# 1. Hilbert spaces. 1.1 Definitions Vector spaces

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1 . Hilbert spaces. Defiitios.. Vector spaces Defiitio. Vector space (*9&)8& "(9/). A vector space over a field F is a set V that has the structure of a additive group. Moreover, a product F V V, deoted (a,x) ax, is defied, satisfyig: Distributivity i V : a(x +y) = ax +ay. Distributivity i F : (a +b)x = ax +bx. Homogeeity i F : a(b x) = (ab)x. Scalar uit elemet: x = x. The elemets of V are called vectors; the elemets of F are called scalars. Throughout this course the field F will be either the field of complex umbers C (V is a complex vector space) or the field of reals R (V is a real vector space). Defiitio.2 Let V be a vector space. A (fiite) set of vectors {x,...,x } V is called liearly idepedet (;*9!**-.**&-; *;-") if the idetity k= a k x k = implies that a k = for all k. Otherwise, this set of elemets is said to be liearly depedet.

2 8 Hilbert spaces Defiitio.3 If a vector space V cotais liearly idepedet vectors ad every + vectors are liearly depedet, the we say that V has dimesio : dimv =. If dimv for every N (for every there exist liearly idepedet vectors) the we say that V has ifiite dimesio. Propositio. Let V be a vector space. Suppose that dimv = ad let (x,...,x ) be liearly idepedet (a basis). The, every y V has a uique represetatio y = a k x k. k= Proof. Obvious. Defiitio.4 Let V be a vector space. A subset Y V is called a vector subspace (*9&)8& "(9/ ;;) (or a liear subspace) if it is a vector space with respect to the same additio ad scalar multiplicatio operatios (the vector space structure o Y is iherited from the vector space structure o V ). Propositio.2 Let V be a vector space. A subset Y V is a vector subspace if ad oly if Y ad for all y,y 2 Y ad a,a 2 F, a y +a 2 y 2 Y, i.e., the subset Y is closed uder liear combiatios. Proof. Easy. Commet. By defiitio, every liear subspace is closed uder vector space operatios (it is algebraically closed). This should ot be cofused with the topological otio of closedess, which is defied oce we edow the vector space with a topology. A liear subspace may ot be closed i the topological sese. Defiitio.5 Let V ad Y be vector spaces over the same field F ; let D V be a vector subspace. A mappig T D Y is said to be a liear trasformatio (;*9!**- %8;3%) if for all x,x 2 D ad a a 2 F : T (a x +a 2 x 2 ) = a T (x )+a 2 T (x 2 ).

3 . Defiitios 9 The set D is called the domai (.&(;) of T. The set Image(T ) = {T (x) x D} Y is called the image (%&/;) of T. If D = V = Y we call T a liear trasformatio o V. If Y = F we call T a liear fuctioal (*9!**- -&*78&5). Commet.2 Liear trasformatios preserve the vector space operatios, ad are therefore the atural isomorphisms i the category of vector spaces. This should be kept i mid, as the atural isomorphisms may chage as we edow the vector space with additioal structure. Iverse trasformatio If T Domai(T ) Image(T ) is oe-to-oe (ijective) the we ca defie a iverse trasformatio such that for all x Domai(T ) ad y Image(T ). T Image(T ) Domai(T ), T (Tx) = x ad T (T y) = y Notatio. I these otes we will use A B to deote ijectios, A B to deote surjectios, ad A B to deote bijectios. Propositio.3 Let V be a vector space ad D V a liear subspace. Let T D Y be a liear trasformatio. The, Image(T ) is a liear subspace of Y. Proof. Sice D, Image(T ) T () =. Let x,y Image(T ). By defiitio, there exist u,v D such that x = T (u) ad y = T (v). By the liearity of T, for every a,b F : Image(T ) T (au+bv) = ax +by. Thus, Image(T ) is closed uder liear combiatios. A digressio o categories: A category is a algebraic structure that comprises objects that are liked by morphisms. A category has two basic properties: the ability to compose the morphisms associatively ad the existece of a idetity morphism for each object. A simple example is the category of sets, whose morphisms are fuctios. Aother example is the category of groups, whose morphisms are homomorphisms. A third example is the category of topological spaces, whose morphisms are the cotiuous fuctios. As you ca see, the chose morphisms are ot just arbitrary associative maps. They are maps that preserve a certai structure i each class of objects.

4 Hilbert spaces..2 Normed spaces A vector space is a set edowed with a algebraic structure. We ow edow vector spaces with additioal structures all of them ivolvig topologies. Thus, the vector space is edowed with a otio of covergece. Defiitio.6 Metric space. A metric space (*9)/ "(9/) is a set X, edowed with a fuctio d X X R, such that Positivity: d(x,y) with equality iff x = y. Symmetry: d(x,y) = d(y,x). Triagle iequality: d(x,y) d(x,z) + d(z,y). Please ote that a metric space does ot eed to be a vector space. O the other had, a metric defies a topology o X geerated by ope balls, B(x,r) = {y X d(x,y) < r}. As topological spaces, metric spaces are paracompact (every ope cover has a ope refiemet that is locally fiite), Hausdorff spaces, ad hece ormal (give ay disjoit closed sets E ad F, there are ope eighborhoods U of E ad V of F that are also disjoit). Metric spaces are first coutable (each poit has a coutable eighborhood base) sice oe ca use balls with ratioal radius as a eighborhood base. Defiitio.7 Norm. A orm (%/9&) over a vector space V is a mappig V R such that Positivity: x with equality iff x =. Homogeeity: ax = ax. Triagle iequality: x +y x+y. A ormed space (*/9& "(9/) is a pair (V, ), where V is a vector space ad is a orm over V. A orm is a fuctio that assigs a size to vectors. Ay orm o a vector space iduces a metric: Propositio.4 Let (V, ) be a ormed space. The d(x,y) = x y

5 . Defiitios is a metric o V. Proof. Obvious. The coverse is ot ecessarily true uless certai coditios hold: Propositio.5 Let V be a vector space edowed with a metric d. If the followig two coditios hold: Traslatio ivariace: d(x + z,y + z) = d(x,y) Homogeeity: d(ax,ay) = ad(x,y), the x = d(x,) is a orm o V. Exercise. Prove Prop Ier-product spaces Vector spaces are a very useful costruct (e.g., i physics). But to be eve more useful, we ofte eed to edow them with structure beyod the otio of a size. Defiitio.8 Ier product space. A complex vector field V is called a ier-product space (;*/*5 %-5,/ "(9/) if there exists a product (, ) V V C, satisfyig: Symmetry: (x,y) = (y,x). Biliearity: (x +y,z) = (x,z)+(y,z). Homogeeity: (ax,y) = a(x,y). Positivity: (x,x) with equality iff x =. A ier-product space is also called a pre-hilbert space. Propositio.6 A ier-product (H,(, )) space satisfies:

6 2 Hilbert spaces (x,y+z) = (x,y)+(x,z). (x,ay) = a(x,y). Proof. Obvious. Propositio.7 Cauchy-Schwarz iequality. Let (H,(, )) be a ierproduct space. Defie = (, ) 2. The, for every x,y H, (x,y) xy. Proof. There are may differet proofs to this propositio 2. For x,y H defie u = xx ad v = yy. Usig the positivity, symmetry, ad biliearity of the ier-product: That is, (u (u,v)v,u (u,v)v) = +(u,v) 2 (u,v) 2 (u,v) 2 = (u,v) 2. (x,y) xy. Equality holds if ad oly if u ad v are co-liear, i.e., if ad oly if x ad y are co-liear. Corollary.8 Triagle iequality. I every ier product space H, x +y x+y. Proof. Applyig the Cauchy-Schwarz iequality x +y 2 = (x +y,x +y) = x 2 +y 2 +2Re(x,y) x 2 +y 2 +2(x,y) x 2 +y 2 +2xy = (x+y) 2. 2 There is a book called The Cauchy-Schwarz Master Class which presets more proofs that you wat.

7 . Defiitios 3 Corollary.9 A ier-product space is a ormed space with respect to the orm: x = (x,x) 2. Proof. Obvious. Thus, every ier-product space is automatically a ormed space ad cosequetly a metric space. The (default) topology associated with a ier-product space is that iduced by the metric (i..e, the ope sets are geerated by ope metric balls). Exercise.2 Show that the ier product (H,(, )) is cotiuous with respect to each of its argumets: ( x,y H)( e > )( d > ) ( z H z x < d)((z,y) (x,y) < e). Exercise.3 Let (H,(, )) be a complex ier-product space. Defie x,y = Re(x,y). Show that (H,, ) is a real ier-product space. Exercise.4 Prove that if a collectio of o-zero vectors {x,...,x } i a ier-product space are mutually orthogoal the they are liearly idepedet. Exercise.5 Prove that i a ier-product space x = iff (x,y) = for all y. Exercise.6 Let (H,(, )) be a ier-product space. Show that the followig coditios are equivalet: (x,y) =. x +ly = x ly for all l C. x x +ly for all l C.

8 4 Hilbert spaces Exercise.7 Cosider the vector space V = C [,] (cotiuouslydifferetiable fuctios over the uit iterval) ad defie the product (, ) V V C: Is (, ) a ier-product? ( f,g) = f (x)g(x)dx. Set V = { f V f () = }. Is (V,(, )) a ier-product? Propositio. Parallelogram idetity (;*-*"8/% &*&&:). I every ierproduct space (H,(, )): x +y 2 +x y 2 = 2x 2 +y 2. (This equatio is called the parallelogram idetity because it asserts that i a parallelogram the sum of the squares of the sides equals to the sum of the squares of the diagoals.) x +y y x y x Proof. For every x,y H: x ±y 2 = x 2 +y 2 ±2 Re(x,y). The idetity follows from addig both equatios. A ier product defies a orm. What about a coverse? Suppose we are give the the orm iduced by a ier product. Ca we recover the ier product? The aswer is positive, as show by the followig propositio:

9 . Defiitios 5 Propositio. Polarizatio idetity (%*7'*9-&5% ;&%'). I a ier-product space (H,(, )), (x,y) = 4 x +y2 x y 2 +ıx +ıy 2 ıx ıy 2. Proof. It is easy to see that Settig y ıy, x +y 2 x y 2 = 4 Re(x,y). x +ıy 2 x ıy 2 = 4 Reı(x,y) = 4 Im(x,y). Multiplyig the secod equatio by ı ad addig it to the first equatio we obtai the desired result. Commet.3 I a real ier-product space: (x,y) = 4 x +y2 x y 2. Defiitio.9 Let (H,(, )) be a ier-product space. x,y H are said to be orthogoal (.*"7*) if (x,y) = ; we deote x y. Propositio.2 Orthogoal vectors i a ier-product space satisfy Pytagoras law: x +y 2 = x 2 +y 2. Proof. Obvious. Exercise.8 Show that a orm over a real vector space V is iduced from a ier-product over V if ad oly if the parallelogram law holds. Hit: set (x,y) = 2 x +y2 x 2 y 2, ad show that it is a ier product ad that the iduced orm is ideed.

10 6 Hilbert spaces Exercise.9 Show that the `p spaces (the spaces of sequeces with the appropriate orms) ca be tured ito a ier-product space (i.e., the orm ca be iduced from a ier-product) oly for p = Hilbert spaces Defiitio. Hilbert space. A complete ier-product space is called a Hilbert space. (Recall: a space is complete (.-:) if every Cauchy sequece coverges.) Commet.4 A ier-product space (H,(, )) is a Hilbert space if it is complete with respect to the metric d(x,y) = (x y,x y) 2. Completeess is a property of metric spaces. A sequece (x ) H is a Cauchy sequece if for all e > there exists a N N such that for every m, > N: x x m < e. Exercise. Let H,...,H be a fiite collectio of ier-product spaces. Defie the space H = H H, alog with coordiate-wise vector space operatios. Defie a product (, ) H H H C: ((x,...,x ),(y,...,y )) H = (x k,y k ) Hk. Show that (, ) H is a ier-product o H. Show that covergece i H is equivalet to compoet-wise covergece i each of the H k. Show that H is complete if ad oly if all the H k are complete. k= We metioed the fact that a ier-product space is also called a pre-hilbert space. The reaso for this omeclature is the followig theorem: ay ier-product space ca be completed caoically ito a Hilbert space. This completio is aalogous to the completio of the field of ratioals ito the field of reals.

11 . Defiitios 7 Theorem.3 Completio. Let (G,(, ) G ) be a ier-product space. The, there exists a Hilbert space (H,(, ) H ), such that: There exists a liear ijectio T G H, that preserves the ier-product, (x,y) G = (Tx,Ty) H for all x,y G (i.e., elemets i G ca be idetified with elemets i H ). Image(T ) is dese i H (i.e., G is idetified with almost all of" H ). Moreover, the iclusio of G i H is uique: For ay liear ier-product preservig ijectio T G H where H is a Hilbert space ad Image(T ) is dese i H, there is a liear isomorphism S H H, such that T = S T (i.e., H ad H are isomorphic i the category of ier-product spaces). I other words, the completio G is uique modulo isomorphisms. Proof. We start by defiig the space H. Cosider the set of Cauchy sequeces (x ) i G. Two Cauchy sequeces (x ) ad (y ) are defied to be equivalet (deoted (x ) (y )) if lim x y =. It is easy to see that this establishes a equivalece relatio amog all Cauchy sequeces i G. We deote the equivalece class of a Cauchy sequece (x ) by [x ] ad defie H as the set of equivalece classes. We edow H with a vector space structure by defiig a[x ]+b[y ] = [ax +bz ]. (It is easy to see that this defiitio is idepedet of represetig elemets.) Let (x ) ad (y ) be Cauchy sequece i G. Cosider the series This limit exists because lim (x,y ) G. (x,y ) G (x m,y m ) G = (x,y ) G (x,y m ) G +(x,y m ) G (x m,y m ) G (triagle ieq.) (x,y y m ) G +(x x m,y m ) G (Cauchy-Schwarz) x G y y m G +x x m y m G, Sice Cauchy sequeces are bouded (easy!), there exists a M > such that for all m,, (x,y ) G (x m,y m ) G M (y y m G +x x m G ). It follows that (x,y ) G is a Cauchy sequece i F, hece coverges (because F is complete).

12 8 Hilbert spaces Moreover, if (u ) (x ) ad (v ) (y ) the, (x,y ) G (u,v ) G = (x,y ) G (x,v ) G +(x,v ) G (u,v ) G from which follows that (x,y v ) G +(x u,v ) G x G y v G +x v G v G, lim (x,y ) G = lim (u,v ) G. Thus, we ca defie uambiguously a product (, ) H H F ; ([x ],[y ]) H = lim (x,y ) G. It remais to show that (, ) H is ideed a ier product (do it). The ext step is to defie the iclusio T G H. For x G let Tx= [(x,x,...)], amely, it maps every vector i G ito the equivalece class of a costat sequece. By the defiitio of the liear structure o H, T is liear. It is preserves the ier-product as (Tx,Ty) H = lim ((Tx),(Ty) ) G = lim (x,y) G = (x,y) G. The ext step is to show that Image(T ) is dese i H. Let h H ad let (x ) be a represetative of h. Sice (x ) is a Cauchy sequece i G, lim Tx h H = lim lim x x k G =. k which proves that Tx h, ad therefore Image(T ) is dese i H. The ext step is to show that H is complete. Let (h ) be a Cauchy sequece i H. For every let (x,k ) be a Cauchy sequece i G i the equivalece class of h. Sice Image(T ) is dese i H, there exists for every a y G, such that It follows that Ty h H = lim k y x,k G. y y m G = Ty Ty m H Ty h H +h h m H +h m Ty m H h h m H + + m, i.e., (y ) is a Cauchy sequece i G ad therefore h = [y ] is a elemet of H.

13 . Defiitios 9 We will show that lim h h H =, which will prove that ay Cauchy sequece i H coverges. By defiitio, Now ad lim h h H = lim lim x,k y k G. k x,k y k G x,k y G +y y k G, lim lim x,k y G lim = ad lim k lim y y k G =. k The last step is to show the uiqueess of the completio modulo isomorphisms. Let h H. Sice Image(T ) is dese i H, there exists a sequece (y ) G, such that lim Ty h H =. It follows that (Ty ) is a Cauchy sequece i H, ad because T preserves the ier-product, (y ) is a Cauchy sequece i G. It follows that (T y ) is a Cauchy sequece i H, ad because the latter is complete (T y ) has a limit i H. This limit is idepedet of the choice of the sequece (y ), hece it is a fuctio of h, which we deote by S(h) = lim T Y. We leave it as a exercise to show that S satisfies the required properties. Exercise. Complete the missig details i the above proof...5 Examples of Hilbert spaces. The space R is a real vector space. The mappig (x,y) is a ier product. The iduced metric i= x i y i 2 d(x,y) = (x i y i ) 2 i= is called the Euclidea metric. It is kow that R is complete with respect to this metric, hece it is a Hilbert space (i fact, ay fiite-dimesioal ormed space is complete, so that the otio of completeess is oly of iterest i ifiite-dimesioal spaces).

14 2 Hilbert spaces 2. The space C is a complex vector space. The mappig (x,y) i= as a ier product. C edowed with this metric is a Hilbert space. 3. Cosider the space of square summable sequeces: `2 = x C N = x i y i x 2 <. It is a complex vector space with respect to poitwise operatios. We defie (x,y) = i= x i y i. This series coverges absolutely as for every fiite, the Cauchy-Schwarz iequality implies: i= x i y i i= 2 x i 2 i= 2 y i 2, ad the right had side is uiformly bouded. It ca be show that `2 is complete. Moreover, it is easy to show that the subset of ratioal-valued sequeces that have a fiite umber of o-zero terms is dese i `2, i.e., `2 is a separable (*-*952) Hilbert space (has a coutable dese subset). Exercise.2 Prove ( by had") that `2 is complete. 4. Let W be a bouded set i R ad let C(W) be the set of cotiuous complexvalued fuctios o its closure 3. This space is made ito a complex vector space by poitwise additio ad scalar multiplicatio: ( f +g)(x) = f (x)+g(x) ad (a f )(x) = a f (x). We defie o C(W) a ier product with the correspodig orm: ( f,g) = W f (x)g(x)dx, f = f (x) 2 2 dx. W 3 The fact that the domai is compact is crucial; we shall see later i this course that such a structure caot be applied for cotiuous fuctios over ope domais.

15 . Defiitios 2 This space is ot complete ad hece ot a Hilbert space. To show that, let x W ad r > be such that B(x,2r) W. Defie the sequece of fuctios, x x r f (x) = +(r x x ) r x x r + x x > +, which are defied for sufficietly large. f = f = x The fuctios f are cotiuous ad coverge poitwise to the discotiuous fuctio x x r f (x) = x x > r. The sequece ( f ) is a Cauchy sequece as for > m, f f m (B(x,r +m)b(x,r)) 2, which teds to zero as m,. Suppose that the space was complete. It would imply the existece of a fuctio g C(W), such that lim f g = lim f (x) g(x) 2 2 =. W By Lebesgue s bouded covergece (%/&2(% ;&2,;%% )5:/) theorem lim f g = lim f (x) g(x) 2 2 =, W i.e., g = f a.e., which is a cotradictio. The completio of C(W) with respect to this metric is isomorphic to the Hilbert space L 2 (W) of square itegrable fuctios. Commet.5 The costructio provided by the completio theorem is ot coveiet to work with. We prefer to work with fuctios rather tha with equivalece classes of Cauchy sequeces of fuctios. TA material. Hilbert-Schmidt matrices. Let M be a collectio of all ifiite matrices over C that oly have a fiite umber of o-zero elemets. For A M we deote by (A) the smallest umber for which A ij = for all i, j > (A). (i)

16 22 Hilbert spaces Show that M is a vector space over C with respect to matrix additio ad scalar multiplicatio. (ii) Defie (A,B) = Tr(AB ) ad show that it is a ier-product. (iii) Show that M is ot complete. (iv) Show that it is possible to idetify the completio H of M with the set alog with the ier-product H = A = (a ij ) i, j= (A,B) = i, j= i, j= A ij B ij. a ij 2 <, This space is kow as the space of Hilbert-Schmidt matrices. Exercise.3 Prove that every fiite-dimesioal ier product space is complete (ad hece a Hilbert space). TA material.2 Sobolev spaces. Edow the space C k (R) with the ierproduct d i f d i g ( f,g) = dx i k R dx i dx i Its completio is deoted H k (R) (or W k,2 (R)). Defie the otio of a weak derivative ad show that if it exists, the it is uique. Show how to idetify H k (R) with the space of fuctios that have square-itegrable k weak derivatives..2 Covexity ad projectio Orthogoality is oe of the cetral cocepts i the theory of Hilbert spaces. Aother cocept, itimately related to orthogoality, is orthogoal projectio (;"7* %-)%). Before gettig to projectios we eed to develop the otio of a covex set. Covexity, is a purely algebraic cocept, but as we will see, it iteracts with the topology iduced by the ier-product..2. Covexity Defiitio. Covex set. Let V be a vector space. A subset C V is called covex (9&/8) if x,y C ad t tx+( t)y C.

17 .2 Covexity ad projectio 23 (The segmet that coects ay two poits i C is i C ). Differetly stated, for all t [,]: tc +( t)c C. A C B Lemma.4 For ay collectio of sets {C a } ad D a ad every t R: t C a = tc a, a A a A ad C a + D a (C a +D a ). a A a A a A Proof. First, Secod, if t C a = {tx x C a a} = tc a. a A a A x C a + D a, a A a A the there is a c C a for all a ad a d D b for all b, such that x = c + d. Now c+d = C a +D a for all a, hece x (C a +D a ). a A Propositio.5 Covexity is closed uder itersectios. Let V be a vector space. Let {C a V a A} be a collectio of covex sets (ot ecessarily coutable). The C = C a a A is covex.

18 24 Hilbert spaces Proof. A obscure" proof relies o Lemma.4. For all t [,]: tc +( t)c = t C a +( t) C a a A a A = tc a + ( t)c a a A a A (tc a +( t)c a ) a A C a = C. a A Now for a more trasparet proof: let x,y C. By defiitio: Sice all the C a are covex: ( a A)(x,y C a ). ( a A)( t )(tx+( t)y C a ). Iterchagig the order of the quatifiers, which implies that ( t )( a A)(tx+( t)y C a ), ( t )(tx+( t)y C ). Propositio.6 Covex sets are closed uder covex liear combiatios. Let V be a vector space. Let C V be a covex set. The for every (x,...,x ) C ad every o-egative (t,...,t ) real umbers that sum up to, i= t i x i C. (.) Proof. Equatio (.) holds for = 2 by the very defiitio of covexity. Suppose (.) were true for = k. Give (x,...,x k+ ) C ad (t,...,t k+ ), defie t = k i= t i. The, k+ i= t i =, k+ i= t i x i = k i= k t i x i +t k+ x k+ = t i= t x i +( t)x k+ t i C C

19 .2 Covexity ad projectio 25 Propositio.7 Let (X, ) be a ormed space ad C X a covex subset. The, The closure C is covex. The iterior C is covex. Commet.6 Iterior ad closure are topological cocepts, whereas covexity is a vector space cocept. The coectio betwee the two stems from the fact that a ormed space has both a topology ad a vector space structure. Proof. Let x,y C. For every e > there are poits x e,y e C with x x e < e ad y y e < e. Let t. The, tx e +( t)y e C ad (tx+( t)y) (tx e +( t)y e ) tx x e +( t)y y e < e, which implies that tx+( t)y C, hece C is covex. Let x,y C. By defiitio of the iterior there exists a r > such that B(x,r) C ad B(y,r) C. Sice C is covex, t [,] tb(x,r)+( t)b(y,r) C, but B(tx+( t)y,r) tb(x,r)+( t)b(y,r), which proves that tx+( t)y C. Hece, C is covex. x tx+( t)y y Examples. Every ope ball B(a,r) i a ormed vector space is covex, for if x,y B(a,r), the for all t : tx+( t)y a = t(x a)+( t)(y a) t x a+( t)y a < r.

20 26 Hilbert spaces Every liear subspace of a vector space is covex, because it is closed uder ay liear combiatios ad i particular, covex oes. For example, let V = L 2 [,] ad let C be the subset of polyomials. C is a liear subspace of V, hece it is covex. Let W R be a domai ad cosider the Hilbert space L 2 (W). The subset of fuctios that are o-egative (up to a set of measure zero) is covex (but it is ot a liear subspace). Exercise.4 Let V be a vector space ad C V. The covex hull (9W/8) of C is defied by Cov(C) = {x V x is a covex combiatios of elemets i C}. Show that Cov(C) is the smallest covex set that cotais C. Exercise.5 Prove Carathéodory s theorem: let A R ad let x Cov(A). The x is a covex combiatio of + poits i A or less. (Hit: suppose that x is a covex combiatio of x,...,x p A, where p > +. Use the fact that {x i x } p i=2 are liearly depedet to show that x ca be writte as a covex sum of p poits). Show that Carathéodory s theorem may fail if the dimesio of the vector space is ifiite..2.2 Orthogoal projectio Defiitio.2 Let (H,(, )) be a ier-product space, ad let S H (it ca be ay subset; ot ecessarily a vector subspace). We deote by S the set of vectors that are perpedicular to all the elemets i S, S = {x H (x,y) = y S}.

21 .2 Covexity ad projectio 27 Propositio.8 Let (H,(, )) be a ier-product space. Let S H. The set S is a closed liear subspace of H, ad S S {}, Proof. We start by showig that S is a liear subspace. Let x,y S, i.e., For all a,b F, z S (x,z) = (y,z) =. z S (ax +by,z) = a(x,z)+b(y,z) =, which implies that ax +by S, i.e., S is a liear subspace. We ext show that S is closed. Let (x ) be a sequece i S that coverges to x H. By the cotiuity of the ier product, z S (x,z) = lim (x,z) =, i.e., x S. Suppose x S S. As a elemet i S, x is orthogoal to all the elemets i S, ad i particular to itself, hece (x,x) =, which by the defiig property of the ier-product implies that x =. Exercise.6 Show that S = {x}. x S Exercise.7 Show that if M ad N are closed subspaces of a Hilbert space H, ad N is fiite dimesioal, the M +N is a closed subspace (hit: iductio o the dimesio of N). Show that M +N may ot be closed if N is ifiite dimesioal.

22 28 Hilbert spaces The followig theorem states that give a closed covex set C i a Hilbert space H, every poit i H has a uique poit i C that is the closest to it amog all poits i C : Theorem.9 Let (H,(, )) be a Hilbert space ad C H closed ad covex. The, x H!y C such that x y = d(x,c ), where d(x,c ) = if x y. y C The mappig x y is called the projectio (%-)%) of x oto the set C ad it is deoted by P C. Commet.7 Note the coditios of this theorem. The space must be complete ad the subset must be covex ad closed. We will see how these coditios are eeded i the proof. A very importat poit is that the space must be a ier-product space. Projectios do ot geerally exist i (complete) ormed spaces. Proof. We start by showig the existece of a distace miimizer. By the defiitio of the ifimum, there exists a sequece (y ) C satisfyig, lim d(x,y ) = d(x,c ). Sice C is covex, 2 (y +y m ) C for all m,, ad therefore, 2 (y +y m ) x d(x,c ). By the parallelogram idetity (which is where the ier-product property eters), a b 2 = 2(a 2 +2b 2 ) a+b 2, ad so y y m 2 = (y x) (y m x) 2 = 2y x 2 +2y m x 2 y +y m 2x 2 2y x 2 +2y m x 2 2d(x,C ) m,. It follows that (y ) is a Cauchy sequece ad hece coverges to a limit y (which is where completeess is essetial). Sice C is closed, y C. Fially, by the cotiuity of the orm, x y = lim x y = d(x,c ), which completes the existece proof of a distace miimizer. Next, we show the uiqueess of the distace miimizer. Suppose that y,z C both satisfy y x = z x = d(x,c ).

23 .2 Covexity ad projectio 29 By the parallelogram idetity, i.e., y+z 2x 2 +y z 2 = 2y x 2 +2z x 2, y+z 2 x2 = d 2 (x,c ) 4 y z. If y z the (y+z)2, which belogs to C is closer to x tha the distace of x from C, which is a cotradictio. x y C (y+z)/2 z TA material.3 Projectios i Baach spaces. The existece of a uique projectio does ot hold i geeral i complete ormed spaces (i.e., Baach spaces). A distace miimizer does exist i fiite-dimesioal ormed spaces, but it may ot be uique)=. I ifiite-dimesioal Baach spaces distace miimizers may fail to exist. TA material.4 Coditioal expectatios. The followig is a importat applicatio of orthogoal projectios. Let (W,F,P) be a probability space, ad let A F be a sub-s-algebra. Let X W C be a radom variable (i.e., a measurable fuctio) satisfyig X <. The radom variable Y W C is called the coditioal expectatio of X with respect to the s-algebra A if (i) Y is A -measurable, ad (ii) for every A A, A YdP= A XdP. Prove that the coditioal expectatio exists ad is uique i L (W,F,P). (Note that L is ot a Hilbert space, so that the costructio has to start with the subspace L 2 (W,F,P), ad ed up with a desity argumet.) Propositio.2 Let H be a Hilbert space. Let C be a closed covex set. The mappig P C H C is idempotet, P C P C = P C. Proof. Obvious, sice P C = Id o C.

24 3 Hilbert spaces Exercise.8 Let (H,(, )) be a Hilbert space ad let P A ad P B be orthogoal projectios o closed subspaces A ad B. Show that if P A P B is a orthogoal projectio the it projects o A B. Show that P A P B is a orthogoal projectio if ad oly if P A P B = P B P A. Show that if P A P B is a orthogoal projectio the P A +P B P A P B is a orthogoal projectio o A+B. Fid a example i which P A P B P B P A. The ext propositio has a geometric iterpretatio: the segmet coectig a poit x C with its projectio P C x makes a obtuse agle with ay segmet coectig P C x with aother poit i C. The propositio states that this is i fact a characterizatio of the projectio. Propositio.2 Let C be a closed covex set i a Hilbert space (H,(, )). The for every x H, z = P C x if ad oly if z C ad y C Re(x z,y z). x z = P C x y C Proof. Suppose first that z = P C x. By defiitio z C. Let y C. Sice C is covex the ty+( t)z C for all t [,], ad sice z is the uique distace miimizer from x i C : > x z 2 x (ty+( t)z) 2 = x z 2 (x z) t(y z) 2 = t 2 y z 2 +2t Re(x z,y z). Thus, for all < t, Re(x z,y z) < 2 ty z2.

25 .2 Covexity ad projectio 3 Lettig t we get that Re(x z,y z). Coversely, suppose that z C ad that for every y C, Re(x z,y z). For every y C, x y 2 x z 2 = (x z)+(z y) 2 x z 2 = y z 2 2Re(x z,y z), which implies that z is the distace miimizer, hece z = P C x. Corollary.22 Projectios are distace reducig. Let C be a closed covex set i a Hilbert space (H,(, )). The for all x,y H, Re(P C x P C y,x y) P C x P C y 2 ad P C x P C y 2 x y 2. y P C y x P C x C Proof. By Propositio.2, with P C y as a arbitrary poit i C, Re(x P C x,p C y P C x). Similarly, with P C x as a arbitrary poit i C Re(y P C y,p C x P C y). Addig up both iequalities: Re((x y) (P C x P C y),p C x P C y), which proves the first assertio. Next, usig the first assertio ad the Cauchy-Schwarz iequality, P C x P C y 2 Re(P C x P C y,x y) (P C x P C y,x y) P C x P C yx y, ad it remais to divide by P C x P C y.

26 32 Hilbert spaces The ext corollary characterizes the projectio i the case of a closed liear subspace (which is a particular case of a closed covex set). Corollary.23 Let M by a closed liear subspace of a Hilbert space (H,(, )) The, y = P M x if ad oly if y M ad x y M. x y m y = P M x y+m M Proof. Let y M ad suppose that x y M. The, for all m M : hece y = P M x by Propositio.2. (x y, m y M ) =, Coversely, suppose that y = P M x ad let m M. By Propositio.2, Re(y x,y m). We may replace m by y m M, hece for all m M : Re(y x,m). Sice we may replace m by ( m), it follows that for all m M : Re(y x,m) =. Replacig m by ım we obtai Im(y x,m) =. Fiite dimesioal case This last characterizatio of the projectio provides a costructive way to calculate the projectio whe M is a fiite-dimesioal subspace (hece a closed subspace). Let = dimm ad let (e,...,e )

27 .2 Covexity ad projectio 33 be a set of liearly idepedet vectors i M (i.e., a basis for M ). Let x H. The, x P M x is orthogoal to each of the basis vectors: (x P M x,e`) = for ` =,...,. Expadig P M x with respect to the give basis: we obtai, P M x = a k e k, k= (x,e`) = a k (e k,e`) = k= for ` =,...,. The matrix G whose etries are G ij = (e i,e j ) is kow as the Gram matrix. Because the e i s are liearly idepedet, this matrix is o-sigular, ad a k = G k` (x,e`), i.e., we have a explicit expressio for the projectio of ay vector: `= P M x = k= `= G k` (x,e`)e k. Exercise.9 Let H = L 2 (R) ad set M = { f H f (t) = f ( t) a.e.}. Show that M is a closed subspace. Express the projectio P M explicitly. Fid M. Exercise.2 What is the orthogoal complemet of the followig sets of L 2 [,]? The set of polyomials. The set of polyomials i x 2. The set of polyomials with a =. The set of polyomials with coefficiets summig up to zero.

28 34 Hilbert spaces Theorem.24 Projectio theorem (%-)%% )5:/). Let M be a closed liear subspace of a Hilbert space (H,(, )). The every vector x H has a uique decompositio x = m+ m M, M. Furthermore, m = P M x. I other words, H = M M. Proof. Let x H. By Corollary.23 x P M x M, hece x = P M x +(x P M x) satisfies the required properties of the decompositio. Next, we show that the decompositio is uique. Assume x = m + = m 2 + 2, where m,m 2 M ad, 2 M. The, M m m 2 = 2 M. Uiqueess follows from the fact that M M = {}. TA material.5 Show that the projectio theorem does ot holds whe the coditios are ot satisfied. Take for example H = `2, with the liear subspace M = {(a ) `2 N > Na = }. This liear subspace it ot closed, ad its orthogoal complemet is {}, i.e., M M = M H. Corollary.25 For every liear subspace M of a Hilbert space H, (M ) = M.

29 .2 Covexity ad projectio 35 Proof. Let m M. By the defiitio of M : which implies that m (M ), i.e., M (m,) =, M (M ). By Propositio.8 ay orthogoal complemet is closed. If a set is cotaied i a closed set, so is its closure (prove it!), M (M ). Let x (M ). Sice M is a closed liear subspace, there exists a (uique) decompositio x = m+, m M (M ). Takig a ier product with, usig the fact that m : Sice M M, the 2 = (x,). (M ) (M ) (the smaller the set, the larger its orthogoal complemet). Thus (M ), ad therefore x. It follows that =, which meas that x M, i.e., This completes the proof. (M ) M. Corollary.26 Let M be a closed liear subspace of a Hilbert space (H,(, )). The, every x H has a decompositio x = P M x +P M x, ad x 2 = P M x 2 +P M x 2. Proof. As a cosequece of the projectio theorem, usig the fact that both M ad M are closed ad the fact that (M ) = M : x = P M x +(x P M x) M M x = (x P M x) M + P M x. M

30 36 Hilbert spaces By the uiqueess of the decompositio, both decompositios are idetical, which proves the first part. The secod idetity follows from Pythagoras law. Corollary.27 A projectio is liear, orm-reducig ad idempotet. Let M be a closed liear subspace of a Hilbert space (H,(, )). The the projectio P M is a liear operator satisfyig P 2 M = P M, ad x H P M x x. Proof. We have already see that P M is idempotet. The orm reducig property follows from x 2 = P M x 2 +P M x 2 P M x 2. It remais to show that P M is liear. Let x,y H. It follows from Corollary.26 that x = P M x +P M x hece y = P M y+p M y, x +y = (P M x +P M y) +(P M x +P M y). M M O the other had, it also follows from Corollary.26 that By the uiqueess of the decompositio, x +y = P M (x +y) +P M (x +y). M M P M (x +y) = P M x +P M y. Similarly, but also ax = a P M x +a P M x, ax = P M (ax)+p M (ax), ad from the uiqueess of the decompositio, P M (ax) = a P M x.

31 .2 Covexity ad projectio 37 The ext theorem shows that the last corollary is i fact a characterizatio of projectios: there is a oe-to-oe correspodece betwee closed subspaces of H ad orthogoal projectios. Theorem.28 Every liear, orm-reducig, idempotet operator is a projectio. Let (H,(, )) be a Hilbert space ad let P H H be a liear, orm reducig, idempotet operator. The P is a projectio o a closed liear subspace of H. Proof. The first step is to idetify the closed subspace of H that P projects oto. Defie M = {x H Px = x} N = {y H Py = }. Both M ad N are liear subspaces of H. If x M N the x = Px =, amely, M N = {}, Both M ad N are closed because for every x,y H : Px Py = P(x y) x y, from which follows that if x M is a sequece with limit x H, the x Px = lim x Px = lim Px Px lim x x =, i.e., Px = x, hece x M. By a similar argumet ee show that N is closed. Let x H. We write By the idempotece of P, x = Px +(Id P)x. Px M ad (Id P)x N. To prove that P = P M it remais to show that N = M (because of the uiqueess of the decompositio). Let Obviously, y N, hece (x,y) =, ad x N ad y = Px x. x 2 Px 2 = x +y 2 = x 2 +y 2 x 2, i.e., y =, i.e., x = Px, i.e., x M. We have just show that N M.

32 38 Hilbert spaces Take ow x M. By the projectio theorem, there exists a uique decompositio, x = u + v. N M N But sice N M, it follows that u is both i M ad i N, i.e., it is zero ad x N, amely M N. Thus N = M ad further N = (N ) = M. This cocludes the proof..3 Liear fuctioals Amog all liear maps betwee ormed spaces stad out the liear maps ito the field of scalars. The study of liear fuctioals is a cetral theme i fuctioal aalysis..3. Boudedess ad cotiuity Defiitio.3 Let (X, ) ad (X 2, 2 ) be ormed spaces ad let D X be a liear subspace. A liear trasformatio T D X 2 is said to be cotiuous if x D limty Tx 2. y x It is said to be bouded if there exists a costat C > such that x D Tx 2 Cx. If T is bouded, the the lowest boud C is called the orm of T : Tx 2 T = sup = sup T x D x x D x x 2 = sup Tx 2. x = (Recall that if X = R the we call T a liear fuctioal.) Commets. We are dealig here with ormed spaces; o ier-product is eeded. As for ow, we call T a orm, but we eed to show that it is ideed a orm o a vector space. If T is bouded the x D Tx T x. All liear operators betwee fiite-dimesioal ormed spaces are bouded. This otio is therefore oly relevat to ifiite-dimesioal cases.

33 .3 Liear fuctioals 39 Propositio.29 Boudedess ad cotiuity are equivalet. Let (X, ) ad (X 2, 2 ) be ormed spaces. A liear operator T D X X 2 is bouded if ad oly if it is cotiuous. Proof. Suppose T is bouded. The, Tx Ty 2 = T (x y) 2 T x y, ad y x implies Ty Tx. Suppose that T is cotiuous at D. The there exists a d > such that y B(,d) Ty 2. Usig the homogeeity of the orm ad the liearity of T : x D Tx 2 = 2 d x T d x 2 2 x 2 d x. which implies that T is bouded, T 2d. Commet.8 We have oly used the cotiuity of T at zero. This meas that if T is cotiuous at zero, the it is bouded, ad hece cotiuous everywhere. Examples.2. A orthogoal projectio i a Hilbert space is bouded, sice P M x x, i.e., P M. Sice P M x = x for x M, it follows that P M = sup P M x x= sup x M x= P M x = sup x M x= x =, P M =. 2. Let H = L 2 [,] ad let D = C[,] L 2 [,]. Defie the liear fuctioal T D R, Tf= f (). This operator is ubouded (ad hece ot-cotiuous), for take the sequece of fuctios, x x f (x) = otherwise. The f, whereas T f = f () =, i.e., T f lim f =.

34 4 Hilbert spaces 3. Cosider the Hilbert space H = L 2 [,] with D the subspace of differetiable fuctios with derivatives i H. Defie the liear operator T D H, (T f)(x) = f (x). This operator is ubouded. Take for example the sequece of fuctios, f (x) = 2 e x22. 2p The f = ad lim Tf =. 4. Importat example! Let H be a Hilbert space. Set y H ad defie the fuctioal: T y = (,y). This fuctioal is liear, ad it is bouded as hece y = (y,y) y (x,y) sup x x T y (x) T = sup = y. x x yx sup = y, x x I other words, to every y H correspods a bouded liear fuctioal T y..3.2 Extesio of bouded liear fuctioals Lemma.3 Give a bouded liear fuctioal T defied o a dese liear subspace D of a Hilbert space (H,(, )), it has a uique extesio T over H. Moreover, T = T. Proof. We start by defiig T. For x H, take a sequece (x ) D that coverges to x. Cosider the sequece Tx. Sice T is liear ad bouded, Tx Tx m T (x x m ) T x x m, which implies that (Tx ) is a scalar-valued Cauchy sequece. The limit does ot deped o the chose sequece: if (y ) D coverges to x as well, the Thus, we ca defie uambiguously lim Tx Ty lim T x y =. Tx= lim Tx.

35 .3 Liear fuctioals 4 For x D we ca take the costat sequece x = x, hece Tx= lim Tx = Tx, which shows that T is ideed a extesio of T : T D = T. Next, we show that T is liear. Let x,y D. Let x,y D coverge to x,y, respectively, the: T (ax +by) = lim T (ax +by ) = a lim Tx +b lim Ty = a Tx+b Ty. It remais to calculate the orm of T : Tx T = sup x x = sup x lim Tx x T lim x sup = T, x x ad sice T is a extesio of T : T = T. The followig theorem is a istace of the Hah-Baach theorem, which we will meet whe we study Baach spaces: Theorem.3 Extesio theorem. Give a bouded liear fuctioal T defied o a liear subspace D of a Hilbert space (H,(, )), it ca be exteded ito a liear fuctioal over all H, without chagig its orm. That is, there exists a liear fuctioal T o H, such that T D = T ad T = T. Proof. By the previous lemma we may assume without loss of geerality that D is a closed liear subspace of H. We defie T = T P D. Sice T is a compositio of two liear operators, it is liear. Also, T D = T. Fially, T = T P D T P D = T. Sice T is a extesio of T it follows that T = T. TA material.6 Hamel basis. Defiitio.4 Let V be a vector space. A set of vectors {v a a A} V is called a Hamel basis (or algebraic basis) if every v V has a uique represetatio as a liear combiatio of a fiite umber of vectors from {v a a A}.

36 42 Hilbert spaces Propositio.32 Every vector space V has a Hamel basis. Exercise.2 Prove it. Hit: use the axiom of choice. Propositio.33 Let (X, ) be a ifiite-dimesioal ormed space. The, there exists a ubouded liear fuctioal o X. Proof. Let {x a a A} be a Hamel basis. Sice X is ifiite-dimesioal there is a sequece (x a ) such that {x a N} is liearly idepedet. For x = a A t a x a defie F(x) = x a t a. = It is easy to see that this is a liear fuctioal. However, it is ot cotiuous. Defie y = x a x a, The, y by F(y ) = for all..3.3 The Riesz represetatio theorem The ext (very importat) theorem asserts that all bouded liear fuctioals o a Hilbert space ca be represeted as a ier-product with a fixed elemet of H : Theorem.34 Riesz represetatio theorem, 97. Let H be a Hilbert space ad T a bouded liear fuctioal o H. The, ad moreover T = y T. (!y T H ) T = (,y T ), Commet.9 The represetatio theorem was proved by Frigyes Riesz (88 956), a Hugaria mathematicia, ad brother of the mathematicia Marcel Riesz ( ). Proof. We start by provig the uiqueess of the represetatio. If y ad z satisfy T = (,y) = (,z), the y z 2 = (y z,y z) = (y z,y) (y z,z) = T (y z) T(y z) =,

37 .3 Liear fuctioals 43 which implies that y = z. Next, we show the existece of y T. Sice T is liear it follows that kert is a liear subspace of H. Sice T is moreover cotiuous it follows that kert is closed, as (x ) kert with limit x H, implies that Tx= lim Tx =, i.e., x kert. If kert = H, the x H Tx= = (x,), ad the theorem is proved with y T =. If kert H, the we will show that dim(kert ) =. Let y,y 2 (kert ), ad set y = T (y 2 )y T(y )y 2 (kert ). By the liearity of T, T (y) =, i.e., y kert, but sice kert (kert ) = {}, it follows that T (y 2 )y = T (y )y 2, i.e., every two vectors (kert ) are co-liear. Take y (kert ) with y =. The, (kert ) = Spa{y }. The, set y T = T (y )y. By the projectio theorem, for every x H, x = (x,y )y +[x (x,y )y ], (kert ) kert ad applyig T, T (x) = (x,y )T (y ) = (x,y T ). Fially, we have already see that T = y T. The space dual to a Hilbert space Cosider the set of all bouded liear fuctioals o a Hilbert space. These form a vector space by the poitwise operatios, (at +bs)(x) = a T (x)+b S(x).

38 44 Hilbert spaces We deote this vector space by H ; it is called the space dual (*-!&\$ "(9/) to H. The Riesz represetatio theorem states that there is a bijectio H H, T y T. H is made ito a Hilbert space by defiig the ier-product ad the correspodig orm over H is (T,S) = (y T,y S ), T = y T coicides with the previously-defied orm" (we ever showed it was ideed a orm) 4. The followig theorem (the Rado-Nikodym theorem restricted to fiite measure spaces) is a applicatio of the Riesz represetatio theorem: Theorem.35 Rado-Nikodym. Let (W,B, µ) be a fiite measure space. If is a fiite measure o (W,B) that is absolutely cotiuous with respect to µ (i.e., every zero set of µ is also a zero set of ), the there exists a o-egative fuctio f L (W), such that B B (B) = B fdµ. (The fuctio f is called the desity of with respect to µ.) Proof. Let l = µ +. Every zero set of µ is also a zero set of l. Defie the fuctioal F L 2 (l) C: F is a liear fuctioal; it is bouded as F(g) = W gd. 2 2 F(g) gd g 2 d W W d W g L 2 (l) ((W)) 2. (This is where the fiiteess of the measure is crucial; otherwise a L 2 fuctios is ot ecessarily i L.) By the Riesz represetatio theorem there is a uique fuctio h L 2 (l), such that W gd = W hgdl. (.2) We ow show that h satisfies the followig properties: 4 Our defiitio of a geuie orm over H is somewhat awkward. Whe we get to Baach spaces it will be made clear that we do ot eed ier-products ad represetatio theorems i order to show that the operator orm is ideed a orm.

39 .3 Liear fuctioals 45. h l-a.e: Because the measures are fiite, idicator fuctios are itegrable. The, settig g = I B, (B) = B hdl, which implies that h l-a.e. 2. h < l-a.e: Settig, B = {x W h(x) }, we get (B) = B hdl l(b) = (B)+ µ(b), which implies that µ(b) = (B) = l(b) =, i.e., h < l-a.e. Sice h <, we may represet h as where f is l-a.e. o-egative. Back to Eq. (.2), h = f + f, gd = f W W + f gd + f W + f gdµ, hece W + f gd = W f + f gdµ. Let ow k W R be o-egative ad bouded. Defie ad g = k(+ f )I Bm. The, B m = {x W k(x)(+ f (x)) m}, Bm kd = Bm kfdµ, Sice k is o-egative we ca let m ad get for all measurable ad bouded k: 5 W kd = W kfdµ. Settig k we get that f is i L (µ). Settig k = I B we get B d = B fdµ, which completes the proof. 5 Here we apply Lebesgue s Mootoe Covergece Theorem, whereby the sequece of itegrads is mootoe ad has a poitwise limit.

40 46 Hilbert spaces.3.4 Biliear ad quadratic forms Defiitio.5 Let (X, ) be a ormed space. A fuctio B X X C is called a biliear form (;*9!**- *" ;*";) if it is liear i its first etry ad ati-liear i its secod etry, amely, B(x,ay+bz) = ab(x,y)+bb(x,z). It is said to be bouded if there exists a costat C such that x,y X B(x,y) Cxy. The smallest such costat C is called the orm of B, i.e., B(x,y) B = sup x,y xy or B = sup B(x,y). x=y= Defiitio.6 Let (X, ) be a ormed space. A mappig Q X C is called a quadratic form (;*3&"*9 ;*";) if Q(x) = B(x,x), where B is a biliear form. It is said to be bouded if there exists a costat C such that x X Q(x) Cx 2. The smallest such C is called the orm of Q, i.e., Q = sup x Q(x), or Q = sup Q(x). x2 x= Propositio.36 Polarizatio idetity. Let (X, ) be a ormed space. For every biliear form B(x,y) with Q(x) = B(x,x), B(x,y) = {Q(x +y) Q(x y)+ı[q(x +ıy) Q(x ıy)]}. 4 Proof. Just follow the same steps as for the relatio betwee the ier-product ad the orm (which is a particular case). By the properties of B, Q(x ±y) = B(x ±y,x ±y) = Q(x)+Q(y)±B(x,y)±B(y,x). Hece Lettig y ıy, Q(x +y) Q(x y) = 2(B(x,y)+B(y,x)). Q(x +ıy) Q(x ıy) = 2ı( B(x,y)+B(y,x)).

41 .3 Liear fuctioals 47 Multiplyig the secod equatio by ı ad addig to the first we recover the desired result. Propositio.37 Let (X, ) be a complex ormed space. Let Q(x) = B(x,x) be a quadratic form over X. The Q is bouded if ad oly if B is bouded, i which case Q B 2Q. If, moreover, B(x,y) = B(y,x) for all x,y X, the Q = B. Proof. First, suppose that B is bouded. The, i.e., Q is bouded ad Q B. Q(x) = B(x,x) Bx 2, Secod, suppose that Q is bouded. Takig the polarizatio idetity, B(x,y) = {Q(x +y) Q(x y)+ı[q(x +ıy) Q(x ıy)]}, 4 ad usig the defiitio of Q: B(x,y) 4 Qx +y2 +x y 2 +x +ıy 2 +x ıy 2. Usig (twice) the parallelogram law: B(x,y) Qx 2 +y 2, hece B = sup x=y= B(x,y) 2Q. Remais the last part of the theorem. Usig agai the polarizatio idetity, ad otig that Q(y+ıx) = Q(ı(x ıy)) = Q(x ıy) we get Q(y ıx) = Q( ı(x +ıy)) = Q(x +ıy) B(x,y)+B(y,x) = {Q(x +y) Q(x y)+ı[q(x +ıy) Q(x ıy)]} 4 + {Q(x +y) Q(y x)+ı[q(y+ıx) Q(y ıx)]} 4 = {Q(x +y) Q(x y)}. 2

42 48 Hilbert spaces Thus, usig oce agai the parallelogram law, B(x,y)+B(y,x) 2 Qx +y2 +x y 2 = Qx 2 +y 2, (.3) If B(x,y) = B(y,x), the there exists a phase a = a(x,y) such that B(y,x) = e ıa(x,y) B(x,y). Note that that is B(y,e ıb x) = e ıa(x,y) ıb B(x,y) = e ıa(x,y) 2ıb B(e ıb x,y) B(e ıb y,x) = e ıa(x,y)+ıb B(x,y) = e ıa(x,y)+2ıb B(x,e ıb y), a(e ıb x,y) = a(x,y) 2b a(x,e ıb y) = a(x,y)+2b. For all x = y =, +e ıa(x,y) B(x,y) 2Q. Ad lettig x e ıb x, we get for all b: +e ı(a(x,y) 2b) B(x,y) 2Q. I particular, for b = 2 a(x,y): B(x,y) Q, i.e., B Q. Example. Cosider the real ormed space X = R 2 edowed with the Euclidea ier-product. Let Clearly B >, however B(x,y) = x y 2 x 2 y. Q(x) = B(x,x) =, hece Q =, i cotradictio to B 2Q. What goig o? The above proof was based o the assumptio that F = C. The propositio does ot hold whe F = R. Example.2 Let (H,(, )) be a ier-product space, ad let T be a bouded liear operator o H. Set B(x,y) = (Tx,y). B is a biliear form. By the Cauchy-Schwarz iequality, B(x,Tx) T = sup Tx = sup sup x= x= Tx x= It follows that B = sup x= y= sup B(x,y) sup sup (Tx,y) T. y= sup B(x,y) = T. x= y=

43 .3 Liear fuctioals 49 The followig theorem states that, i fact, every bouded biliear form o a Hilbert space ca be represeted by a bouded liear operator (this theorem is very similar to the Riesz represetatio theorem). Please ote that like for the Riesz represetatio theorem, completeess is essetial. Theorem.38 To every bouded biliear form B over a Hilbert space (H,(, )) correspods a uique bouded liear operator o H, such that x,y H B(x,y) = (Tx,y). Furthermore, B = T. Commet. The Riesz represetatio theorem asserts that This theorem asserts that H H. H H H H. Proof. We start by costructig T. Fix x H ad defie the fuctioal: F x is liear ad bouded, as F x = B(x, ). F x (y) Bxy, i.e., F x Bx. It follows from the Riesz represetatio theorem that there exists a uique z x H, such that F x = (,z x ) = B(x, ). Deote the mappig x z x by T, thus for every x H : (,Tx) = B(x, ) i.e., for every x,y H : B(x,y) = (Tx,y). Next, we show that T is liear. By defiitio of T ad by the biliearity of both B ad the ier-product: (T (a x +a 2 x 2 ),y) = B(a x +a 2 x 2,y) = a B(x,y)+a 2 B(x 2,y) = a (Tx,y)+a 2 (Tx 2,y) = (a Tx +a 2 Tx 2,y).

44 5 Hilbert spaces Sice this holds for all y H it follows that T (a x +a 2 x 2 ) = a Tx +a 2 Tx 2. T is bouded, as we have already proved that T = B. It remais to prove that T is uique. Suppose T ad S are both bouded liear operators satisfyig ( x,y H ) B(x,y) = (Tx,y) = (Sx,y). The, hece ( x,y H ) ((T S)x,y) =, ( x H ) (T S)x 2 =, ad hece T = S..3.5 The Lax-Milgram theorem Defiitio.7 Let (X, ) be a ormed space. A biliear form B o X is called coercive (39-/ %/&2() if there exists a costat d > such that x H B(x,x) dx 2. The followig theorem is a cetral pillar i the theory of partial differetial equatios: Theorem.39 Lax-Milgram, 954. Let B be a bouded ad coercive biliear form o a Hilbert space H. The, there exists a uique bouded liear operator S o H such that x,y H (x,y) = B(Sx,y). Furthermore, S exists, it is bouded, ad S d where d is the coercivity parameter. ad S = B, Commet. This theorem is amed after Peter Lax (926 ) ad Arthur Milgram (92 96). Proof. By Theorem.38 there is a uique bouded liear operator T such that x,y H B(x,y) = (Tx,y),

45 .3 Liear fuctioals 5 ad T = B. If we succeed to show that T is ivertible with bouded iverse S = T ad S d the we are doe, as ad we kow that T = B. (x,y) = (TSx,y) = B(Sx,y), Settig x = y ad usig the coercivity of B ad the Cauchy-Schwarz iequality, we get dx 2 B(x,x) = (Tx,x) Txx, i.e., x H Tx dx. (.4) It follows that kert = {}, hece T is ijective. We ext show that Image(T ) is a closed liear subspace. If Image(T ) x = Tu x H, the u u m d Tu Tu m = d x x m, which implies that (u ) is a Cauchy sequece with limit which we deote by u. By the cotiuity of T, we have x = Tu. Next, we show that T is surjective. Let z (ImageT ), the d z 2 B(z,z) = (Tz,z) =, which implies that z =, i.e., (Image(T )) = {}, ad sice Image(T ) is closed it is equal to H. We coclude that T is a bijectio. Hece, T exists. It is liear because the iverse of a liear operator is always liear. It is bouded because by Eq. (.4) T x d T (T x) d x, from which we coclude that T d. This completes the proof. Example.3 We will study a typical (relatively simple) applicatio of the Lax- Milgram theorem. Let k C[,] satisfy x [,] < c k(x) c 2, ad cosider the boudary value problem d dx du k = f u() = u() =, (.5) dx

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