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 l2 ). Most completeness proofs follow these steps: ! 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 ∈l2 for some !n. As x ∈l2, this implies that x = x −(x − x)∈l2 ! ! ! 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 x n x x n x 2 , hence x n x , and x − = i 1 i − i ≤ ε − ∈l2 ∈l2 ! ∑ = ! ! 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. (1 p ) is separable: The set M of sequences of the form x {x , ,x ,0,0, } for lp ≤ < ∞ = 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 . δ 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 δ δ δ δ δ 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).