1 Inner Product Space

Transcription

1 Ch - Hilbert Space 1 4 Hilbert Space 1 Inner Product Space Let E be a complex vector space, a mapping (, ) : E E C is called an inner product on E if i) (x, x) 0 x E and (x, x) = 0 if and only if x = 0; ii) (x, ) is linear on E for each x E; and iii) (x, y) = (y, x) for all x, y in E. With such an inner product E is called an inner product space. If we write x for (x, x), then is a norm on E and hence E is a normed vector space. For this fact we show first Schwarz Inequality (x, y) x y, for x, y E. Proof. For t R we have 0 (x + ty, x + ty) = (x, x + ty) + (ty, x + ty) = x 2 +2t Re(x, y) + t 2 y 2, from which we necessarily have Re(x, y) 2 x 2 y 2, and hence Re(x, y) x y. In the above, replace y by θy for some θ C with θ = 1 and Re(x, θy) = (x, y), then (x, y) = Re(x, θy) x θy = x y.

2 Ch - Hilbert Space 2 From Schwarz inequality, it follows that x + y 2 = x 2 +2 Re(x, y)+ y 2 x 2 +2 x y + y 2 = ( x + y ) 2, or x + y x + y, i.e. triangular inequality holds for. Hence is a norm on E. For an inner product space E, the norm on E is the norm so defined unless stated otherwise. Examples i) Let E = C n. For z = (z 1,, z n ) and z = (z 1,, z n) in C n let (z, z ) = n z j z j. ii) Let E = l 2 (N) = {(z 1, z 2, ) : { z j 2 } j N is summable }. For z = (z 1, z 2, ), and z = (z 1, z 2, ) l 2 (N) let (z, z ) = j N z j z j. The space E = l 2 (N) will be hereafter simply denoted by l 2. iii) Let E = L 2 (Ω, Σ, µ). For f, and g in L 2 (Ω, Σ, µ), define (, ) := fgdµ. Ω

3 Ch - Hilbert Space 3 Exercise 1.1. i) For z, z l 2 (N), show that { z j z j } j N is summable and hence (z, z ) defined above is absolutely convergent. ii) Show that l 2 (N) is complete. An inner product space is called a Hilbert space if it is complete. Both C n and l 2 (N) are Hilbert space. Since L 2 (Ω, Σ, µ) is complete with respect to the metric defined by L 2 -norm and since L 2 -norm is given by the inner product defined in Example iii), L 2 (Ω, Σ, µ) is a Hilbert space of which both C n and l 2 (N) are special case. Exercise 1.2. Define real inner product space and real Hilbert space. 2 Geometry in Hilbert Space Theorem 2.1. Let E be an inner product space, and M a complete convex subset of E. Suppose x E, then the following are equivalent: 1) y M satisfies x y = min z M x z ; 2) y M satisfies Re(y x, y z) 0 z M. Furthermore, there is unique y M satisfying 1) and 2). Proof. 1) 2): For z M and 0 < θ 1, let f(θ) = x {(1 θ)y + θz} 2 = x y + θ(y z) 2 = x y 2 + θ 2 x z 2 +2θRe(x y, y z).

4 Ch - Hilbert Space 4 Since f(θ) f(0) = x y 2 for 0 < θ 1, we have f(θ) f(0) lim θ 0 θ = 2Re(x y, y z) 0. 2) 1) :Re(y x, y z) = Re(x y, y x + x z) = x y 2 Re(x y, x z) 0, hence x y 2 Re(x y, x z) x y x z, and consequently x y x z for all z M. That there exists at most one such y follows from 2), for if both y 1 and y 2 satisfy 2) for all z M, then 0 y 1 y 2 2 = (y 1 y 2, y 1 y 2 ) = (y 1 x, y 1 y 2 ) + (y 2 x, y 2 y 1 ) = Re(y 1 x, y 1 y 2 ) + Re(y 2 x, y 2 y 1 ) 0, so y 1 = y 2. To show that there exists y satisfying 1), let α = inf z M x z. Consider then a sequence {z n } M satisfying α 2 x z n 2 α n.

5 Ch - Hilbert Space 5 We claim that {z n } is a Cauchy sequence. We have z n z m 2 = (z n x) (z m x) 2 = z n x 2 + z m x 2 2Re(z n x, z m x) ; 4 z n + z m 2 x 2 = z n x 2 + z m x 2 +2Re(z n x, z m x), consequently z n z m 2 = 2 z n x 2 +2 z m x 2 4 z n + z m 2 x 2 2(α n ) + 2(α2 + 1 m ) 4α2 = 2( 1 n + 1 m ), which shows that {z n } is a Cauchy sequence. Now since M is complete, there is y M with y = lim z n. Obviously x y = lim x z n = α. The map t : E M defined by tx = y, where y is the unique element in M which satisfies 1) and 2) of Theorem 1 is called the projection from E onto M and is more precisely denoted by t M if necessary. Theorem 2.1 is usually applied in the special case when M is a closed convex subset of a Hilbert space. Corollary 2.1. Let M be a closed convex subset of a Hilbert space E, then t = t M has the following properties: i) t 2 = t (t is idempotent); ii) tx ty x y (t is contractive); and iii) Re(tx ty, x y) 0 (t is monotone).

6 Ch - Hilbert Space 6 Proof. i) is obvious. ii) : From Re(tx x, tx ty) 0 and Re(ty y, ty tx) 0 we obtain Re(x y (tx ty), tx ty) 0, hence tx ty 2 Re(x y, tx ty) x y tx ty, from which tx ty x y follows. iii) : Again from Re(x y (tx ty), tx ty) 0 we have 0 tx ty 2 Re(x y, tx ty). Exercise 2.1. If M is a closed convex cone of a Hilbert space E and x E, then y = tx if and only if Re(x y, y) = 0 and Re(x y, z) 0 for all z M. Note : A convex set M in a vector space is called a convex cone if αx M for x M and α > 0. Exercise 2.2. Let M be a closed convex cone in a Hilbert space E and let N = {y E : Re(y, x) 0 x M}. Put t = t M and s = t N. Show that i) s = I t, I being the identity map of E. ii) t(λx) = λtx if λ 0 (t is positively homogeneous), iii) x 2 = tx 2 + sx 2, x E, iv) N = {x E : tx = 0},M= { x E : sx = 0}.

7 Ch - Hilbert Space 7 v) Re(tx, sx) = 0 and x = tx+sx ; conversely if x = y+z, y M, z N and Re(y, z) = 0, then y = tx, z = sx. In the remaining part of the exercise, suppose that M is a closed vector subspace of E. Show that vi) N = M := {y E : (y, x) = 0 x M}. vii) both t and s are continuous and linear. viii) M = te = ker s ; N = ker t = se. ix) (tx, y) = (x, ty) x, y E. x) tx and sx are the unique elements y M and z M such that x = y + z. 3 Linear transformation In this section we consider a linear transformation T from a normed vector space X into a normed vector space Y over the same field R or C. Exercise 3.1. Show that T is continuous on X if and only if it is continuous at one point. Theorem 3.1. T is continuous if and only if there is C 0 such that T x C x (3.1) for all x X.

8 Ch - Hilbert Space 8 Proof. If there is C 0 such that (3.1) holds for all x X, then T is obviously continuous at x = 0 and hence by Exercise 3.1 it is continuous on X. Conversely, suppose that T is continuous on X, and is hence continuous at x = 0. There is then δ > 0 such that if x δ, then T x 1. Let now x X and x 0, then δ δ x = δ, so T ( x) 1. Thus x x T x 1 x. If we choose C = 1, then (3.1) holds for x 0. But when δ δ x = 0, (3.1) holds always. From this theorem it follows that if T is a continuous linear transformation from X into Y, then T := T x sup x X, x 0 x < + (3.2) and is the smallest C for which (3.1) holds. T is called the norm of T. Of course, T can be defined for any linear transformation T from X into Y, and T is continuous if and only if T < +. Hence a centinuous linear transformation is also called a bounded linear transformation. Exercise 3.2. Show that T = sup T x. x X, x =1 Exercise 3.3. Let L(X, Y ) be the space of all bounded linear transformaitons from X into Y. Show that it is a normed vector space with norm given by (3.2). Theorem 3.2. If Y is a Banach space, then L(X, Y ) is a Banach space.

9 Ch - Hilbert Space 9 Proof. We will show that L(X, Y ) is complete. Let {T n } be a Cauchy sequence in L(X, Y ). Since T n x T m x = (T n T m )x T n T m x, {T n x} is a Cauchy sequence in Y for each x X. Put T x = lim T n x. T is obviously a linear transformation from X into Y. We claim now T L(X, Y ). Since {T n } is Cauchy T n C for some C > 0, and for all n. Now T x = lim T n x lim inf T n x ( ) sup n T n x C x for each x X. Hence T is a bounded linear transformation. We show next lim T n T = 0. Given ε > 0, there is n 0 such that T n T m < ε if n, m n 0. Let n n 0, we have T n T = sup T n x T x x X, x =1 = sup x X, x =1 this shows that lim T n T = 0, or lim T nx T m x m = sup lim inf T n T m x x X, x =1 m sup ε x = ε, x X, x =1 lim T n = T.

10 Ch - Hilbert Space 10 Thus the sequence {T n } has a limit in L(X, Y ). L(X, C), or L(X, R), depending on whether X is a complex or a real vecter space, is called the topological dual of X and is denoted by X. X is a Banuch space. Theorem 3.3 (Riesz Representation Theorem). Let X be a Hilbert space and l X, then there is y 0 X such that l(x) = (y 0, x) for x X. Furthermore, the mapping l y 0 is conjugate linear and l = y 0. Proof. We may assume l 0. Let M = ker l, then M is one dimensional. For x X, x can be expressed as x = v + λx 0, where x 0 is a fixed non-zero element of M, v M and λ a scalar. We have l(x) = l(v) + λl(x 0 ) = λl(x 0 ) and (x 0, x) = (x 0, v + λx 0 ) = λ x 0 2. Hence if we let y 0 = l(x 0) x 0 2 x 0, then (y 0, x) = λl(x 0 ) = l(x). All the other assertions are obvious.

11 Ch - Hilbert Space 11 4 Lebesque-Nikodym Theorem Let (Ω, Σ, µ) be a measure space and f a Σ-measurable function on Ω. Suppose that fdµ has a meaning, then the set function ν defined by Ω ν(a) = A fdµ, A Σ is called the indefinite integral of f. Then ν( ) = 0 and ν is σ-additive i.e. if {A n } Σ is a disjointed sequence, then ν( n A n ) = n ν(a n ). It enjoys also the following property: ν(a) = 0 whenever A Σ and µ(a) = 0. This fact suggests the following definition of absolute continuity of a measure with respect to another measure. Let (Ω, Σ, µ) and (Ω, Σ, ν) be measure spaces, ν is said to be absolutely continuous w.r.t. µ if ν(a) = 0 whenever A Σ and µ(a) = 0. Theorem 4.1 (Lebesque-Nikodym Theorem). Let (Ω, Σ, µ) and (Ω, Σ, ν) be measure spaces with µ(ω) < + and ν(ω) < +. Suppose ν is absolutely continious w.r.t. to µ, then there is a unique h L 1 (Ω, Σ, µ) such that ν(a) = hdµ, A Σ. Furthermore, h 0 µ a.e. A Proof. Let ρ = µ + ν, then ρ is a finite measure on Σ. Consider the real Hilbert space L 2 (Ω, Σ, ρ) and consider the linear functional l on L 2 (Ω, Σ, ρ)

12 Ch - Hilbert Space 12 defined by l(f) = fdν. Since l(f) ( f 2 dν ) 1/2 ( 1dν ) 1/2 ν(ω) 1/2 [ f 2 dρ ] 1/2 = ν(ω) 1/2 f L 2 (ρ), l is a bounded linear functional on L 2 (Ω, Σ, ρ). By Riesz Representation Theorem there is unique g L 2 (Ω, Σ, ρ) such that fdν = fgdρ = fgdµ + fgdν for all f L 2 (Ω, Σ, ρ), or f(1 g)dν = fgdµ (4.1) for all f L 2 (Ω, Σ, ρ). Claim 1 0 g(x) < 1 for ρ a.e. x on. Let A 1 = {x Ω : g(x) < 0} and A 2 = {x Ω : g(x) 1}. If we let f = χ A1 in (4.1), then 0 ν(a 1 ) A 1 (1 g)dν = A 1 gdµ, from which follows that µ(a 1 ) = 0 and hence ν(a 1 ) = 0. Thus ρ(a 1 ) = 0. Now in (4.1) choose f = χ A2, we have 0 A 2 (1 g)dν = A 2 gdµ µ(a 2 ). This implies µ(a 2 ) = 0, hence ν(a 2 ) = 0. Consequently, ρ(a 2 ) = 0. Thus Claim 1 is established.

13 Ch - Hilbert Space 13 Claim 2 (4.1) holds for all Σ-measurable and ρ a.e. non-negative functions f. For each positive integer n, let f n = f n. Since 1 g > 0 and g 0 ρ a.e., 0 f n (1 g) f(1 g), and 0 f n g fg, then from Monotone Convergence Theorem and (4.1) it follows that f(1 g)dν = lim f n (1 g)dν = lim f n gdµ = fgdµ, which proves the claim. For a Σ-measurable and ρ a.e. 0 function z, choose f = z 1 g in (4.1), then zdν = g z 1 g dµ = zhdµ, (4.2) where h = g 1 g. If for A Σ we take z = χ A in (4.2), then ν(a) = χ A dν = A hdµ. Since ν(ω) < +, we know hdµ < +, hence h L 1 (Ω, Σ, µ). That such an h is unique is obvious. That h 0 µ a.e. is also obvious. Exercise 4.1. A measure space (Ω, Σ, µ) is said to be σ-finite if there are A 1, A 2, such that A n = Ω and µ(a n ) < +, n = 1, 2,. Show that Lebesque-Nikodym Theorem holds if both (Ω, Σ, µ) and (Ω, Σ, ν) are σ-finite. But in this case h may not be µ-integrable.

14 Ch - Hilbert Space 14 5 The Lax-Milgram Theorem Let X be a Hilbert space. For definiteness, let X be a complex Hilbert space. B(, ) : X X C is called sesquilinear if for x, x 1, x 2 in X and λ 1, λ 2 C the following equalities hold B(x, λ 1 x 1 + λ 2 x 2 ) = λ 1 B(x, x 1 ) + λ 2 B(x, x 2 ), B(λ 1 x 1 + λ 2 x 2, x) = λ 1 B(x 1, x) + λ 2 B(x 2, x). B is said to be bounded if there is r > 0 such that B(x, y) r x y for all x and y in X ; and B is said to be positive definite if there exists ρ > 0 such that B(x, x) ρ x 2 for all x in X. Exercise 5.1. Suppose that B is a bounded, positive definite and sesquilinear function on X X and assume that B(x, y) = B(y, x) for all x and y in X. let ((, )) = B(, ), then ( X, ((, )) ) is a Hilbert space which is equivalent to ( ) X, (, ) as Banach space. Theorem 5.1 (The Lax-Milgram Theorem). Let X be a Hilbert space and B a bounded, positive definite and sesquilinear functional on X X. Then there is a unique bounded linear operator S : X X such that (x, y) = B(Sx, y) for all x and y in X and S ρ 1. Furthermore S 1 exists and is bounded with S 1 r.

15 Ch - Hilbert Space 15 Proof. Let D = {x X : x X such that (x, y) = B(x, y) y X}. D is not empty, for 0 D. Also x is uniquely determined by x. For, if B(x 1, y) = B(x 2, y) = (x, y) y X, then B(x 1 x 2, y) = 0 y X, and hence 0 = B(x 1 x 2, x 1 x 2) ρ x 1 x 2 2, implying x 1 x 2 =0, or x 1 = x 2. For x D, let Sx = x. Since B is sesquilinear, D is a vector subspace of X and S is linear on D. Furthermore, from ρ Sx 2 B(Sx, Sx) = (x, Sx) x Sx, it follows that Sx ρ 1 x for x D. Thus S is bounded on D with S ρ 1. We proceed to show that D = X. For this we show first that D is closed. Let {x n } n=1 D with lim x n = x for some x X, then (x, y) = lim (x n, y) = lim B(Sx n, y), for all y X. Since S is bounded on D, Sx n is Cauchy in X, and hence has a limit z X. From this and the fact that B is bounded, it follows that (x, y) = lim B(Sx n, y) = B(z, y), for all y X. Hence x D and z = Sx. So D is closed. Now if D X, there is y 0 D, y 0 0. Consider the linear functional l defined on X by l(x) = B(y 0, x), x X. As B is bounded, l is a bounded linear functional on X, and hence by Riesz Representation Theorem there is x 0 X such that B(y 0, x) = (x 0, x), x X.

16 Ch - Hilbert Space 16 Thus x 0 D and ρ y 0 2 B(y 0, y 0 ) = (x 0, y 0 ) = 0. Hence y 0 = 0, this contradicts the fact that y 0 0. Therefore D = X. Thus S is a bounded linear operater on X and S ρ 1. As Sx = 0 implies (x, y) = B(Sx, y) = 0 y X and hence x = 0, S is an one-to-one map. Applying Riesz Representation Theorem again, as in the last paragraph, for each x in X, there is x X such that (x, y) = B(x, y) y X, i.e. x = Sx. Thus S is an onto map. Hence S 1 exists. But from S 1 x 2 = (S 1 x, S 1 x) = B(x, S 1 x) r S 1 x x, it follows that S 1 r. 6 Gram-Schmidt Orthogonalization Procedure A family {x α } α I of non-zero elements of a Hilbert space X is said to be orthogonal if (x α, x β ) = 0 whenever α β. An orthogonal family is obviously linearly independent. A finite or countable orthogonal family is usually refered to as an orthogonal system in X. From a linearly independent system {x n } in X, one can construct an orthogonal system {y n } in the following way. Let E k = < x 1,, x k >, the vector subspace of X generated by {x 1,, x k }, and let t k = t Ek be the orthogonal projection of X onto E k. {y n } is defined inductively as follows. Let y 1 = x 1 ; suppose y 1,, y k have

17 Ch - Hilbert Space 17 been defined, let y k+1 = x k+1 t k x k+1. Exercise 6.1. Show that (i) < y 1,, y n > = < x 1,, x n >, n = 1, 2, 3, ; (ii) {y n } is an orthogonal system in X. This procedure of obtaining an orthogonal system from a linearly independent system is called the Gram-Schmidt orthogonalization procedure. An orthogonal family {x α } α I is called an orthonormal family if (x α, x β ) = δ αβ, where δ αβ = 1 or 0 according as α = β or α β. If {y n } is the orthogonalized system of {x n } through Gram-Schmidt procedure, the system {e n }, e n = yn y n, n = 1, 2, 3,, is called the Gram-Schmidt orthonormalized system of {x n }. 7 Bessel Inequality and Parseval Relation Let {e n } be an orthonormal system in a Hilbert space X, U be the closed vector subspace of X generated by {e n }, and t U be the orthogonal projection of X onto U. For each positive integer k, let E k = < e 1,, e k > and t k = t Ek, the orthogonal projection of X onto E k. The following propositions hold: 1) t k x = k (e j, x)e j, x X.

18 Ch - Hilbert Space 18 Proof. Let t k x = k λ j e j, λ j C. Then k (e i, t k x) = (e i, λ j e j ) = = λ i, 1 i k. k λ j (e i, e j ) But (e i, x) = ( e i, t k x + (1 t k )x ) = (e i, t k x) = λ i, 1 i k, where the fact that I t k is the orthogonal projection of X onto E k (See Exercise 2.2) has been used, hence t k x = k (e j, x)e j. 2) For x X, lim k t k x = t U x. Proof. For y U and ε > 0, there is a finite linear combination z = such that z y < ε. For k N, z E k, and hence t k z = z, then N λ j e j t k y y = t k y t k z + z y t k (y z) + z y 2 y z < 2ε. Consequently, lim k t k y = y. Now for x X, t U x U, thus from what is proved above lim t k (t U x) = t U x. But t k t U = t k. k k 3) For each k and x, y in X, (t k x, t k y) = (e j, x)(e j, y).

19 Ch - Hilbert Space 19 Proof. The proposition follows easily from 1). 4) For x, y in X, (t U x, t U y) = (e j, x)(e j, y). Proof. Since (t U x, t U y) (t k x, t k y) = (t U x t k x, t U y) + (t k x, t U y t k y), we have (t U x, t U y) (t k x, t k y) t U y t U x t k x + t k x t U y t k y y t U x t k x + x t U y t k y, and hence by 2) and 3) (t U x, t U y) = lim k (t k x, t k y) = lim k k (e j, x)(e j, y) = (e j, x)(e j, y). 5) (e j, x) 2 x 2, x X. (Bessel inequality) Proof. From 4), (e j, x) 2 = t U x 2 x 2. 6) (e j, x) 2 = x 2 for all x X if and only if U = X. Proof. If U = X, then t U = I and hence from 4) (e j, x) 2 = t U x 2 = x 2 for all x X. On the other hand, if U X, there is x X such that x t U x, hence x 2 = t U x 2 + (1 t U )x 2 > t U x 2 = (e j, x) 2.

20 Ch - Hilbert Space 20 An orthonormal system {e n } in X is called complete if U = X. Exercise 7.1. Show that an orthonormal system {e n } is complete if and only if from the fact that (e n, x) = 0 for all n it follows that x = 0. A Hilbert space is called separable if it contains a countable dense subset. Theorem 7.1. A separable Hilbert X is isometrically isomorphic either to C n for some n or to l 2. Proof. Let {z k } k=1 be a sequence of elements which is dense in X. By a wellknown elementary selection procedure in linear algebra, one can extract from {z k } an independent subsequence {x k } such that < {x k } > = < {z k } >. If {x k } is finite, then the proof that X is isometrically isomorphic to C n, where n is the cardinality of {x k }, is an easy imitation of that of the case when {x k } is infinite. Hence we assume that {x k } is infinite. From Gram-Schmidt orthonormalization procedure we construct from {x k } an orthonormal system {e k } k=1 such that < {x k} > = < {e k } >. As before, let U =< {e k } >, then U = X; thus for x, y in X we have from 4) (x, y) = (e k, x)(e k, y), (7.1) k=1 because t U = I. Define a map τ : X l 2 by letting, for x X, τx = (α k ) k=1, where α k = (e k, x). Since x 2 = (e k, x) 2 by 6), τx is in l 2 and k=1

21 Ch - Hilbert Space 21 x = τx l 2, so τ is an isometry. That τ is linear is obvious. We show now that τ is onto l 2. Let (α k ) k=1 l2. For each positive integer n, let x n = n α k e k. k=1 We claim that {x n } is a Cauchy sequence in X. For n > m we have x n x m 2 = n k=m+1 α k 2, which tends to 0 as m. Hence {x n } is a Cauchy sequence. Let x = lim x n. Then (e k, x) = lim (e k, n α j e j ) = α k and hence τx = (α k ) k=1. Therefore τ is onto. That τ is an isomorphism follows from (7.1). The equality (e j, x) 2 = x 2 for x X and (7.1) are called Parseval relations.