HILBERT SPACE GEOMETRY

Transcription

1 HILBERT SPACE GEOMETRY

2 Defiitio: A vector space over is a set V (whose elemets are called vectors) together with a biary operatio +:V V V, which is called vector additio, ad a eteral biary operatio : V V, which is called scalar multiplicatio, such that (i) (V,+) is a commutative group (whose eutral elemet is called zero vector) ad (ii) for all λ,µ,,y V: λ(µ)=(λµ), =, λ(+y)=(λ)+(λy), (λ+µ)=(λ)+(µ), where the image of (,y) V V uder + is writte as +y ad the image of (λ,) V uder is writte as λ or as λ. Eercise: Show that the set together with vector additio ad scalar multiplicatio defied by y + y + = y + y λ ad λ =, λ respectively, is a vector space.

3 Remark: Usually we do ot distiguish strictly betwee a vector space (V,+, ) ad the set of its vectors V. For eample, i the et defiitio V will first deote the vector space ad the the set of its vectors. Defiitio: If V is a vector space ad M V, the the set of all liear combiatios of elemets of M is called liear hull or liear spa of M. It is deoted by spa(m). By covetio, spa( )={0}. Propositio: If V is a vector space, the the liear hull of ay subset M of V (together with the restrictio of the vector additio to M M ad the restrictio of the scalar multiplicatio to M) is also a vector space. Proof: We oly eed to prove that spa(m) cotais the zero vector ad that it is closed uder vector additio ad scalar multiplicatio: M= spa(m)={0} 0 spa(m) M M: 0 =0 spa(m),y spa(m) +y= + y spa(m) spa(m), λ λ spa(m) The other properties of a vector space are satisfied for all elemets of V ad therefore also for all elemets of M V. Defiitio: If a subset M of a vector space V is also a vector space, it is called a liear subspace of V.

4 Defiitio: A ier product space is a vector space V together with a fuctio :V V (called ier product) satisfyig the followig aioms: For all,y,z V, λ (i) (ii) (iii) (iv) (v), y = y,, + y, z =, z + y, z, λ, y =λ, y,, 0,, =0 =0. A semi-ier product satisfies (i) (iv), but zero if 0., ca be Eercise: Show that the ier product aioms (i)-(iii) imply that for all,y,z,u V, λ,µ,ν,ξ λ + µ y, νz + ξu =λν, z +λξ, u +µν y, z +µξ y, u. Eercise: Show that the vector space fuctio defied by together with the is a ier product space. y, y = y + y 3

5 Defiitio: The orm (semiorm) of a elemet of a ier product space (semi-ier product space) is defied by =,. Cauchy-Schwarz Iequality: If ad y are elemets of a ier product space, the, y y. Proof: 0 y ± y, y ± y = y, ± y, y + = y ± y, y = y ( y ±, y ) y, y 0 y ±, y ±, y y Eercise: Let V be a semi-ier product space. Show that for all,y,z V, λ (i) + y + y, (ii) λ = λ, (iii) 0, ad, if V is a ier product space, also (iv) =0 =0. 4

6 Lemma: The triagle iequality + y + y implies that for all ad y y y. Proof: = ( y) + y y + y y y = ( y ) + y + y - y y - Cotiuity of the Norm: If the sequece ( ) of elemets of a ier product space V coverges i orm to V, the the sequece coverges to, i.e.,. 0. Proof: 0 0 Cotiuity of the Ier Product: If the sequeces ( ) ad (y ) of elemets of a ier product space V coverge i orm to V ad y V, respectively, the the sequece, coverges to, y, i.e., y 0, y y 0, y, y. Proof: 0, y, y =, y y +, y, y y +, y y y + y 0 0 5

7 Defiitio: A ier product space H is called a Hilbert space, if it is complete i the sese that every Cauchy sequece ( ) of elemets of H coverges to some elemet H, i.e.,, m H, 0 as m, H: 0. m Eample: That the ier product space ca be see as follows. m m ( m - ) 0, ( m - ) 0 =( m - ) +( m - ) 0 is a Hilbert space, :, (by the completeess of ) =( - ) +( - ) 0 Defiitio: A liear subspace S of a Hilbert space is said to be a closed subspace, if S, 0 S. 6

8 Eercise: Show that the itersectio I i I Si of a family of closed subspaces of a Hilbert space is also a closed subspace. Defiitio: The closed spa of a subset M of a Hilbert space is defied as the itersectio of all closed subspaces which cotai all elemets of M. It is deoted by spa(m). Defiitio: Two elemets ad y of a ier product space are said to be orthogoal ( y), if, y =0. Propositio: The orthogoal complemet M ={ H: y y M} of ay subset M of a Hilbert space H is a closed subspace. Proof: M is a liear subspace, because z M 0, z =0 0 z,,y M, z M + µ, z =, z + y, z =0 +y z, M, λ, z M λ, z =λ, z =0 λ z. Moreover, M is eve a closed subspace, because M, 0, z M, z =0 for all, z =lim, z =0. 7

9 Projectio Theorem: If S is a closed subspace of a Hilbert space H, the each H ca be uiquely represeted as =ˆ +u, where ˆ S ad u S. Furthermore, ˆ (which is called the projectio of oto S) satisfies for ay other elemet y S. ˆ < y Defiitio: Let S be a closed subspace of a Hilbert space H. The mappig P S ()=ˆ, H, where ˆ is the projectio of oto S, is called the projectio mappig of H oto S. Properties of Projectio Mappigs: If S, S, S are closed subspaces of a Hilbert space H,,y, H, ad λ,µ the: (i) P S (λ+µy)= λp S ()+µp S (y) (ii) S P S ()= (iii) S P S ()=0 (iv) =P S ()+ P S () (v) S S P S ( P S ())= P S () (vi) 0 PS ( ) PS () 0 8

10 Propositio: If M={u,,u } is a set of mutually orthogoal elemets of a Hilbert space H ad 0 M, the,u P spa(m) ()= k u k u Proof: For each u j M we have k H. Thus - -,u u k,u u k k u k, j k u k u =,u j -,u u k,u j =,u j - u j =0., λk u k =0 k u u k, u j j, u j for each liear combiatio λk u k spa(m)= spa (M). 9