UNIT 4 HILBERT SPACE AND ORTHONORMALITY Structure Page No. 4.1 Introduction Objectives 4.2 Complete Inner Product Spaces 4.3 Orthonormal Sets 4.4 Projection and Riesz Representation Theorems 4.5 Summary 4.6 HintsISolutions 4.1 INTRODUCTION In this unit, we provide geometric structure to a linear space. This leads us to Hilbert spaces. Hilbert spaces are special kind of Banach spaces, those which possess an additional structure in which we can tell when two elements are orthogonal (i.e. perpendicular). In Sec. 1, we will be concerned with the geometric implications of this additional structure. You will observe that this geometry is very much like the familiar two- and three-dimensional Euclidean geometry. In Sec. 2, we shall see some applications of orthogonality to function spaces. Sec. 3 deals with two important theorem for Hilbert spaces, namely, the Riesz representation theorem which gives a characterization of bounded linear functionals on Hilbert spaces and Projection theorem. Objectives After studying this unit, you should be able to a define an inner product space and give examples; a identify inner product spaces which are complete and are called Hilbert spaces; a explain the geometrical features of Hilbert spaces; a explain the orthogonality property; a state and prove: i) the projection theorem ii) the Riesz representation theorem 4.2 COMPLETE INNER-PRODUCT SPACES In this section, we study an inner product space which is new structure on a linear space. This sh-ucture helps us to study perpendicularity of vectors. You will observe that the theory of inner product space does not depend on angles but yet one can talk about perpendicularity READL r Chapter VI Section 21 beginning upto Page 38 1 Line 12 1 Functional Analysis NOTES 1) Page 368 Last line. The inner product is conjugate-linear in the second variable. For (x, y + z) = (y + z, x) (conjugate-symmetry) = (y, x) + (2,x) (linearity in the first variable) -- = (y, x) + (2, x) (property of conjugate) = (x, y) + (x, z) (conjugate-symmetry) Also (x, ky ) = (ky ,x) = k(Y, x) = k(y,x) - = k(x,y) 2) Page 369 Line 15 It is easy to check that the right-side of the desired identity (i.e. 21.l(a)) Textbook L reduces to 4(x, y). To see this by successive use of linearity in the first variable and conjugate-linearity in the second variable, we have (X+ Y,X f Y) = (x,x) + (x,Y)+ (~4)+ (Y,Y) (x - YlX - Y) = (x,x) - (x,Y)- (~9)+ (Y~Y) i(x + iy, x + iy) = i{(x, x) - i(x, y) + i(y, x) + (y, y)) D = i(, x) + (x,Y)- (Y,x(+~(Y,Y) (because (x , iy ) = -i(x , y) ) 3) Page 369 Line 15 It is easy to check that the right side of the desired identity reduces to 4( This can be verified as follows: The right-side of the identity is equal to (x + y, x + y) - (x - y, x - y) + i(x + iy, x + iy) - (x - iy, x - iy). Now (x + Y, x + Y) = (x, x + Y) + (Y, x + Y) (by linearity in the first variable) = (x,x)+ (x,Y)+ (Y,x)+ (Y,Y) (by linearity in the second variable) Sirnilarly (x - Y),(x - Y) = (x,Y)- (x,Y) - (Y,x)+ (Y,Y) Also (x + iy, x + iy) = (x, iy) + (x, x) + (iy, x) + (iy, iy) = -i(x, y) + (x, x) + i(y, x) - i2(y,y) = -i(x, Y) + (x, x) + i(Y, x) + (Y,Y) and Substituting (4), (3, (6) on the right side of the identity we get 4(x, y) Hilbert Space and' Orthonormality 1 1 i(x - iy, x - iy) = i(x, x) - (x, y) + (y, x) + i(y, y) (because i(x, -iy) = - (x, y)) (8) Substituting (I), (2), (3) and (4) in the right side of 21.l(a) of Textbook L we get 4(x, Y). 4) Page 370 Lines 6 and 7: We see that if y = kx, then (x, x>(Y? Y) = (x, x) 0% kx) = kE(x, x)~ = (k(2(~,~)2 (If k is a complex number then kE = lkI2) = (E(2(x,x)~ = I(x1 kx) l2 = I(x, y>I2. 5) Page 371 Line 26 or Line 4 from bottom For any real number a, we have that a 5 la1 and a2 5 a2 + b2 for any real b. Hence la\ = &? 5 dm,i.e. a 5 la1 5 la + ibl. In other words, real part of a + ib is less equals Ja+ ibJ. 6) Page 372 Line 9 Since {((x,}} is a bounded sequence we see that (x,, y,) - (x, y). Since IJx, - XI( -+ 0, (xn) is a convergent sequence and hence it is bounded. So there exists a real number cr such that JJx,JJ< cr for all n. Now for a given E > 0, let N be an integer 2 1 such that E E ll~n- YI( < and (lxn - xJI 5 - ally ll provided y # 0. (The case y = 0 can be argued separately.) You may be aware that one can choose a common stage for two convergent sequences. 7) Page 372 Line 17 or Line 7 from below IIx - y 11 2 E. Then E 5 2. This follows by triangle inequality of the norm 6 I IIx - yll 5 llxll+ ll~llI 1 + 1 = 2. 8) Page 372 Line 21 or Lines 1,2 and 3 from below If 11 1) is a norm on linear space X which satisfies the parallelogram law (21.2(a)in Textbook L), i.e. we define for x, y, E X, 1 ('1~) = q(llx + yIl2 - IIx - yIl2 + illx + iyII2 - illx - iy1I2), then it can be shown that (, ) is an inner product on X. This essentially implies that a linear space is an inner product space if and only if it satisfies the parallelogram law. The proof of this result which is due to mathematician Jordan and Von Neumann, is somewhat tedious as you will see here. We will give you the important steps and leave them as exerdses to you as they are routine calculations. We are given that for all x, y E X ((x+ y(I2+ ((x- yl12 = 211~11~+ 211yl12(~arallelogramlaw) (9) Let us define 1 (x, y) = -(((~+~((~-((~-~((~+i~~x+i~~~~-i~~x-i~((~)(Polarization identity) (10) e 4 Functional Analysis Firstly let us prove that For this, replace y by x throughout in (9). By using the property /lax11 = la1 llx 11 and simplifying you will see that (11) holds. Now to show conjugate symmetry of ( ), take the complex conjugate of both sides of (10). Since IIx + iY1I2,IIx - iy1I2, IIx + ~(1~and IIx - y(12,they do not change under conjugation on simplifying you will see that To prove that (X+ Y14 = (~1~1+ (~7') (13) replace x by (x + y) and y by z in (2); you will get 4(x+y,z) = (~~(x+y)+z~~2-(l(x+y)-z(i2+i~((x+y)-iz((2-i(((x+y)-iz))2)(14) Now let us replace x by x + z in (I), to get II(x + 2) + y1I2 + Il(x + z) - yH2 = 211~+ z1I2 + 211~11~ (15) Again by (I), you can see that lib+ 2) - y1I2 + Ilk - Y)+ x1I2 = 211~- z1I2 + 211x1I2 - II(x +Y) - z1I2 (16) Substituting the value of 11 (x + z) 7 Y(12 from (16) in (15) you will see that II(x+y)+z(J2- II(~+y)-z11~= IIx+z~~~- II~-z11~+lly+zH~-lly-z11~(17) Interchanging x and y in (17) and adding it with (17), you will get that ll(x+~)+zlJ~-I~(x+Y) -412 = IIx+z1I2 - llx-z112+ IIY +z1I2- (IY -z112 (18) Now we replace z by iz in (18) and multiply both sides by i, to get ill(~+~)+izll~-ill(x+y)-iz1(~= i(~x+i~((~-i~(x-iz(~~+i~~~+iz~~~-i~~~-iz~(19) Using (14), you can see that (18) becomes 4(x + y, 2) = 4(x, 2) + 4(y, 2). Now comes the most tedious part of our calculations. We still have to show that (ax, y) = a(x,y) for all complex a and for all x, y in X. We prove this by looking at different values that a can take Firstly suppose (i) cr is a pvsitive integer Replacing z by x in (13), you see that Thus, the result is true for cr = 2. Now arguing by induction, you c& see that (nx, y) = n(x, y) for all positive integers n ' (20) (ii) a is a negative integer Hilbert Space and For this replace x by -x in (lo), to see that Orthonormality If & is a negative integer then we can write a = -p where ,f3 is a positive integer. (In fact ,f3 = -a.) We have then (iii) a is a rational number. Let us write a = p/q where p, q are integers and q # 0. Then X X Also (9.-, y) = q(-, y) because q is an integer. 9 9 X 1 x Therefore (- , y) = - (q. - , y) 4 9 and hence iv) cu is a real number. You know then that there exists a sequence (an) of rationals which converges to cr (The set of rationals is dense in reals!) Now (ax, Y)= (( lim a;l)x, Y) n--too = lim (a,x, y) (This follows from 21.2(a) of Textbook L) n-cc = lirn a, (x, y) (By (iii)) n--cc = (x, y) lim a, n-cc = (x,Y)Q. V) a is a complex number. Firstly replace x by ix in (10) and simplify to see that (ix, y) = i(x, y) Now write a = a1 + ia2 where a1 and a2are real and use (iv) to see that (ax, Y)= 4x7 Y) We have, thus, proved that X is an inner product space.