HILBERT AND BANACH SPACES

• Completeness

• The Space L (a,b) ! 2

• The Closest Point Property

• Banach Fixed Point Theorem

EL3370 Hilbert and Banach Spaces - 1

COMPLETENESS

A property that many important metric spaces share is completeness: a sequence that seems to converge, actually does. This, however, does not always hold (recall the l ! 0 example from last lecture). This property is what distinguishes  from !

Definition {x } is a Cauchy sequence in a metric space (X ,d) if, for every ε > 0, there is an N ∈ such ! n ! ! ! that for all n,m > N , d(x ,x )< ε ! ! n m !(X ,d) is a complete metric space if every Cauchy sequence in !X is convergent

Observation Every convergent sequence in a metric space is a Cauchy sequence (exercise!)

EL3370 Hilbert and Banach Spaces - 2

COMPLETENESS (CONT.)

Example 1  and  are complete, while  is not: take x = 1+1 1!+1 2!++1 n!; {x } is a Cauchy ! n ! n sequence in , since it is convergent in  (and hence Cauchy in  ), but its limit in  is !e , which is not rational (see bonus slides for proof)

Example 2 1 Consider C[0,1], with inner product (x, y)= x(t)y(t)dt ! ! ∫0 This space is not complete: consider the sequence {x } in ! n the figure. This sequence is Cauchy, since d(x ,x )< ε when ! n m n,m >1 ε . However, {x } is not convergent: for every x ∈X , ! ! n !

1 2 2 1 2+1 n 2 1 2 d2(x ,x)= x(t) dt + x (t)− x(t) dt + 1− x(t) dt ! n ∫0 ∫1 2 n ∫1 2+1 n d(x ,x)→ 0 means that x(t)= 0 for t <1 2 and x(t)= 1 for ! n ! ! ! !t >1 2, but then !x cannot be continuous!

EL3370 Hilbert and Banach Spaces - 3

COMPLETENESS (CONT.)

Theorem n ! and l are complete metric spaces ! 2

Proof (for l ). Most completeness proofs follow these steps: ! 2 n 1) Take a Cauchy sequence {x } 2) Postulate a candidate limit !x 3) Show that !x belongs to the metric space 4) Show that x n → x

Steps: n n n n 1) Pick a Cauchy sequence {x } in l , where x = {x ,x ,…} ! ! 2 ! 1 2

n 2) Consider a fixed index !k . Since !{x } is Cauchy, for a given !ε > 0 there is an !N > 0 such 2 n m ∞ n m n m n that x − x ≤ x − x = d(x ,x )< ε when n,m > N . Hence, {x } is Cauchy in , ! k k ∑i=0 i i ! ! k n so it converges to, say, x . Therefore, consider the candidate limit x = {x ,x ,…} ! k ! 1 2

EL3370 Hilbert and Banach Spaces - 4

COMPLETENESS (CONT.)

Proof (cont.) n n n n 3) We will show that x − x ∈l for some !n. As x ∈l , this implies that x = x −(x − x)∈l ! 2 ! 2 ! 2 n n m Since !{x } is Cauchy, given !ε > 0 there is an !N > 0 such that !d(x ,x )< ε for !n,m > N . Fix !M ∈ . Then, for !n,m > N ,

M 2 ∞ 2 x n − x m ≤ x n − x m = d2(x n ,x m )< ε 2 !∑i=1 i i ∑i=1 i i

M 2 Taking the limit !m→ ∞, we get x n − x ≤ ε 2 , and taking !M → ∞ we finally obtain !∑i=1 i i 2 2 n ∞ n 2 n x − x = x − x ≤ ε , hence x − x ∈l , and x ∈l ! ∑i=1 i i ! 2 ! 2 n n 4) We have shown in Step 3 that there is an !N > 0 such that ! x − x ≤ ε for !n > N , so !x → x

We have proven that every Cauchy sequence in l is convergent. This means that l is ! 2 ! 2 complete ■

EL3370 Hilbert and Banach Spaces - 5

COMPLETENESS (CONT.)

Definition Hilbert Space: Inner product space which is complete (as a metric space) Banach Space: Normed space which is complete (as a metric space)

Every Hilbert space is a Banach space, but not conversely. E.g., l is a Banach space, but ∞ not a Hilbert space, since the l -norm does not satisfy the parallelogram law (exercise!) ∞

Examples 1. The l space (for 1≤ p ≤ ∞) are Banach spaces. Of them, only l is a Hilbert space ! p ! ! 2 2. Every finite-dimensional normed space is Banach (prove it! [this follows from Exercise Set n 2]). In particular, ! is a Banach space for every !p-norm (!1≤ p ≤ ∞)

EL3370 Hilbert and Banach Spaces - 6

THE SPACE L (a,b) ! 2

1 We have seen that C[0,1] with inner product (x, y)= x(t)y(t)dt is not complete. The ! ! ∫0 problem lies in requiring that the elements of !C[0,1] should be continuous functions Removing this requirement leads to

Definition Let −∞ ≤ a < b ≤ ∞. L (a,b) is the vector space of Lebesgue measurable functions ! ! 2 b 2 b f :(a,b)→  such that f (t) dt < ∞, with inner product (x, y)= x(t)y(t)dt ! !∫a ! ∫a

Subtlety b 2 b 2 (x,x)= x(t) dt = 0 for some !x ≠ 0! (e.g., x(t) dt = 0 if x(t)≠ 0 only for finite # of !t ’s) ! ∫a !∫a ! b 2 To solve this, define an equivalence relation on L (a,b): x  y iff x(t)− y(t) dt = 0 ! 2 ! !∫a This relation is compatible with the operations in L (a,b) (i.e., if x  x and y  y , then ! 2 ! 1 2 ! 1 2 x + y  x + y , λx  λx and (x , y )= (x , y )) ! 1 1 2 2 ! 1 2 ! 1 1 2 2

EL3370 Hilbert and Banach Spaces - 7

THE SPACE L (a,b) (CONT.) ! 2

Digression (Density and separability) Let X be a metric space, and D ⊆ X . D is dense in X if D = X , i.e., if every point in X can be approximated arbitrarily well by points in D. E.g., the set of real-valued polynomials in [a,b] is dense in C[a,b] (this will be proven in Lecture 5)

A metric space is separable if it contains a countable subset which is dense in it. Many proofs in functional analysis can be simplified if the underlying space is separable

Examples 1. ! is separable: a countable dense set in ! is {x + iy : x, y ∈!}. Thus, ! is separable too 2. l is not separable: Let A ⊆ l be the set of sequences x = {x ,x ,…} where x ∈{0,1}. ∞ ∞ 1 2 k Each x ∈A can be mapped to a number xˆ = x /2+ x /22 + x /23 +!∈[0,1]. The mapping 1 2 3 is bijective (except for the countable set of numbers with finite binary expansion), and [0,1] is uncountable, so A is uncountable. If x, y ∈A and x ≠ y , then x − y = 1, hence ∞ every dense set in l must be uncountable (why?) ∞ 3. l (1≤ p < ∞) is separable: The set M of sequences of the form x = {x ,…,x ,0,0,…} for p 1 n some n∈!, and x ∈!, is countable and dense in l (exercise!) k p

EL3370 Hilbert and Banach Spaces - 8

THE SPACE L (a,b) (CONT.) ! 2

It can be proven that L (a,b) is a Hilbert space, and that C[a,b] is a dense subspace of ! 2 ! L (a,b) (i.e., C[a,b]⊂ L (a,b), and the closure of C[a,b] is L (a,b)) ! 2 ! 2 ! ! 2

Similarly, we can define the Banach spaces L (a,b) (1≤ p ≤ ∞) of Lebesgue measurable ! p ! b p functions f :(a,b)→  such that f (t) dt < ∞ (for p < ∞) or esssup f (t) < ∞ (for ! !∫a ! ! t∈[a,b] !p = ∞ ), with norm

1 p ⎧ b p ⎪ x(t) dt , 1≤ p < ∞ x p := ⎨ ()∫a ⎪ esssup x(t) , p = ∞ ! ⎩ t∈[a,b]

EL3370 Hilbert and Banach Spaces - 9

THE CLOSEST POINT PROPERTY

Remember the “projection theorem”? (shortest distance from a point to a plane is achieved n by the perpendicular) This result is fundamental in ! , and it also holds in Hilbert spaces!

Definition A subset !A of a vector space is convex if, for all !x, y ∈A and !0≤ λ ≤1, !λx +(1− λ)y ∈A.

Definition In an inner product space !V , !x, y ∈V are orthogonal, !x ⊥ y , if !(x, y)= 0. !x is orthogonal to a subset !S ⊆ V , !x ⊥ S , if !x ⊥ s for all !s ∈S

Lemma (Pythagorean theorem) 2 2 2 If !x, y are orthogonal in an inner product space, then ! x + y = x + y (Exercise!)

EL3370 Hilbert and Banach Spaces - 10

THE CLOSEST POINT PROPERTY (CONT.)

x Theorem (Closest point property) Let !M be a non-empty closed convex set in a Hilbert space !V . For every !x ∈V there is a unique point !y ∈M which is closer to !x than any other point of !M (i.e., x − y = inf x − m ) y ! m∈M M

Why closed and convex? x x x

M M M

(No y) (A unique y) (Infinitely may y’s)

In a Banach space there may be infinitely many closest points, or none at all

EL3370 Hilbert and Banach Spaces - 11

THE CLOSEST POINT PROPERTY (CONT.)

Proof 1) Existence: Let δ = inf x − m , which is finite (because M ≠ ∅). Then, for each n∈ ! m∈M ! ! 2 there is a y ∈M such that x − y ≤δ 2 + n−1 . We will show that { y } is Cauchy: For n,m, ! n ! n ! n ! 2 2 2 2 (x − yn )−(x − ym ) + (x − yn )+(x − ym ) = 2 x − yn +2 x − ym (parallelogram!law)

2 ⎛ 1 1 ⎞ < 4δ +2⎜ + ⎟ ! ⎝ n m⎠ 2 2 2 ⎛ 1 1 ⎞ or!! y − y < 4δ +2⎜ + ⎟ − 4 x −( y + y )/2 n m ⎝ n m⎠ n m

2 ⎛ 1 1 ⎞ 2 < 4δ +2⎜ + ⎟ − 4δ (because!M!is!convex,!so!(y + y ) 2∈M) ⎝ n m⎠ n m ⎛ 1 1 ⎞ < 2⎜ + ⎟ ! ⎝ n m⎠ so { y } is Cauchy, and it converges, say, to y ∈M (since M is closed). Taking the limit ! n ! ! 2 !n→ ∞ in x − y ≤δ 2 +1 n, we obtain x − y ≤δ , so x − y = inf x − m ! n ! ! m∈M

EL3370 Hilbert and Banach Spaces - 12

THE CLOSEST POINT PROPERTY (CONT.)

Proof (cont.) 2) Uniqueness: Let y , y ∈M be such that x − y = x − y = δ . Then ( y + y ) 2∈M , so ! 1 2 ! 1 2 ! 1 2 x −( y + y ) 2 ≥δ . By the parallelogram law, ! 1 2

2 2 2 2 y1 + y2 2 22 2 y1 − y2 = 2 x − y1 +2 x − y2 − 4 x − ≤ 2δ +2δ − 4δ = 0 ! 2

Therefore, y = y ■ ! 1 2

EL3370 Hilbert and Banach Spaces - 13

THE CLOSEST POINT PROPERTY (CONT.)

Corollary (Projection Theorem) If, in the theorem, !M is a closed linear subspace, then !y ∈M is the minimizer of ! x − m over all !m∈M iff !x − y ⊥ M

Proof. Assume that !y is the minimizer, but that there is an !m∈M such that 2 !(x − y,m)= η ≠ 0. Then, if !y′:= y +ηm m ,

2 2 2 2 m 2 η ⎛ m ⎞ 2 η 2 x − y′ = x − y −η = x − y + −2 x − y,η = x − y − < x − y 2 2 ⎜ 2 ⎟ 2 ! m m ⎝ m ⎠ m

This contradiction shows that !x − y ⊥ M Conversely, if !x − y ⊥ M , then the Pythagorean theorem gives, for every !m∈M,

2 2 2 2 ! x − m = x − y + y − m = x − y + y − m so ! x − m > x − y for all !m ≠ y ■

EL3370 Hilbert and Banach Spaces - 14

BANACH FIXED POINT THEOREM

Let (X ,d) be a metric space. A mapping A: X → X is a contraction with rate 0≤α <1 if d(Ax,Ay)≤αd(x, y) for all x, y ∈ X

Theorem (Banach Fixed Point Theorem) Every contraction mapping A on a complete metric space X has a unique fixed point, i.e., Ax = x has a unique solution in X ∗ Also, if Y is another metric space, and λ ! A(λ) is a weak -continuous, where A(λ) is a contraction with rate , i.e., for every x X and Y , lim d(T( )x,T( )x) 0, then α ∈ λ0 ∈ λ→λ λ λ0 = 0 the fixed point of A(λ) is continuous in λ

Proof 1) Existence: Take x ∈ X , and define by induction x = Ax = Anx for every n ∈ ! . If m≥ n, 0 n n−1 0

m−n−1 m−n−1 α nd(x ,x ) d(x ,x )= d(Anx ,Amx )≤α nd(x ,x )≤α n d(x ,x )≤α nd(x ,x ) α i ≤ 0 1 n m 0 0 0 m−n ∑ i i+1 0 1 ∑ 1 i=0 i=0 −α hence {x } is Cauchy, so it converges to, say, x ∈ X . Now, a contraction is continuous n (why?), so Ax = A(lim x )= lim Ax = lim x = x , thus x is a fixed point of A n n n+1

EL3370 Hilbert and Banach Spaces - 15

BANACH FIXED POINT THEOREM (CONT.)

Proof (cont.) 2) Uniqueness: Assume that Ax = x and Ay = y for some x, y ∈ X . Then d(x, y)= d(Ax,Ay)≤αd(x, y), where α <1, so d(x, y)=0, i.e., x = y 3) Continuity: Note that

d(x(λ),x(λ0 ))= d(A(λ)x(λ),A(λ0 )x(λ0 ))

≤ d(A(λ)x(λ),A(λ)x(λ0 ))+ d(A(λ)x(λ0 ),A(λ0 )x(λ0 )) ≤αd(x(λ),x(λ ))+ d(A(λ)x(λ ),A(λ )x(λ )) 0 0 0 0

so d(x(λ),x(λ ))≤(1−α )−1d(A(λ)x(λ ),A(λ )x(λ ))→ 0 as λ → λ ■ 0 0 0 0 0

EL3370 Hilbert and Banach Spaces - 16

BANACH FIXED POINT THEOREM (CONT.)

Application: Value iteration in dynamic programming

Consider a Markov chain {x } on a finite set X such that P{x = i|x = j,a = a}= p (a), k k+1 k k ij where {a } is a sequence of actions on a finite set A. The problem is to find a , as a function k k ∞ of x , such that E α kR(x ) is maximized, where R: X → !+ is a reward function k {∑k=1 k } 0

∞ To solve this, define the value function V(x) max R(x) kE{R(x )}, which = {a ,a ,…} +∑ α k 1 2 k=2 satisfies the dynamic programming (DP) equation

V(x)= R(x)+αmax V(i)p (a) a∈A ∑ ix i

Once V is found, the optimal action is aopt (x)= argmax V(i)p (a) a∈A ∑i ix

To solve the DP equation, one can use the value iteration algorithm:

1) Let V (x)=0 0 2) For n=1,2,…, let V (x)= R(x)+αmax V (i)p (a) n a∈A ∑i n−1 ix

EL3370 Hilbert and Banach Spaces - 17

BANACH FIXED POINT THEOREM (CONT.)

Application: Value iteration in dynamic programming (cont.)

The value iteration algorithm yields a sequence of value functions {V } that converges to n the unique solution of the DP equation, since it is defined by a contraction mapping!

Indeed, define the Bellman operator A:C(X) C(X) as A(V )(x) R(x) max V(i)p (a). → = +α ix a∈A ∑i Then, for V ,V ∈ C(X), 1 2

A(V )(x)− A(V )(x) =α max V (i)p (a)−max V (i)p (a) 1 2 a∈A ∑i 1 ix a∈A ∑i 2 ix max V (i)p (a) V (i)p (a) ≤α ∑ 1 ix −∑ 2 ix a∈A i i ≤αmax V (i)−V (i) p (a) a∈A ∑i 1 2 ix ≤αmax V (i)−V (i) so A is a contraction i∈X 1 2

EL3370 Hilbert and Banach Spaces - 18

BONUS: IRRATIONALITY OF e

∞ Euler’s number is defined as e:= ∑1 n!. We will show that e is irrational n=0 Assume, to the contrary, that e is rational, and in particular that e = a/b where a,b ∈!. Let

b x = b! e − 1/n! ( ∑n=0 )

Notice that: ∞ (1) x = b! 1/n!> 0 ∑n=b+1 ⎛ a b ⎞ b (2) x = b! − 1/n! = a(b−1)!− b!/n! is an integer ⎜ ∑n=0 ⎟ ∑n=0 ⎝ b ⎠ b! 1 1 (3) For n b 1, , so ≥ + = ≤ n−b n! (b+1)!n (b+1)

∞ b! ∞ 1 ∞ 1 1 (b+1) 1 x = < = = = <1 ∑ ∑ n−b ∑ n n=b+1 n! n=b+1 (b+1) n=1 (b+1) 1−1 (b+1) b which is a contradiction, since there is no integer between 0 and 1, so e is irrational ■

EL3370 Hilbert and Banach Spaces - 19

BONUS: COMPLETION OF METRIC SPACES

Let (X ,d) be a metric space. If X is not complete, it can be “embedded” into a complete ! ! ! ! ! metric space (X ,d), i.e., there exists a complete metric space (X ,d) and a mapping T : X → X ! which is an isometry, that is, such that d(T(x),T( y))= d(x, y) for all x, y ∈ X , and such that ! ! ! R(T) is dense in X . Such an (X ,d) is a completion of (X ,d)

Theorem. Every metric space (X ,d) has a unique completion, up to isometry, i.e., if X! and Xʹ are two completions of X , they are isometric

Proof 1) Existence: Let {x } and {xʹ } be two Cauchy sequences in X . {x } and {xʹ } are said to be n n n n equivalent, {x } ∼ {xʹ }, if d(x ,xʹ )→0 as n→∞. Let X! consist of all equivalence classes of n n n n Cauchy sequences in X , and define the metric d! as d!({x },{ y })= lim d(x , y ). This n n n→∞ n n metric is well defined since d(x , y )−d(x , y ) ≤ d(x ,x )+d( y , y )→0 as n,m→∞, n n m m n m n m and if {x } ∼ {xʹ } and { y } ∼ { yʹ }, then d(x , y )−d(xʹ , yʹ ) ≤ d(x ,xʹ )+d( y , yʹ )→0 as n n n n n n n n n n n n n→∞. d! is a metric because (i) if d!({x },{ y })=0 then d(x , y )→0, so {x } ∼ { y }, and n n n n n n (ii) d(x , y )≤ d(x ,z )+d(z , y ), so n→∞ gives d!({x },{ y })≤ d!({x },{z })+d!({z },{ y }) n n n n n n n n n n n n

EL3370 Hilbert and Banach Spaces - 20

BONUS: COMPLETION OF METRIC SPACES (CONT.)

Proof (cont.) An isometry T : X → X! is given by T(x)= {x } with x = x for all n. Clearly, n n d!(T(x),T( y))= d(x, y). Now, take an {x } ∈ X! , and let ε >0. There is an N ∈ ! such that n d(x ,x )<ε for all n≥ N ; then, { y } ∈R(T) given by y = x satisfies n N n n N d!({x },{ y })= lim d(x ,x )≤ε , so R(T) is dense in X! n n n→∞ n N

To prove the completeness of X! , consider a Cauchy sequence {x! n } in X! , where x! n = {x! n } . k k Since R(T) is dense in X! , for every x! n there is a y! n = { y! n } ∈R(T), with y! n = y , such k k k n ! n n ! n m ! n n ! n m ! m m ! n m that d(x! , y! )<1 n, and d( y! , y! )≤ d( y! ,x! )+d(x! ,x! )+d(x! , y! )<1 n+d(x! ,x! )+1 m, so { y! n } is Cauchy. Consider x! = { y }; since d( y , y )= d!( y! n , y! m ), x! is Cauchy, so x! ∈ X! , n n m and d!(x! n ,x!)≤ d!(x! n , y! n )+d!( y! n ,x!)<1 n+lim d( y , y )→0 as n→∞, so x! n → x! and X! is k→∞ n k complete

EL3370 Hilbert and Banach Spaces - 21

BONUS: COMPLETION OF METRIC SPACES (CONT.)

Proof (cont.) 2) Uniqueness: If (Xʹ,dʹ) is another completion of (X ,d), with Tʹ being an isometry of X n n into Xʹ, then for every xʹ, yʹ ∈ Xʹ there are sequences {xʹ },{ yʹ } in R(Tʹ) such that n n n n n n xʹ → xʹ and yʹ → yʹ. Hence, dʹ(xʹ, yʹ)−dʹ(xʹ , yʹ ) ≤ dʹ(xʹ,xʹ )+dʹ( yʹ, yʹ )→0 as n→∞. n n Since R(Tʹ) and R(T) are isometric to X , there are Cauchy sequences {x! },{ y! } in R(T), and their limits x!, y! are uniquely defined for each xʹ, yʹ ∈ Xʹ and satisfy ! ! d(x!, y! )= dʹ(xʹ, yʹ), so X and Xʹ are isometric ■

Application to ! as a completion of ! The previous theorem yields a constructive means to complete a metric space. In particular, the set of rational numbers ! can be completed by constructing the set of all Cauchy sequences in !. This is a representation (model) of the set of real numbers, !

All operations in ! can be defined similarly to its metric: given x = {x }, y = { y } ∈ ! , n n

d(x, y)= x − y = lim x − y n→∞ n n x ± y = lim x ± y , x ⋅ y = lim x y , etc. n→∞ n n n→∞ n n

EL3370 Hilbert and Banach Spaces - 22