Chapter 3 Inner Product Spaces. Hilbert Spaces

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1 Chapter 3 Ier Product Spaces. Hilbert Spaces 3. Ier Product Spaces. Hilbert Spaces 3.- Defiitio. A ier product space is a vector space X with a ier product defied o X. A Hilbert space is a complete ier product space. A ier product o X is a mappig of X X ito the scalar field of X such that for all x y ad z i X ad ay scalar we have (Ip) < x + y z > = < x z > + < y z >. (Ip) < x y > = < x y >. (Ip 3) < x y > = y x the complex cojugate of < y x >. (Ip4) < x x > 0 ad < x x > = 0 if ad oly if x = 0. otes. () A ier product o X defies a orm o X give by x = x x > ad a metric o X give by d( x y ) = x y = x y x y >. () The ier product spaces are ormed spaces Hilbert spaces are Baach spaces. (3) For all x y ad z i a ier product space ad ay scalar a) < x + y z > = < x z > + < y z >. b) < x y > = < x y >. c) < x y + z > = < x y > + < x z >. The the ier product is sesquiliear ( liear i the first factor ad cojugate liear i the secod ). 4) Parallelogram equality. If X is a ier product space the for all x yx x + y + x y = ( x + y ). ( How?) ( From 4) we have: if parallelogram equality is ot satisfied i a ormed space X the X is ot a ier product space.) 3.-3 Defiitio. A elemet x i a ier product space X is said to be orthogoal to yx if < x y > = 0 a we write x y. If A B are subsets of X the x A if x a for all aa ad A B if a b for all aa ad all bb. Examples Euclidea space R is a Hilbert spaces with ier product defied by < x y > = i i. This ier product iduces the orm x = ( i ) i ad the metric d( x y ) = ad y = (... ). i i i where x = i ( ( ) ) (... )

2 3.-5 Uitary space C is a Hilbert spaces with ier product defied by < x y > = i. This ier product iduces the orm x = ( ) i i i ad the metric d( x y ) = ad y = (... ). i i i where x = i ( ) (... ) 3.-6 Space L [a b]. The completio of the space of all cotiuous real-valued fuctios o [a b] with orm defied by x = b ier product < x y > = x ( t ) y ( t ) dt. a b x t dt ad the ( ( ) ) ote. The fuctio x(t) ca be exteded to be complex valued o [a b] ad the correspodig ier product is < x y > = x = b b a x ( t ) y ( t ) dt ad the orm is x t dt. The L [a b] becomes as the completio of the ( ( ) ) a space of all all cotiuous complex-valued fuctios o [a b] correspodig to this orm. I each case real or complex L [a b] is a Hilbert space Hilbert sequece space product defied by < x y > = a. This space is a Hilbert spaces with ier i i. This ier product iduces the orm x = ( i ) where x = ( i ) ad y = ( i ). i p 3.-8 Space. This space with p is ot a ier product space hece is ot a Hilbert spaces. Proof. Cosider x = ( ) ad y = ( ). The x y p ad x + y + x y = ( ) + ( ) = 8. But ( x + y ) = ( ( ) + ( ) p p ) = [ ( ) + ( ) ] = 4( p ) 8 whe p. Therefore x + y + x y ( x + y ). p Hece by ote 4) above the space with p is ot a ier product space hece is ot a Hilbert spaces i

3 3.-9 Space C[a b]. This space is ot a ier product space hece is ot a Hilbert spaces. t a Proof. Cosider x(t) = ad y(t) = o the closed iterval J b a ta = [a b]. The x y C[a b] ad x = y = max = ta ta bt x + y = max = ad x y = max = max =. tj ba Hece ( x + y ) = 4 but x + y + x y = 5. The x + y + x y ( x + y ). Hece by ote 4) above the space C[a b] is ot a ier product space hece is ot a Hilbert spaces 3.-0 Remars. () For ay x y i a real ier product space tj < x y > = ( x + y 4 - x y ). () For ay x y i a complex ier product space a) Re < x y > = ( x + y 4 - x y ). a) Im < x y > = ( x + iy 4 - x iy ). ba tj tj ba ba H. W. -9. H.W.*

4 3. Further Properties of Ier Product Spaces. 3.- Lemma ( Schwarz iequality triagle iequality). If x ad y are elemets i a ier product space the a) < x y > x y ( Schwarz iequality) where the equality sig holds if ad oly if { x y } is a liearly depedet set. b) x + y x + y ( Triagle iequality) where the equality sig holds if ad oly if y =0 or x = cy for some c 0. Proof. a) If y = 0 the < x y > = 0 ad the result is trivial. Let y 0 ad be a scalar we have 0 x - y = < x - y x - y > = < x x > - < x y > - [ < y x > - < y y > ]...() Put yx = yy we have 0 < x x > - yx yy < x y > = x - xy. Hece < x y > x y ad the the result follows. y If { x y } is liearly depedet the there is a scalar such that x = y ad < x y > = < y y > = y = y y = x y. Coversly if equality holds the by () either y = 0 or x - y = 0. Hece { x y } is liearly depedet. b) By usig Schwarz iequality we have x + y = < x + y x + y > = x + < x y > + < y x > + y x + < x y > + < y x > + y x + x y + y = ( x + y ). Hece x + y x + y. If equality holds the < x y > + < y x > = x y. However < x y > + < y x > = Re< x y >. The Re< x y > = x y < x y >. Sice the real part ca t exceed the modulas the Re< x y > = x y = < x y > 0 the by a) { x y } is liearly depedet ad so y = 0 or x = cy. Sice 0 < x y > = < cy y > = c y the c 0. Coversly if y = 0 or x = cy the it is a easy calculatio for gettig x + y = x + y 3.- Lemma ( Cotiuity of ier product). If i a ier product space x x ad y y the < x y > < x y >. H. W H.W.* 8. 4

5 3.3 Orthogoal Complemets ad Direct Sums Remar. A subset M of a vector space X is said to be covex if for all x ym ad all [0 ] x + (-)ym. Hece every subspace of X is covex ad the itersectio of covex sets is covex Theorem. Let X be a ier product space ad M a o empty covex subset which is complete. The for every xx there exists a uique ym such that = if x y = x y. ym Proof. a) Existece. Sice = if x y the there is a sequece (y ) ym i M such that where = x y. We show that (y ) is Cauchy. Let y x = v. The = v ad v + v m = y + y m x = (y + y m ) x where (y + y m ) M because M is covex. Furthermore y y m = v v m. By usig parallelogram equality we have y y m = v v m = v + v m + ( v + v m ) () + ( + m ). As m y y m 0. Hece for every > 0 there is sufficietly large such that y y m < for all m >. Hece (y ) is a Cauchy sequece i M. But M is complete the (y ) coverges say y ym. The x y. However x y x y + y y = + y y. The x y. Therefore x y =. b) Uiqueess. Suppose that there are y y 0 M with x y = ad x y 0 =. By parallelogram equality y y 0 = (y x) (y 0 x) = y x + y 0 x (y x) + (y 0 x) = + 4 (y + y 0 ) x...() Sice M is covex the (y + y 0 )M which implies that (y + y 0 ) x. By this ad () y y = 0. Hece y y 0 = 0 that is y = y 0. This proves the uiqueess 3.3- Lemma. I Theorem 3.3- let M be a complete subspace Y ad xx fixed. The z = x y is orthogoal to Y. Proof. Suppose that z = x y is ot orthogoal to Y. The there is y Y such that < z y > = 0. Clearly y 0 otherwise < z y > = 0. Furthermore for ay scalar z y = < z z > < z y > [ < y z > < y y > ] = < z z > [ < y y > ] Choose = y y to get z y = z -. However y z = x y = ad 0 the z y < z =..() Sice Y is a subspace the ( y y ) Y ad so z y = x ( y y ). This is a cotradictio with (). Therefore z = x y is orthogoal to Y 5

6 3.3-3 Defiitio. A vector space X is said to be the direct sum of two subspaces Y ad Z of X writte X = Y Z if each xx has a uique represetatio x = y + z yy ad zz. The Z is called the algebraic complemet of Y i X Theorem ( Projectio Theorem ). Let Y be ay closed subspace of a Hilbert space H. The H = Y Z where Z = Y = { zh : z Y}. Proof. Sice Y is a closed subspace of a Hilbert space H the Y is complete. Sice Y is covex the Theorem 3.3- ad lemma 3.3- imply that for ay xx there is yy such that z = x yy. The x = y + z yy ad zy = Z..() Suppose that there are y y Y ad z z Z with x = y + z = y + z. The y y = z z. However y y Y ad z z Z. The y y YY = {0}. Hece y = y ad z = z. This proves the uiqueess Remar. From () a bove we ca defie a mappig ( called the orthogoal projectio of H oto Y ) P: H Y defied by Px = y where x = y + z yy ad zz. This mappig has the followig properties: () P is liear ad bouded. () P maps H oto Y. (3) P(Y) = Y. (4) P(Y ) = {0}. (5) P = P ad the restrictio of P o Y is the idetity operator o Y Lemma. The orthogoal complemet Y of a closed subspace Y of a Hilbert space H is the ull space (P) of the orthogoal projectio P of H oto Y. Remar 3. If M is a oempty subset of a ier product space X the M is a closed vector subspace of X ad M is a subset of M Lemma. If Y is a closed subspace of a Hilbert space H the Y = Y. Proof. By Remar 3 Y Y. Coversely let xy. By Theorem there is a uique yy such that x = y + z. However Y Y ad Y is a vector space the z = x y Y that is z Y. Sice Y is a closed subspace of a Hilbert space the it is complete ad so by Lemma 3.3- z Y. Hece z z ad so z = 0. So that x = y ad xy. Therefore Y Y. Hece Y = Y Remar 4. If Y is a closed subspace of a Hilbert space H ad Z =Y the Z =Y = Y H = Z + Z ad P z x = z defies a projectio P z :HZ. 6

7 3.3-7 Lemma. For ay o empty subset M of a Hilbert space H the spa of M is dese if ad oly if M = {0}. Proof. Let xm ad assume that V = spa M to be dese i H. The x V = H the there is a sequece (x ) of elemets i V such that x x. Sice xm the for all mm < x m > = 0 which implies that < x v > = 0 for all vv i particular < x x > = 0 for all. By the cotiuity of the ier product < x x >< x x >. Hece < x x > = 0 ad so x =0. Therefore M = {0}. Coversely suppose that M = {0}. Sice M V the V M = {0}. Hece V = {0}. By Theorem H = V V = V {0} = V. Hece V is dese i H H. W H.W.* 8 Remars 3 ad Lemma

8 3.4 Orthoormal Sets ad Sequeces 3.4- Defiitio. A orthogoal set M i a ier product space X is a subset M of X whose elemets are pairwise orthogoal. A orthoormal subset M of X is a orthogoal set i X whose elemets have orm. That is 0 if x y for all x ym < x y > =. if x = y otes. a) If a orthogoal or a orthoormal set M is coutable we ca arrage it i a sequece (x ) ad call it a orthogoal or a orthoormal sequece respectively. b) A family (x ) I is called orthogoal if x x for all I ad. The family is orthoormal if it is orthogoal ad all x have orm 0 if so that for all I < x x > =. if = Remar. For a orthogoal elemets x ad y we have x + y = x + y ( Pythagorea Relatio ). I geeral if {x x. x } is a orthogoal set the x + x +. + x = x + x +. + x Lemma. A orthoormal set is liearly idepedet. Examples: Euclidea space R. The stadard basis of R forms a orthoormal set. ( How?) Hilbert sequece space orthoormal sequece i. The Schauder basis (e ) of. ( How?) forms a Cotiuous fuctios. Let X be the ier product space of all cotiuous real-valued fuctios o [0 ] with < x y > = x ( t ) y ( t ) dt. The (cost ) = 0.. ad ( sit ) are orthogoal sequeces. Moreover ( cost cos t si t. ) ad ( ) are orthoormal sequeces. 0 8

9 3.4-6 Theorem ( Bessel's Iequality). Let (e ) be a orthoormal sequece i a ier product space X. The for every xx xe x. Proof. Cosider the fiite subset { e e. e } from ( e ). The 0 x x ei ei = < x i x < x x ei ei x i x e e > = x e e > < x ei ei x > + < i x e e > = x x e x e i i x ei ei i x ei ei x + x ei x e ei e = x x e x e + x e x e = x By lettig we get xe. Hece xe x xe x. otes. a) The ier product < x e > above is called the Fourier coefficiets of x with respect to the orthoormal sequece (e ). b) If dim(x) is fiite the ay orthoormal set i X must be fiite because it is liearly idepedet. Gram-Schmidit process for orthoormalizig a liearly idepedet sequece (xj) i a ier product space X. x x st step. The first elemet of (e ) is e = d step. Let v = x < x e > e. Sice (x j ) is liearly idepedet the v 0. Also v e where < v e > = < x e > < x e >< e e > = 0. So we ca tae e = v. v 3 rd step. Let v 3 = x 3 < x 3 e > e < x 3 e > e. As above v 3 0 v 3 e ad v 3 e (how?). So we ca tae e 3 = v 3. th step. Let v = x x e e 3. v. As above v 0 ad v e i for all i = -. So we ca tae e =. Therefore we have (e ) as a v orthoormal sequece. v H. W H.W.*

10 3.5 Series Related to Orthoormal Sets ad Sequeces 3.5- Theorem. Let (e ) be a orthoormal sequece i a Hilbert space H. The for ay scalars 3. a) b) If e coverges ( i the orm of H ) if ad oly if coverges. e coverges the 's are the Fourier coefficiets < x e > where x deotes the sum x = c) For ay xh e ad so x e e coverges. e = x = x e e. Proof. a) Let S = e + e +.+ e ad σ = Sice (e ) is orthoormal the for > m S S m = m+ e m+ + m+ e m+ +.+ e = < m+ e m+ + m+ e m+ +.+ e m+ e m+ + m+ e m+ +.+ e > = m+ e m+ + m+ e m+ +.+ e = m+ + m+ +.+ = σ σ m. Therefore (S ) is Cauchy i H if ad oly if (σ ) is Cauchy i R. However both H ad R are complete the (S ) coverges if ad oly if (σ ) coverges. Hece e coverges (i the orm of H) if ad oly if b) Let S x; that is x = e coverges i R.. Let be fixed the for a fixed j = ad we have < S e j > = < e + e +.+ e e j > = < j e j e j > = j. By the cotiuity of the ier product we have j = <S e j ><x e j > as. The we ca tae ( ) as large as we please. Hece j = lim = < x e j > for all j = 3. Therefore x = b) Let x be ay elemet i H. By Bessel s iequality Hece xe coverges ad so by a) e = x e e. xe x. x e e coverges 3.5- Lemma. Ay x i a ier product space X ca have at most accoutably may o zero Fourier coefficiets < x e > with respect to a orthoormal family (e ) I i X. Remar. a) The Bessel iequality holds i the case of (e ) I as a orthoormal family that it is the parseval relatio. H.W.* 4. xe x. If the equality holds we say j 0

11 3.6 Total Orthoormal Sets ad Sequeces 3.6- Defiitio. A total set (fudametal set) i a ormed space X is a subset M of X whose spa is dese i X. A total orthoormal set i X is a orthoormal set which is total. Remar. a) I every Hilbert space H {0} there exists a total orthoormal set. b) All total orthoormal sets i a give Hilbert space H {0} have the same cardiality where the cardiality of the total orthoormal set i a Hilbert space H {0} is called the Hilbert dimesio or orthogoal dimesio of H. If H = {0} the Hilbert dimesio is defied to be zero Theorem. Let M be a subset of a ier product space X. The a) If M is total i X the xx ad x M implies x = 0. b) If X is complete ad if the coditio ( xx ad x M implies x = 0 ) is satisfied the M is total i X. Proof. a) Let H be the completio of X. The X ca be regarded as a subspace of H which is dese i H. However M is total i X the spam is dese i X the it is dese i H. Hece by Lemma3.3-7 M = {0}. Therefore xx ad x M implies x = 0. b) Use Lemma3.3-7 ad the defiitio of total set to get the result Theorem. A orthoormal set M i a Hilbert space H is total i H if ad oly if xe = x for all xh. Proof. If M is ot total the by Theorem3.6- b) there is xh x 0 ad x M. The < x e > = 0 for all e M. Hece xe = 0 x. Therefore if xe = x for all xh the M is total i H. Coversely suppose that M is total i H. Let x be ay elemet i H. ad arrage all its ozero Fourier coefficiets i a sequece < x e > < x e >. or writte i some defiite order if there are oly fiitely may of them. Defie y by y = x e e () Sice M is orthoormal the for every e j occurig i () we have < x y e j > = < x e j > < y e j > = < x e j > x e e ej = < x e j > < x e j > = 0. But for all vm ot cotaied i () we have < x v > = 0. So that < x y v > = < x v > x e e v = 0. Hece x y M. However M is total i H the by Lemma3.3-7 M = {0} ad so x y = 0 that is x = y. Therefore x = < y y > = < x e e = x e x em e em = m xe x em em > m

12 3.6-4 Theorem. Let H be a Hilbert space. The a) If H is separable the every orthoormal set i H is coutable. b) If H cotais a orthoormal sequece which is total i H the H is separable. Proof. a) Let H be separable B ay dese set i H ad M ay orthoormal set. Sice M is orthoormal the for ay x ym with x y we have x y = < x y x y > = < x x > + < y y > =. Hece spherical eighborhoods x of x ad y of y of radius 3 are disjoi ( why?). Sice B is dese i H the for ay xm x B. Hece there is a x B ad b y B. Therefore a b. If M is ucoutable the we have uaccoutably may pair wise disjoit spherical eiborhoods so that B would be ucoutable. Sice B was ay dese set this maes that H would ot cotai a coutable dese set which cotradicts the separability of H. Therefore M must be coutable. a) Let (e m ) be a total orthoormal sequece i H ad A = { ( ) e : = ( ) ( ) a + i ( ) b ( ) a ( ) b Q }. ( b = 0 whe H is real ). A is coutable (how?). We show that A is dese i H. Let x be ay fixed elemet i H. Sice (e m ) is total i H the spa( e m) = H. The for every > 0 there is wspa(e m ) such that x w <. Hece wy = spa{e e e } for some. By Lemma3.3- there is yy such that x y Y ad x y x w <. By (8a&b) i 3.4[ see the text boo] y ca be writte as y = x e e.the x dese i R the for ay < x e > there is such that ( ) [ x e ] e ( ) ( ) x e e <. Sice Q is = ( ) a + ib ( ) <. Hece there is v = that satisfies x v = x ( ) x e e e < dese i H + ( ) e x a ( ) b ( ) e ( ) Q A x e e + =. Hece v B(x; ) A. Thus A is

13 3.6-5 Theorem. Two Hilbert spaces H ad H both real or both complex are isomorphic if ad oly if they have the same Hilbert dimesio. Proof. Suppose that H is isomorphic with H the there is a bijective liear mappig T :H H that satisfies < Tx Ty > = < x y > for all x yh. Hece orthoormal elemets i H have orthoormal images uder T. However T is bijective the T maps every total orthoormal set i H oto a total orthoormal set i Hilbert dimesio. Coversly suppose that H ad H ( how?) Therefore H ad H have the same H have the same Hilbert dimesio. The case that H = {0} ad H = {0} is trivial. Let H {0} the H {0} ad ay total; orthoormal sets M i H ad M i H have the same cardiality. So we ca idex them by the same idex set {} ad write M = (e ) ad M = ( e ). ow defie T:H H by Tx= e x e. This is well defied because for all xh we have x = x e e ad by Bessel s iequality xe coverges. The by Theorem 3.5- e x e coverges so that Tx H. Let x = x e e ad y = y e e be ay elemets i H ad ay scalar T(x+ y) = x y e e = x e e + y e e = Tx +Ty. Hece T is liear. Sice ( e ) is orthoormal the < Tx Tx > = < x e e x e e > = m xe = x. For ay x y H (if H is real) < Tx Ty > = 4 x e x em e em = ( Tx + Ty - Tx Ty ) = ( T(x + y) 4 - T(x y) ) = ( x + y 4 - x y ) = < x y >. Similarly for the complex case. Hece T preserves the ier product which implies that T is -. Give ay x = By Bessel's iequality e H. coverges (how?) ad so e is a fiite sum or a series which coverges to xh by Theorem3.5- ad = <x e > by the same theorem. Hece x = e x e = Tx. Thus T is oto. Therefore T is a isomorphism so H ad H are isomorphic H. W H.W.*

14 3.8 Represetatio of Fuctioals o Hilbert Spaces 3.8- Riesz's Theorem. Every bouded liear fuctioal f o a Hilbert space H ca be represeted i terms of ier product amely f(x) = < x z > where z depeds o f is uiquely determied by f with orm z = f. Proof. The case f = 0 is a trivial case (why?). Let f 0. The (f) H. However (f) is a closed subspace of H the by Theorem H = (f) (f). Hece (f) {0}. Let z 0 0 ad z 0 (f). Set v = z 0 f(x) xf(z 0 ) where x is arbitrary elemet i H. Sice f is liear the f(v) = f(z 0 )f(x) f(x)f(z 0 ) = 0 ad so v(f). Hece z 0 v ad 0 = < v z 0 > = < z 0 f(x) xf(z 0 ) z 0 > = f(x)< z 0 z 0 > f(z 0 ) < x z 0 >. The f(x) = f ( z0 ) z z < x z 0 > = < x 0 0 the by taig z = f ( z0 ) z z 0 0 f ( z0 ) z z 0 0 z 0 we get f(x) = < x z >. z 0 >. Sice x was arbitrary To prove the uiqueess of z. Suppose that there are z ad z such that f(x) = < x z > = < x z > for all xh. The < x z - z > = 0 for all xh ad for x = z - z we have < z - z z - z > = 0. Hece z = z ad the uiqueess is proved 3.8- Lemma. If < v w > = < v w > for all w i a ier product space X the v = v. I particular < v w > = 0 for all wx implies that v = Defiitio. Let X ad Y be vector spaces over a field. A sesquiliear form (sesquiliear fuctioal) h o X Y is a mappig h : X Y such that for all x x x X y y y Y ad all scalars it satisfies a) h(x + x y) = h(x y) + h(x y) b) h(x y + y ) = h(x y ) + h(x y ) c) h(x y) = h(x y) d) h(x y) = h(x y). otes. ) If = R the h above is biliear. ) If X ad Y are ormed spaces ad there is c > 0 such that for all xx ad all yy h(x y) c x y the h is bouded ad the umber h = sup{ h ( x y ) x y : xx\{0} & yy\{0} } = sup{h(x y): x = y = } is called the orm of h the h(x y) h x y for all x y. 3) The ier product is sesquiliear ad bouded (how?) 4

15 3.8-4 Theorem (Riesz's Represetatio). Let H ad H be Hilbert spaces over a field ad h : H H a bouded sesquiliear form. The h has a represetatio h(x y) = < Sx y > where S : H H is a bouded liear operator. Moreover S is uiquely determied by h ad has the orm S = h. Proof. Cosider h ( x y ) which is liear i y. ow for ay fixed xh we apply Theorem 3.8- to get a uique zh that depeds o xh such that h( x y ) = < y z > for all yh ad so h(x y) = < z y >. Hece S : H H give by Sx = z is a operator ad h(x y) = < Sx y >. For all x x H y H ad all scalars < S(x + x ) y > = h(x + x y) = h(x y) + h(x y) = <Sx y > +< Sx y > = <Sx + Sx y >. The by Lemma3.8- S(x + x ) = Sx +Sx so S is liear. If S=0 the it is bouded. If S 0 the Sx y Sx Sx h = sup{ : xh \{0} & yh \{0}} sup{ : xh \{0}} = sup{ Sx x y x Sx x : xh \{0}} = S. Hece S is bouded ad h S. Sx y Sx y But h = sup{ x y : xh \{0} & yh \{0}} sup{ x y : xh \{0} & yh \{0}} = S. Therefore h = S. To prove the uiqueess Suppose that there is a liear operator T : H H such that h(x y) = <Sx y > = <Tx y > for all xh ad yh. The by Lemma3.8- Sx = Tx for all xh. This proves the uiqueess of S H. W H.W.* 0. 5

16 3.9 Hilbert adjoit operator 3.9- Defiitio. Let H ad H be two Hilbert spaces ad T:H H a bouded liear operator. The Hilbert adjoit operator T* of T is the operator T*:H H such that for all xh ad all yh < Tx y > = < x T*y > Theorem. The Hilbert adjoit operator T* of T i Defiitio 3.9- exists is uique is liear ad is bouded with orm T* = T. Proof. Cosider h(y x) = < y Tx > which defies a sesquiliear form o H H ( The details is left to the reader ). By Schwarz iequality h(y x) = < y Tx > y Tx y x T this implies that h is bouded ad h T. But h = sup{ h ( y x ) y x : yh \{0} & xh \{0} } = sup{ y Tx : yh \{0} & xh \{0}} sup{ Tx Tx y x Tx x : xh \{0}} = T. Therefore h = T. Sice h is a bouded sesquiliear form the by Theorem3.8-4 there exists a bouded liear operator call it T* that is uiquely determied by h ad h(y x) = < T*y x > with T* = h. Hece T* = T. However h(y x) = <y Tx> the < y Tx > = < T*y x > ad so < Tx y > = < x T*y > Lemma. Let X ad Y be ier product spaces ad Q:XY a bouded liear operator. The a) Q = 0 if ad oly if < Qx y > = 0 for all x X ad all yy. b) If Q:XX X is complex ad < Qx x > = 0 for all xx the Q =0. Proof. a) Left to the reader. b)from assumptio < Qv v > = 0 for all v = x + yx; that is 0 = < Q(x + y) x + y > = < Qx x > + < Qy y > + < Qx y > + < Qy x > = < Qx y > + < Qy x >. Put = ad the = i to get < Qx y > + < Qy x > = 0 ad < Qx y > - < Qy x > = 0 respectively. By additio < Qx y > = 0 ad so Q = 0 follows from a) Remar. If X is real the b) above eed ot holds. To see this cosider the mappig Q : R R give by Q( ) = ( -). It is clear that for ay x = ( )R < Qx x > = ( -).( ) = 0 but Q Theorem. Let H ad H be Hilbert spaces S T:H H bouded liear operators ad ay scalar. The we have a) < T*y x > = < y Tx > for all xh ad yh. b) (S +T)* = S* + T* ad (T )* = T*. c) ( T* )* = T ad i the case H = H (ST)* = T*S*. d) T*T = TT* = T. e) T*T = 0 if ad oly if T = 0. Proof. We prove d) ad left the proof of the other parts to the reader. First ote that T*T:H H but TT*:H H. The Tx = <Tx Tx> = < T*Tx x > T*Tx x T*T x. The T = sup{ Tx : x = xh } T*T T* T = T T = T. Hece T*T = T. Replacig T by T* to get T**T* = T*. However T**T* = TT* ad T = T* Therefore T*T = TT* = T H. W.-8. H.W.*

17 3.0 Self- adjoit Uitary ad ormal Ooperators 3.0- Defiitio. A bouded liear operator T : H H o a Hilbert space H is said to be ) self-adjoit or Hermitia if T* = T. ) Uitary if T is bijective ad T* = T -. 3) ormal if TT* = T*T Remar. a) If T is self-adjoit the < Tx y > = < x Ty > for all x ad y. b) If T is self-adjoit or uitary the T is ormal. c) ormal operators eed ot be self-adjoi or uitary. To see this cosider the idetity operator I : H H o a Hilbert space H. It is easy to see that T = ii is ormal but it is either self-adjoit or uitary Theorem. Let T : H H be a bouded liear operator o a Hilbert space H. The a) If T is self-adjoit the < Tx x > is real for all xh. b) If H is complex ad <Tx x > is real for all xh the T is self-adjoit. Proof. a) Suppose that T is self-adjoit. The for all xh Tx x = < x Tx > = <Tx x >. Hece <Tx x > is real. b) Suppose that H is complex ad <Tx x > is real for all xh. The < Tx x > = Tx x = x T * x = < T*x x > ad 0 = < Tx x > - < T*x x > = < (T-T*)x x >. The by Lemma3.9-3(b) T-T* = 0. Hece T is self-adjoit Theorem. The product of two bouded self-adjoit liear operators S ad T o a Hilbert space H is self adjoit if ad oly if ST = TS Theorem. Let ( T ) be a sequece of bouded self-adjoit liear operators T : H H o a Hilbert space H. Suppose that T T that is T - T 0 where this orm is the orm o the space B(H H). The T is a bouded self-adjoit liear operator o H Theorem. Let the operators U V : H H be uitary o a Hilbert space H. The : a) U is isometric thus Ux = x for all xh. b) U = provided that H {0}. c) U - ad UV are uitary. d) U is ormal. Furthermore c) A bouded liear operator T o a complex Hilbert space H is uitary if ad oly if T is isometric ad oto Remar. Isometric operators eed ot be uitary. Proof. Cosider the bouded liear operator T: give by T(.) = (0.). It is easy to see that T is a isometric but T is ot oto where there is (.) ad 0 but there is o x with Tx = (.). Hece we have a isometric which is ot uitary H. W H.W.*

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