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.
At this point you will probably agree to grant that completeness is a very desirable property for spaces to have. Every inner product space is a medlinear space given by 11 x 11 = (x, X) and hence is a metric space. If this metric space is complete then the space is called a Hilbert space. As you will see the Hilbert spaces have "richer" geometric structure than the inner product spaces. So it is useful to note carefully in the rest of this chapter which theorems are true for inner product spaces and which for Hilbert spaces. Functional Analysis 9) Page 373 Line 13 Note that an inner product space X can be completed into a Hilbert spa& H. You are aware that every metric space can be embedded as a subspace of a complete metric space. In a similar manner an inner product space can be embedded as a subspace of a Hilbert space. This is left as an exercise to you (E3). 10) Page 374 Line 15 The function - (x, YJM = x(l)y(l) + x(2)y(2) +x(3)y(3) - x(4)x(4) is not an inner product because for x = (0,0,01) in K*,(x, X)M= -1 which is not positive. 11) Note that X = coo is dense in H = C2. However X # H. Hence X is not closed in H. This is because if x = (x(l), x(2), x(3), . . .) is in C2 then the sequence
lies in coo and converges to x in C2. Therefore coo is dense in 12.Since coo # e2 and coo = C2, coo cannot be closed in 12,for otherwise = coo = C2 which is not true. 12) Page 375 Lines 15,16 and 17 We find that among all the b-spaces 1 < p < co,only l2is an inner product space. If x = (1,O: 0,. . .) and y = (0,1,0,1,. . .) are elements of Cp for p # 2 then the parallelogram law yields that 2 = 22/p which is not true for p # 2.
Let us see some examples.
Example 1: Let us show that b with p-norm is an inner product space unless p = 2. Solution: If the norm is induced by an inner product then it must obey the parallelogram law. If el, e2 are the first two unit vectors in b,then
el +e2 = (l,l,O,.. .),el -e2 = (1,-1,0,. . .) Ilel + en11 = (1 + I)'/'' = 2l/p, //el- ezll = (1 + ].)l/p = 21/p. The parallelogram law gives
that is s2/p + 22/p = 2 + 2. This is possible only if 2/p = 1.This proves the result.
Example 2: Let {x,) be a sequence in an inner product space X. Let us show that if
(x", X) -+ (x, X) and llxnll -+ IIxII, then xn + x. Solution: Since
llxn - x(I2= (xn - X,xn - X) = (xn, xn) - (xn, X) - (x, xn) + (x, x),
(xn,xn) = IIxnI12 + IIxI12, (xtxo) = (xncx) + (x,x), we see that
Ilx, - x1I2 -+ llx112 - (x, x) - (x, x) + (x, x) = 0. That is, x, + x. Hilbert Space and * * * Orthonormality You can now try the following exercises.
El) Let X be an inner product space and let x, y E X. Suppose (a,) is a sequence of scalars converging to a scalar a prove that (a,x, y) -+ (ax,y) in X. E2) Prove that a subspace of a Hilbert Space is Hilbert under the induced inner product if and only if the subspace is closed. (It is useful to recall a similar result for complete metric spaces here.) E3) Prove that the quotient of a Hilbert space by one of its closed subspaces is again a Hilbert space. (It will be worthwhile to look up Theorem 8.2(a) of Textbook L on page 127.)
ORTHONORMAL SETS As we mentioned earlier the geometry of Hilbert spaces is roughly speaking a generalization of Euclidean geometry. The concept of inner product allows us to introduce perpendicularity of vectors, i.e. orthogonality of vectors. This is motivated by the fact that in Euclidean spaces, two vectors are orthogonal if and only if their dot product is 0. In this section, orthogonal vectors and orthonormal sets in an inner product space are defined and we look at some properties and also some examples.
You can now READL : Chapter VI Section 22 beginning upto page 401 last line. I
NOTES I) Page 386 Line 15
n=l For
n=l k= 1 m = (x,~n) (m) (u., uk)(by linearity and conjugate linearity) n,k=l m
(because {un)being orthogonal(u,, uk) = 0 for all n # k) m = x(x,un)(m) (because(u,, un) = 1 for all n) n=l m Functional Analysis By similar calculations you can see that
2) Page 386 Line 17
This is because
But (x,xm) = (xm, X) = (xm, xm) Therefore
3) Page 387 Line 15 The orthonormality of the set {ul, u2, . . .) shows that (x, u,) = k, for each n. This, of course needs some proof. In fact, we have that
= ( lim knun un) m-w C
= lim ((Cknun, u,) (By Theorem 21.2(a) of L) m-oo
, = !im k, (u,, u,) (because(u,, um) = 0 for n # m) m-w = k,( because (u,, u,) = 1
4) Page 388 Line 23 or Line 5 from below For any vector space basis exist, you may recall here that by a basis we mean a linearly independent set which also generates the vector space. Banach spaces and Hilbert spaces are vector spaces and hence they contain bases in the sense of vector spaces. These bases are known as Hamel bases. If B is a Hamel basis, then every element in the space is expressible uniquely as a finite linear combination of the basis elements. However, with topological structure present in Banach and Hilbert spaces, we have new opportunities open to us. It is now possible to attach w meaning to infinite linear combinations of the form C knun.Thus, we have the n= 1 possibility of introducing bases which involve topological as well as algebraic structure. One such useful concept is that of an orthonormal basis in a Hilbert space. You will see, in what follows, the orthonormal basis behave in a manner similar to bases in a fink dimensional vector space. 5) Page 389 Line 5 It is easy to see that every totally orde~edsubset of E has an upper bound in E. Let T be any totally ordered subset of E, say T = {A,\& E A). We shall show that the set theoretic union A = U A, is an orthonormal set in E. For this let x, y c A (YEA he dictinrt Then thprp pyict n R in A c~lrhthat Y A. and v F A Riit T ic totally ordered. Therefore either A, Ap or Ap C A,. If A, C Ai3 then Hilbert Space and Orthonormality x, y E Ap. But Ap is an orthonormal set. Hence x Iy and llx 11 = 1 = 11 y 11. Similar argument holds if Ap C A,. This shows that A is an orthonormal set clearly A > E. Therefore A E E. Moreover A being a set theoretic union it is an upper bound for T in e. 6) Page 389 Line 20 If x = 0, there is nothing to prove. For if x = 0 then (x, u,) = 0 for all a and hence the set Ex is empty. 7) Page 390 Line 2 1 ix, ua) I Ex is the union of all the Ejs. For if u, E Ex then (x, u,) # 0 and hence llxll 1. is a positive real number. But -+ 0 as j 4a. Therefore there is a j for which J
- < '(X'Ua)' i.e. u, E E, far some j. j llxll 8) Page 390 Line 2 Ex is a countable set because Ex is a union of all Ej's for j = 1,2, . . . , where each Ei has atmost j2 elements. This means that Ex is a countable union of finite sets. 9) Page 391 Line 9 rn Condition (ii) imples condition (iv), because x(x,u,)un being a finite linear n=l a combination of the un7sbelongs to span {u,). But by (ii), x = x(x,un)un. This n= 1 m means that x is a limit of sequence of partial sums x(x,un)u,, and each such sum n=l belongs to span {u,). In other words, x is a limit of a sequence of elements of span {u,} i.e. x E span{u,}. 10) Page 391 Line 11: (iv) implies (v) (x, u,) = 0 for all a implies that (x. x,) = 0 for all x, E span{u,}. 11) Page 391 Line 14 I.et E be an orthonormal set in H containing {u,). Such an orthonormal sets because of Zorn's Lemma. 12) Page 397 Line 2 from below If {ul, u2, . . .) is a finite set having n elements then F is a linear map from H to Kn. Since the inner product is linear in the first variable, F(ax + py) = aF(x) + PF(y).
Example 3: Let X be the real space C[-l ,1] of all real-valued continuous functions on [- 1, I] with the inner product given by
If Y is the set of all odd functions in X, find its orthogonal complement Y~.
Solution: A function x is odd if x(-t) = -x(t) for all t; it is even if x(-t) = x(t) for all t. The orthogonal complement of Y is
Y'- = {x EX: (x,y) = 0 forally E Y}. Let x be any even function and y E Y. Then x(t)y(t) is an odd function oft, and hence Functional Analysis Thus, (x, y) = 0 for all y E Y. Therefore, x E Y'.
On the other hand, let x E Y', and y(t) = x(t) - x(-t). Then, y(-t) = x(-t) - x(t) = -y(t), and so y E Y. Hence, (x, y) = 0. Since
a we get
Since y is continuous this implies that y(t) = 0 for all t, that is, x(-t) = x(t). So x is an even function if x E yL.Thus, Y' is the set of all even functions.
Example 4: Let xl(t) = t2, x2(t) = t and xs(t) = 1. Let us orthonormalise XI,x2, x3 in this order, on the interval [-I, I.] with respect to the inner product
Solution: We find orthonormal functions ul, u2, us as in Problem 5.2.3.
Since
we take
Let
b Y2 = Xz - (x2,ul)ul.
Then (yz, ul)'= 0. Since
we have y2 = xz,
So we take
Let
Then, (ys, uz) = 0 = (y3, ~1).Since
. fl ,*- Hilbert Space and Orthonormality we get
Take
Thus,
We conclude by leaving some exercises for you to try.
E4) Let u, be the sequence in l2with 1 in the nth place and zeros elsewhere. Prove that the set {u,) is an orthonormal basis for 12.
E5) Let X = Coo, u, = (0, . . .O, 1,0,0, . . .O) where only the nth entry is 1 and let 1 k, = - for n = 1,2, . . .. Prove that the Riesz-Fischer theorem does not hold good n for this space.
E6) Let X = e2, for n = 1,2, . . . let x, = (1, . . .l,0,O,. . .) where 1 occurs only in the first n places. Prove that the Gram-Schmidt orthogonalization yields the sequences {u,) where u, = (0, . . . ,0,1,0, . . .O, . . .) where 1 occurs only in the nth place. E7) Letx,(t) = tnforn = 0,1,2.. . and-1 5 t < 1. i) Show that {xo , x 1, . . .) is a linearly independent set in L2[- l,l]. ii) Use the Gram-Schmidt process on 1x0,XI,. . .) to determine the first three vectors in the orthonormal set.
In the next section, we shall give a characterisation of bounded linear functionals on Hilbert spaces.
4.4 PROJECTION AND RIESZ-REPRESENTATION THEOREMS
You were introduced to bounded (continuous) linear functionals on normed spaces in Section 1.4. In this section we discuss the linear functionals on inner product spaces. Mainly we wish to discuss the relationship between inner product spaces and continuous linear functionals on them. It will be seen that every continuous linear functional on a Hilbert space (completeness of the space is important) has a particularly simple representation, so simple that you can Literally lay your hands on them.
Now you can read what the Textbook L says
0 READL : From the beginning of Section 24 Chapter VI Page 420 upto Page 431 Line 16 Functional Analysis NOTES 1) Page 420 Line 2 Following example illustrates E'- Let H = l2 and E = {el, e2, es, e4) where e~=(1,0,0,0,~~~),e~=(0,1,0,0,~~~),e~=(0,0,1,0,~~~)and 'e4 = (0, 0, 0, 1,0. . . An element x = (xl, x2, . . . ) belongs to E~ if and only (x,ei) = 0 for each i = 1,2,3and 4. But (x, el) = ((XI,~2,. . . ), (1,0,0,. . . )) = (XI, 0, 0, . . .), (x,e2) = (O,x~,O,~~~),(x,e~)= (O,O,x~,O,~~~)and(x,e~) = (O,~,O,X~,O...) . Therefore, x must be of the form (O,O, 0,0, xs, xs, . . . ) The following theorem shows that E'- is a closed linear subspace for any subset E(# 4). We have that - = 6 for any x # 0. Therefore, since (f(x)( _< 1 for any x for 11 11 6 which 1JxJ(5 6, 1 ($) / _< i. By linearity off, this implies that -- If(x) 1 5 1 IIxIl i.e. (f(x)I I llx 11 16. Observe that this is true even when x = 0. 2) Page 423 Line 8 This proves that the number of elements in any finite subset of Ej is atmost j2 and hence Ej has to be a finite set and the number of elements in it is of course, atmost j2. 3) Page 424 Line 7 from below The unique y E H such that
depends on f. As f varies, y also will vary. 4) Page 426 Line 5 For vm = uno,(vm, ~n) = (uno,u,) = 0 for all n # no, and (u,, , u,,) 1 because {u,) is orthonormal. 5) Page 426 Line 6 If V, # un for any n, then v, 6 {u, : (x,u,) # 0) i.e. (v,,u,) = 0 for all a. 6) Page 428 Line 1 f # 0 becabse there is a non-zero element, for example x = (1,0,0,. . . ) which is mapped into 1.
Let us see some more properties.
Theorem 1: Let us prove the following. a) B'- c A'- b) A C A'-'- ) All'- = A'- Give examples of situations where we have equalities and strict inclusions in (a) and (b).
Solution: (The proof of (a), (b) and (c) are left as an exercise for you to try since these can be verified directly.)
Here we shall show examples to show that the inclusion in (a), (b) and (c) are strict.
Let B be a nonzero subspace of X and A = (0). Since (x, 0) = 0 for any x, we see that x E A'- for any x. Thus, A'- = X. If x E B and x E B'-, then (x, x) = 0 and so x = 0. Hence, if x # 0 and x E B. then x $! B'. So, B' 5 X = A'. Thus, we have a situation Hilbert Space and where the inclusion in (a) is strict. Orthonormality
Next, as above, let B be a nonzero subspace of X and A be the set obtained by removing a nonzero element u from B. Then A B. If x E A',
since -u E A. Thus, we see that (x, y) = 0 for all y E B, that is x E B'. Hence A' c B'. So by (a), equality holds in (a). If x E BL, then (x, u) = 0 sir= u E B. Thus,
u 1x for all x E B' = A'
Hence, u E A" while u $! A. So we have a situation where the inclusion in (b) is strict.
Finally, if A = E' for some E c X then
by (c). Thus, equality holds in (b).
'Theorem 2: Let us prove the following: a) If S is a nonempty subset of an inner product space X, and Y = span M, then yl = s'. b) If M is a nonempty subset of a Hilbert space H, then M" = span M. Solution: a) Let span S = Z. Then S C Z C Y, and hence yL C S' by Theorem 1. For the reverse inclusion, let x E s'. Then, (x, u) = 0 for all u E S. If z E Z,
and hence (x, z) = 0. Let y E Y. Then there is a sequence {z,) in Z such that Z, --+ y. So (x, z,) --+ (x, y). But (x, 2,) = 0 for all R. Hence, (x, y) = 0. This is true for any y E Y. This means that x E Y'. Thus, S' C yL.This proves (a). b) Let Y = span M. Then Y' = M' by (a). Since Y is a closed subspace of a Hilbert space we see that Y is complete. So yLi = Y. Thus MIL = Y'~= Y. This completes the proof. Theorem 3: Let F be a closed subspace of H and a E H, a $! F. Let us show that there is a unique f E HI such that f(a) = d(a,F), f(x) = 0 for all x E F and ((f/J= 1.
Solution: There exist b, c such that
Since a - b = c E FI, we see that b is the best approximation to a from F. Further, c # 0 since a $! F. Let
Then,f E HI and Jlf/l= IlylJ = 1. Ifx EF, thenx E H, and hence
f(x) = (x, 2)= 0 IlcIJ Thus. f(x) = 0 for all x E F, Further, Functional Analysis Since d(a, F) = Ila - bll = IIcII, we see that f(a) = d(a, F). This completes the proof. , You can try some exercise.
E8) Suppose that F1 and Fz are closed subspaces of a Hilbert space H. Show that i) (F1 n F2)lequals the closure of F: + F;; ii) if, F1 I F2, then F1 + F2 is a closed subspace. Show also that F1 I FZ,need not be closed even though F1 and Fa are closed subspaces.
E9) Let F be a finite dimensional subspace of an inner product space. Then show that x =~+F'and~" = F. ElO) Let X be an inner product space. Then show that the projection theorem holds if and only if Riesz representation theorem holds iff X is complete.
4.5 SUMMARY
In this unit, we have covered the following points: 1. You were introduced to inner product spaces and Hilbert spaces. You have seen that these spaces behave, in many respects, like finite dimensional spaces. 2. The notion of orthogonality of two elements was discussed. You saw how this lead to a theory of orthonormal sets and bases. 3. We discussed the relationship between inner product spaces and bounded linear functionals on them. 4. We discussed two important theorems: i) Projection Theorem ii) Riesz Representation Theorem.
El) Hint: Try it by yourself by applying the definitions. E2) Try it by yourself E3) Try it by yourself. E4) Hit: Verify that (u,, u,) = 0 when n # m and 1111, (1 = 1. E5) Note that
CJk"12< m
60 Ck.u.= (1,O ...0)+ (0,- ;,o ...> + (O,o,;...) n=l 11 = (1, 5,i,. . .)
= (b) E t2 E6) Hint: Apply 22.2 Theorem on Page 383 of Textbook L. Hilbert Space and Orthonormality E7) i) By construction {u,) is an orthonormal sequence in L2[- 1,1] such that
Yn = span{xo, xl, . . . , x,) = span{uo, ul, . . . , u,).
Since x,(t) = tn, we see that Y, consists of polynomials of degree not exceeding n. To prove that {u,) is an orthonormal basis for L2[- 1, I.], let x E L2[-I, 11 and (x, u,) = 0 for all n 2 0. It follows that (x, y) = 0 if y E Ym for some m.
Let y be any continuous function on [-1,1]. Then there is a sequence {y,) of polynomials converging uniformly to y on [- 1,1]. So, y, --t y in the norm of L2[- 1, I.]. Hence (x, y,) -+ (x, y) . But each polynomial y, lies in some Y,, and so (x, y,) = 0 for all n. Thus, (x, y) = 0. Since continuous functions are dense in L2[-I, 11, there is a sequence {z,) of continuous functions on [- 1, I} such that z, --t x in the norm of L2[- 1,1]. Hence, (x, z,) --t (x, x). But (x, z,) = 0 for all n. Thus, (x, x) = 0 and x = 0. This proves (a). ii) The first three terms uo, ul , u2 of the desired orthonormal sequence are determined as follows: Since xo(t) = 1 for all t,
So we take
Let
Then, (yl, uo) = 0. Since
l1 (XI,uO) = 11 t-dtJZ = 0, we haveyl = xl, and so
Take
Let
Y2 = x2 - (~2,~1)~l - (x2, u0)uo.
Then (y2, ul) = O(y2,uo). Since Functional Analysis we get
JZ 1 y2 = X2 - UO= X2 - - 3 3' 1 2 8 11~211~= 11(~2(t))2dt= Ll(t2 - 113) dt = -.45
Take
u2 = 2.-= g (x2 - i). llv2ll Thus,
5 3t2 - 1 uo(O = 8,u1(t) = fit, u2(t) = & (T)
i) Since F1 n F2 C F1 and F1 n Fz C F2, we have by Theorem 1,
F: C (Fl n ~2)l,F; C (Fl n ~2)~.
Thus, (F1 n ~2)~is a closed subspace containing F: and F;, and hence F: + F:. So if F is the closure of F: + F;, then
(F~n F~)~ F
On the other hand,
F: C F, F; C F;
FfL>F', F;'>F'. But F1, F2,F are closed subspaces of a Hilbert space H. So, by Projection theorem,
F:~ = F1, F;~= F2,FL1 = F.
Hence,
F1 IF', F2 IF', FlnF2 >FL,
(Fl n ~2)'~C FL1 = F. Hence the result. ii) Hint: Let {x,) be a sequence in F1 + F2. Show that {x,) is convergent in F1 + F2. Hint: Proof of theorem 24.1 Page 420 of Textbook L Hint: Proof of theorem 24.1 and 24.2 of Textbook L