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CHAPTER 9 Inner Product Spaces and Hilbert Spaces BY Prof. M. Saha Professor of Mathematics The University of Burdwan West Bengal, India E-mail : [email protected] Chapter 9 Inner Product Spaces and Hilbert Spaces 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 di- mensional 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) 2 X × X there corresponds a scalar denoted by hx; yi called inner product of x and y of X such that the following properties hold. (IP.1) hx; yi = hy; xi where (x; y) 2 X × X and ` ' denotes the conjugate of the complex number. (IP.2) for all α 2 C; hαx; yi = αhx; yi; 8 (x; y) 2 X × X. (IP.3) for all x; y; z 2 X hx + y; zi = hx; zi + hy; zi. (IP.4) hx; xi ≥ 0 and hx; xi = 0 iff x = θ. Then (X; h : i) 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 2 C; hαx + βy; z) = αhx; zi + βhy; zi 8 x; y; z 2 X. (ii) for all α 2 C hx; αyi =α ¯hx; yi 8 x; y 2 X (iii) for all α; β 2 C 8 x; y; z 2 X. hx; αy + βzi =α ¯hx; yi + β¯hx; zi. 3 Chapter 9 Inner Product Spaces and Hilbert Spaces 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 2 X p Define jjxjj = + hx; xi (9.1.1) Now, jjxjj ≥ 0 as hx; xi ≥ 0 by (IP.4). Also, jjxjj = 0 iff x = θ by (IP.4) Take α 2 C so, jjαxjj2 = hαx; αxi; α 2 X = αα¯hx; xi; x 2 X = jαj2 jjxjj2; x 2 X so jjαxjj = = jαj jjxjj; x 2 X To prove the triangle inequality we first state and prove Cauchy Schwarz Inequality. Cauchy Schwarz Inequality: jhx; yij ≤ jjxjj jjyjj 8 x; y 2 X (9.1.2) Proof: If y = θX then the result follows trivially. Let y =6 θX . Then for every scalars λ 2 C; hx + λy; x + λyi ≥ 0 =) hx; xi + hx; λyi + hλy; xi + hλy; λyi ≥ 0 (By (IP.3)) =) hx; xi + λ¯hx; yi + λhy; xi + jjλyjj2 ≥ 0 =) jjxjj2 + λ¯hx; yi + λhx; yi + jλj2jjyjj2 ≥ 0 hx; yi Take λ = − hy; yi hx; yihx; yi hx; yihx; yi jhx; yij2 So, jjxjj2 − − + jjyjj2 ≥ 0: jjyjj2 jjyjj2 jjyjj4 jhx; yij2 jhx; yij2 jhx; yij2 So; jjxjj2 − − + ≥ 0 jjyjj2 jjyjj2 jjyjj2 4 Chapter 9 Inner Product Spaces and Hilbert Spaces jhx; yij2 So jjxk2 − ≥ 0 jjyjj2 =) jjxjj2jjyjj2 ≥ jhx; yij2 =) jjxjj jjyjj ≥ jhx; yij So jhx; yij ≤ jjxjj jjyjj Sometimes this inequality is abbreviated as C-S inequality. We see that equality sign will hold if and only if in above derivation hx + λy; x + λyi = 0 =) jjx + λyjj2 = 0 =) x + λy = θX , i.e x and y are linearly dependent. We shall now prove triangle inequality for norm. Now 8 x; y 2 X. Now; jjx + yjj2 = hx + y; x + yi = hx; xi + hx; yi + hy; xi + hy; yi = jjxjj2 + jjyjj2 + hx; yi + hy; xi So; jjx + yjj2 = jjxjj2 + jjyjj2 + hx; yi + hy; xi ≤ jjxjj2 + jjyjj2 + hx; yi + hy; xi ≤ jjxjj2 + jjyjj2 + 2jjxjj jjyjj ( ) 2 = jjxjj + jjyjj So; jjx + yjj ≤ jjxjj + jjyjj: 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 2 X define d : X × X ! R by p d(x; y) = jjx − yjj = + hx − y; x − yi (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) = hx; yi; 8 x; y 2 X then f is continuous). Proof: Let X be an inner product space. Define f : X × X ! C by f(x; y) = hx; yi; 8 x; y 2 X. Now take fxng and fyng be a sequence in X such that xn ! x as 5 Chapter 9 Inner Product Spaces and Hilbert Spaces n ! 1 and yn ! y as n ! 1. So, jjxn − xjj ! 0 as n ! 1 and jjyn − yjj ! 0 as n ! 1. As, xn ! x as n ! 1 then, jjxnjj ! jjxjj as n ! 1. So, fjjxnjjg are bounded. So, there exists a constant M > 0 such that jjxnjj ≤ M; 8 n: Now, jhxn; yni − hx; yij = jhxn; yni − hxn; yi + hxn; yi − hx; yij = jhxn; yn − yi + hxn − x; yij ≤ jhxn; yn − yij + jhxn − x; yi ≤ jjxnjj jjyn − yjj + jjxn − xjj jjyjj [By C-S inequality (9.1.2)] ≤ Mjjyn − yjj + jjxn − xjj jjyjj ! 0 as n ! 1 i.e hxn; yni ! hx; yi as n ! 1, implying that f(xn; yn) ! f(x; y) as n ! 1. So, f is continuous. Theorem 9.1.2 (Parallelogram Law): Let X be an inner product space and let x; y 2 X. Then, ( ) jjx + yjj2 + jjx − yjj2 = 2 jjxjj2 + jjyjj2 Proof: jjx + yjj2 = hx + y; x + yi = hx; xi + hx; yi + hy; xi + hy; yi = jjxjj2 + jjyjj2 + hx; yi + hy; xi (9.1.4) and jjx − yjj2 = hx − y; x − yi = hx; xi + hx; −yi + ⟨−y; xi + ⟨−y; −yi = jjxjj2 + jjyjj2 − hx; yi − hy; xi (9.1.5) Adding (9.1.4) and (9.1.5) we get ( ) jjx + yjj2 + jjx − yjj2 = 2 jjxjj2 + jjyjj2 Theorem 9.1.3 (Polarization Identity): Let X be an inner product space, let x; y 2 X. Then h i 1 hx; yi = jjx + yjj2 − jjx − yjj2 + ijjx − iyjj2 − ijjx − iyjj2 (9.1.6) 4 6 Chapter 9 Inner Product Spaces and Hilbert Spaces Proof: Now; jjx + yjj2 = jjxjj2 + jjyjj2 + hx; yi + hy; xi (9.1.7) jjx − yjj2 = jjxjj2 + jjyjj2 − hx; yi − hy; xi (9.1.8) Replacing y by iy in (9.1.7) and (9.1.8) jjx + iyjj2 = jjxjj2 + jjiyjj2 + hx; iyi + hiy; xi = jjxjj2 + jjyjj2 − ihx; yi + ihy; xi (9.1.9) jjx − iyjj2 = jjxjj2 + jjiyjj2 − hx; iyi − hiy; xi = jjxjj2 + jjyjj2 + ihx; yi − ihy; xi (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 fxng and fyng are two Cauchy sequences in X then fhxn; ynig is also a Cauchy sequence in C and hence convergence in C. Proof: (i) Let, fxng be a Cauchy sequences in X. Then for " = 1 there exists a positive integer N such that, jjxn − xmjj < 1, whenever, n; m ≥ N. In particular, jjxn − xN jj < 1, whenever n ≥ N. jj jj ≤ jj − jj jj jj jj jj 8 ≥ Now, xn n xn xN + xN < 1 + xN on N. Let, M = max jjx1jj; jjx2jj; ··· ; jjxN−1jj; jjxN jj + 1 so, jjxnjj ≤ M 8 n so, fxng is bounded. (ii) Let fxng, fyng be two Cauchy sequences in X. So, jjxn − xmjj ! 0 as n; m ! 1 and jjyn − ymjj ! 0 as n; m ! 1. Also, by (i) jjxnjj ≤ M for all n and for some M > 0. Similarly jjynjj ≤ K, for some K > 0 and 8 n. Now; jhxn; yni − hxm; ymij = jhxn; yni − hxn; ymi + hxn; ymi − hxm; ymij = jhxn; yn − ymi + hxn − xm; ymij ≤ jhxn; yn − ymij + jhxn − xm; ynij 7 Chapter 9 Inner Product Spaces and Hilbert Spaces ≤ jjxnjj jjyn − ymjj + jjxn − xmjj jjynjj (By C-S inequality) ≤ Mjjyn − ymjj + Kjjxn + xmjj ! 0 as n; m ! 1: So, fhxn; ynig is a Cauchy sequences of scalars in C.
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  • Ruler and Compass Constructions and Abstract Algebra

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    Ruler and Compass Constructions and Abstract Algebra Introduction Around 300 BC, Euclid wrote a series of 13 books on geometry and number theory. These books are collectively called the Elements and are some of the most famous books ever written about any subject. In the Elements, Euclid described several “ruler and compass” constructions. By ruler, we mean a straightedge with no marks at all (so it does not look like the rulers with centimeters or inches that you get at the store). The ruler allows you to draw the (unique) line between two (distinct) given points. The compass allows you to draw a circle with a given point as its center and with radius equal to the distance between two given points. But there are three famous constructions that the Greeks could not perform using ruler and compass: • Doubling the cube: constructing a cube having twice the volume of a given cube. • Trisecting the angle: constructing an angle 1/3 the measure of a given angle. • Squaring the circle: constructing a square with area equal to that of a given circle. The Greeks were able to construct several regular polygons, but another famous problem was also beyond their reach: • Determine which regular polygons are constructible with ruler and compass. These famous problems were open (unsolved) for 2000 years! Thanks to the modern tools of abstract algebra, we now know the solutions: • It is impossible to double the cube, trisect the angle, or square the circle using only ruler (straightedge) and compass. • We also know precisely which regular polygons can be constructed and which ones cannot.
  • 3. Hilbert Spaces

    3. Hilbert Spaces

    3. Hilbert spaces In this section we examine a special type of Banach spaces. We start with some algebraic preliminaries. Definition. Let K be either R or C, and let Let X and Y be vector spaces over K. A map φ : X × Y → K is said to be K-sesquilinear, if • for every x ∈ X, then map φx : Y 3 y 7−→ φ(x, y) ∈ K is linear; • for every y ∈ Y, then map φy : X 3 y 7−→ φ(x, y) ∈ K is conjugate linear, i.e. the map X 3 x 7−→ φy(x) ∈ K is linear. In the case K = R the above properties are equivalent to the fact that φ is bilinear. Remark 3.1. Let X be a vector space over C, and let φ : X × X → C be a C-sesquilinear map. Then φ is completely determined by the map Qφ : X 3 x 7−→ φ(x, x) ∈ C. This can be see by computing, for k ∈ {0, 1, 2, 3} the quantity k k k k k k k Qφ(x + i y) = φ(x + i y, x + i y) = φ(x, x) + φ(x, i y) + φ(i y, x) + φ(i y, i y) = = φ(x, x) + ikφ(x, y) + i−kφ(y, x) + φ(y, y), which then gives 3 1 X (1) φ(x, y) = i−kQ (x + iky), ∀ x, y ∈ X. 4 φ k=0 The map Qφ : X → C is called the quadratic form determined by φ. The identity (1) is referred to as the Polarization Identity.