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DUAL

• Linear Functionals

• Duals of Some Common Banach Spaces

• Hahn-Banach Theorem (Extension of Linear Functionals)

• The Dual of C[a,b]

• Second Dual

• Alignment and Orthogonal Complements

• Minimum Norm Problems

• Hahn-Banach Theorem (Geometric Form)

EL3370 Dual Spaces - 1

LINEAR FUNCTIONALS

Definition In a real !V , !f :V →  is a linear if for all !x, y ∈V , !α ,β ∈, !f (α x + β y)= α f (x)+ β f ( y)

Examples n n 1. In ! , every linear functional is of the form f (x)= α x , where x = (x ,…,x ) ! ∑k=1 k k ! 1 n 2. In !C[0,1], !f (x)= x(1 2) is a linear functional b 3. In L (a,b), f (x)= y(t)x(t)dt , for any y ∈L (a,b), is a linear functional ! 2 ! ∫a ! 2

EL3370 Dual Spaces - 2

LINEAR FUNCTIONALS (CONT.)

Definition. A linear functional !f in a normed space is bounded if there is an !M > 0 such that ! f (x) ≤ M x for all !x ∈V . The smallest such !M is the norm of !f , ! f

Theorem. Let !f be a linear functional on a normed space !V . The following are equivalent: (1) !f is continuous (2) !f is continuous at !0 (3) !f is bounded

Proof (1)⇒(2): By definition (2)⇒(1): Let x → x , where x ,x ∈V . If f is continuous at 0, then f (x )− f (x) = f (x − x) → f (0) =0, n n n n hence f is continuous at x (2)⇒(3): Let f be continuous at 0. There is a δ >0 such that f (x) ≤1 for all x ∈V such that x ≤δ . Then, by linearity, f (x) ≤(1 δ) x for all x ∈V (3)⇒(2): If f is bounded and x is an arbitrary convergent sequence to 0, then there is an M >0 such n that f (x ) ≤ M x →0. Then, f is continuous at 0 ■ n n

EL3370 Dual Spaces - 3

LINEAR FUNCTIONALS (CONT.)

Example In l (the space of sequences with finitely non-zero entries), with norm equal to the ! 0 maximum of the absolute value of its components, define for x = (x ,…,x ,0,0,…), ! 1 n n f (x)= kx . This linear functional is unbounded ! ∑k=1 k

Linear functionals on !V form a vector space, called algebraic dual of !V , by defining

( f + f )(x)= f (x)+ f (x) 1 2 1 2 ! (λ f )(x)= λ f (x), x ∈V

EL3370 Dual Spaces - 4

LINEAR FUNCTIONALS (CONT.)

Definition The space of bounded linear functionals on a normed space !V is the normed dual of !V , denoted as !V ∗ . The norm of an f ∈V ∗ is f = sup f (x) ! ! x ≤1

Theorem. !V ∗ is a

Proof (for real normed spaces). We only need to show that V ∗ is complete. Take a Cauchy sequence {x * } in V ∗ . For every x ∈V , {x *(x)} is Cauchy in !, since x *(x)− x * (x) ≤ x * − x * x , so there is an n n n m n m x *(x)∈ such that x∗(x)→ x *(x). The functional x * defined in this way is linear, because ! n x∗(αx +β y)= lim x∗(αx +β y)=αlim x∗(x)+β lim x∗( y)=αx∗(x)+βx∗( y). In addition, for a given ε >0, n n n since {x * } is Cauchy, there is an N ∈ ! such that x∗ − x∗ <ε for all n,m≥ N , or x∗(x)− x∗ (x) <ε x for n n m n m all x ∈ X , hence taking n→∞ gives x∗(x)− x∗ (x) <ε x . Thus, x∗(x) ≤ x∗(x)− x∗ (x) + x∗ (x) < m m m ε + x∗ x , so x∗ is bounded, i.e., x∗ ∈V ∗ . Finally, as x∗(x)− x∗ (x) <ε x , we have x∗ → x∗ . Therefore, ( m ) m m V ∗ is complete ■

EL3370 Dual Spaces - 5

DUALS OF SOME COMMON BANACH SPACES

n • ! (with Euclidean norm): itself!

Take the linear functional

n f (x)= α x , x = (x ,…,x ) ! ∑k=1 k k ! 1 n

n n From Cauchy-Schwarz, f (x) = α x ≤ α 2 x , so every f is bounded, and since ! ∑k=1 k k ∑k=1 k ! n equality is achieved for x = (α ,…,α ), we have f = α 2 ! 1 n ! ∑k=1 k

Conversely, if f ∈(n )∗ , take the standard basis vectors e . Every x = (x ,…,x ) can be ! ! i ! 1 n n n n written as x = x e , so f (x)= x f (e )= α x , with α = f (e ) ! ∑k=1 k k ! ∑k=1 k k ∑k=1 k k ! k k

n Hence the of ! is itself, in the sense that all bounded functionals are of the n n form f (x)= α x , with f = α 2 ! ∑k=1 k k ! ∑k=1 k

EL3370 Dual Spaces - 6

DUALS OF SOME COMMON BANACH SPACES (CONT.)

• l (1≤ p < ∞): l ! ! p ! ! q

The dual of l is l , where 1 p+1 q = 1, in the sense that every bounded linear functional ! p ! q ! ∞ in l can be written as f (x)= α x , with α = {α }∈l , and f = α q ! p ! ∑k=1 k k ! n q !

The dual of l is not l : see bonus slides for proof ∞ 1

• L (a,b) (1≤ p < ∞): L (a,b)! ! p ! ! q

Similarly to l , the dual of L (a,b) is L (a,b), with 1 p+1 q = 1, since every bounded linear ! p ! p ! q ! b functional is of the form f (x)= x(t)y(t)dt , with y ∈L (a,b), and f = y ! ∫a ! q ! q

EL3370 Dual Spaces - 7

DUALS OF SOME COMMON BANACH SPACES (CONT.)

• Dual of a !H

A particular linear functional in !H is !f (x)= (x, y) for !y ∈H . By Cauchy-Schwarz, ! f (x) ≤ x y , and taking !x = y shows that ! f = y . Conversely,

∗ Theorem (Riesz-Fréchet). If !f ∈H , there is a unique !y ∈H such that !f (x)= (x, y) for all !x ∈H , and ! f = y

Proof. If f =0, the theorem is trivial. Otherwise, the set N = {x ∈ H : f (x)=0} is closed (why?) and not ⊥ ⊥ equal to H . Since H = N ⊕ N , take a z ∈ N \{0}, scaled so that f (z)=1. We will show that y is a multiple of z . Given x ∈ H , we have x − f (x)z ∈ N , since f (x − f (x)z)=0. As z ⊥ N , we have 2 (x − f (x)z,z)=0, or (x,z)= f (x)(z,z), hence y = z z . If (x, y )=(x, y ) for all x ∈ H then y = y , 1 2 1 2 which proves the uniqueness of y ■

EL3370 Dual Spaces - 8

HAHN-BANACH THEOREM (EXTENSION OF LINEAR FUNCTIONALS)

Definitions 1. Let !f be a linear functional defined on a subspace !M of a vector space !V . An extension !F of !f (from !M to !N ) is a linear functional defined on N ⊇ M , !N ≠ M , such that !F M = f

2. A real valued function p defined on !V is a sublinear functional if a) p(x + x )≤ p(x )+ p(x ), for all x ,x ∈V ! 1 2 1 2 ! 1 2 b) !p(α x)= αp(x), for all !α ≥ 0 and !x ∈V

Theorem (Hahn-Banach; extension form) Let !V be a real vector space, and !p a sublinear functional on !V . Let !f be a linear functional defined on a subspace M ⊆V satisfying !f (x)≤ p(x) on !M . Then there is an extension !F of !f from !M to !V such that !F(x)≤ p(x) on !V

EL3370 Dual Spaces - 9

HAHN-BANACH THEOREM (EXTENSION OF LINEAR FUNCTIONALS) (CONT.)

Proof Let us define a partial order ≺ on the set E of extensions of f satisfying the conditions of the theorem: say that F ≺ G if G is an extension of F . For every chain C ⊆ E , according to this partial order, define Fˆ as Fˆ(x)= F(x) if x ∈D(F) for some F ∈C . Fˆ is a linear functional with domain ∪ D(F), F∈C and it is an upper bound for the chain C . Thus, by Zorn’s lemma, E has a maximal element F

The linear functional F should have domain equal to V , since otherwise there is an element y ∈V \D(F ). All x ∈ lin{D(F )+ y} are of the form x = m+α y with m ∈D(F ) and α ∈ ! . An extension of F from D(F ) to lin{D(F )+ y} has the form g(x)= F(m)+αg( y), and we can choose g( y) so that g(x)≤ p(x) on lin{D(F )+ y} (see details in the bonus slides), thus contradicting the maximality of F . This contradiction shows that F is the sought extension of f to V ■

Remark In general it is not possible to avoid the use of Zorn’s lemma (or the axiom of choice), since in fact it can be shown that the Hahn-Banach theorem is equivalent to the axiom of choice

EL3370 Dual Spaces - 10

HAHN-BANACH THEOREM (EXTENSION OF LINEAR FUNCTIONALS) (CONT.)

Corollary 1 Let !f be a bounded linear functional defined on a subspace !M of a normed space !V . Then there is an extension !F of !f to !V such that ! F = f

Proof. Take p(x)= f x in Hahn-Banach theorem ■

Corollary 2 ∗ Let !x ∈V . Then there is a nonzero !F ∈V on a normed space !V such that !F(x)= F x

∗ Proof. Assume x ≠0 (otherwise, take any F ∈V ). On lin{x}, define the bounded functional ∗ f (αx)=α x , which has norm 1. By Corollary 1, extend f to an F ∈V with norm 1 ■

The Hahn-Banach theorem, particularly as Corollary 1, can be used as an existence theorem for minimization problems

EL3370 Dual Spaces - 11

THE DUAL OF !C[a,b]

Definition !x :[a,b]→  is of bounded variation if there is a !K > 0 such that for every partition of ![a,b] n (i.e., a finite set {t ,…,t } such that a = t

A consequence of Hahn-Banach theorem is that the dual of !C[a,b] is (almost) !BV[a,b]:

Theorem (Riesz Representation Theorem) ∗ Let !f ∈C[a,b] . Then there is !ν ∈BV[a,b] such that

b f (x)= x(t)dν(t), x ∈C[a,b] ! ∫a

∗ and ! f = TV(ν). Conversely, every !ν ∈BV[a,b] defines an !f ∈C[a,b] in this way

EL3370 Dual Spaces - 12

THE DUAL OF !C[a,b] (CONT.)

Proof b Given f ∈ C[a,b]∗ , we need to find a ν ∈ BV[a,b] such that f (x)= x(t)dν(t) ∫a A natural idea is to set x =u ( s ∈[a,b]), where u (t)=1 for t ∈[a,s] and 0 otherwise, so s s

b s f (u )= u (t)dν(t)= dν(t)=ν(s)−ν(a) s ∫a s ∫a and we can set ν(a)=0 and ν(s)= f (u ) s

Unfortunately, u ∉ C[a,b]. However, u belongs to the space B of bounded functions on [a,b], and s s C[a,b]⊆ B , so by Hahn-Banach it is possible to obtain an extension F of f from C[a,b] to B , preserving its norm. This will allow us to define ν(s)= F(u )! s

b The rest of the proof is to show that: ν ∈ BV[a,b], f (x)= x(t)dν(t), and f =TV(ν) (see bonus ∫a slides for the details) ■

EL3370 Dual Spaces - 13

THE DUAL OF !C[a,b] (CONT.)

Riesz Representation Theorem does not provide a unique !ν ∈BV[a,b], since, e.g., !f (x)= x(1/2) can be described by

⎧ 0, 0≤t <1 2 ⎪ v(t)= ⎨ α , t = 1 2 ⎪ 1, 1 2

Definition !NBV[a,b]⊆ BV[a,b], normalized space of functions of bounded variation, consists of those !ν ∈BV[a,b] such that !ν(a)= 0 and are right continuous on !(a,b). Here ! v = TV(v)

Then, !NBV[a,b] is the dual of !C[a,b]

EL3370 Dual Spaces - 14

SECOND DUAL SPACE

∗ ∗ ∗ ∗ Let !x ∈V . For any !x ∈V , denote !< x,x >:= x (x) (resembling an inner product) ∗ This notation also allows to define a functional !f on !V (for given !x ∈V ):

∗ ∗ ∗ ∗ !f (x )=< x,x >, x ∈V

∗ ∗ ∗ This !f is linear, and since ! f (x ) = < x,x > ≤ x x , so ! f ≤ x , and by Hahn-Banach, ∗ ∗ ∗ ∗ there is an !x ∈V such that !< x,x > = x x , so ! f = x

Definition ∗∗ ∗ ∗ ∗ !V = (V ) , second dual of a normed space !V , is the set of bounded linear functionals on !V

∗∗ ∗∗ As we saw, there is a natural mapping !ϕ :V →V given by !ϕ(x)= x , where ∗ ∗∗ ∗ !< x ,x >=< x,x >. This mapping is linear and “norm-preserving”: !ϕ(x) = x . However, ϕ is not always surjective, i.e., there may be an !x∗∗ ∈V ∗∗ not represented by elements of !V

Definition. A normed space !V is reflexive if ϕ is surjective. If so, we write !V = V ∗∗

EL3370 Dual Spaces - 15

SECOND DUAL SPACE (CONT.)

Examples

1. l and L (1< p < ∞) are reflexive, since (l )∗∗ = (l )∗ = l , where 1 p+1 q = 1 ! p ! p ! ! p q p !

2. l and L are not reflexive ! 1 ! 1

3. Every Hilbert space is reflexive

EL3370 Dual Spaces - 16

ALIGNMENT AND ORTHOGONAL COMPLEMENTS

∗ ∗ ∗ ∗ ∗ For !x ∈V and !x ∈V , !< x,x >≤ x x . In Hilbert spaces, we have equality iff !x is ∗ represented by a nonnegative multiple of !x , i.e., !< x,x >= (x,α x) for some !α ≥ 0. In a normed space !V , we define

∗ ∗ ∗ ∗ Definition. !x ∈V is aligned with !x ∈V if !< x,x >= x x

Example Let !x ∈C[a,b], let !Γ := {t ∈[a,b]: x(t) = x } ≠ ∅ b An x∗(x)= x(t)dν(t) is aligned with !x iff ! ∫a ν only varies on Γ so that ν is nondecreasing at !t if !x(t)> 0 and nonincreasing if !x(t)< 0 (exercise!)

EL3370 Dual Spaces - 17

ALIGNMENT AND ORTHOGONAL COMPLEMENTS (CONT.)

∗ ∗ ∗ Definition. !x ∈V and !x ∈V are orthogonal if !< x,x > = 0

In Hilbert spaces, this coincides with the original definition of

⊥ ∗ ∗ Definition. The orthogonal complement of !S ⊆ V , !S , is the set of all !x ∈V such that ∗ !< x,x > = 0 for all !x ∈S

For subsets of !V ∗ , we have a “dual” definition:

* ⊥ Definition. The orthogonal complement of !U ⊆ V , ! U , is the set of all !x ∈V such that ∗ ∗ !< x,x > = 0 for all !x ∈U

⊥ ⊥ ∗ ⊥ ∗∗ ⊥ Notice that !U ≠ U (!U ⊆ V ) in general, unless !V is reflexive, since !U ⊆ V , while ! U ⊆ V

⊥ ⊥ Theorem. If !M is a closed subspace of !V , then ! [M ]= M (see proof at the end)

EL3370 Dual Spaces - 18

MINIMUM NORM PROBLEMS

Goal: Extend the projection theorem to minimum norm problems in normed spaces

The situation is more difficult: possibly several optima, given by nonlinear equations, ...

Example Let !V = 2 with norm ( y , y ) = max{ y , y }, and let M = {(α ,0): α ∈}. Then ! 1 2 1 2 ! min (2,1)− y = 1, but the optimum is not unique! ! y∈M

EL3370 Dual Spaces - 19

MINIMUM NORM PROBLEMS (CONT.)

Theorem Let !x ∈V , where !V is a real normed space, and let !d be the distance from !x to a subspace !M . Then,

d = inf x − m = max < x,x∗ > m∈M x∗ ≤1 ! x∗∈M⊥ where the max is achieved for some x∗ ∈M ⊥ . Also, if the inf is achieved for some m ∈M , ! ! 0 ! ! 0 then x∗ is aligned with x − m ! 0 ! 0

Proof (≥) Take ε >0, and m ∈ M such that x − m ≤ d + ε . Then, for every x∗ ∈ M ⊥ , x∗ ≤1: ε ε

< x,x∗ >= < x −m ,x∗ > ≤ x −m x∗ ≤ d +ε ε ε

∗ ∗ ⊥ ∗ and since ε was arbitrary, we have < x,x >≤ d for all x ∈ M , x ≤1

EL3370 Dual Spaces - 20

MINIMUM NORM PROBLEMS (CONT.)

Proof (cont.) (≤) Let N = lin{x +M}. If n ∈ N , then n=αx +m for some m ∈ M , α ∈ ! . Define the functional f on N as f (n)=αd . Then

f (n) α d d d f = sup = sup = sup = =1 n∈N n m∈M αx +m m∈M x +m α inf x +m α m∈M

By Hahn-Banach, let x∗ be a norm-preserving extension of f to V . Then x∗ =1 and x∗ = f on N . 0 0 0 By construction, x∗ ∈ M ⊥ and < x,x∗ >= d 0 0

Now, let m ∈ M be such that x −m = d , and take x∗ ∈ M ⊥ such that x∗ =1, and < x,x∗ > = d . Then, 0 0 0 0 0 < x −m ,x∗ > = < x,x∗ > = d = x −m x∗ , so x∗ is aligned with x −m ■ 0 0 0 0 0 0 0

EL3370 Dual Spaces - 21

MINIMUM NORM PROBLEMS (CONT.)

Corollary (generalized projection theorem) Let !x ∈V and !M be a subspace of !V . Then m ∈M satisfies x − m ≤ x − m for all !m∈M iff ! 0 ! 0 there is an x∗ ∈M ⊥ aligned with x − m ! 0 ! 0

The following is a “dual” version of the previous theorem:

Theorem Let !M be a subspace of !V , and !x∗ ∈V ∗ , which is at a distance !d from !M ⊥ . Then

d = min x∗ − m∗ = sup < x,x∗ > ∗ ⊥ m ∈M x∈M ! x ≤1 where the min is achieved for m∗ ∈M ⊥ . If the sup is achieved for some x ∈M , then x∗ − m∗ ! ! 0 ! ! ! 0 is aligned with x ! 0

EL3370 Dual Spaces - 22

MINIMUM NORM PROBLEMS (CONT.)

Example (control problem, again) Consider the DC motor problem again, where we want to find the current !u:[0,1]→  to drive a motor governed by

  !θ(t)+θ(t)= u(t) from θ(0)=θ(0)= 0 to θ(1)= 1, θ(1)= 0, so as to minimize max u(t) ! ! ! ! t∈[0,1]

We can formulate this problem in a dual space, by forcing u∈(L )∗ = L , so that we seek a ! 1 ∞ u∈(L )∗ = L of minimum norm ! 1 ∞

As in the previous example, the constraints are

1 t−1 t−1 e u(t)dt = < y1 ,u > = 0, y1(t)= e ∫0 1 (1− et−1 )u(t)dt = < y ,u > = 1, y (t)= 1− et−1 !∫0 2 2

EL3370 Dual Spaces - 23

MINIMUM NORM PROBLEMS (CONT.)

Example (cont.) Therefore, the optimization problem is

L = min u = min u! −u = min u! −u u L u L ⊥ ∈ ∞ !∈ ∞ u!∈M < y1 ,u>=0 < y1 ,u!>=0 < y ,u>=1 < y ,u!>=0 ! 2 2 where !u is any function in L satisfying the constraints: < y ,u > = 0, < y ,u > = 1, and ! ∞ ! 1 ! 2 M = lin{ y , y } ! 1 2

From the theorems:

L = min u = sup < x,u > = sup < ay + by ,u > = sup b u L 1 2 ∈ ∞ x M ∈ ay1+by2 ≤1 ay1+by2 ≤1 < y1 ,u>=0 x ≤1 < y ,u>=1 ! 2

1 and the constraint ay + by ≤1, in L , corresponds to (a− b)et−1 + b dt ≤1 ! 1 2 ! 1 !∫0

EL3370 Dual Spaces - 24

MINIMUM NORM PROBLEMS (CONT.)

Example (cont.) This means that the optimal value is equal to

sup b 1 (a−b)et−1+b dt≤1 !∫0 which is attained and can be solved explicitly (since it is simple and finite dimensional!)

t−1 Qualitatively, the optimal !x(t)= (a− b)e + b is aligned with the optimal !u, i.e., !< x,u > = x u L x Since the right side is the maximum of the left side t over all u∈L with u = L, !u has to take only values 0 1 ! ∞ ! !±L, depending on !sgn(x(t)): “Bang-bang” control! u L As !x is the sum of an exponential and a constant, it can change sign at most once, so !u has to change sign at most once

EL3370 Dual Spaces - 25

;#;E<8#E#(;%=;)CI)D%>G)CD)=IH(%/CID@

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!"##$%&'()*&+,)-./&0&2S&

HAHN-BANACH THEOREM (GEOMETRIC FORM)

By definition, a hyperplane !H is either closed or dense in !V , since H ⊆ H ⊆V , and H is also a linear variety (why?), so either !H = H or !H = V , because H is maximally proper (i.e., it cannot be inside a larger but proper linear variety)

Theorem. Let f ≠ 0 be a linear functional on !V . Then, for every !c , !H = {x : f (x)= c} is closed iff !f is continuous

−1 Proof. If f is continuous, then H = f ({c}) is closed since {c} is closed for every c ∈! Conversely, as H = {x : f (x)= c} is a hyperplane, there is a y ∈V \H , and every x ∈V fulfills x =α y +h, with α ∈ ! and h∈ H . Then, consider a sequence {x } in V , with x =α y +h (α ∈ !, h ∈ H ), such n n n n n n that x → x =α y +h∈V (α ∈ ! , h∈ H ). Since H is closed, d = inf y −h >0 (from Exercise set 1), so n h∈H x − x = (α −α)y +h −h ≥ α −α d , which implies that α →α , hence f (x )=α f ( y)→α f ( y)= f (x) , n n n n n n n so f is continuous ■

See bonus slides for an example of a linear functional for which H is dense

EL3370 Dual Spaces - 27

HAHN-BANACH THEOREM (GEOMETRIC FORM) (CONT.)

A hyperplane !H = {x : f (x)= c} generates the half-spaces

!{x : f (x)≤ c}, !{x : f (x)≥ c}, !{x : f (x)< c}, !{x : f (x)> c}

If !f is continuous, the first !2 are closed and the last two are open

Our goal is to establish a geometric Hahn-Banach theorem: Given a convex set K such that intK ≠ ∅, and x ∉intK , ! ! ! 0 x0 there is a closed hyperplane containing x but disjoint from ! 0 K !intK

If !K were !B(0,1), the theorem is easy, since Hahn-Banach assures the existence of an x∗ aligned with x : if x ∈intB(0,1), ! 0 ! 0 ! < x,x∗ > ≤ x x ∗ < x x ∗ = < x ,x∗ >, so H = {x : < x,x∗ > = < x ,x∗ >} is enough ! 0 0 0 0 0 0 ! 0 0 0

We thus need to generalize the unit ball !B(0,1) to arbitrary convex sets!

EL3370 Dual Spaces - 28

HAHN-BANACH THEOREM (GEOMETRIC FORM) (CONT.)

Definition (Minkowski functional) Let !K be a convex set in a normed space !V , and assume !0∈intK . The Minkowski functional p:V → !+ of !K on !V is 0

p(x)= inf r >0: x ∈ rK { }

Properties (assuming !0 ∈intK ) (1) !0≤ p(x)< ∞ for all !x ∈V (2) !p(α x)= αp(x) for !α > 0 (3) p(x + x )≤ p(x )+ p(x ) ! 1 2 1 2 (4) !p is continuous (5) !K = {x : p(x)≤1}, !intK = {x : p(x)<1}

!p is then a continuous sublinear functional

(see bonus slides for proofs of these properties)

EL3370 Dual Spaces - 29

HAHN-BANACH THEOREM (GEOMETRIC FORM) (CONT.)

Theorem (Mazur’s Theorem, Geometric Hahn-Banach) Let !K be a convex set in a real normed space !V such that !intK ≠ ∅. Let !W be a linear variety such that !W ∩intK = ∅. Then there is a closed hyperplane containing !W but no ∗ ∗ ∗ interior points of !K , i.e., there is an x ∈V \{0} and !c ∈ such that !< w,x > = c for !w ∈W , ∗ ∗ < k,x > ≤ c for k ∈ K , and !< k,x > < c for !k ∈intK

Proof. Idea: use the Minkowski functional p in the Hahn-Banach theorem! Assume that 0∈intK . Let M = linW . Then, W is a hyperplane in M which does not contain 0, so there is a linear functional !f on M such that W = {x : f (x)=1}

Let p be the Minkowski functional of K . Since W has no interior points of K , f (x)=1≤ p(x) for x ∈W . Every point in M can be written as αx , where α ∈ ! and x ∈W , so if α >0, f (αx)=α ≤ p(αx) (by homogeneity), and if α <0, f (αx)≤0≤ p(αx), so f (x)≤ p(x) in M

By Hahn-Banach, there is an extension F of f to V such that F(x)≤ p(x). Let H = {x : F(x)=1}. Since F(x)≤ p(x) on V , and p is continuous, so is F (because −p(−x )≤ F(x )≤ p(x ), so if x → 0, then n n n n F(x )→ 0), and F(x)<1 on intK , while F(x)≤1 on K . Therefore, H is the desired hyperplane ■ n

See bonus slides for example of why !intK ≠ ∅ is needed in Mazur’s theorem

EL3370 Dual Spaces - 30

HAHN-BANACH THEOREM (GEOMETRIC FORM) (CONT.)

Some corollaries

1. (Supporting Hyperplane Theorem). If x ∉intK ≠ ∅, there is a closed hyperplane !H containing !x such that !K lies on one side of !H (If !x ∈∂K , !H is a supporting hyperplane) Proof. A special case of Mazur’s theorem, where W = {x} ■

2. (Eidelheit Separating Hyperplane Theorem). If K and K are convex, ! 1 ! 2 such that intK ≠ ∅ and K ∩intK = ∅. Then there is closed ! 1 ! 2 1 hyperplane separating K and K (i.e., an x∗ ∈V ∗ such ! 1 ! 2 ! that sup x,x∗ inf x,x∗ ) x∈K < > ≤ x∈K < > ! 1 2 Proof. Let K = K +(−K ). intK ≠ ∅ , and 0∉intK , so apply Mazur with W = {0} ■ 1 2

3. If !K is closed and convex, and !x ∉K , there is a closed halfspace that contains !K but not !x Proof. d = inf x − y > 0, so B(x,d /2) does not intersect K . Apply y∈K Corollary 2 with K = K and K = B(x,d /2) ■ 1 2

4. (Dual formulation of convex sets). A closed convex set is equal to the intersection of all the closed half-spaces that contain it Proof. Follows directly from Corollary 3 ■

EL3370 Dual Spaces - 31

BONUS: DETAILS OF PROOF OF HAHN-BANACH THEOREM

To show that !g( y) can be chosen so that !g(x)≤ p(x) on !lin{M + y}, notice that for every m ,m ∈M , f (m )+ f (m )= f (m + m )≤ p(m + m )≤ p(m − y)+ p(m + y), or ! 1 2 ! 1 2 1 2 1 2 ! 1 2

f (m )− p(m − y)≤ p(m + y)− f (m ) ! 1 1 2 2 so sup [ f (m)− p(m− y)]≤ inf [p(m+ y)− f (m)]. Take a !c ∈ between these bounds. ! m∈M m∈M For !x = m+α y , we then define !g(x)= f (m)+αc . If !α > 0,

⎡ ⎛ m⎞ ⎤ ⎡ ⎛ m⎞ ⎛ m ⎞ ⎛ m⎞ ⎤ ⎛ m ⎞ f (m)+αc = α ⎢ f ⎜ ⎟ + c⎥ ≤α ⎢ f ⎜ ⎟ + p⎜ + y⎟ − f ⎜ ⎟ ⎥ = αp⎜ + y⎟ = p(m+α y) ! ⎣ ⎝ α ⎠ ⎦ ⎣ ⎝ α ⎠ ⎝ α ⎠ ⎝ α ⎠ ⎦ ⎝ α ⎠ and if !α = −β < 0,

⎡ ⎛ m⎞ ⎤ ⎡ ⎛ m⎞ ⎛ m ⎞ ⎛ m⎞ ⎤ ⎛ m ⎞ f (m)− βc = β ⎢ f ⎜ ⎟ − c⎥ ≤ β ⎢ f ⎜ ⎟ + p⎜ − y⎟ − f ⎜ ⎟ ⎥ = βp⎜ − y⎟ = p(m− β y) ⎣ ⎝ β ⎠ ⎦ ⎣ ⎝ β ⎠ ⎝ β ⎠ ⎝ β ⎠ ⎦ ⎝ β ⎠

Thus !f (m)+αc ≤ p(m+α y) for all α , and !g is an extension of !f to !lin{M + y}

EL3370 Dual Spaces - 32

⊥ ⊥ BONUS: PROOF THAT ! [M ] = M

⊥ ⊥ It is clear that M ⊆ [M ]

Let x ∉ M . On lin{x +M}, define the linear functional f (αx +m)=α for all m ∈ M . Then,

f (x +m) 1 f = sup = m∈M x +m inf x +m m∈M

∗ * and since M is closed, f <∞. Thus, by Hahn-Banach, we can extend f to an x ∈V . Since f (x)=0 on ∗ ⊥ ∗ ⊥ ⊥ M , x ∈ M , but also < x,x > =1, so x ∉ [M ] ■

EL3370 Dual Spaces - 33

BONUS: PROOF THAT l IS NOT THE DUAL OF l ! 1 ∞

Let c be the subspace of l consisting of sequences x = {x } such that x →0. We will first show that 0 ∞ n n ∞ c∗ = l . First, let z ∈l and define ϕ :c → ! so that ϕ (x)= z x . ϕ is linear, and bounded, since 0 1 1 z 0 z ∑n=1 n n z ∞ ϕ (x) = z x ≤ z 1 x ∞ , and if x = sgn(z ), then ϕ (x) = z 1 x ∞ , so ϕ = z 1 . Conversely, if z ∑n=1 n n n n z z N f :c → ! is a bounded linear functional, define z = {z } by z = f (e ); then, if v = sgn(z )e , 0 n n n ∑n=1 n n N z = f (v) ≤ f v ∞ = f , so taking N →∞ we conclude that z ∈l . Furthermore, for every x ∈ c , ∑n=1 n 1 0 ∞ N N ∞ f (x)= f x e = f lim x e = lim x f (e )= z x =ϕ (x) (by continuity of f ), so (∑n=1 n n ) ( N→∞ ∑n=1 n n ) N→∞ ∑n=1 n n ∑n=1 n n z there is a one-to-one correspondence between c∗ and l 0 1

Now we show that there are elements of l∗ which are not of the form ϕ with z ∈l . Since c ⊆ l , ∞ z 1 0 ∞ and x ∈l , x =1, does not belong to c , Mazur’s theorem with K = B(x,1/2) and W = c implies the ∞ n 0 0 existence of a f ∈l∗ such that f (c )=0 and f (x)=1; this f cannot be written in the form ϕ , since ∞ 0 z otherwise z =0 and f ≡0, which is not true ■

EL3370 Dual Spaces - 34

BONUS: DETAILS OF DERIVATION OF DUAL OF !C[a,b]

(1) ν ∈ BV[a,b]: Let a =t

n n ⎛ n ⎞ n

v(ti )−v(ti 1 ) = εi[v(ti )−v(ti 1 )]= F⎜ εi[ut −ut ]⎟≤ F εi[ut −ut ] = F = f ∑ − ∑ − ∑ i i−1 ∑ i i−1 i 1 i 1 ⎝ i 1 ⎠ i 1 = = = != ##"##$ =1 so ν ∈ BV[a,b], with TV(v)≤ f

b n (2) f (x) x(t)d (t): For x C[a,b], let z x(t )[u u ] B . Then, z x = ν ∈ =∑ i−1 t − t ∈ − B = ∫a i=1 i i−1 max max x(t ) x(t ) 0 as max t t 0, since x is uniformly continuous. Also, continuity i t ≤t≤t i−1 − i → i i − i−1 → i−1 i n b of F yields F(z)→ F(x)= f (x), but F(z)= x(t )[v(t − v(t )]→ x(t)dv(t), by definition of the ∑i=1 i−1 i i−1 ∫a b Riemann-Stieltjes integral, so f (x)= x(t)dv(t) ∫a

n (3) f =TV(ν): Since F(z) = x(t )[v(t −v(t )] ≤ x TV(v), since F(z)→ F(x)= f (x), we obtain ∑i=1 i−1 i i−1 f (x) ≤ x TV(v), i.e., f ≤TV(v). Combining this with (1) yields f =TV(v) ■

EL3370 Dual Spaces - 35

BONUS: EXAMPLE OF A DENSE HYPERPLANE

It is easy to build examples of closed hyperplanes, since these are the contour surfaces of bounded functionals. Finding dense hyperplanes is a bit harder. Here is an example:

Consider the space c ⊆ l of sequences that converge to 0, and let {e } be the standard 0 ∞ n n∈! basis of these spaces. Let v = {1,1 2,1 3,1 4,…}∈c . The set {v ,e ,e ,…} is linearly 0 0 0 1 2 independent (recall that linear independence refers to finite linear combinations); this set can be complemented with a set {b } ⊆ c to obtain a Hamel basis, i.e., a linearly i i∈I 0 independent set {v ,e ,e ,…}∪{b } such that every x ∈c can be written as a finite linear 0 1 2 i i∈I 0 combination of elements of this set (recall the application of Zorn’s Lemma from Lecture 1)

Define a functional f :c → ! by 0

∞ f α v + α e + β b = α ( 0 0 ∑n=1 n n ∑i∈I i i ) 0

Note that f ≠ 0 (because f (v )= 1≠ 0), and since e ∈Ker f for every n∈!, Ker f = 0 n {x ∈c : f (x)= 0} is a dense hyperplane in c . Exercise: prove that f is unbounded 0 0

EL3370 Dual Spaces - 36

BONUS: PROOF OF PROPERTIES OF MINKOWSKI FUNCTIONAL

(1) 0≤ p(x)< ∞ for all x ∈V : By definition, p(x)≥ 0. Also, since 0∈intK , there is an δ > 0 such that B(0,δ )⊆ K , so for every ε >0, x ∈( x δ + ε)B(0,δ )⊆ ( x δ + ε)K , hence p(x)≤ x δ + ε for all ε > 0, so p(x)≤ x δ < ∞

(2) p(αx)= αp(x) for α > 0: p(αx)= inf{r >0:αx ∈ rK }= inf{αr! >0: x ∈ r!K }=αp(x)

(3) p(x + x )≤ p(x )+ p(x ): Given ε > 0, let p(x )≤ r ≤ p(x )+ ε so x /r ∈K (i = 1,2). Then, 1 2 1 2 i i i i i (x + x )/(r + r )= [r /(r + r )](x /r )+ [r /(r + r )](x /r )∈K , since this is a convex combination of 1 2 1 2 1 1 2 1 1 2 1 2 2 2 x /r and x /r , and K is convex. Hence, p(x + x )≤ r + r ≤ p(x )+ p(x )+2ε , and since ε > 0 was 1 1 2 2 1 2 1 2 1 2 arbitrary, p is sublinear

(4) p is continuous: By (1), if x → 0, 0≤ p(x )≤ x δ → 0, so p is continuous at 0. Also, given x, y ∈V , n n n p(x)= p(x − y + y)≤ p(x − y)+ p( y) and p( y)≤ p( y − x)+ p(x), so −p( y − x)≤ p(x)− p( y)≤ p(x − y). Thus, if x → x , −p(x − x)≤ p(x)− p(x )≤ p(x − x ), so p(x )→ p(x) and p is continuous n n n n n

(5) K = {x : p(x)≤1}, intK = {x : p(x)<1}: If x ∈K , p(x)≤1, so if x ∈K → x ∈V , by (4), p(x)≤1, thus n K ⊆ {x :p(x)≤1}, and if x ∈intK , there is an ε > 0 such that (1+ ε)x ∈K , so intK ⊆ {x :p(x)<1}. Conversely, if x ∉K , there is an ε ∈(0,1) such that (1−δ)x ∉ K for all 0≤δ ≤ε , hence p(x)>1 and −1 K = {x :p(x)≤1}, and if p(x)<1, x ∈K , so p ((0,1)) is a nbd of x , thus intK = {x :p(x)<1} ■

EL3370 Dual Spaces - 37

8CE"&S%[;Y%!C%[)%E))!%! " !" %HE%D#\"IW&%=;)CI)D]%

! " = ! !"#$ !" !" !" & !" 3;&!"#$ 9&:7&:/&487&,8//:D*.&:4&E.4.5)*&78&;:4B&)&>=,.5,*)4.& &-847):4:4E& &/(->&7>)7& & (30+2,.<,?-7:4,%*:./&84&84.&/:B.&8;& & & !" = ! " = # # # ! ! # = " = # # # ! ! # = # = " ! ( ! ( $ & & ! " = !( !" ! $# =& "& $ & 34& 9&*.7& "# % % ' ) *+&)4B& "# % % ' ) *+9&)/&/>8@4&:4& ! ! !" !" =# ! !" 7>.& ;:E(5.& D.*8@<& a87.& 7>)7& !"#$ <& O>:*.& ! "#$ 9& 7>.5.& :/& 84*=& 84.& >=,.5,*)4.& -847):4:4E&7>.&,*)4.& &H4)?.*=&:7/.*;G& I9&)4B& &*:./&84&D87>&/:B./&8;&:7& & & "

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!"##$%&'()*&+,)-./&0&#V&

BONUS: WEAK∗ CONVERGENCE

A sequence {x } in a normed space V is said to (strongly) converge to x ∈V if x − x →0 n n

Another notion of convergence in V is: {x } converges weakly to x ∈V if < x ,x∗ >→< x,x∗ > n n as n→∞ for all x∗ ∈V ∗ . If x → x strongly, then < x ,x∗ > − < x,x∗ > ≤ x∗ x − x →0, so n n n x → x weakly too n

Example. In l , consider the sequence {e } with e = {0,…,0,1,0,…}, with 1 in the n-th ! 2 n n place. For every x∗ =(η ,η ,…), < e ,x * >=η →0 as n→∞, so e →0 weakly. However, 1 2 n n n e −0 =1 for every n, so e does not converge to 0 strongly n n

In the dual space V ∗ , we can similarly define strong and weak convergence (since it is a normed space), but an even weaker notion is available:

{x ∗ } converges in weak∗ -sense to x∗ ∈V ∗ if < x,x∗ >→< x,x∗ > as n→∞ for all x ∈V n n

Weak convergence in V ∗ implies weak ∗ convergence, since in the former one requires that < x ,x∗ >→< x,x∗ > for all x ∈V ∗∗ , where V ∗∗ ⊇V n

EL3370 Dual Spaces - 39

BONUS: WEAK∗ CONVERGENCE (CONT.)

Example. For V = c ⊂ l , the set of sequences that converge to 0, V ∗ = l and V ∗∗ = l 0 ∞ ! 1 ∞ (why?). In V ∗ = , e →0 weak ∗ , but not weakly, since < x ∗ ,x∗∗ >→/ 0 for x∗∗ =(1,1,…,)∈V ∗∗ !l1 n

Weak ∗ convergence corresponds to point-wise convergence for linear functionals, so it has the relative topology on V ∗ of the product topology of !V = !; in particular, the ∏ x∈V weak ∗ topology is Hausdorff but not first-countable in general. Also,

Theorem (Banach-Alaoglu). The closed unit ball in V ∗ , K , is weak ∗ -compact Proof. Since K = {x∗ ∈V ∗ : x∗(x) ≤ x for all x ∈V }=V ∗ ∩ [− x , x ]=:V ∗ ∩D, and D is compact by ∏ x∈V Tychonoff, it suffices to show that K is closed. If f ∈ K ∩D, pick x, y ∈V , α,β ∈ ! , and ε >0, and let g ∈ K such that f (x)− g(x) <ε , f ( y)− g( y) <ε and f (αx +β y)− g(αx +β y) <ε . Then, f (αx +β y)−α f (x)−β f ( y) ≤ f (αx +β y)− g(αx +β y) + α g(x)− f (x) + β g( y)− f ( y) <(1+ α + β )ε and f (x) ≤ f (x)− g(x) + g <ε +1, and since ε,x, y,α,β were arbitrary, f is linear and f ≤1, so f ∈ K and K is closed ■

Weak*-compactness is useful for establishing existence of solutions of optimization problems involving a weak∗ -continuous functional on a dual space

EL3370 Dual Spaces - 40

BONUS: WEAK∗ CONVERGENCE (CONT.)

∗ ∗ ∗ ∗ The weak* topology of V is the weakest in which functionals f (x )=< x,x > on V , for some x ∈V , are continuous (i.e., every other such topology contains the open sets of the ∗ ∗ weak* topology). This is because for f (x ) = < x,x > <ε to hold, the topology should ∗ ∗ ∗ contain open sets of the form {x ∈V : < x,x > <ε} for all x ∈V , as well as finite intersections of these sets, which generate the weak* topology. Conversely,

∗ ∗ ∗ Theorem. Every weak*-continuous linear functional on V has the form f (x )=< x,x > for some x ∈V

Proof. Let f be such a functional. Then, for every ε >0, there is a set U = {x∗ : < x ,x∗ > <δ,…, < x ,x∗ > <δ} where x ,…,x ∈V , such that if x∗ ∈U , f (x∗ ) <ε . Thus, by 1 n 1 n linearity, f (x∗ ) ≤(ε /δ)max { < x ,x∗ > }, so if < x ,x∗ >=0 for all i , then f (x∗ )=0; assume now i∈{1,…,n} i i w.l.o.g. that {x ,…,x } is l.i. Let F(x∗ )=(< x ,x∗ >,…,< x ,x∗ >, f (x∗ )). Since there is a nbd of (0,…,0,1) 1 n 1 n n+1 T ∗ ∗ not intersecting R(F), by Mazur’s theorem there is a λ ∈ ! \{0} such that λ F(x )=0 for all x , i.e., n < λ x ,x∗ >+λ f (x∗ )=0 for all x∗ , and since {x ,…,x } are l.i., λ ≠0, so ∑i=1 i i n+1 1 n n+1 n f (x∗ )=< (−λ /λ )x ,x∗ > ■ ∑i=1 i n+1 i

EL3370 Dual Spaces - 41

BONUS: WEAK∗ CONVERGENCE (CONT.)

∗ ∗ Let B,B be the closed unit balls in V ,V , respectively. The following is a converse of the Banach-Alaoglu theorem:

∗ ∗ Theorem (Krein-Smulian). If E ⊆V is convex and such that E ∩(rB ) is weak*-compact for every r >0, then E is weak*-closed

Proof Firstly note that E is norm-closed, since if {x∗ } is a sequence in E that converges to x∗ ∈V ∗ , then {x∗ } n n is bounded, i.e., x∗ ≤ M for some M > 0, thus x∗ ∈ E ∩(M B∗ ) for all n, so x∗ ∈ E ∩(M B∗ )⊆ E . This n n ∗ means that if x ∈V \E , there is a ball centered at x which does not intersect E ; thus, by translation and scaling, the theorem will be proven by showing that “if E ∩B∗ = ∅, then there exists an x ∈V such ∗ ∗ ∗ ∗ ∗ that < x,x >≥1 for every x ∈E ”: since {x : < x,x >≥1} is a weak -closed half-space, E is the intersection of closed half-spaces, and hence it is weak ∗ -closed

∗ ∗ Next, given a subset F ⊆ X , define its polar P(F):= {x : < x,x > ≤1 for all x ∈F}; notice that the −1 ∗ intersection of all sets P(F) as F ranges over all finite subsets of r B is exactly rB

EL3370 Dual Spaces - 42

BONUS: WEAK∗ CONVERGENCE (CONT.)

Proof (cont.) To establish the proposition, let F = {0}, and assume that finite sets F ,…,F have been chosen such 0 0 k−1 that F ⊆ i−1B and P(F )∩!∩P(F )∩E ∩kB∗ = ∅; note that this holds for k = 1. Let Q = i 0 k−1 P(F )∩!∩P(F )∩E ∩(k+1)B∗ ; if P(F)∩Q ≠ ∅ for every finite set F ⊆ k−1B , the weak ∗ -compactness 0 k−1 ∗ of Q implies (via the finite intersection property) that kB ∩Q ≠ ∅ , which contradicts the properties of F ,…,F . Hence, there is a finite set F ⊆ k−1B such that P(F )∩Q = ∅, which yields the sequence {F } 0 k−1 k k k

By construction, {F } satisfies E ∩∩∞ P(F )= E ∩P(∪F )= ∅ . Now, arrange the elements of ∪F in a k k=1 k k k sequence {x }; note that x →0. Define T :V ∗ → c (the space of sequences converging to 0) by k k 0 Tx∗ = {< x ,x∗ >}. T(E) is a convex subset of c , and, due to E ∩P(∪F )= ∅ , Tx∗ = sup < x ,x∗ > ≥1 for n 0 k n n every x∗ ∈E . Thus, since c∗ = l (see previous bonus slides), by the separating hyperplane theorem 0 1 ∞ ∞ there is a y ∈l (i.e., y < ∞ ) such that = y < x ,x∗ > ≥1 for all x∗ ∈E . The vector 1 ∑n=1 n ∑n=1 n n ∞ x = y x is well defined (why?) and it satisfies the condition < x,x∗ >≥1 for every x∗ ∈E ■ ∑n=1 n n

EL3370 Dual Spaces - 43

8CE"&S%ZI)HE

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BONUS: KREIN-MILMAN AND CARATHÉODORY THEOREMS (CONT.)

The Krein-Milman theorem is one of the cornerstones of , as it extends to very general vector spaces some of the intuitive results of Euclidean geometry shown in the previous slide. To state it, we first need some definitions:

Definitions - : a vector space V over F with a Hausdorff topology such that the addition +:V 2 →V and scalar multiplication ⋅:F ×V →V operations are continuous. Every normed space induces a topological vector space

- Locally convex topological (LCT) vector space: a real topological vector space V such that for every x ∈V and every nbd U of x , there is a convex nbd W ⊆U of x

- V ∗ (dual of a topological vector space V ): set of continuous linear functionals on V

- Extreme point of a convex set K : a point x ∈ K such that if x = λ y +(1−λ)z with y,z ∈ K and λ ∈[0,1], then either λ =0 or λ =1

- conv(X) (convex hull of a subset X of a real vector space): set of all convex combinations (i.e., expressions α x +!+α x where α ,…,α ≥0, α =1 and n ∈ !) of points in X 1 1 n n 1 n ∑i i

EL3370 Dual Spaces - 45

BONUS: KREIN-MILMAN AND CARATHÉODORY THEOREMS (CONT.)

In locally convex topological vector spaces, linear functionals can separate points:

Theorem (separation of points) ∗ If V is an LCT vector space, V separates points in V (i.e., if x, y ∈V , x ≠ y , there is an ∗ f ∈V such that f (x)≠ f ( y))

Proof. Since V is an LCT vector space, all properties of Minkowski functional of a convex set hold. In particular, if U is a convex set and 0∈ int U , then p is finite, because scalar multiplication is continuous, so for every z ∈V there is an α ∈ ! such that αz ∈U

Therefore, Mazur’s theorem applies to LCT vector spaces, as well as its 4 corollaries. Corollary 3, in particular implies this result, by taking K = { y} ■

Corollary (separation of a convex set and a point) ∗ If K is a closed, convex set in an LCT vector space V , and y ∈V \K , then there is an f ∈V such that f (x)≤ c for all x ∈ K , and f ( y)> c

EL3370 Dual Spaces - 46

BONUS: KREIN-MILMAN AND CARATHÉODORY THEOREMS (CONT.)

Theorem (Krein-Milman) Let V be an LCT vector space, and K ⊆V a convex, compact and non-empty set. Then: (i) K has at least one extreme point (ii) K is the closure of the convex hull of its extreme points

Proof. (i) Call a set E ⊆ K extreme if it is convex, compact, non-empty and such that if x ∈ E is a convex combination of y,z ∈ K , then both y,z ∈ E . Since K is extreme, the collection F of all extreme subsets of K is not empty. Partially order {E } by set inclusion; we will show that every subchain {E } j j of F has a lower bound in F , ∩E . ∩E is clearly convex, compact (as it is the intersection of compact j j sets), and if x ∈∩E is a convex combination of y,z ∈ K , then for every j , x ∈ E , so both y,z ∈ E , j j j and hence y,z ∈∩E . It remains to show that ∩E is non-empty: if ∩E = ∅ , then ∪E C = K , but since K j j j j C C C is compact, there is a finite subcollection {E j } of {E j } whose union is K ; however, since {E j } is k k C totally ordered, there is a largest extreme set E j = K , i.e., E j = ∅, which is a contradiction k k By Zorn’s lemma, K has a minimal extreme subset E . If E had at least two points, by the theorem on ∗ separation of points, there is an f ∈V that separates them

EL3370 Dual Spaces - 47

BONUS: KREIN-MILMAN AND CARATHÉODORY THEOREMS (CONT.)

Proof (cont.) Since E is compact and f is continuous and non-constant, it achieves its maximum, say, f , on some max proper subset M of E , which is closed (and hence compact) due to the continuity of f . M is convex, and if x ∈ M satisfies x = λ y +(1−λ)z for some y,z ∈ K and λ ∈[0,1], then both y,z ∈ E and furthermore f = f (x)= λ f ( y)+(1−λ)f (z), so f ( y)= f (z)= f and y,z ∈ M ; thus, M is a smaller max max extreme subset of K than E , a contradiction, so E has only one point, which is an extreme point of K

(ii) Let K be the set of all extreme points of K , and C = conv(K )⊆ K . If x ∉ C , by the corollary on the e e e e separation of a convex set and a point, there is an f ∈V ∗ such that f ( y)≤ c for all y ∈ C and f (x)> c . e As before, f achieves its maximum over K , f , on some closed non-empty set E ⊆ K . If y ∈ E max satisfies y = λz+(1−λ)w for some z,w ∈ K and λ ∈[0,1], then f = f ( y)= λ f (z)+(1−λ)f (w), so max f (z)= f (w)= f and z,w ∈ E , hence E is extreme. By part (i), E should contain at least one extreme max point, p, which belongs to K ⊆ C , hence f = f (p)≤ c . Therefore, since f (x)> c ≥ f , x ∉ K . Thus, e e max max C ⊇ K , which implies that C = K ■ e e

EL3370 Dual Spaces - 48

BONUS: KREIN-MILMAN AND CARATHÉODORY THEOREMS (CONT.)

Example: c is not the dual space of a Banach space 0 By Banach-Alaoglu, the closed unit ball of a dual space is weak∗ -compact, as well as convex and non-empty, hence by Krein-Milman, it should contain at least one extreme point However, the closed unit ball of c does not have extreme points: indeed, let x ∈c , x ≤1. 0 0 By definition, x = {x }, with x → 0, so there is an N ∈! such that x <1/2 for all n ≥ N . n n n Let y1 , y2 ∈c be such that y1 = y2 = x for n ≤ N and y1 = x +2−n , y2 = x −2−n for n > N . 0 n n n n n n n 1 2 1 2 1 2 Then, y ≤1, y ≤1, x = (1/2)( y + y ), but y ≠ x , y ≠ x , so x is not an extreme point

EL3370 Dual Spaces - 49

BONUS: KREIN-MILMAN AND CARATHÉODORY THEOREMS (CONT.)

To apply Krein-Milman, the following lemma is often useful:

Lemma. If V is an LCT vector space, K ⊆ V a compact and convex set, and F ⊆ K a compact set such that conv(F)= K , then the extreme points of K are contained in F

Proof. Let x be an extreme point of K not in F . Since F is compact, there is a nbd U of 0 such that 0 (x +U )∩F = ∅ , and a convex nbd U of 0 such that U −U ⊆ U , so (p+U)∩(F +U)= ∅, hence p ∉F +U . 0 0 The family { y +U} is an open cover of F , so by compactness, { y +U} is a finite subcover. Let y∈F i i=1,…,n Q = conv([ y +U]∩K )⊆ y +U ; note that Q is closed, and hence a compact subset of K , therefore i i i i

K = conv(Q ∪!∪Q )= conv(Q ∪!∪Q ) 1 n 1 n

(This result is proven by induction on n: for n = 2, the mapping ϕ :[0,1]×Q ×Q → conv(Q ∪Q ) given 1 2 1 2 by ϕ(α , y , y )= α y +(1−α )y is continuous, and [0,1]×Q ×Q is compact, hence ϕ([0,1]×Q ×Q )= 1 2 1 2 1 2 1 2 conv(Q ∪Q ) is compact, so conv(Q ∪Q )= conv(Q ∪Q )) 1 2 1 2 1 2

n Since x ∈K , x = α x for x ∈Q , α ≥ 0, α +!+α = 1. But x is an extreme point, so x = x for ∑k=1 k k k k k 1 n k some k , which implies that x ∈Q ⊆ y +U ⊆ F +U , a contradiction ■ k k

EL3370 Dual Spaces - 50

BONUS: KREIN-MILMAN AND CARATHÉODORY THEOREMS (CONT.)

In finite dimensions, the Krein-Milman theorem can be made more explicit:

Theorem (Carathéodory) N Let X ⊆ ! . Every point x ∈ conv(X) can be written as a convex combination of at most N +1 points from X

Proof. Let x ∈ conv(X). Then, x = λ x +!+λ x , where λ ,…,λ >0, λ +!+λ =1 and x ,…,x ∈ X . If 1 1 n n 1 n 1 n 1 n n> N +1, the points x − x ,…,x − x are linearly dependent, i.e., µ (x − x )+!+µ (x − x )=0 for some 2 1 n 1 2 2 1 n n 1 µ ,…,µ ∈ ! , not all zero. Defining µ = −µ −!−µ , we have that µ x +!+µ x =0 and µ +!+µ =0. 2 n 1 2 n 1 1 n n 1 n Since not all µ ’s are zero, there is at least one µ >0. Then, x = λ x +!+λ x −α(µ x +!+µ x )= i 1 1 n n 1 1 n n (λ −αµ )x +!+(λ −αµ )x . Pick α as the minimum of λ /µ , over the indices j for which µ >0; 1 1 1 n n n j j j note that α >0 and λ −αµ ≥0 for all j , but λ −αµ =0 for at least one index. Thus, j j j j x =(λ −αµ )x +!+(λ −αµ )x , where each coefficient is non-negative, their sum is one, and one of 1 1 1 n n n them is zero. Applying this procedure iteratively one can write x as a convex combination of at most N +1 points ■

EL3370 Dual Spaces - 51

BONUS: KREIN-MILMAN AND CARATHÉODORY THEOREMS (CONT.)

Application: Optimal multisine spectrum In experimental design, one searches for an input cumulative spectrum Φopt (a non- decreasing function such that Φ(0)=0) such that the information matrix I of a parameter F d π θ ∈ ! is maximized in some sense, under a power constraint Φ(π )= dΦ(ω)=1. Now, ∫0

π I (Φ)≈ F(ω)dΦ(ω)∈ !d×d , F(ω): symmetric positive semi-definite matrix F ∫0

Note that, under the power constraint, I (Φ) lies in the convex hull of {F(ω):ω ∈[0,π ]}, F which has dimension d(d +1)/2 (since F(ω) is a symmetric matrix). Thus, by Carathéodory’s Theorem, I (Φopt ) is a convex combination of at most n= d(d +1)/2+1 F matrices F(ω ),…,F(ω ), i.e., for some λ ,…,λ ≥0, 1 d(d+1)/2+1 1 n

opt π ⎡ ⎤ IF (Φ )= λ1F(ω1 )+!+λnF(ωn )= F(ω)⎣λ1δ(ω −ω1 )+!+λnδ(ω −ωn )⎦dω ∫0 where δ is the Dirac distribution. Therefore Φopt can be replaced by a “multisine” spectrum with frequencies at ω ,…,ω to obtain the same maximal information matrix! 1 n

EL3370 Dual Spaces - 52

BONUS: POSITIVE LINEAR FUNCTIONALS

Let X be a topological space. A linear functional F on C(X) is positive, denoted F ≥0, if F( f )≥0 whenever f (x)≥0 for all x ∈ X . Positive linear functionals are called Borel measures on X , and those of unit norm are called Borel probability measures on X

Lemma (Cauchy-Schwarz inequality for positive linear functionals) 2 2 2 If F :C(X)→ ! is a positive linear functional, and f , g ∈ C(X), then [F( fg)] ≤ F( f )F(g ) 2 2 2 2 Proof. Since F[(λx + y) ]= λ F(x )+2λF(xy)+F( y )≥0 for all λ ∈ !, the discriminant 2 2 2 4[F( fg)] −F( f )F(g ) is non-positive ■

Lemma (characterization of positive linear functionals) A linear functional F :C(X)→ ! is positive iff F(1)= F , where 1: x ∈ C(X)!1 Proof. (⇒) If F ≥0, then F(1)≤ F , since 1 =1, and, by Cauchy-Schwarz, for every f ∈ C(X) of unit 2 2 2 2 2 2 norm, [F( f )] =[F(1 f )] ≤ F(1 )F( f )≤ F(1) F , since f ≤ f ≤1; taking the supremum over all f of 2 unit norm, F ≤ F(1) F , so in conclusion F(1)= F (⇐) If f ∈ C(X) satisfies f (x)≥0 for all x , and f ≤1, then also 1(x)− f (x)≥0, and 1− f ≤1, so F −F( f )= F(1)−F( f )= F(1− f )≤ F , so F( f )≥0. By linearity, F is positive ■

EL3370 Dual Spaces - 53

BONUS: POSITIVE LINEAR FUNCTIONALS (CONT.)

Lemma (Jordan decomposition for linear functionals) Every l ∈ C(X)∗ can be written as l = l −l , where l ,l ∈ C(X)∗ are positive, and l = l + l + − + − + − ∗ ∗ Proof. Let S = {l ∈ C(X) :l(1)= l ≤1} and U = {l ∈ C(X) : l ≤1}; by Banach-Alaoglu, S and U are weak ∗ -compact. We will first show that U equals K = conv(S∪(−S)). If l ∈ K , then l =αl −(1−α)l , 1 2 with l ,l ∈ S and α ∈[0,1] (since S is convex, every convex combination in S∪(−S) can be reduced to 1 2 this form); hence l ≤α l +(1−α ) l ≤1, so K ⊆ U . Furthermore, K is weak ∗ -compact: indeed, define 1 2 ϕ :S × S ×[0,1]→U as ϕ(l ,l ,α)= αl −(1−α)l ; this map is continuous and S × S ×[0,1] is the product of 1 2 1 2 compact sets, hence it is compact, and K =R(ϕ) is compact. Suppose now that l ∈U \K . By the ∗ corollary on the separation of a convex set and a point, there is a weak -continuous functional f on ∗ C(X) such that f (l)> c and f (m)≤ c for all m ∈ K . This f is of the form f (m)= m(x) for some x ∈ C(X), and since K is symmetric (i.e., m ∈ K iff −m ∈ K ), m(x) ≤ c for all m ∈ K ˆ ∗ ˆ ˆ Let t ∈ X such that x(t) = max{ x(τ ) :τ ∈ X}, and define m ∈ C(X) as m(x)= x(t), hence m≥0 and mˆ =1, so x = mˆ (x) = sup m(x) . Thus, x ≤ c but l(x)> c , contradicting that l ≤1; hence K =U m∈S Now, take an l ∈ C(X)∗ ; assume w.l.o.g. that l =1. As l ∈U = conv(S∪(−S)), l =αl −(1−α)l for some 1 2 l ,l ∈ S and α ∈[0,1]. Define the positive functionals l =αl and l =(1−α)l , and note that l + l = 1 2 + 1 − 2 + − α l +(1−α) l ≤1= l = l −l ≤ l + l ■ 1 2 + − + −

EL3370 Dual Spaces - 54

BONUS: POSITIVE LINEAR FUNCTIONALS (CONT.)

Lemma (Extreme points of Borel probability measures) The extreme points of the set of Borel probability measures on X , Δ(X), are the delta measures δ for x ∈X , where δ ( f )= f (x) for all f ∈C(X) x x

Proof (Barvinok, 2002, and Dunford&Schwartz, 1958) Let x ∈ X . Assume that δ =αl +(1−α)l , where l ,l ∈ Δ(X) and α ∈(0,1). If f ∈ C(X) satisfies f (x)=1 x 1 2 1 2 and f ( y)≤1 for all y ∈ X , then l ( f )= l (1−(1− f ))= l (1)−l (1− f )≤1 for i =1,2, but δ ( f )= f (x)=1, so i i i i x l ( f )=1. Hence, l ,l agree with δ on every function with its maximum at x . Now, take any f ∈ C(X), i 1 2 x and write it as f = f − f , where f ( y)= min{ f (x), f ( y)} and f ( y)= min{0, f (x)− f ( y)}. Then, f , f 1 2 1 2 1 2 attain their maxima at x , so l ( f )= l ( f )−l ( f )= f (x)− f (x)= f (x)= δ ( f ), i.e., δ is extreme i i 1 i 2 1 2 x x Now, let A = {δ : x ∈X}⊆ Δ(X). Since Δ(X) is convex and weak ∗ -closed, conv(A)⊆ Δ(X) (the left side is x ∗ ∗ the weak -closure). If l ∈C(X) \conv(A), by the corollary on the separation between a convex set and a point, there is an f ∈C(X) such that l( f )> c but δ ( f )= f (x)≤ c for all x ∈X , so l ∉Δ(X) . Therefore, x ∗ ∗ ∗ conv(A)= Δ(X). On the other hand, A is weak -closed in C(X) (why?), and hence weak -compact by Banach-Alaoglu, so the lemma on slide 50 implies that all extreme points of Δ(X) are in A ■

Corollary. The extreme points of U = {l ∈ C(X)∗ : l ≤1} are δ and −δ for x ∈X x x

EL3370 Dual Spaces - 55

BONUS: CONVEX FUNCTIONS AND JENSEN’S INEQUALITY

Definitions Let X be a normed space, and f : X → ! - f is convex if f (λx +(1−λ)y)≤ λ f (x)+(1−λ)f ( y) for all x, y ∈ X and λ ∈[0,1] - epi f = {(x, y)∈ X ×!: y ≥ f (x)}: epigraph of f : X → !

A function is convex iff its epigraph is a convex set, so it can be represented as the intersection of closed half-spaces that contain it. Now, for all a,b ∈ ! and x∗ ∈ X ∗,

a ≤< x,x∗ >+by whenever y ≥ f (x)

∗ ⇔ a ≤< x,x >+bf (x)

∗ If H = {(x, y)∈ X ×!:a ≤< x,x >+by} contains epi f , ∗ then b cannot be zero; otherwise f would be undefined when a >< x,x >. Thus, every ∗ closed half-space containing epi f has the form H = {(x, y)∈X × !:c+ < x,x >≤ y}, and

f (x)= sup c+ < x,x∗ > (c ,x∗ ): c+≤ f (x! ) for all x!∈X

EL3370 Dual Spaces - 56

BONUS: CONVEX FUNCTIONS AND JENSEN’S INEQUALITY (CONT.)

Jensen’s Inequality Let X be a set, and E a Borel probability measure on a set of functions f : X → !. Also, let g: X → ! be such that E(g) is well-defined, and let f :! → ! be a convex function. Then

E( f ! g)≥ f (E(g)) if the left hand side is well-defined

Proof. Take a,b ∈ ! such that f ( y)≥ ay +b for all y ∈ ! . Then f (g(x))≥ ag(x)+b for all , so applying E yields E{ f ! g}≥ aE{g}+b. Taking the supremum over all a,b ∈ ! such that f ( y)≥ ay +b for every y ∈ ! finally gives

E{ f ! g}≥ sup aE{g}+b = f (E{g}) ■ (a,b): a+by≤ f ( y ) for all y∈X

Remark. The Borel probability measure E can be, e.g., the Riemann/Lebesgue integral, or the mathematical expectation operator with respect to a probability measure

EL3370 Dual Spaces - 57

BONUS: STONE-WEIERSTRASS THEOREM

Weierstrass’ Theorem, on the approximation of continuous functions on a real interval by polynomials, was considerably generalized by M.H. Stone in two directions: - it applies to continuous functions on an arbitrary compact Hausdorff space, and - instead of polynomials, it allows more general algebras of functions

We will provide a functional-analytic proof of this result, based on the following lemma:

Lemma (spanning criterion) ∗ Let V be a normed space, x ∈V , and Y ⊆V . Then, x ∈ clin Y iff every f ∈V that vanishes on Y also vanishes on x (i.e., if f ( y)=0 for all y ∈Y , then f (x)=0)

Proof (⇒) If f ( y)=0 for all y ∈Y , then by linearity f ( y)=0 for all y ∈ lin Y , and by continuity f (x)=0 since x ∈ clin Y (⇐) Assume x ∉ clin Y , and define the functional f on lin{clin Y + x} by f ( y +αx)=α for all y ∈ clin Y and α . Let d = inf x − y , which is >0 since clin Y is closed. Then, y +αx = α α −1 y + x ≥ d α , so y∈clin Y −1 ∗ f ≤ d , and Corollary 1 of Hahn-Banach provides an extension F ∈V of f that vanishes on Y but not on x ■

EL3370 Dual Spaces - 58

BONUS: STONE-WEIERSTRASS THEOREM (CONT.)

Theorem (Stone-Weierstrass) Let X be a compact Hausdorff space, and let E be a subalgebra of C(X), i.e., a of C(X) such that if f , g ∈ E , also fg ∈ E . In addition, assume that 1 ∈ E , and that E separates points of X , i.e., for every pair x, y ∈ X , x ≠ y , there is an f ∈ E such that f (x)≠ f ( y). Then, E is dense in C(X)

Proof (adapted from version due to Louis de Branges) ∗ By the spanning criterion, E is dense in C(X) iff for every l ∈ C(X) , l(x)=0 for all x ∈ E implies that ∗ l =0. Assume the contrary, and let U be the set of l ∈ C(X) of norm ≤1 that vanish on E ; by Banach- Alaoglu, U is weak ∗ -compact. By Krein-Milman, U has an extreme point, say, l , which should have unit norm (why?)

∗ Let g ∈ E be such that 0< g(t)<1 for all t ∈ X . Since E is an algebra, the functionals lʹ,lʹʹ ∈ C(X) defined by lʹ( f )= l(gf ) and lʹʹ( f )= l((1− g)f ) also vanish in E . Write l as l = l −l , where l ,l ≥ 0, and + − + − also lʹ and lʹʹ as lʹ( f )= l(gf )= l (gf )−l (gf )=: lʹ( f )−lʹ( f ) and lʹʹ( f )= l ((1− g)f )− l ((1− g)f )=: + − + − + − l′′( f )− l′′( f ), where l′ ,l′ ,l′′,l′′≥ 0, and note that l = l + l = l (1)+l (1)= lʹ(1)+lʹʹ(1)+lʹ(1)+lʹʹ(1)= + − + − + − + − + − + + − − lʹ + lʹʹ + lʹ + lʹʹ , while l = lʹ+lʹʹ ≤ lʹ + lʹʹ ≤ lʹ + lʹʹ + lʹ + lʹʹ , so combining these inequalities we see + + − − + + − − that they should be all equalities, and in particular l = lʹ + lʹʹ

EL3370 Dual Spaces - 59

BONUS: STONE-WEIERSTRASS THEOREM (CONT.)

Proof (cont.) Let a = lʹ ≥ min g(x) l >0 and similarly b = lʹʹ >0, so a+b = l =1. Then, l = a(lʹ/a)+b(lʹʹ/b) shows t∈X that l is a convex combination of lʹ/a and lʹʹ/b, both of which have unit norm and vanish in E . Since l is an extreme point, we have that l = lʹ/a = lʹʹ/b, i.e., l ([1− g/a]f )=0 for all f ∈ C(X), or ± l (g) l ( f )= l (gf ), where l ( f )= l ( f )+ l ( f ) + −

Using l (g) l ( f )= l (gf ), we will now show that all functions in the nullspace of l in E (which is a closed hyperplane) have a common zero. Indeed, every g ∈Ker l ∩E should have a zero in X , because −1 −1 −1 otherwise g ∈C(X), so taking f = g gives l (g) l (g )= l (1)> 0, which contradicts the assumption that l (g)= 0. Now, if g ,…, g ∈Ker l ∩E , then they share a common zero, since otherwise 1 n g = g2 +!+ g2 ∈E is strictly positive, while l (g)= l (g2 )+!+ l (g2 )=[ l (g )]2 +!+[ l (g )]2 = 0, which 1 n 1 n 1 n is a contradiction. Finally, for every g ∈E , let X = {x ∈X : g(x)= 0}; X is a closed set in X , and every g g finite intersection X X is non-empty, so by the compactness of X , X , i.e., all g ∩!∩ g ∩ g Ker l g ≠ ∅ 1 n ∈ functions in Ker l ∩E share a common zero

EL3370 Dual Spaces - 60

BONUS: STONE-WEIERSTRASS THEOREM (CONT.)

Proof (cont.) ∗ The of any h∈E for which h( fg)= h( f )h(g) (called a homomorphism, such as l ) satisfies another important property: it is an ideal in E , i.e., it is a linear subspace of E such that if f ∈E and g ∈Ker h, then fg ∈Ker h (since h( fg)= h( f )h(g)= 0); furthermore, it is a maximal ideal, that is, a proper subset of E which is not contained in a larger proper ideal of E (since Ker h has co-dimension 1, so adding an extra dimension would yield E )

Other homomorphisms on E include δ for every x ∈X . Since E separates points on X , x E Ker δ = {g ∈E : g(x)= 0} ≠ Ker δ for x ≠ y (indeed: pick a function g ∈E such that g(x)= 0 and x E y E g( y)= 1; this g belongs to Ker δ but not to Ker δ ) x E y E

Therefore, Ker l ∩E ⊆ Ker δ for some x ∈X corresponding to the common zero of the functions in x E Ker l ∩E , but since Ker l is a maximal ideal in E , we have that Ker l ∩E = Ker δ , i.e., x E l ( f )= αδ ( f )= α f (x) for all f ∈E , with α = 1 due to the positivity of l x

Since δ is an extreme point of the unit ball of C(X)∗ , δ = l = l + l implies that l = α δ , so l = cδ . On x x + − ± ± x x the other hand, 1∈E , so cδ (1)= l(1)= 0, thus c 0 and l = 0, which contradicts the assumption that x = l = 1, hence E is dense in C(X) ■

EL3370 Dual Spaces - 61