= 1−θ1

+ θ2

and 1s′

= 1−θ∞ + θ

2. Note that

‖A‖`1→`∞ = maxjk|Ajk|.

16 Clarkson’s Inequality 20210303

Clarkson’s inequality =⇒ Uniform convexity =⇒ Lp is reflexive =⇒ L∗p = Lp′

We want to use the Clarkson’s inequality (proof ref. Boa) to prove uniform convexity

of Lp.

Theorem 16.1 (Reformulation of Riesz-Thorin). Let A be a matrix. Consider

F (x, y) = log ‖A‖`1/x→`1/y . Then F is a convex function.

Now let A =

(1 1

1 −1

): `2p(C)→ `2q(C). Thus ‖A‖`s→`r ≤ 21− 1

s for all s ≤ r ≤ s′.

We have seen

36

• ‖A‖`22→`22 ≤√

2, and U =

(1√2− 1√

21√2− 1√

2

)is unitary (preserves inner product).

• ‖A‖`2∞→`2∞ = 2.

• ‖A‖`21→`2∞ = 1.

Remark 16.2. ‖A‖`2p→`2q = ‖A‖`2q′→`

2p′

.

Figure 1: Picture taken from here, p.1366

Explanation of the picture: by the value at a point, I mean the power of 2. (If I call

the value α, then 2α is an upper bound for ‖A‖`2p→`2q .)• (Region III) The point (1

p, 1q) = (1

2, 12) corresponds to ‖A‖`22→`22 and has value

log2(√

2) = 12.

• (Region IIIa) The point (1p, 1q) = (1, 1) corresponds to ‖A‖`2∞→`2∞ and has value

log2(2) = 1. By the remark above ‖A‖`21→`21 = ‖A‖`2∞→`2∞ = 2, so the point

(1p, 1q) = (1, 1) also has value 1.

• (Region IIIa) Using convexity, for 2 < p < ∞, point (1p, 1q) on the line y = x

has value 1q.

37

• (Region IIIb) The point (1p, 1q) = (1,∞) corresponds to ‖A‖`2∞→`2∞ and has value

log2(1) = 0.

• (Region IIIb) Using convexity, for 2 < p <∞, point (1p, 1q) on the line y = 1−x

has value 1q. Vertical lines between the lines y = x and y = 1− x has value 1

q.

• (Region II) For 1 ≤ s ≤ 2 we have

‖A‖`s→`s′ ≤ ‖A‖1−θ`1→`∞ · ‖A‖

θ`2→`2 ≤ 21− 1

s = 21s′ .

So (1p, 1q) = (1

s, 1s′

) has value 1− 1s.

‖A‖`s′→`s′ = ‖A‖`s→`s ≤ ‖A‖1−θ`1→`∞ · ‖A‖θ`2→`2 ≤ 21− 1

s .

For s ≤ r ≤ s′ (???)

‖A‖`s→`r = ‖A‖1−θ`s→`s · ‖A‖θ`s→`s′ ≤ (21− 1

s′ )1−θ · (21− 1s′ )θ = 21− 1

s′ .

Theorem 16.3 (Minkowski’s inequality). Let Lp(`q) and `q(Lp) be the space

of functions with the norm

‖f‖Lp(`q) =

(∫ (∑k

|fk(ω)|q) pq

dµ

) 1p

,

‖f‖`q(Lp) =

(∑k

(∫|fk(ω)|p dµ

) qp

) 1q

.

If p ≤ q, then Lp(`q) ⊂ `q(Lp) and `p(Lq) ⊂ Lq(`p).

Proof. We want to show ‖f‖`q(Lp) ≤ ‖f‖Lp(`q), i.e.(∑k

(∫|fk(ω)|p dµ

) qp

) 1q

≤

(∫ (∑k

|fk(ω)|p) pq

dµ

) 1p

.

Let p ≤ q and r = qp≥ 1. The continuous version of triangle inequality says

‖∫g dµ‖r ≤

∫‖g‖r dµ. (Prove this first for simple function and approximation.)

Define g(ω) = |fk(ω)|q, then∥∥∥∥∫ g(ω) dµ

∥∥∥∥`r

≤∫‖g(ω)‖`r dµ

38

By definition of ‖ · ‖`r(∑k

(∫|fk(ω)|p dµ

)r) 1r

≤∫ (∑

k

|fk(ω)|pr) 1r

dµ,

so (∑k

(∫|fk(ω)|q dµ

) qp

) pq

≤∫ (∑

k

|fk(ω)|q) pq

dµ.

Taking q-th root on both sides gives the first inclusion. The second inclusion is proved

using triangle inequality in `p.

17 Uniform convexity of Lp 20210305

Generalize the scalar valued inequality to function valued inequality.

Theorem 17.1. For f , g ∈ Lp and r ≤ p ≤ s then

(‖f + g‖rp + ‖f − g‖rp)1r ≤ 21− 1

s · (‖f‖sp + ‖g‖sp)1s .

Proof. Recall (Minkowski inequality or generalized Fubini Theorem).

Lp(`r) ⊂ `r(Lp) if p ≤ r and (2)

`s(Lp) ⊂ Lp(`s) if s ≤ p (3)

Let f , g ∈ Lp(Ω,Σ, µ) then

LHS = (‖f + g‖rp + ‖f − g‖rp)1r

≤(∫|f(ω) + g(ω)|r + |f(ω)− g(ω)|r)

pr dµ

) 1p

(by (2))

≤ 21− 1s ·(∫|f(ω)|s + |g(ω)|s)

1s·p dµ

) 1p

(by Corollary (15.5))

≤ 21− 1s ·(∫

(|f(ω)|p)sp + (|g(ω)|p)

sp dµ

) 1s

= RHS. (by (3))

39

Now we show the above theorem implies uniform convexity.

Theorem 17.2. The space Lp is uniformly convex for 1 < p <∞. In particu-

lar, Lp is reflexive.

We need to show ∀ε > 0, ∃δ > 0 with ‖f‖p ≤ 1, ‖g‖p ≤ 1 and ‖f − g‖p > ε then

‖f+g2‖p ≤ 1− δ.

Example 17.3. When p = 2, X = L2(Ω,R). Fixing ε > 0, if we take δ = ε8

then

(‖f + g‖22 + ‖f − g‖22)12 ≤√

2 · (‖f‖22 + ‖g‖22)12 ≤√

2 ·√

2

and ‖f + g‖22 + ‖f − g‖22 > ‖f + g‖22 + ε2.

So ‖f + g‖22 + ε2 ≤ 4, i.e. ‖f+g2‖2 ≤

√1− ε2

4≤ 1− ε

8= 1− δ.

Proof. Assume p ≥ 2, s = min(p, p′) and r = max(p, p′), so that s ≤ p ≤ r. Fixing

ε > 0, and assume ‖f‖p ≤ 1, ‖g‖p ≤ 1 and ‖f − g‖p > ε. Then Theorem 17.1 gives

(‖f + g‖rp + ‖f − g‖rp)1r ≤ 21− 1

s · (‖f‖sp + ‖g‖sp)1s ≤ 21− 1

s · 21s = 2.

Same as previous example(∥∥∥∥f + g

2

∥∥∥∥rp

+( ε

2

)r) 1r

<

(∥∥∥∥f + g

2

∥∥∥∥rp

+

∥∥∥∥f − g2

∥∥∥∥rp

) 1r

≤ 1.

So we can choose δ (δ = O( ε2)r). Note that when p→∞, ( ε

2)r → 0.

Example 17.4. For 1 < p, q,∞, the Sobolov space

Wmp,q =

f ∈ C(R)

∣∣∣∣∣ ‖f‖ =

(∫ ( m∑k=1

∣∣f (k)(x)∣∣q) pq) l

q

<∞

is uniformly convex. Uniformly convex and reflexive properties pass to subspaces.

(Uniform convexity is a property of two points.) We can embeds Wmp,q into Lp(`

mq ) = Y

and show Y is uniformly convex.

40

Our goal is to find r, s so that(‖F +G‖rY + ‖F −G‖rY

) lr ≤ 21− 1

s ·(‖F‖sY + ‖G‖sY

) 1s .

We need the inclusions Lp(`q(`r)) ⊂ `r(Lp(`q)) and Ls(`p(`q)) ⊂ Lp(`q(`s)). These

require p, q ≤ r and s ≤ p, q. Hence s = min(p, q, p′, q′) and r = max(p, q, p′, q′).

Check this gives the above inequality.

18 Uniform Boundedness and Open Mapping 20210308

Theorem 18.1 (Uniform boundedness principle). Let X be a Banach space

and Y a normed vector space. Suppose that F is a collection of continuous

linear operators from X to Y . If F is pointwise bounded:

supT∈F‖T (x)‖Y <∞,∀x ∈ X

then F is norm-bounded:

supT∈F‖T‖B(X,Y ) = sup

T∈F ,‖x‖=1

‖T (x)‖Y <∞.

Application: If Tn ⊂ L(X, Y ) is a sequence such that limn Tn(x) = y exists for all

x, then supn ‖Tn‖ <∞.

Proof. Ref. Let F be a family and start with a subset (not a subspace)

Xn = x | supT∈F‖Tx‖ ≤ n ⊂ X.

Claim 18.2. Xn is closed.

Assume xα → x and we have ‖Txα‖ ≤ n for all α and T ∈ F . Since T is continuous,

lim ‖Tx‖ = lim supα ‖Txα‖ ≤ n (not clear what the first limit is taking with respect

to), and

‖Tx‖ = ‖T limαxα‖ = lim

α‖Txα‖ ≤ lim sup

α‖Txα‖ ≤ n.

41

Note that ∪nXn = X by assumption, and X1 ⊂ X2 ⊂ · · · ⊂ Xn.

Assume that the int(Xn) = ∅ for all n, then On = Xcn is dense for all n. Baire’s Cate-

gory Theorem gives ∩nOn is dense, in particular nonempty. But ∩nOn = (∪nXn)c = ∅gives a contradiction. So there exists n such that int(Xn) 6= ∅.

Take x0 ∈ X, δ > 0 and ‖y‖ < δ be such that make Bδ(x0) ⊂ Xn. Then ‖T (x0+y)‖ ≤n for all T ∈ F . Therefore

‖T (y)‖ =

∥∥∥∥T (x0 + y)− T (x0 − y)

2

∥∥∥∥ ≤ ‖T (x0 + y)‖+ ‖T (x0 − y)‖2

≤ n.

and

‖T (y)‖ =

∥∥∥∥T( y

‖y‖· δ

2

)∥∥∥∥ · 2‖y‖δ≤ n · 2‖y‖

δ

implies ‖T‖ ≤ 2nδ

=⇒ supT∈F ‖T‖ ≤ 2nδ

This argument also works for convex maps with values in another space.

A famous example is the following.

Example 18.3. Consider X = C[−π, π]. Define the truncation of Fourier series

Pn(f) =n∑

k=−n

f(k)eikt, where f(k) =1

2π

∫ π

−πf(t)e−ikt dt.

Note that Pn ∈ L(X,X). Recall in L2, ‖f‖2 =(∑

k |f(k)|2) 1

2 and Pn(f)→ f in L2.

If we had that Pn(f) → f uniformly: limn→∞

Pn(f) = f for all f ∈ X. That is, if Pnwere pointwise bounded: supn ‖Pn(f)‖ <∞, then uniform boundedness would imply

supn ‖Pn‖ < ∞. We will prove later that ‖Pn‖ ≥ C(1 + lnn) (see Theorem 19.1

below).

This gives a contradiction. So there exists a continuous f such that limn→∞

n∑k=−n

f(k)eikt

42

diverges. Another fact says that the space of trigonometric polynomials p(t) =n∑

k=−n

akeikt

are dense, and Pn(p)→ p uniformly. The partial Fourier series converges almost ev-

erywhere.

Let X be a Banach space, D ⊂ X. What does it mean to be bounded? Two answers

1. ∃ r such that D ⊂ RBX

2. D is weakly bounded: ∀x∗ ∈ X∗, x∗(D) ⊂ (−Rx, Rx) (or z | |z| ≤ Rx in

complex case)

With respect to the weak topology, weak bounded implies norm bounded.

Corollary 18.4. Let X be a Banach space, D ⊂ X such that x∗(D) is bounded in Kfor all x∗ ∈ X∗. Then D is bounded.

Proof. Let F = ϕx | x ∈ D ⊂ L(X∗,K), where ϕx(x∗) = x∗(x). We know

supx∈D

ϕx(x∗) = sup

x∈D|x∗(x)| <∞,∀x.

Uniform boundedness principle implies supx∈D‖ϕx‖ ≤ C. Then D is bounded, because

supx∈D‖x‖ = sup

x∈Dsup‖x∗‖≤1

|x∗(x)| = supx∈D‖ϕx‖ ≤ C.

Theorem 18.5 (Open mapping theorem). Let X and Y be Banach spaces and

T : X → Y be linear and surjective. Then T is open.

Proof. Step 1. Let ε > 0 and Yn = T (BX(0, nε)). Then Y = ∪nYn. Uniform

boundedness principle implies one of the Yn’s has nonempty interior. So there exists

x and δ > 0 such that, BY (x, δ) ⊂ Yn. WLOG we can assume x = 0, so BY (0, δ) ⊂ Yn.

Hence for some δ′ > 0, we have BY (0, δ) ⊂ T (BX(0, ε)). Our goal is to remove this

closure.

43

Step 2. Choose εk so that∑εk < ε. According to the previous step, we know that

there exists δk such that BY (0, δk) ⊂ T (BX(0, εk)) for all k. WLOG we can assume

δk → 0 because we can always take smaller value for δ’s.

Now let y ∈ Y with ‖y‖ < δ0. Since BY (0, δ0) ⊂ T (BX(0, ε0)) we can find x0 in

BX(0, ε0) such that ‖y− T (x0)‖ < δ1. Call y1 = y− T (x0). Then we can find we can

find x1 in BX(0, ε1) such that ‖y − T (x0)− T (x1)‖ = ‖y1 − T (x1)‖ < δ2. Iterate this

step and we have a sequence of xk such that

‖y − T (x0)− T (x1)− · · · − T (xk)‖ < δk. (4)

Since δk → 0, y =∑

k T (xk) by construction. Moreover∑

k xk converges to some

point x ∈ X because ‖∑

k xk‖ ≤∑

k ‖xk‖ ≤∑

k εk < ε < ∞ (completeness of

the Banach space). Note that ‖x‖ < ε and passing limit of inequality (4) gives

‖y− T (x)‖ = 0. So y = T (x) ∈ T (BX(0, ε)). This proves for all ε, there exists δ such

that BY (0, δ) ⊂ T (BX(0, ε)). ( trick Write x and y as converging sequences and use

the completeness of Banach spaces).

Step 3. Take O an open set.

Example 18.6.

• There exists a map T : `∞ → `2 which is linear and onto.

• There is no map T : `∞ → `4/3.

19 ‖Pn‖ is Unbounded 20210310

Theorem 19.1. Let X = C[−π, π] and let

Pn(f) =n∑

k=−n

f(k)eikt, where f(k) =1

2π

∫ π

−πf(t)e−ikt dt.

Then ‖Pn‖ ≥ C(1 + lnn).

44

Lemma 19.2. If T : C(K)→ C(K), then

‖T‖ = supx∈K‖T ∗(δx)‖C(K)∗ ,

where δx ∈ C(K)∗ is defined by δx(f) = f(x).

Proof. Certainly

‖T‖ = ‖T ∗‖ = supϕ∈C(K)∗,‖ϕ‖≤1

‖T ∗(ϕ)‖ ≥ supx∈K‖T ∗(δx)‖.

It remains to show “≤”.

Step 1. Take ϕ =∑

x αxδx, we first prove ‖ϕ‖ =∑

x |αx|. One one hand ‖ϕ‖ ≤∑x |αx| because

|ϕ(f)| =∣∣∣∑

x

αxf(x)∣∣∣ ≤∑

x

|αx| · |f(x)| ≤∑x

|αx| · ‖f‖∞.

To show the other direction, we need to find f(xj) = εj, with |ε| = 1, for a compact

topological space K. Recall Urysohn’s lemma.

Lemma 19.3 (Urysohn’s lemma). A topological space (X, τ) is normal if and only

if for every pair of disjoint nonempty closed subsets C,D ⊂ X there is a continuous

function f : X → [0, 1] such that f(x) = 0 for all x ∈ C and f(x) = 1 for all x ∈ D.

More generally, for Oi ⊂ X disjoint open subsets, and xi ∈ Oi, we can find a function

positive function f ∈ C(K) such that fi(xi) = 1, supp fi ⊂ Oi and∑

i fi = 1. Here

we only need K = [0, 1], fi(xi) = 1, supp fi ⊂ Oi and∑

i fi ≤ 1.

Define f(x) =∑

j εjfj(x), with |εj| = 1. Then

|ϕ(f)| =∣∣∣∑

j

αjδxj

∣∣∣ =∣∣∣∑

j

εjαjfj(xj)∣∣∣ =

∑j

|εj| · |αj| =∑j

|αj|.

45

Existence of such an f gives ‖ϕ‖ ≥∑

j |αj|. This shows that for disjoint xj’s,∥∥∥∑j αjδxj

∥∥∥ =∑

j |αj|.

Step 2. For an arbitrary ϕ ∈ C(K)∗. Recall we have the following extension

C(K) `∞(K)

K

ι

ϕ∃ϕ

Then there exists a family ϕα ⊂ `∞(K) with

ϕα(f) =∑j

λj(α)f(xj) and ‖ϕα‖∞ =∑j

|λj(α)| = 1.

Denote ϕ(f) = limα ϕα(f). Then ϕα → ϕ in σ(C(K)∗, C(K))-topology. This implies

for T : C(K1)→ C(K2),

‖T ∗(ϕ)‖ = sup‖f‖C(K1)

≤1|T ∗(ϕ)(f)| = sup

‖f‖C(K1)≤1|ϕ(T (f))|

= sup‖f‖C(K1)

≤1| limαϕα(T (f))| ≤ sup

‖f‖C(K1)≤1

lim supα|ϕα(T (f))|.

Note that

|ϕα(T (f))| =∣∣∣∑

j

λj(α) · (T (f))(xj)∣∣∣ =

∣∣∣∑j

λj(α) · T ∗(δxj)(f)∣∣∣

≤∑j

|λj(α)| · ‖T ∗(δxj)‖ ≤∑j

|λj(α)| · ‖f‖∞ ≤ supxj

‖T (δxj)‖.

This gives ‖T‖ = ‖T ∗‖ ≤ ‖T ∗(δx)‖ and thus the equality.

In short, we could use the fact that the convex hull of the δ measures are weak*-dense

in the unit ball of C(K)∗.

Lemma 19.4. Let K = [0, 2π], µ be a measure on K and F (x, y) be a continuous

functional in two variables. Define an integral operator T : C(K)→ C(K) by

TF (h)(x) =

∫K

F (x, y)h(y) dµ(y).

46

Then ‖TF‖ = supy∫K|F (x, y)| dµ(x).

Proof. We know ‖TF‖ = supx ‖T ∗F (δx)‖ and T ∗F (δx)(f) =∫KF (x, y)f(y) dµ(y). This

is given by integration against h(y) = F (x, y). Note that

‖T ∗F (δx)‖C(K)∗ = ‖h‖L1(µ) =

∫K

|F (x, y)| dµ(x).

Often this lemma is used for groups: G is a compact group and µ a measure on G.

We can prove the existence of Haar measure, which means there exists µ such that∫f(gh) dµ(h) =

∫f(h) dµ(h) for all g. The integral is invariant under translation.

In the case of K = [−π, π], this is the Lebesgue measure λ.

Lemma 19.5. Let f : G→ G be a continuous map, and define a translation invariant

operator T : C(G)→ C(G) by

Tf1(f2)(g) =

∫K

f1(gh−1)f2(h) dµ(h).

Then ‖Tf1‖ =∫|f1(h)| dµ(h).

The norm does not see translation by g and thus the supremum disappears. In

particular, on K = [−π, π],

Tf1(f2)(s) =

∫K

f1(s− t)f2(t)1

2πdt.

Proof. Let F (g, h) = f1(gh−1), and Tf1 as above. Then the previous lemma and right

invariant suggests

‖Tf1‖ = supg

∫K

|f1(gh−1)| dµ(x) =

∫K

|f1(h−1)| dµ(x).

47

Proof of Theorem 19.1. Consider Pn(f) =n∑

k=−n

f(k)eikt, where f(k) =1

2π

∫ π

−πf(t)e−ikt dt.

By substitution

Pn(f) =n∑

k=−n

f(k)eikt =n∑

k=−n

∫ π

−πf(s)e−iks

1

2πds · eikt =

∫ π

−π

n∑k=−n

eik(t−s)f(s)1

2πds.

Thus we take f1(s) =∑n

k=−n eik(t−s). By previous Lemma,

‖Pn‖ =

∫ π

−π

∣∣∣∣∣n∑

k=−n

eik(t−s)

∣∣∣∣∣ ds.Sum of geometric series gives for s 6= 0,

n∑k=−n

eiks =e−ins − ei(n+1)s

1− eis.

Multiplying both sides by e−is/2 we get∣∣∣∣e−i(n+1/2)s − ei(n+1/2)s

eis/2 − e−is/2

∣∣∣∣ =

∣∣∣∣sin((n+ 1/2)s)

sin(s)

∣∣∣∣.Note that sin(s) ∼ s when s ∼ 0, and | sin(ns)| ∼ 1 when s ∼ π

2n+ 2lπ, l ∈ N.

So there are sj’s such that on the interval Ij = s | |s − sj| ≤ 14πn, sj ∼ jπ

2nand

| sin(ns)| ≥ 14. This implies∫ ∣∣∣∣sin(ns)

sin(s)

∣∣∣∣ ds ≥ n∑j=1

1

4

∫Ij

dλn

j=

n∑j=1

1

4|Ij|

n

j∼ const.

n∑j=1

1

j.

So the integral is unbounded.

Remark 19.6.

1.∫ π−π fn(s) ds = 0. (fn should be refering to the oscillating function sin(ns) but

I’m not sure).

2. The kernel K(t, s) =∑n

k=−n hk(t)hk(s) appears a lot in solutions of PDEs.

48

20 Krein–Milman Theorem 20210312

Lecture recording missing. I’ll try to type the lecture notes later.

Let X be a locally convex topological space. Recall that a set C ⊂ X is convex if

and only if for all x, y ∈ C, 0 ≤ λ ≤ 1, we have λx+ (1− λ)y ∈ C.

Definition 20.1. Let C be convex set. A point x ∈ C is called extreme if x =

λy + (1 − λ)z with y, z ∈ C implies λ ∈ 0, 1 or y = z = x. We denote the set of

extreme points of C as Ext(C).

Warning: the set of extreme points need not to be closed.

Remark 20.2. Let x1, x2, · · · , xm ∈ Rn. Then

Ext( conv(xi) | 1 ≤ i ≤ m ) ⊂ x1, x2, · · · , xm.

Theorem 20.3 (Krein-Milman Theorem). Let X be a locally convex topological vector

space and let C be a nonempty, convex, compact subset of X. Then C is equal to the

closure of the convex hull of the extreme points of C, i.e. C = conv(Ext(C)). In

particular, C contains the extreme points.

21 Krein–Milman Theorem Cont. 20210315

Applications of Krein–Milman Theorem

Example 21.1. Consider K = T ∈ L(`n1 , `n1 ) | ‖T‖ ≤ 1 . Krein-Milman Theorem

gives K = conv(Ext(K)). The map T has an associated matrix A : Rn → Rn.

Compute

‖T‖ = sup∑j |αj |≤1

∑i

∣∣∑j

aijαj∣∣ ≤∑

i

∑j

|aij| · |αj| ≤ maxj

∑i

|aij|.

49

This implies K = B(L(`n1 , `n1 )) = B(`n∞(`n1 )) =

∏B(`n1 ) (the RHS is the product of n

copies of B(`n1 )). Recall

Ext(K1, K2, · · · , Kn) = (x1, x2, · · · , xn) | xj ∈ Ext(Kj) .

So Ext(B(`n1 )) = ±ek | 1 ≤ k ≤ n implies

Ext(K) = (ε1ek1 , ε2ek2 , · · · , εnekn) | kj ∈ 1, · · · , n and |ε| = 1 .

Write this as a matrix

A =

a11 a12 · · · a1n

a21 a22 · · · a2n...

.... . .

...

an1 an2 · · · ann

,

where aij ∈ 0, εj. The extreme points are matrices with exactly one entry of

absolute value one in each column (reputation in rows is allowed).

Example 21.2. Let K to be the set of bistochastic matrices. That is,

K = A = (aij)1≤i;j,≤n | aij ≥ 0,∑i

aij = 1,∀i, and∑j

aij = 1,∀j ,

A =

1 1 1 1

1 a11 a12 · · · a1n

1 a21 a22 · · · a2n

1...

.... . .

...

1 an1 an2 · · · ann

.

These matrices are contained in the set

S = A = (aij) | ‖T‖ ≤ 1 and ‖T t‖ ≤ 1 .

Clearly the identity matrix and more generally all permutation matrices are extreme

points of S. By Birkhoff’s Theorem (which we will not prove), the extreme points of

S are exactly the permutation matrices.

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For n = 2, there is a nice decomposition for the permutation matrices, namely(a b

b a

)= a

(1 0

0 1

)+ b

(0 1

1 0

).

The general case is proved by the Hall’s Marriage Theorem.

Example 21.3. Let K be the set of non-increasing convex functions. That is,

K = f | f ′ ≤ 0, f ′′ ≥ 0 and f(0) = 1 .

The extreme points are in Ext(K) = e−λx | λ ≥ 0 .

The prove is not so simple. One needs first to show these exponential functions are

extreme points and then use the fact that every function can be writen as a convex

combination f(x) =∫g(x, y)e−xy dµ(y) with

∫g(y) dµ(y) = 1.

Lemma 21.4 (Contraction lemma). Let C be a subset of a locally convex and Haus-

dorff topological space X. Let x ∈ int(C), y ∈ ∂C and λ ∈ [0, 1), then z =

(1− λ)x+ λy ∈ int(C).

Proof. Assume X is a normed space. (The proof for a general case is similar using

semi-norms.) Fix λ ∈ [0, 1) WLOG assume x = 0. For any y ∈ int(C), there exists

δ > 0 so that the set

B = (1− λ)0 + λ(y +Bδ) = λy + λBδ = γ = λ(y + a) | ‖a‖ ≤ δ ⊂ C.

Now choose a sequence yn → y.

Claim 21.5. There exists n ∈ N so that λy + λ δn2B ⊂ C.

Let zn = (1− λ)0 + λyn, then

‖z − zn‖ = ‖λ(y − yn)‖ = λ‖y − yn‖ → 0.

So for some n, ‖z−zn‖ < λ δn2

, z ∈ zn+λ δn2Bδn ⊂ C and z+λ δn

2Bδn ⊂ zn+λδnBδn ⊂ C.

So z ∈ int(C).

The following proof was done in previous lecture.

51

Proof of Theorem 20.3

We had

F = L | L nonempty proper convex open subset of C .

Step 1. Lmax = L in the family F .

Step 2. Define the affine map T xλ (y) = (1− λ)x+ λy (preserves convex combination).

T xλ (C) ⊂ L for all x ∈ C and λ ∈ [0, 1).

• Observe that if L is maximal, then L = C. Using contradiction: T xλ (L) ⊂L =⇒ L(⊂ T xλ )−1(L) which is open and convex.

• T xλ (L) ⊂ int(L) by contraction lemma.

Step 3. If O is open, then O ∪ L = C or O ⊂ L. If O ⊂ L done. Otherwise O ∪ L is

open convex. Take a, b ∈ O then aλ + b(1 − λ) ∈ O ⊂ O ∪ L, similarly for a, b ∈ L.

If a ∈ O, b ∈ L and 0 < λ < 1, then aλ + b(1− λ) = T bλ(a) ∈ L because a ∈ C = L.

Maximally says O ∪ L is not a subset so O ∪ L = C.

Step 4. C \ L has at most one point. Take two disjoint neighborhoods V1, V2 and

apply the alternative to L ∪ V1. If L ∪ V1 = C, then V2 ⊂ L ∪ V1, which gives a

contradiction.

Step 5.

Lemma 21.6. Let l : X → R be a continuous real linear functional. Then

supx∈C

l(x) = supx∈Ext(C)

l(x).

Proof. By compactness the above supreme is attained at some x0 ∈ C. Consider the

set F = x ∈ C | l(x) = l(x0) . This F is a face, because C \ F = x ∈ C | l(x) <

l(x0) is open. Also C \ F is convex: x1, x2 ∈ C \ F , then l(λx1 + (1 − λ)x2) =

λl(x1) + (1−λ)l(x2) < l(x0). So C \F ∈ F . By Zorn’s lemma, there exists Lmax such

that C \ F ⊂ Lmax and we know L = C \ x1 for some x1 ∈ Ext(C). This means

x1 6∈ C \ F . So l(x1) = l(x0).

52

Now we can finish the proof of Krein-Milman Theorem: C = conv(Ext(C)).

Suppose this does not hold, that is K := conv(Ext(C)) is contained properly in C.

Then there exists x0 ∈ C \K. By Hahn-Banach, there exists a linear functional l so

that l(x0) > supx∈K l(x) ≥ supy∈Ext(C) l(y). But l(x0) = l(x1) for some x1 ∈ Ext(C).

This gives a contradiction.

Corollary 21.7. For a convex continuous function f ,

supx∈C

f(x) = supx∈Ext(C)

f(x).

Proof. This is because we can write x = limα

∑i λ

αi xi, and estimate

f(x) ≤ limα

∑i

λαi f(xi) ≤ supx∈Ext(C)

f(x).

53