CHAPTER 9 BY Prof. M. Saha Professor of Mathematics The University of Burdwan West Bengal, India E-mail : mantusaha.bu@gmail.com
Introduction and Objectives In the preceding chapters, we discussed normed linear spaces and Banach spaces. These spaces has linear properties as well as metric properties. Although the norm on a linear space generalizes the elementary concept of the length of a vector, but the main geometric concept other than the length of a vector is the angle between two vectors, In this chapter, we take the opportunity to study linear spaces having an inner product, a generalization of the usual dot product on finite dimensional linear spaces. The concept of an inner product in a linear space leads to an inner product space and a complete inner product space which is called a Hilbert space. The theory of Hilbert Spaces does not deal with angles in general. Most interestingly, it helps us to introduce an idea of perpendicularity for two vectors and the geometry deals in various fundamental aspects with Euclidean geometry. The basics of the theory of Hilbert spaces was given by in 1912 by the work of German mathematician D. Hilbert (1862-1943) on integral equations. However, an axiomatic basis of the theory was given by famous mathematician J. Von Neumann (1903-1957). However, Hilbert spaces are the simplest type of infinite dimensional Banach spaces to tackle a remarkable role in functional analysis. This chapter is divided into four modules. Module 1 ideals with the introduction of inner product spaces and Hilbert spaces with some examples. Some basic properties are given in it. Module 2 is concerned with the conception of orthogonal and orthonormal vectors in a Hilbert space. Some basic result are shown.. Module 3 is devoted mainly with very fundamental results on inner product spaces. Module 4 is discussed on some fundamental results on the Hilbert spaces particularly on approximation theory of convex subsets of a Hilbert spaces. Module 5 is emphasized on infinite series are those convergence in Hilbert spaces. Also with an introduction of isometric isomorphism property between two infinite dimensional Hilbert spaces an important theorem namely Reisz Fischer s Theorem has been stated. 2
Module-1: Inner Product Spaces Definition 9.1.1: Let, X be a linear space over a field of complex numbers. If for every pair (x, y) X X there corresponds a scalar denoted by x, y called inner product of x and y of X such that the following properties hold. (IP.1) x, y = y, x where (x, y) X X and complex number. denotes the conjugate of the (IP.2) for all α C, αx, y = α x, y, (x, y) X X. (IP.3) for all x, y, z X x + y, z = x, z + y, z. (IP.4) x, x 0 and x, x = 0 iff x = θ. Then (X,. ) is called an inner product space or pre-hilbert space. Remark 9.1.1: The following properties hold in an inner product space. Let X be an inner product space then, (i) for all α, β C, αx + βy, z) = α x, z + β y, z x, y, z X. (ii) for all α C x, αy = ᾱ x, y x, y X (iii) for all α, β C x, y, z X. x, αy + βz = ᾱ x, y + β x, z. 3
Remark 9.1.2: Inner product induces a norm. For this we proceed as follows: Proof: Let X be an inner product space. Take x X Define x = + x, x (9.1.1) Now, x 0 as x, x 0 by (IP.4). Also, x = 0 iff x = θ by (IP.4) Take α C so, αx 2 = αx, αx, α X = αᾱ x, x, x X = α 2 x 2, x X so αx = = α x, x X To prove the triangle inequality we first state and prove Cauchy Schwarz Inequality. Cauchy Schwarz Inequality: Proof: If y = θ X then the result follows trivially. Let y θ X. Then for every scalars λ C, x + λy, x + λy 0 x, y x y x, y X (9.1.2) = x, x + x, λy + λy, x + λy, λy 0 (By (IP.3)) = x, x + λ x, y + λ y, x + λy 2 0 = x 2 + λ x, y + λ x, y + λ 2 y 2 0 Take x, y λ = y, y So, x 2 x, y x, y x, y x, y x, y 2 + y 2 0. y 2 y 2 y 4 So, x 2 x, y 2 y 2 x, y 2 y 2 + x, y 2 y 2 0 4
So x 2 x, y 2 0 y 2 = x 2 y 2 x, y 2 = x y x, y So x, y x y Sometimes this inequality is abbreviated as C-S inequality. We see that equality sign will hold if and only if in above derivation x + λy, x + λy = 0 = x + λy 2 = 0 = x + λy = θ X, i.e x and y are linearly dependent. We shall now prove triangle inequality for norm. Now x, y X. Now, x + y 2 = x + y, x + y = x, x + x, y + y, x + y, y = x 2 + y 2 + x, y + y, x So, x + y 2 = x 2 + y 2 + x, y + y, x x 2 + y 2 + x, y + y, x = So, x + y x + y. x 2 + y 2 + 2 x y ( ) 2 x + y Hence, inner product induces a norm and consequently every inner product space is a normed linear space. Remark 9.1.3: So every inner product space is a metric space and the metric induced by inner product is defined as follows: for all x, y X define d : X X R by d(x, y) = x y = + x y, x y (9.1.3) Theorem 9.1.1: Every inner product function is a continuous function. (Equivalently, if f : X X C defined by f(x, y) = x, y, continuous). x, y X then f is Proof: Let X be an inner product space. Define f : X X C by f(x, y) = x, y, x, y X. Now take {x n } and {y n } be a sequence in X such that x n x as 5
n and y n y as n. So, x n x 0 as n and y n y 0 as n. As, x n x as n then, x n x as n. So, { x n } are bounded. So, there exists a constant M > 0 such that x n M, n. Now, x n, y n x, y = x n, y n x n, y + x n, y x, y = x n, y n y + x n x, y x n, y n y + x n x, y x n y n y + x n x y [By C-S inequality (9.1.2)] M y n y + x n x y 0 as n i.e x n, y n x, y as n, implying that f(x n, y n ) f(x, y) as n. So, f is continuous. Theorem 9.1.2 (Parallelogram Law): Let X be an inner product space and let x, y X. Then, x + y 2 + x y 2 = 2 ( x 2 + y 2) Proof: x + y 2 = x + y, x + y = x, x + x, y + y, x + y, y = x 2 + y 2 + x, y + y, x (9.1.4) and x y 2 = x y, x y = x, x + x, y + y, x + y, y = x 2 + y 2 x, y y, x (9.1.5) Adding (9.1.4) and (9.1.5) we get x + y 2 + x y 2 = 2 ( x 2 + y 2) Theorem 9.1.3 (Polarization Identity): Let X be an inner product space, let x, y X. Then x, y = 1 ] [ x + y 2 x y 2 + i x iy 2 i x iy 2 (9.1.6) 4 6
Proof: Now, x + y 2 = x 2 + y 2 + x, y + y, x (9.1.7) x y 2 = x 2 + y 2 x, y y, x (9.1.8) Replacing y by iy in (9.1.7) and (9.1.8) x + iy 2 = x 2 + iy 2 + x, iy + iy, x = x 2 + y 2 i x, y + i y, x (9.1.9) x iy 2 = x 2 + iy 2 x, iy iy, x = x 2 + y 2 + i x, y i y, x (9.1.10) (9.1.7) (9.1.8) + i(9.1.9) i(9.1.10), we get (9.1.6). Hence the result. Theorem 9.1.4: Let X be an inner product space. Then (i) Every Cauchy sequence is bounded (ii) If {x n } and {y n } are two Cauchy sequences in X then { x n, y n } is also a Cauchy sequence in C and hence convergence in C. Proof: (i) Let, {x n } be a Cauchy sequences in X. Then for ε = 1 there exists a positive integer N such that, x n x m < 1, whenever, n, m N. In particular, x n x N < 1, whenever n N. Now, x n x n x N + x N < 1 + x N n N. { } Let, M = max x 1, x 2,, x N 1, x N + 1 so, x n M bounded. (ii) Let {x n }, {y n } be two Cauchy sequences in X. So, x n x m 0 as n, m and y n y m 0 as n, m. n so, {x n } is Also, by (i) x n M for all n and for some M > 0. Similarly y n K, for some K > 0 and n. Now, x n, y n x m, y m = x n, y n x n, y m + x n, y m x m, y m = x n, y n y m + x n x m, y m x n, y n y m + x n x m, y n 7
x n y n y m + x n x m y n (By C-S inequality) M y n y m + K x n + x m 0 as n, m. So, { x n, y n } is a Cauchy sequences of scalars in C. As C is complete, { x n, y n } is convergent in C. Definition 9.1.2: A complete inner product space is called a Hilbert space i.e. an inner product space X which is complete with respect to a metric d : X X R induced by the inner product, on X X i.e. d(x, y) = x y, x y 1/2 x, y X. Theorem 9.1.5: A Banach space X is a Hilbert space if and only if parallelogram law holds in it. Proof: We know that every Hilbert space X is a Banach space where parallelogram law holds in it. Conversely suppose that X is a Banach space where parallelogram law holds. Without loss of generality we can assume a function, whose range is R. For all x, y X. Define, : X X R by x, y = 1 ] [ x + y 2 x y 2 4 (9.1.11) i.e. for real inner product space we start with (9.1.11) and sometimes we write R x, y = 1 4[ x + y 2 x y 2], x, y X. Clearly x, y = y, x as x, y is real. Also x, x 0 x X and x, x = 0 iff x = 0. So, (IP.1) and (IP.4) holds. Now, for u, v, w X u + v + w 2 + u + v w 2 = 2 ( u + v 2 + w 2) (9.1.12) u v + w 2 + u v w 2 = 2 ( u v 2 + w 2) (9.1.13) By (9.1.12) (9.1.13) we get u + v + w 2 + u + v w 2 u v + w 2 + u v w 2 = 2 ( u + v 2 u v 2) = 4 [ u + w, v + u w, v ] = 2.4 u, v = u + w, v + u w, v = 2 u, v (9.1.14) Put, u = w 2u, v = 2 u, v (9.1.15) 8
Again put, x 1 = u + w, x 2 = u w, x 3 = v then from (9.1.14) we get So, (IP.3) holds. x 1, x 3 + x 2, x 3 = 2 u, v = 2u, v (by (9.1.15)) From (9.1.16) for x 1, x 2, x 3, x 4 X, = x 1 + x 2, x 3 (9.1.16) x 1 + x 2 + x 3, x 4 = x 1 + x 2, x 4 + x 3, x 4 = x 1, x 4 + x 2, x 4 + x 3, x 4 Put x 1 = x 2 = x 3 = x, x 4 = y i.e. 3x, y = 3 x, y. So, by Principle of Mathematical Induction for any positive integer n nx, y = n x, y x, y X (9.1.17) Now, x, y = 1 ] [ x + y 2 x y 2 4 = 1 [ ] x + y 2 + y x 2 = x, y (9.1.18) 4 Take, n = m (m > 0) nx, y = mx, y = m( x), y = m x, y (by (9.1.17)) = m x, y (by (9.1.18)) = n x, y So (9.1.17) is also true for any negative integer. Thus λx, y = λ x, y, when λ is either positive integer or a negative integer. Take λ = p/q = a rational number where,, gcd(p, q) = 1 and p and q are integer. p Now λx, y = q x, y p So q λx, y = q q x, y = px, y = p x, y = λx, y = p x, y q p = q x, y = p x, y (9.1.19) q 9
Let, λ be any real. So there exists a sequence of rationals {r n } such that r n λ as n. So, r n x, y = r n x, y λ x.y as n Now, r n x, y λx, y = (r n λ)x, y (r n λ)x y (by C-S inequality (9.1.2)) = r n λ x y 0 as n r n x, y λx, y as n. So, λx, y = λ x, y x, y X. So, (IP.2) holds. So, X is an inner product space with respect to (9.1.11) consequently, X is a Hilbert space. Note: For a complete inner product space we start with Rl x, y = 1 [ x + 4 y 2 x y 2 ], Im x, y = 1 [ x + 4 iy 2 x iy 2 ], x, y X and the proof is similar to the proof of real inner product space. (Readers can verify it) Example 9.1.1: The Euclidean space R n is a Hilbert space. Solution: Let x = (ζ 1, ζ 2,, ζ n ) and y = (η 1, η 2,, η n ) be two element of R n. We define the inner product of x and y by (x, y) = ζ 1 η 1 + ζ 2 η 2 + + ζ n η n. Then x = x, x = (ζ1 2 + ζ2 2 + + ζn) 2 1 2. It may be easily verified that all the inner product axioms are satisfied in R n and the Euclidean metric d is obtained by d(x, y) = n x y = x y, x y 1 2 = (ζ i η i ) 2. With respect to this metric we can at once see that R n is complete so as to make R n, a Hilbert space. i=1 Example 9.1.2: The Euclidean space C n is a Hilbert space. Solution: Let x = (ζ 1, ζ 2,, ζ n ) and y = (η 1, η 2,, η n ) be two elements of C n. We define the inner product of x and y by x, y = ζ 1 η 1 + ζ 2 η 2 + + ζ n η n. Then x = x, x = ( ζ 1 2 + ζ 2 2 + + ζ n 2 ) 1 2. It may be easily verified that all the inner product axioms are satisfied in C n and the Euclidean metric d is obtained by n d(x, y) = x y = x y, x y 1 2 = (ζ i η i ) 2. With respect to this metric we can at once see that C n is complete so as to make C n, a Hilbert space. i=1 Example 9.1.3: The space l 2 is a Hilbert space. 10
Solution: Let x = {ζ i } and y = {η i } be elements of l 2. We define the inner product of x and y by x, y = ζ i η i convergence of the series on the right hand side follows i=1 from the fact that x l 2 and ζ i η i ζ i 2 + η i 2. 2 2 Then x = ( ) 1 2 x, x = ζ i. It can be easily shown that all the inner product i=1 axioms (IP.1) -(IP.4) are satisfied in l 2. The metric d of l 2 is defined by d(x, y) = ( n ) 1 x y = x y, x y 1 2 = ζ i η i 2 2. With respect to this metric we can at i=1 once see that l 2 is complete so as to make l 2 a Hilbert space. But for 1 p <, following example. l p (p 2 is not a Hilbert space. It can be shown by the Example 9.1.4: For 1 p <, l p (p 2) is not an inner product space and hence not a Hilbert space. Solution: Let x = (1, 1, 0, 0, ) l p and y = (1, 1, 0, 0, ) l p. Then x = y = 2 1 p does not hold. and x + y = x y = 2. Now we see that if p 2, the parallelogram law Hence 1 p < l p (p 2) is not an inner product space and consequently it is not a Hilbert space. Example 9.1.5: The space c[a, b] of all real valued continuous in the closed interval [a, b] is not an inner product space with respect to sup norm and hence not a Hilbert space. Solution: Here the norm defined by x = sup x(t). Take x(t) = 1, t [a, b] and a t b y(t) = t a, t [a, b]. Then x = 1, y = 1, x + y = 2, x y = 1. By simple b a calculations we see that parallelogram law does not hold in it. Hence c[a, b] is not a Hilbert space. Example 9.1.6: The space L 2 [a, b], the space of all square integrable functions over [a, b] is a Hilbert space. Solution: Define the inner product on L 2 [a, b] by x, y = b x(t)y(t) dt, x, y a b L 2 [a, b] and the norm on L 2 [a, b] is given by x = z x(t) 2 dt. Also with respect 11
to this norm it can be shown that L 2 [a, b] is complete with respect to a metric defined by So L 2 [a, b] is a Hilbert space. [ b d(x, y) = a ] 1 x(t) y(t) 2 2 dt. 12