FUNCTIONAL ANALYSIS - University Of Pittsburgh

FUNCTIONAL ANALYSISPIOTR HAJLASZ1. Banach and Hilbert spacesIn what follows K will denote R of C.Definition. A normed space is a pair (X, k · k), where X is a linear spaceover K andk · k : X [0, )is a function, called a norm, such that(1) kx yk kxk kyk for all x, y X;(2) kαxk α kxk for all x X and α K;(3) kxk 0 if and only if x 0.Since kx yk kx zk kz yk for all x, y, z X,d(x, y) kx ykdefines a metric in a normed space. In what follows normed paces will alwaysbe regarded as metric spaces with respect to the metric d. A normed spaceis called a Banach space if it is complete with respect to the metric d.Definition. Let X be a linear space over K ( R or C). The inner product(scalar product) is a functionh·, ·i : X X Ksuch that(1)(2)(3)(4)(5)hx, xi 0;hx, xi 0 if and only if x 0;hαx, yi αhx, yi;hx1 x2 , yi hx1 , yi hx2 , yi;hx, yi hy, xi,for all x, x1 , x2 , y X and all α K.As an obvious corollary we obtainhx, y1 y2 i hx, y1 i hx, y2 i,Date: February 12, 2009.1hx, αyi αhx, yi ,

2PIOTR HAJLASZfor all x, y1 , y2 X and α K.For a space with an inner product we definepkxk hx, xi .Lemma 1.1 (Schwarz inequality). If X is a space with an inner producth·, ·i, then hx, yi kxk kyk for all x, y X .Proof. We can assume that hx, yi 6 0. For x, y X and t R we have0 hx ty, x tyi hx, xi t[hx, yi hy, xi] t2 hy, yi hx, xi 2t re hx, yi t2 hy, yi .We obtained a quadratic function of a variable t which is nonnegative andhence0 4(re hx, yi)2 4hx, xihy, yi ,(re hx, yi)2 hx, xihy, yi .If α 1, then replacing y by αy we obtainhx, xihy, yi hx, xihαy, αyi (re hx, αyi)2 (re (αhx, yi))2 .In particular for α hx, yi/ hx, yi we havehx, xihy, yi re!!2hx, yihx, yi hx, yi 2 . hx, yi This completes the proof.2Corollary 1.2. If X is a space with an inner product h·, ·i, thenpkxk hx, yiis a norm.Proof. The properties kαxk α kxk and kxk 0 if and only if x 0 areobvious. To prove the last property we need to apply the Schwarz inequality.kx yk2 hx y, x yi hx, xi 2 re hx, yi hy, yi kxk2 2kxk kyk kyk2 (kxk kyk)2 .Definition. A space with an inner product h·, ·i is called a Hilbert space ifit is a Banach space with respect to the normpkxk hx, xi .Proposition 1.3 (The Polarization Identity). Let h·, ·i be an inner productin X.

FUNCTIONAL ANALYSIS3(1) If K R, thenhx, yi 1(kx yk2 kx yk2 ) ,4for all x, y X.(2) If K C, then 1 1hx, yi kx yk2 kx yk2 i kix yk2 kix yk244for all x, y X.Proof is left as an easy exercise.Theorem 1.4. Let (X, k · k) be a normed space over K ( R or C). Thenthere is an inner product h·, ·i such thatpkxk hx, xiif and only if the norm satisfies the Parallelogram Law, i.e.kx yk2 kx yk2 2kxk2 2kyk2for all x, y X .The Parallelogram Law has a nice geometric interpretation.Proof. The implication follows from a direct computation. To prove theother implication we define the inner product using the Polarization Identities and we check that it has all the required properties. We leave the detailsas an exercise.21.1. Examples. 1. Cn with respect to each of the following norms is aBanach spacenXkxk1 xi ,i 1kxk kxkp where x (x1 , . . . , xn ) pn .nXmax xi ,i 1,2,.,n!1/p xi pi 1nC . The,1 p ,Banach space (Cn , k · kp ) is denoted by

4PIOTR HAJLASZ2. Cn with the inner producthx, yi nXxi yii 1is a Hilbert space. Note that the corresponding norm is!1/2nXp2hx, xi xi kxk2i 1so 2nis a Hilbert space. The metric associated with the norm isvu nuXd(x, y) t xi yi 2 ,i 1i.e. it is the Euclidean metric.3. The spaces pn were defined over K C, but we can also do the sameconstruction for K R by replacing Cn by Rn . The resulting space is alsodenoted by pn , but in each situation it will be clear whether we talk aboutthe real or complex space pn so there is no danger of a confusion.4. Consider p2 over K R. Then the shape of the unit ball {x : kxk 1} is5. A subset of the Euclidean space Rn is called an ellipsoid if it is the imageof the unit ball in Rn under a nondegenerate linear mapping L : Rn Rn(i.e. det L 6 0).For every ellipsoid E in Rn there is an inner product in Rn such thatE is the unit ball in the associated norm. Indeed, if L : Rn Rn is anisomorphism such that E L(B n (0, 1)), then it suffices to define the innerproduct ashx, yi (L 1 x) · (L 1 y)where · stands for the standard inner product in Rn .More precisely, if L : Rn Rn is non-degenerate, then according to thepolar decomposition theorem (P. Lax, Linear Algebra, p. 139)L RU

FUNCTIONAL ANALYSIS5where U is unitary and R is positive self-adjoint. The mapping R can becomputed explicitly LLT RU U T RT R2 , R LLT .According to the spectral theorem there is an orthonormal basis v1 , . . . vn inRn (with respect to the standard inner product) such that λ10 λ2 R . .0λnin this basis. That means the mapping L has the following structure. First werotate (the mapping U ) and then we apply R which has a simple geometricmeaning of extending the length of vectors v1 , . . . , vn by factors λ1 , . . . , λn .Now if B n (0, 1) is the unit ball, then U (B n (0, 1)) is also the unit ball, soL(B n (0, 1)) R(B n (0, 1)) is an ellipsoid with semi-axes λ1 v1 , . . . , λn vn ofthe lengths being eigenvalues of R LLT . These numbers are called singular values of L.Now the ellipsoid E L(B n (0, 1)) is the unit ball for the inner productnnnXXXai biai vi ,bi vi .λ2ii 1i 1i 1We will see later (Corollary 5.12) any real inner product space space Hof dimension n is isometrically isomorphic to 2n , i.e. Rn with the standardinner product, so if L : 2n H is this isometric isomorphism, the unit ballin H is L(B n (0, 1)), so it is an ellipsoid. Thus we proved.Theorem 1.5. A convex set in Rn is a unit ball for a norm associated withan inner product if and only if it is an ellipsoid.6. , the space of all bounded (complex, real) sequences x (an ) n 1 withthe normkxk sup xn nis a Banach space. This is very easy to check.

6PIOTR HAJLASZ7. c1 , the space of all (complex, real) convergent sequences with the normk · k is a Banach space.8. c0 , the space of all (complex, real) sequences that converge to zero withthe norm k · k is a Banach space.9. Note that c0 c and both c0 and c are closed linear subspaces of with respect to the metric generated by the norm.Exercise. Prove that , c and c0 are Banach spaces.Exercise. Prove that the spaces c and c0 are separable, while is not.10. p , 1 p is the space of all (complex, real) sequences x (xn ) n 1such that!1/p Xpkxkp xn .n 1It follows from the Minkowski inequality for sequences that k · kp is a normand that p is a linear space. We will prove now that p is a Banach space,pi.e. that it is complete. Let xn (ani ) i 1 be a Cauchy sequence in , i.e. forevery ε 0 there is N such that for all n, m N!1/p Xnm pkxn xm kp ai ai ε.i 1Hence for each i the sequence (ani ) n 1 is a Cauchy sequence in K ( C ornR). Let ai limn ai . Fix an integer k. Then for n, m N we havekXpp ani ami εi 1and passing to the limit as m yieldskX ani ai p εp .i 1Now taking the limit as k we obtain X ani ai p εp ,i 1i.e.kxn xm kp ε where x (ai ) i 1 .ppThis proves that x and xn x in . The proof is complete.In particular the space 2 is a Hilbert space because its norm is associatedwith the inner product Xhx, yi xi yi .i 1

FUNCTIONAL ANALYSIS7Exercise. Prove that p , 1 p is separable.11. We will prove that p for p 6 2 is not an inner product space. Letx (1, 0, 0, . . .), y (0, 1, 0, 0, . . .). If 1 p , thenkx yk2p kx yk2p 21 2/p ,2kxk2p 2kyk2p 4 ,and thence the Parallelogram Law is violated. The same example can alsobe used in the case p . 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 , p 6 2, along the space generated by the first two coordinates is not anellipse.12. If X is equipped with a positive measure µ, then for 1 p , Lp (µ)is a Banach space with respect to the norm Z 1/pp.kf kp f dµXL (µ)Alsois a Banach space with the norm being the essential supremumof f . For the proofs see the notes from Analysis I.For p 2 the space L2 (µ) is a Hilbert space with respect to the innerproductZhf, gi f g dµ .X13. If X {1, 2, 3, . . .} and µ is the counting measure (i.e. µ(A) #A),then Lp (µ) p . In particular this gives another proof that p is a Banachspace. However the proof given above is much more elementary.14. If X is a compact metric space, then the space of continuous functionson X with respect to the normkf k sup f (x) x Xis a Banach space. This space is denoted by C(X). The resulting metric inC(X) is the metric of uniform convergence.15. Let H 2 be the class of all holomorphic functions on the unit disc D {z C : z 1} such thatZ 2π 1/21kf kH 2 sup f (reiθ ) 2 dθ .0 r 1 2π 0This space is called Hardy space H 2 . We will prove now that it is a Hilbertspace and we will find an explicit formula for the inner product.

8PIOTR HAJLASZSince every holomorphicfunction in D can be represented as a TaylorPnpolynomial f (z) azn 0 n , for z 1 we haveZ 2πZ 2π X X 11iθ 2n inθ f (re ) dθ an r eam rm eimθ dθ2π 02π 0n 0m 0Z 2π X1 ei(n m)θ dθan am rn m2π 0n,m 0 {z}1 if n m, 0 if n 6 m Xr2n an 2 .n 0Hencekf kH 2 limr 1 Xr2n an 2 1/2 Xn 0 an 2 1/2.n 0This proves that the space H 2 is isometrically isomorphic with 2 and thatthe norm in H 2 is associated with the inner productZ 2π X1f (reiθ )g(reiθ ) dθ ,an bn limhf, gi r 1 2π 0n 0P Pnwhere f n 0 an z n , g(z) n 0 bn z .2. Linear operatorsIf X and Y are normed spaces, then linear functions L : X Y will becalled (linear) operators of (linear) transforms. Also we will often write Lxinstead of L(x). We say that a linear operator L : X Y is bounded if thereis a constant C 0 such thatkLxk Ckxkfor all x X.Theorem 2.1. Let L : X Y be a linear operator between normed spaces.Then the following conditions are equivalent:(1) I is continuous;(2) L is continuous at 0;(3) L is bounded.Proof. The implication (1) (2) is obvious. (2) (3) For if notthere would exist a sequence xn X such that kLxn k nkxn k. ThenkL(xn /(nkxn k))k 1, but on the other hand kL(xn /(nkxn k))k 0, because xn /(nkxn k) 0 which is an obvious contradiction. (3) (1) Letxn x. Then Lxn Lx. Indeed,kLx Lxn k kL(x xn )k Ckx xn k 0 .

FUNCTIONAL ANALYSISThis completes the proof.92For a linear operator L : X Y we define its norm bykLk sup kLxk .kxk 1ThenkLxk kLk kxkfor all x X.Indeed, xx kxk L kxk kLk .kLxk L kxkkxkkxkThus L is bounded if and only if kLk . Moreover kLk is the smallestconstant C for which the inequalitykLxk Ckxkfor all x Xis satisfied. kLk is called the operator norm.The class of bounded operators L : X Y is denoted by B(X, Y ). Wealso write B(X) B(X, X). Clearly B(X, Y ) has a structure of a linearspace.Lemma 2.2. B(X, Y ) equipped with the operator norm is a normed space.Proof is very easy and left as an exercise.Theorem 2.3. If X is a normed space and Y is a Banach space, thenB(X, Y ) is a Banach space.Proof. Let {Ln } be a Cauchy sequence in B(X, Y ). Then(2.1)kLn x Lm xk kLn Lm k kxk .Since the right hand side converges to 0 as n, m , we conclude that{Ln x} is a Cauchy sequence for every x X and hence {Ln x} has a limit inY . We denote it by Lx limn Ln x. Because {Ln } is a Cauchy sequencekLn Lm k ε for all sufficiently large n and m and passing to the limit in(2.1) as n yields(2.2)kLx Lm xk εkxkfor all sufficiently large m .This giveskLxk kLx Lm xk kLm xk (ε kLm k)kxk .Hence L B(X, Y ) and Lm L in B(X, Y ) because of (2.2).2Definition. By a (continuous linear) functional we mean an arbitrarybounded operatorL : X K. The space X B(X, K) is called the dual space of X. Elements of X willusually be denoted by x and we will write hx , xi instead of x (x).