FUNCTIONAL ANALYSIS - University of Pittsburgh

1 FUNCTIONAL ANALYSISPIOTR HAJ and Hilbert spacesIn what followsKwill is a pair (X, ), whereXis a linear spaceoverKand :X [0, )is a function, called anorm, such that(1) x+y x + y for allx,y X;(2) x =| | x for allx Xand K;(3) x = 0 if and only ifx= x y x z + z y for allx,y,z X,d(x,y) = x y defines a metric in a normed space . In what follows normed paces will alwaysbe regarded as metric spaces with respect to the metricd. A normed spaceis called aBanach spaceif it is complete with respect to the a linear space overK(=RorC). Theinner product(scalar product) is a function , :X X Ksuch that(1) x,x 0;(2) x,x = 0 if and only ifx= 0;(3) x,y = x,y ;(4) x1+x2,y = x1,y + x2,y ;(5) x,y = y,x ,for allx,x1,x2,y Xand all an obvious corollary we obtain x,y1+y2 = x,y1 + x,y2 , x, y = x,y ,Date: February 12, HAJ LASZfor allx,y1,y2 Xand a space with an inner product we define x = x,x.]

2 Lemma (Schwarz inequality).IfXis a space with an inner product , , then| x,y | x y for allx,y X . can assume that x,y 6= 0. Forx,y Xandt Rwe have0 x+ty,x+ty = x,x +t[ x,y + y,x ] +t2 y,y = x,x + 2tre x,y +t2 y,y .We obtained a quadratic function of a variabletwhich is nonnegative andhence0 = 4(re x,y )2 4 x,x y,y ,(re x,y )2 x,x y,y .If| |= 1, then replacingyby ywe obtain x,x y,y = x,x y, y (re x, y )2= (re ( x,y )) particular for = x,y /| x,y |we have x,x y,y (re( x,y | x,y | x,y ))2=| x,y | completes the a space with an inner product , , then x = x,y is a properties x =| | x and x = 0 if and only ifx= 0 areobvious. To prove the last property we need to apply the Schwarz inequality. x+y 2= x+y,x+y = x,x + 2 re x,y + y,y x 2+ 2 x y + y 2= ( x + y ) space with an inner product , is called aHilbert spaceifit is a Banach space with respect to the norm x = x,x.

3 Proposition (The Polarization Identity).Let , be an inner ANALYSIS3(1)IfK=R, then x,y =14( x+y 2 x y 2),for allx,y X.(2)IfK=C, then x,y =14( x+y 2 x y 2) 14i( ix+y 2 ix y 2)for allx,y is left as an easy (X, )be a normed space overK(=RorC). Thenthere is an inner product , such that x = x,x if and only if the norm satisfies theParallelogram Law, x+y 2+ x y 2= 2 x 2+ 2 y 2for allx,y Parallelogram Law has a nice geometric implication follows from a direct computation. To prove theother implication we define the inner product using the Polarization Iden-tities and we check that it has all the required properties. We leave the detailsas an respect to each of the following norms is aBanach space x 1=n i=1|xi|, x = maxi=1,2,..,n|xi|, x p=(n i=1|xi|p)1/p,1 p < ,wherex= (x1.)

4 ,xn) Cn. The Banach space (Cn, p) is denoted by` HAJ the inner product x,y =n i=1xiyiis a Hilbert space . Note that the corresponding norm is x,x =(n i=1|xi|2)1/2= x 2so`2nis a Hilbert space . The metric associated with the norm isd(x,y) = n i=1|xi yi|2, it is the Euclidean spaces`pnwere defined overK=C, but we can also do the sameconstruction forK=Rby replacingCnbyRn. The resulting space is alsodenoted by`pn, but in each situation it will be clear whether we talk aboutthe real or complex space `pnso there is no danger of a `p2overK=R. Then the shape of the unit ball{x: x 1} subset of the Euclidean spaceRnis called anellipsoidif it is the imageof the unit ball inRnunder a nondegenerate linear mappingL:Rn Rn( detL6= 0).For every ellipsoidEinRnthere is an inner product inRnsuch thatEis the unit ball in the associated norm.

5 Indeed, ifL:Rn Rnis anisomorphism such thatE=L(Bn(0,1)), then it suffices to define the innerproduct as x,y = (L 1x) (L 1y)where stands for the standard inner product precisely, ifL:Rn Rnis non-degenerate, then according to thepolar decomposition theorem (P. Lax,Linear Algebra, p. 139)L=RUFUNCTIONAL ANALYSIS5whereUis unitary andRis positive self-adjoint. The mappingRcan becomputed explicitlyLLT=RUUTRT=R2, R= to the spectral theorem there is an orthonormal basisv1,..vninRn(with respect to the standard inner product) such thatR= 10 n in this basis. That means the mappingLhas the following structure. First werotate (the mappingU) and then we applyRwhich has a simple geometricmeaning of extending the length of vectorsv1,..,vnby factors 1,.., ifBn(0,1) is the unit ball, thenU(Bn(0,1)) is also the unit ball, soL(Bn(0,1)) =R(Bn(0,1)) is an ellipsoid with semi-axes 1v1.

6 , nvnofthe lengths being eigenvalues ofR= LLT. These numbers are calledsin-gular the ellipsoidE=L(Bn(0,1)) is the unit ball for the inner product n i=1aivi,n i=1bivi =n i=1aibi will see later (Corollary ) any real inner product space spaceHof dimensionnis isometrically isomorphic to`2n, the standardinner product, so ifL:`2n His this isometric isomorphism, the unit ballinHisL(Bn(0,1)), so it is an ellipsoid. Thus we convex set inRnis a unit ball for a norm associated withan inner product if and only if it is an ` , the space of all bounded (complex, real) sequencesx= (an) n=1withthe norm x = supn|xn|is a Banach space . This is very easy to HAJ , the space of all (complex, real) convergent sequences with the norm is a Banach , the space of all (complex, real) sequences that converge to zero withthe norm is a Banach thatc0 c ` and bothc0andcare closed linear subspaces of` with respect to the metric generated by the that` ,candc0are Banach that the spacescandc0are separable, while` is `p, 1 p < is the space of all (complex, real) sequencesx= (xn) n=1such that x p=( n=1|xn|p)1 follows from the Minkowski inequality for sequences that pis a normand that`pis a linear space .

7 We will prove now that`pis a Banach space , that it is complete. Letxn= (ani) i=1be a Cauchy sequence in`p, forevery >0 there isNsuch that for alln,m > N xn xm p=( i=1|ani ami|p)1/p< .Hence for eachithe sequence (ani) n=1is a Cauchy sequence inK(=CorR). Letai= limn ani. Fix an integerk. Then forn,m > Nwe havek i=1|ani ami|p< pand passing to the limit asm yieldsk i=1|ani ai|p taking the limit ask we obtain i=1|ani ai|p p, xn xm p wherex= (ai) i= proves thatx `pandxn xin`p. The proof is particular the space `2is a Hilbert space because its norm is associatedwith the inner product x,y = i= that`p,1 p < is will prove that`pforp6= 2 is not an inner product space . Letx= (1,0,0,..),y= (0,1,0,0,..). If 1 p < , then x+y 2p+ x y 2p= 21+2/p,2 x 2p+ 2 y 2p= 4,and thence the Parallelogram Law is violated.

8 The same example can alsobe used in the casep= . In the real case this result can also be seen as aconsequence of the fact that the two dimensional section of the unit ball in`p,p6= 2, along the space generated by the first two coordinates is not equipped with a positive measure , then for 1 p < ,Lp( )is a Banach space with respect to the norm f p=( X|f|pd )1 ( ) is a Banach space with the norm being the essential supremumof|f|. For the proofs see the notes from ANALYSIS 2 the spaceL2( ) is a Hilbert space with respect to the innerproduct f,g = Xfg d . {1,2,3,..}and is the counting measure ( (A) = #A),thenLp( ) =`p. In particular this gives another proof that`pis a Banachspace. However the proof given above is much more a compact metric space , then the space of continuous functionsonXwith respect to the norm f = supx X|f(x)|is a Banach space .

9 This space is denoted byC(X). The resulting metric inC(X) is the metric of uniform the class of all holomorphic functions on the unit discD={z C:|z|<1}such that f H2=(sup0<r<112 2 0|f(rei )|2d )1/2< .This space is calledHardy spaceH2. We will prove now that it is a Hilbertspace and we will find an explicit formula for the inner HAJ LASZS ince every holomorphic function inDcan be represented as a Taylorpolynomialf(z) = n=0anzn, for|z|<1 we have12 2 0|f(rei )|2d =12 2 0( n=0anrnein )( m=0amrmeim )d = n,m=0anamrn+m12 2 0ei(n m) d 1 if n=m, 0 ifn6=m= n=0r2n|an| f H2=(limr 1 n=0r2n|an|2)1/2=( n=0|an|2)1 proves that the spaceH2is isometrically isomorphic with`2and thatthe norm inH2is associated with the inner product f,g = n=0anbn= limr 1 12 2 0f(rei )g(rei )d ,wheref= n=0anzn,g(z) = n= operatorsIfXandYare normed spaces, then linear functionsL:X Ywill becalled (linear)operatorsof (linear)transforms.

10 Also we will often writeLxinstead ofL(x). We say that a linear operatorL:X Yisboundedif thereis a constantC >0 such that Lx C x for allx :X Ybe a linear operator between normed the following conditions are equivalent:(1)Iis continuous;(2)Lis continuous at0;(3)Lis implication (1) (2) is obvious. (2) (3) For if notthere would exist a sequencexn Xsuch that Lxn > n xn . Then L(xn/(n xn )) >1, but on the other hand L(xn/(n xn )) 0, be-causexn/(n xn ) 0 which is an obvious contradiction. (3) (1) Letxn x. ThenLxn Lx. Indeed, Lx Lxn = L(x xn) C x xn ANALYSIS9 This completes the a linear operatorL:X Ywe define its norm by L = sup x 1 Lx .Then Lx L x for allx , Lx = L( x x x ) = x Lx x x L .ThusLis bounded if and only if L < . Moreover L is the smallestconstantCfor which the inequality Lx C x for allx Xis satisfied.