MAT641- MSC Mathematics, MNIT Jaipur Functional Analysis-Inner Product Space and orthogonality Dr. Ritu Agarwal Department of Mathematics, Malaviya National Institute of Technology Jaipur April 11, 2017 1 Inner Product Space Inner product on linear space X over field K is defined as a function h:i : X × X −! K which satisfies following axioms: For x; y; z 2 X, α 2 K, i. hx + y; zi = hx; zi + hy; zi; ii. hαx; yi = αhx; yi; iii. hx; yi = hy; xi; iv. hx; xi ≥ 0, hx; xi = 0 iff x = 0. An inner product defines norm as: kxk = phx; xi and the induced metric on X is given by d(x; y) = kx − yk = phx − y; x − yi Further, following results can be derived from these axioms: hαx + βy; zi = αhx; zi + βhy; zi hx; αyi = αhx; yi hx; αy + βzi = αhx; yi + βhx; zi α, β are scalars and bar denotes complex conjugation. Remark 1.1. Inner product is linear w.r.to first factor and conjugate linear (semilin- ear) w.r.to second factor. Together, this property is called `Sesquilinear' (means one and half-linear). Date: April 11, 2017 Email:[email protected] Web: drrituagarwal.wordpress.com MAT641-Inner Product Space and orthogonality 2 Definition 1.2 (Hilbert space). A complete inner product space is known as Hilbert Space. Example 1. Examples of Inner product spaces n n i. Euclidean Space R , hx; yi = ξ1η1 + ::: + ξnηn, x = (ξ1; :::; ξn), y = (η1; :::; ηn) 2 R . n n ii. Unitary Space C , hx; yi = ξ1η1 + ::: + ξnηn, x = (ξ1; :::; ξn), y = (η1; :::; ηn) 2 C . 2 R b iii. Function space L [a; b], hx; yi = a x(t)y(t)dt, x(t) and y(t) are complex valued functions (May be real valued also). 2 P1 2 P1 iv. l = fx = (ξ1; ξ2; :::): i=1 jξij < 1g is an IPS for hx; yi = i=1 ξiη¯i. 2 Norm and Parallelogram equality To prove triangle inequality for normed space, first we are required to prove the Schwarz inequality. Theorem 2.1 (Schwarz inequality). jhx; yij ≤ kxkkyk (2.1) where, equality holds if and only if fx; yg is linearly dependent set. Proof. If y = 0, then (2.1) holds since hx; 0i = 0. Let y 6= 0. For every scalar α, we have 0 ≤ kx − αyk2 = hx − αy; x − αyi (2.2) = hx; xi − α¯hx; yi − α[hy; xi − α¯hy; yi] hy;xi Choosing α = hy;yi so that the expression in brackets vanish. We get hy; xi jhx; yij2 0 ≤ hx; xi − hx; yi = kxk2 − (Usinghy; xi = hx; yi) (2.3) hy; yi kyk2 Rearranging the terms, we get (2.1). Equality holds if kx − αyk = 0 i.e. x = αy, fx; yg are linearly dependent. Theorem 2.2. All inner product spaces are normed spaces for the norm defined by kxk = phx; xi. Also, show that the triangle inequality for norm can be obtained from inner product as kx + yk ≤ kxk + kyk (2.4) where, equality holds if and only if fx; yg is linearly dependent set x = cy, c ≥ 0 or y = 0. MAT641-Inner Product Space and orthogonality 3 Proof. (i) From axiom (iv) for inner product space, we have kxk = phx; xi ≥ 0 and kxk = phx; xi = 0 iff x = 0. (ii) kαxk = phαx; αxi = pαα¯hx; xi = pjαj2hx; xi = jαj kxk. (iii) Triangle inequality kx + yk2 = hx + y; x + yi = hx; xi + hx; yi + hy; xi + hy; yi (2.5) = kxk2 + hx; yi + hy; xi + kyk2 Using Schwarz inequality (2.1) kx + yk2 ≤ kxk2 + 2jhx; yij + kyk2 ≤ kxk2 + 2kxk kyk + kyk2 = (kxk + kyk)2 (2.6) Taking square root on both sides, we get(2.4). Further, equality holds in this derivation if and only if hx; yi + hy; xi = 2kxk kyk i.e. hx; yi + hy; xi = 2<hx; yi = 2kxk kyk. But Schwarz inequality says that kxk kyk ≥ jhx; yij, so that <hx; yi ≥ jhx; yij. Since the real part of a complex number cannot exceed the absolute value, we must have equality, which implies linear dependence from (2.1), say y = 0 or x = cy. With the equality sign, we have <hx; yi = jhx; yij, and hence imaginary part must be zero. If x = cy, <(hx; yi) = <(hcy; yi) = <(chy; yi) ≥ 0 ) c > 0. Remark 2.3. Every Hilbert space is Banach space. However converse is not necessarily true. 2.1 Parallelogram equality A norm on an inner product space satisfies the important Parallelogram equality kx + yk2 + kx − yk2 = 2(kxk2 + kyk2) (2.7) Proof. LHS = kx + yk2 + kx − yk2 = hx + y; x + yi + hx − y; x − yi = hx; xi + hx; yi + hy; xi + hy; yi + hx; xi − hx; yi + hy; yi − hy; xi (2.8) = 2(hx; xi + hy; yi) = 2(kxk2 + kyk2) MAT641-Inner Product Space and orthogonality 4 Examples of normed spaces which are not inner product spaces: Example 2. Sequence space lp, p 6= 2 Consider x = (1; 1; 0; 0; 0; ::::) and y = (1; −1; 0; 0; 0; :::) 2 lp and calculate kxk = kyk = 21=p; kx + yk = kx − yk = 2 Parallelogram equality is not satisfied. Hence lp, p 6= 2 is a Banach space but not Hilbert space. Example 3. Space C[a,b] is not an inner product space for kxk = maxt2J jx(t)j, J = [a; b]. Consider x(t) = 1 and y(t) = (t − a)=(b − a). We have, kxk = 1, kyk = 1 and t−a t−a kx(t) + y(t)k = k1 + b−ak = 2, kx(t) − y(t)k = k1 − b−ak = 1. Thus, Parallelogram equality is not satisfied. Theorem 2.4 (Polarization identity). The inner product in term of induced norm is given by 1. For a real IPS X, 1 hx; yi = (kx + yk2 − kx − yk2) (2.9) 4 2. For a complex IPS 1 <hx; yi = (kx + yk2 − kx − yk2) (2.10) 4 and 1 =hx; yi = (kx + iyk2 − kx − iyk2) (2.11) 4 Proof is left as an exercise to the reader. 3 Continuity in IPS Theorem 3.1 (Continuity). If in an inner product space xn ! x and yn ! y, then hxn; yni ! hx; yi. Proof. Consider jhxn; yni − hx; yij = jhxn; yni − hxn; yi + hxn; yi − hx; yij = jhxn; yn − yij + jhxn − x; yij Applying Schwarz inequality jhxn; yni − hx; yij ≤ kxnk kyn − yk + kxn − xk kyk ! 0 as n ! 1 Hence inner product is a continuous function. MAT641-Inner Product Space and orthogonality 5 Definition 3.2 (Isomorphism). An isomorphism T of an inner product space (X; h:i) ~ onto an IPS (X; h:i0) over the same field K of scalars is a bijective linear operator ~ T : X ! X which preserves the values of the inner products i.e. hT x; T yi0 = hx; yi. Exercise 3.3. Let Y be a subspace of a Hilbert space H. 1. Y is complete iff Y is closed in H. 2. If Y is a finite dimensional subspace then Y is complete. 3. Every subspace of a separable space is separable. Theorem 3.4 (Minimizing Vector). Let X be an inner product space and Y 6= φ is a convex subset which is complete (in the metric induced by the inner product) subset. Then for every given x 2 X, there is an unique y 2 Y such that δ = inf kx − y0k = kx − yk: y02Y 0 Proof: (Existence of y): Given δ = infy02Y kx − y k. By definition of infimum, there exists a sequence (yn) in Y such that δn = kx − ynk δn ! δ as n ! 1 We will show that (yn) is a Cauchy sequence in Y. Let x − yn = vn. Then (yn + ym) kvn + vmk = k2x − (yn + ym)k = 2 x − 2 But yn; ym 2 Y ,(yn + ym)=2 2 Y as Y is convex. Hence kvn + vmk ≥ 2δ: (3.1) 2 2 2 2 2 Now kyn − ymk = kvn − vmk = 2kvnk + 2kvmk − kvn − vmk (by Parallelogram law). 2 2 2 2 2 2 2 ) kyn − ymk = 2kx − ynk + 2kx − ymk − kvn − vmk ≤ 2δn + 2δm − 4δ (using (3.1)). As n; m ! 1, δn ! δ and δm ! δ. Thus kyn − ymk ! 0 as n; m ! 1 i.e. (yn) is a Cauchy sequence. Since Y is complete, (yn) ! y 2 Y . Thus existence of y is guaranteed. (Uniqueness): Let y0 and y1 2 Y such that δ = kx − y0k = kx − y1k 2 2 2 . Now ky1 − y0k = k(y1 − x) − (y0 − x)k ≤ 2ky1 − xk + 2ky0 − xk − ky1 − x + y0 − xk (by Parallelogram law) 2 2 2 ) ky1 − y0k ≤ 2δ + 2δ − 4δ ! 0. But ky1 − y0k ≥ 0, we have ky1 − y0k = 0 i.e. y1 = y0. MAT641-Inner Product Space and orthogonality 6 4 Orthogonality Definition 4.1 (Orthogonality). An element x of an inner product space X is said to be orthogonal to an element y 2 X if hx; yi = 0 and we write x?y. Theorem 4.2 (Orthogonality). Let X be an inner product space and Y 6= φ is a convex complete (in the metric induced by the inner product) subset. Then for a fixed x 2 X, z = x − y is orthogonal to Y where kx − yk = δ = inf kx − y0k; y 2 Y: y02Y Proof: Let z ? Y is false. Then there exist a y0 2 Y such that hz; y0i = β 6= 0 Clearly, y0 6= 0 since otherwise hz; y0i = 0. Consider the norm kz − αy0k2 = hz − αy0; z − αy0i = hz; zi − αhz; y0i − α(hz; y0i − αhy0; y0i) hz;y0i β Choose α = hy0;y0i = ky0k2 so that ββ kz − αy0k2 = kzk2 − < kzk2 = δ2 ky0k2 We claim, that kz − αy0k2hδ2 can never happen.