1. Hilbert spaces. 1.1 Definitions Vector spaces


 Leslie Clarke
 2 years ago
 Views:
Transcription
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 oetooe (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 Ierproduct 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 ierproduct 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 ierproduct space is also called a prehilbert space. Propositio.6 A ierproduct (H,(, )) space satisfies:
6 2 Hilbert spaces (x,y+z) = (x,y)+(x,z). (x,ay) = a(x,y). Proof. Obvious. Propositio.7 CauchySchwarz 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 ierproduct: 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 coliear, i.e., if ad oly if x ad y are coliear. Corollary.8 Triagle iequality. I every ier product space H, x +y x+y. Proof. Applyig the CauchySchwarz 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 CauchySchwarz Master Class which presets more proofs that you wat.
7 . Defiitios 3 Corollary.9 A ierproduct space is a ormed space with respect to the orm: x = (x,x) 2. Proof. Obvious. Thus, every ierproduct space is automatically a ormed space ad cosequetly a metric space. The (default) topology associated with a ierproduct 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 ierproduct space. Defie x,y = Re(x,y). Show that (H,, ) is a real ierproduct space. Exercise.4 Prove that if a collectio of ozero vectors {x,...,x } i a ierproduct space are mutually orthogoal the they are liearly idepedet. Exercise.5 Prove that i a ierproduct space x = iff (x,y) = for all y. Exercise.6 Let (H,(, )) be a ierproduct 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 ierproduct? ( f,g) = f (x)g(x)dx. Set V = { f V f () = }. Is (V,(, )) a ierproduct? 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 ierproduct 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 ierproduct space: (x,y) = 4 x +y2 x y 2. Defiitio.9 Let (H,(, )) be a ierproduct space. x,y H are said to be orthogoal (.*"7*) if (x,y) = ; we deote x y. Propositio.2 Orthogoal vectors i a ierproduct 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 ierproduct 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 ierproduct space (i.e., the orm ca be iduced from a ierproduct) oly for p = Hilbert spaces Defiitio. Hilbert space. A complete ierproduct space is called a Hilbert space. (Recall: a space is complete (.:) if every Cauchy sequece coverges.) Commet.4 A ierproduct 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 ierproduct spaces. Defie the space H = H H, alog with coordiatewise 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 ierproduct o H. Show that covergece i H is equivalet to compoetwise 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 ierproduct space is also called a prehilbert space. The reaso for this omeclature is the followig theorem: ay ierproduct 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 ierproduct space. The, there exists a Hilbert space (H,(, ) H ), such that: There exists a liear ijectio T G H, that preserves the ierproduct, (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 ierproduct 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 ierproduct 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 (CauchySchwarz) 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 ierproduct 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 ierproduct, (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 fiitedimesioal ormed space is complete, so that the otio of completeess is oly of iterest i ifiitedimesioal 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 CauchySchwarz 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 ratioalvalued sequeces that have a fiite umber of ozero 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. HilbertSchmidt matrices. Let M be a collectio of all ifiite matrices over C that oly have a fiite umber of ozero 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 ierproduct. (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 ierproduct 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 HilbertSchmidt matrices. Exercise.3 Prove that every fiitedimesioal 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 squareitegrable 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 ierproduct..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 oegative (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 oegative (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 ierproduct 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 ierproduct 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 ierproduct 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 ierproduct 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 ierproduct 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 fiitedimesioal ormed spaces, but it may ot be uique)=. I ifiitedimesioal 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 subsalgebra. 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 salgebra 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 CauchySchwarz 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 fiitedimesioal 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 osigular, 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, ormreducig 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 oetooe correspodece betwee closed subspaces of H ad orthogoal projectios. Theorem.28 Every liear, ormreducig, 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 ierproduct 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 fiitedimesioal ormed spaces are bouded. This otio is therefore oly relevat to ifiitedimesioal 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 otcotiuous), 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 scalarvalued 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 HahBaach 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 ifiitedimesioal ormed space. The, there exists a ubouded liear fuctioal o X. Proof. Let {x a a A} be a Hamel basis. Sice X is ifiitedimesioal 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 ierproduct 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 coliear. 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 ierproduct ad the correspodig orm over H is (T,S) = (y T,y S ), T = y T coicides with the previouslydefied orm" (we ever showed it was ideed a orm) 4. The followig theorem (the RadoNikodym theorem restricted to fiite measure spaces) is a applicatio of the Riesz represetatio theorem: Theorem.35 RadoNikodym. 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 oegative 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 ierproducts ad represetatio theorems i order to show that the operator orm is ideed a orm.
39 .3 Liear fuctioals 45. h la.e: Because the measures are fiite, idicator fuctios are itegrable. The, settig g = I B, (B) = B hdl, which implies that h la.e. 2. h < la.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 < la.e. Sice h <, we may represet h as where f is la.e. oegative. 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 oegative 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 oegative 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 atiliear 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 ierproduct 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 ierproduct. 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 ierproduct space, ad let T be a bouded liear operator o H. Set B(x,y) = (Tx,y). B is a biliear form. By the CauchySchwarz 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 ierproduct: (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 LaxMilgram 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 LaxMilgram, 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 CauchySchwarz 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
SAMPLE QUESTIONS FOR FINAL EXAM. (1) (2) (3) (4) Find the following using the definition of the Riemann integral: (2x + 1)dx
SAMPLE QUESTIONS FOR FINAL EXAM REAL ANALYSIS I FALL 006 3 4 Fid the followig usig the defiitio of the Riema itegral: a 0 x + dx 3 Cosider the partitio P x 0 3, x 3 +, x 3 +,......, x 3 3 + 3 of the iterval
More informationConvexity, Inequalities, and Norms
Covexity, Iequalities, ad Norms Covex Fuctios You are probably familiar with the otio of cocavity of fuctios. Give a twicedifferetiable fuctio ϕ: R R, We say that ϕ is covex (or cocave up) if ϕ (x) 0 for
More informationMetric, Normed, and Topological Spaces
Chapter 13 Metric, Normed, ad Topological Spaces A metric space is a set X that has a otio of the distace d(x, y) betwee every pair of poits x, y X. A fudametal example is R with the absolutevalue metric
More informationChapter 5: Inner Product Spaces
Chapter 5: Ier Product Spaces Chapter 5: Ier Product Spaces SECION A Itroductio to Ier Product Spaces By the ed of this sectio you will be able to uderstad what is meat by a ier product space give examples
More informationLecture 5: Span, linear independence, bases, and dimension
Lecture 5: Spa, liear idepedece, bases, ad dimesio Travis Schedler Thurs, Sep 23, 2010 (versio: 9/21 9:55 PM) 1 Motivatio Motivatio To uderstad what it meas that R has dimesio oe, R 2 dimesio 2, etc.;
More informationSequences and Series
CHAPTER 9 Sequeces ad Series 9.. Covergece: Defiitio ad Examples Sequeces The purpose of this chapter is to itroduce a particular way of geeratig algorithms for fidig the values of fuctios defied by their
More informationOur aim is to show that under reasonable assumptions a given 2πperiodic function f can be represented as convergent series
8 Fourier Series Our aim is to show that uder reasoable assumptios a give periodic fuctio f ca be represeted as coverget series f(x) = a + (a cos x + b si x). (8.) By defiitio, the covergece of the series
More information3. Covariance and Correlation
Virtual Laboratories > 3. Expected Value > 1 2 3 4 5 6 3. Covariace ad Correlatio Recall that by takig the expected value of various trasformatios of a radom variable, we ca measure may iterestig characteristics
More informationx(x 1)(x 2)... (x k + 1) = [x] k n+m 1
1 Coutig mappigs For every real x ad positive iteger k, let [x] k deote the fallig factorial ad x(x 1)(x 2)... (x k + 1) ( ) x = [x] k k k!, ( ) k = 1. 0 I the sequel, X = {x 1,..., x m }, Y = {y 1,...,
More informationLecture 13. Lecturer: Jonathan Kelner Scribe: Jonathan Pines (2009)
18.409 A Algorithmist s Toolkit October 27, 2009 Lecture 13 Lecturer: Joatha Keler Scribe: Joatha Pies (2009) 1 Outlie Last time, we proved the BruMikowski iequality for boxes. Today we ll go over the
More information1 Set Theory and Functions
Set Theory ad Fuctios. Basic De itios ad Notatio A set A is a collectio of objects of ay kid. We write a A to idicate that a is a elemet of A: We express this as a is cotaied i A. We write A B if every
More informationIn nite Sequences. Dr. Philippe B. Laval Kennesaw State University. October 9, 2008
I ite Sequeces Dr. Philippe B. Laval Keesaw State Uiversity October 9, 2008 Abstract This had out is a itroductio to i ite sequeces. mai de itios ad presets some elemetary results. It gives the I ite Sequeces
More informationThe Field Q of Rational Numbers
Chapter 3 The Field Q of Ratioal Numbers I this chapter we are goig to costruct the ratioal umber from the itegers. Historically, the positive ratioal umbers came first: the Babyloias, Egyptias ad Grees
More informationClass Meeting # 16: The Fourier Transform on R n
MATH 18.152 COUSE NOTES  CLASS MEETING # 16 18.152 Itroductio to PDEs, Fall 2011 Professor: Jared Speck Class Meetig # 16: The Fourier Trasform o 1. Itroductio to the Fourier Trasform Earlier i the course,
More informationLinear Algebra II. 4 Determinants. Notes 4 1st November Definition of determinant
MTH6140 Liear Algebra II Notes 4 1st November 2010 4 Determiats The determiat is a fuctio defied o square matrices; its value is a scalar. It has some very importat properties: perhaps most importat is
More informationProperties of MLE: consistency, asymptotic normality. Fisher information.
Lecture 3 Properties of MLE: cosistecy, asymptotic ormality. Fisher iformatio. I this sectio we will try to uderstad why MLEs are good. Let us recall two facts from probability that we be used ofte throughout
More informationSequences II. Chapter 3. 3.1 Convergent Sequences
Chapter 3 Sequeces II 3. Coverget Sequeces Plot a graph of the sequece a ) = 2, 3 2, 4 3, 5 + 4,...,,... To what limit do you thik this sequece teds? What ca you say about the sequece a )? For ǫ = 0.,
More information1. MATHEMATICAL INDUCTION
1. MATHEMATICAL INDUCTION EXAMPLE 1: Prove that for ay iteger 1. Proof: 1 + 2 + 3 +... + ( + 1 2 (1.1 STEP 1: For 1 (1.1 is true, sice 1 1(1 + 1. 2 STEP 2: Suppose (1.1 is true for some k 1, that is 1
More informationAnalysis Notes (only a draft, and the first one!)
Aalysis Notes (oly a draft, ad the first oe!) Ali Nesi Mathematics Departmet Istabul Bilgi Uiversity Kuştepe Şişli Istabul Turkey aesi@bilgi.edu.tr Jue 22, 2004 2 Cotets 1 Prelimiaries 9 1.1 Biary Operatio...........................
More informationFunctional Analysis Basics c 2007 Ulrich G. Schuster SVN Revision: 2109, December 11, 2007 Vector Space Concepts
Fuctioal Aalysis Basics c 2007 Ulrich G. Schuster SVN Revisio: 2109, December 11, 2007 Vector Space Cocepts icomplete liear spaces ormed spaces ier product spaces metric spaces Basis ad Dimesio [2, 1].
More informationLecture 4: Cauchy sequences, BolzanoWeierstrass, and the Squeeze theorem
Lecture 4: Cauchy sequeces, BolzaoWeierstrass, ad the Squeeze theorem The purpose of this lecture is more modest tha the previous oes. It is to state certai coditios uder which we are guarateed that limits
More informationTheorems About Power Series
Physics 6A Witer 20 Theorems About Power Series Cosider a power series, f(x) = a x, () where the a are real coefficiets ad x is a real variable. There exists a real oegative umber R, called the radius
More informationModule 4: Mathematical Induction
Module 4: Mathematical Iductio Theme 1: Priciple of Mathematical Iductio Mathematical iductio is used to prove statemets about atural umbers. As studets may remember, we ca write such a statemet as a predicate
More information7. Sample Covariance and Correlation
1 of 8 7/16/2009 6:06 AM Virtual Laboratories > 6. Radom Samples > 1 2 3 4 5 6 7 7. Sample Covariace ad Correlatio The Bivariate Model Suppose agai that we have a basic radom experimet, ad that X ad Y
More informationTrigonometric Form of a Complex Number. The Complex Plane. axis. ( 2, 1) or 2 i FIGURE 6.44. The absolute value of the complex number z a bi is
0_0605.qxd /5/05 0:45 AM Page 470 470 Chapter 6 Additioal Topics i Trigoometry 6.5 Trigoometric Form of a Complex Number What you should lear Plot complex umbers i the complex plae ad fid absolute values
More informationAlgebra Vocabulary List (Definitions for Middle School Teachers)
Algebra Vocabulary List (Defiitios for Middle School Teachers) A Absolute Value Fuctio The absolute value of a real umber x, x is xifx 0 x = xifx < 0 http://www.math.tamu.edu/~stecher/171/f02/absolutevaluefuctio.pdf
More information2 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS
2 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS 1. The group rig k[g] The mai idea is that represetatios of a group G over a field k are the same as modules over the group rig k[g]. First I
More informationDistributions of Order Statistics
Chapter 2 Distributios of Order Statistics We give some importat formulae for distributios of order statistics. For example, where F k: (x)=p{x k, x} = I F(x) (k, k + 1), I x (a,b)= 1 x t a 1 (1 t) b 1
More informationOverview of some probability distributions.
Lecture Overview of some probability distributios. I this lecture we will review several commo distributios that will be used ofte throughtout the class. Each distributio is usually described by its probability
More informationTHE REGRESSION MODEL IN MATRIX FORM. For simple linear regression, meaning one predictor, the model is. for i = 1, 2, 3,, n
We will cosider the liear regressio model i matrix form. For simple liear regressio, meaig oe predictor, the model is i = + x i + ε i for i =,,,, This model icludes the assumptio that the ε i s are a sample
More informationAsymptotic Growth of Functions
CMPS Itroductio to Aalysis of Algorithms Fall 3 Asymptotic Growth of Fuctios We itroduce several types of asymptotic otatio which are used to compare the performace ad efficiecy of algorithms As we ll
More informationChapter 7 Methods of Finding Estimators
Chapter 7 for BST 695: Special Topics i Statistical Theory. Kui Zhag, 011 Chapter 7 Methods of Fidig Estimators Sectio 7.1 Itroductio Defiitio 7.1.1 A poit estimator is ay fuctio W( X) W( X1, X,, X ) of
More informationInfinite Sequences and Series
CHAPTER 4 Ifiite Sequeces ad Series 4.1. Sequeces A sequece is a ifiite ordered list of umbers, for example the sequece of odd positive itegers: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29...
More informationAn example of nonquenched convergence in the conditional central limit theorem for partial sums of a linear process
A example of oqueched covergece i the coditioal cetral limit theorem for partial sums of a liear process Dalibor Volý ad Michael Woodroofe Abstract A causal liear processes X,X 0,X is costructed for which
More informationFIBONACCI NUMBERS: AN APPLICATION OF LINEAR ALGEBRA. 1. Powers of a matrix
FIBONACCI NUMBERS: AN APPLICATION OF LINEAR ALGEBRA. Powers of a matrix We begi with a propositio which illustrates the usefuless of the diagoalizatio. Recall that a square matrix A is diogaalizable if
More informationDepartment of Computer Science, University of Otago
Departmet of Computer Sciece, Uiversity of Otago Techical Report OUCS200609 Permutatios Cotaiig May Patters Authors: M.H. Albert Departmet of Computer Sciece, Uiversity of Otago Micah Colema, Rya Fly
More informationDiscrete Mathematics and Probability Theory Spring 2014 Anant Sahai Note 13
EECS 70 Discrete Mathematics ad Probability Theory Sprig 2014 Aat Sahai Note 13 Itroductio At this poit, we have see eough examples that it is worth just takig stock of our model of probability ad may
More informationThe Euler Totient, the Möbius and the Divisor Functions
The Euler Totiet, the Möbius ad the Divisor Fuctios Rosica Dieva July 29, 2005 Mout Holyoke College South Hadley, MA 01075 1 Ackowledgemets This work was supported by the Mout Holyoke College fellowship
More informationDivide and Conquer. Maximum/minimum. Integer Multiplication. CS125 Lecture 4 Fall 2015
CS125 Lecture 4 Fall 2015 Divide ad Coquer We have see oe geeral paradigm for fidig algorithms: the greedy approach. We ow cosider aother geeral paradigm, kow as divide ad coquer. We have already see a
More information12. FRAMES IN HILBERT SPACES
1. FRAMES IN HILBERT SPACES 81 1. FRAMES IN HILBERT SPACES Frames were itroduced by Duffi ad Schaeffer i the cotext of oharmoic Fourier series [DS5]. They were iteded as a alterative to orthoormal or Riesz
More information4.1 Sigma Notation and Riemann Sums
0 the itegral. Sigma Notatio ad Riema Sums Oe strategy for calculatig the area of a regio is to cut the regio ito simple shapes, calculate the area of each simple shape, ad the add these smaller areas
More information4.3. The Integral and Comparison Tests
4.3. THE INTEGRAL AND COMPARISON TESTS 9 4.3. The Itegral ad Compariso Tests 4.3.. The Itegral Test. Suppose f is a cotiuous, positive, decreasig fuctio o [, ), ad let a = f(). The the covergece or divergece
More informationPART TWO. Measure, Integration, and Differentiation
PART TWO Measure, Itegratio, ad Differetiatio Émile FélixÉdouardJusti Borel (1871 1956 Émile Borel was bor at SaitAffrique, Frace, o Jauary 7, 1871, the third child of Hooré Borel, a Protestat miister,
More information1 Introduction to reducing variance in Monte Carlo simulations
Copyright c 007 by Karl Sigma 1 Itroductio to reducig variace i Mote Carlo simulatios 11 Review of cofidece itervals for estimatig a mea I statistics, we estimate a uow mea µ = E(X) of a distributio by
More informationCS103X: Discrete Structures Homework 4 Solutions
CS103X: Discrete Structures Homewor 4 Solutios Due February 22, 2008 Exercise 1 10 poits. Silico Valley questios: a How may possible sixfigure salaries i whole dollar amouts are there that cotai at least
More informationWHEN IS THE (CO)SINE OF A RATIONAL ANGLE EQUAL TO A RATIONAL NUMBER?
WHEN IS THE (CO)SINE OF A RATIONAL ANGLE EQUAL TO A RATIONAL NUMBER? JÖRG JAHNEL 1. My Motivatio Some Sort of a Itroductio Last term I tought Topological Groups at the Göttige Georg August Uiversity. This
More informationNotes on exponential generating functions and structures.
Notes o expoetial geeratig fuctios ad structures. 1. The cocept of a structure. Cosider the followig coutig problems: (1) to fid for each the umber of partitios of a elemet set, (2) to fid for each the
More informationHere are a couple of warnings to my students who may be here to get a copy of what happened on a day that you missed.
This documet was writte ad copyrighted by Paul Dawkis. Use of this documet ad its olie versio is govered by the Terms ad Coditios of Use located at http://tutorial.math.lamar.edu/terms.asp. The olie versio
More informationExample 2 Find the square root of 0. The only square root of 0 is 0 (since 0 is not positive or negative, so those choices don t exist here).
BEGINNING ALGEBRA Roots ad Radicals (revised summer, 00 Olso) Packet to Supplemet the Curret Textbook  Part Review of Square Roots & Irratioals (This portio ca be ay time before Part ad should mostly
More informationTHE ABRACADABRA PROBLEM
THE ABRACADABRA PROBLEM FRANCESCO CARAVENNA Abstract. We preset a detailed solutio of Exercise E0.6 i [Wil9]: i a radom sequece of letters, draw idepedetly ad uiformly from the Eglish alphabet, the expected
More informationPermutations, the Parity Theorem, and Determinants
1 Permutatios, the Parity Theorem, ad Determiats Joh A. Guber Departmet of Electrical ad Computer Egieerig Uiversity of Wiscosi Madiso Cotets 1 What is a Permutatio 1 2 Cycles 2 2.1 Traspositios 4 3 Orbits
More informationGregory Carey, 1998 Linear Transformations & Composites  1. Linear Transformations and Linear Composites
Gregory Carey, 1998 Liear Trasformatios & Composites  1 Liear Trasformatios ad Liear Composites I Liear Trasformatios of Variables Meas ad Stadard Deviatios of Liear Trasformatios A liear trasformatio
More informationThe second difference is the sequence of differences of the first difference sequence, 2
Differece Equatios I differetial equatios, you look for a fuctio that satisfies ad equatio ivolvig derivatives. I differece equatios, istead of a fuctio of a cotiuous variable (such as time), we look for
More informationSection 11.3: The Integral Test
Sectio.3: The Itegral Test Most of the series we have looked at have either diverged or have coverged ad we have bee able to fid what they coverge to. I geeral however, the problem is much more difficult
More informationCooleyTukey. Tukey FFT Algorithms. FFT Algorithms. Cooley
Cooley CooleyTuey Tuey FFT Algorithms FFT Algorithms Cosider a legth sequece x[ with a poit DFT X[ where Represet the idices ad as +, +, Cooley CooleyTuey Tuey FFT Algorithms FFT Algorithms Usig these
More informationNOTES ON INEQUALITIES FELIX LAZEBNIK
NOTES ON INEQUALITIES FELIX LAZEBNIK Order ad iequalities are fudametal otios of moder mathematics. Calculus ad Aalysis deped heavily o them, ad properties of iequalities provide the mai tool for developig
More informationMARTINGALES AND A BASIC APPLICATION
MARTINGALES AND A BASIC APPLICATION TURNER SMITH Abstract. This paper will develop the measuretheoretic approach to probability i order to preset the defiitio of martigales. From there we will apply this
More information{{1}, {2, 4}, {3}} {{1, 3, 4}, {2}} {{1}, {2}, {3, 4}} 5.4 Stirling Numbers
. Stirlig Numbers Whe coutig various types of fuctios from., we quicly discovered that eumeratig the umber of oto fuctios was a difficult problem. For a domai of five elemets ad a rage of four elemets,
More informationINFINITE SERIES KEITH CONRAD
INFINITE SERIES KEITH CONRAD. Itroductio The two basic cocepts of calculus, differetiatio ad itegratio, are defied i terms of limits (Newto quotiets ad Riema sums). I additio to these is a third fudametal
More informationI. Chisquared Distributions
1 M 358K Supplemet to Chapter 23: CHISQUARED DISTRIBUTIONS, TDISTRIBUTIONS, AND DEGREES OF FREEDOM To uderstad tdistributios, we first eed to look at aother family of distributios, the chisquared distributios.
More information4 n. n 1. You shold think of the Ratio Test as a generalization of the Geometric Series Test. For example, if a n ar n is a geometric sequence then
SECTION 2.6 THE RATIO TEST 79 2.6. THE RATIO TEST We ow kow how to hadle series which we ca itegrate (the Itegral Test), ad series which are similar to geometric or pseries (the Compariso Test), but of
More informationJoint Probability Distributions and Random Samples
STAT5 Sprig 204 Lecture Notes Chapter 5 February, 204 Joit Probability Distributios ad Radom Samples 5. Joitly Distributed Radom Variables Chapter Overview Joitly distributed rv Joit mass fuctio, margial
More information3 Basic Definitions of Probability Theory
3 Basic Defiitios of Probability Theory 3defprob.tex: Feb 10, 2003 Classical probability Frequecy probability axiomatic probability Historical developemet: Classical Frequecy Axiomatic The Axiomatic defiitio
More informationChapter 6: Variance, the law of large numbers and the MonteCarlo method
Chapter 6: Variace, the law of large umbers ad the MoteCarlo method Expected value, variace, ad Chebyshev iequality. If X is a radom variable recall that the expected value of X, E[X] is the average value
More informationOutput Analysis (2, Chapters 10 &11 Law)
B. Maddah ENMG 6 Simulatio 05/0/07 Output Aalysis (, Chapters 10 &11 Law) Comparig alterative system cofiguratio Sice the output of a simulatio is radom, the comparig differet systems via simulatio should
More informationSoving Recurrence Relations
Sovig Recurrece Relatios Part 1. Homogeeous liear 2d degree relatios with costat coefficiets. Cosider the recurrece relatio ( ) T () + at ( 1) + bt ( 2) = 0 This is called a homogeeous liear 2d degree
More informationNOTES ON PROBABILITY Greg Lawler Last Updated: March 21, 2016
NOTES ON PROBBILITY Greg Lawler Last Updated: March 21, 2016 Overview This is a itroductio to the mathematical foudatios of probability theory. It is iteded as a supplemet or followup to a graduate course
More information1.3 Binomial Coefficients
18 CHAPTER 1. COUNTING 1. Biomial Coefficiets I this sectio, we will explore various properties of biomial coefficiets. Pascal s Triagle Table 1 cotais the values of the biomial coefficiets ( ) for 0to
More informationTHIN SEQUENCES AND THE GRAM MATRIX PAMELA GORKIN, JOHN E. MCCARTHY, SANDRA POTT, AND BRETT D. WICK
THIN SEQUENCES AND THE GRAM MATRIX PAMELA GORKIN, JOHN E MCCARTHY, SANDRA POTT, AND BRETT D WICK Abstract We provide a ew proof of Volberg s Theorem characterizig thi iterpolatig sequeces as those for
More informationNUMBERS COMMON TO TWO POLYGONAL SEQUENCES
NUMBERS COMMON TO TWO POLYGONAL SEQUENCES DIANNE SMITH LUCAS Chia Lake, Califoria a iteger, The polygoal sequece (or sequeces of polygoal umbers) of order r (where r is r > 3) may be defied recursively
More information1.3. VERTEX DEGREES & COUNTING
35 Chapter 1: Fudametal Cocepts Sectio 1.3: Vertex Degrees ad Coutig 36 its eighbor o P. Note that P has at least three vertices. If G x v is coected, let y = v. Otherwise, a compoet cut off from P x v
More information. P. 4.3 Basic feasible solutions and vertices of polyhedra. x 1. x 2
4. Basic feasible solutios ad vertices of polyhedra Due to the fudametal theorem of Liear Programmig, to solve ay LP it suffices to cosider the vertices (fiitely may) of the polyhedro P of the feasible
More informationGibbs Distribution in Quantum Statistics
Gibbs Distributio i Quatum Statistics Quatum Mechaics is much more complicated tha the Classical oe. To fully characterize a state of oe particle i Classical Mechaics we just eed to specify its radius
More informationTAYLOR SERIES, POWER SERIES
TAYLOR SERIES, POWER SERIES The followig represets a (icomplete) collectio of thigs that we covered o the subject of Taylor series ad power series. Warig. Be prepared to prove ay of these thigs durig the
More informationSEQUENCES AND SERIES
Chapter 9 SEQUENCES AND SERIES Natural umbers are the product of huma spirit. DEDEKIND 9.1 Itroductio I mathematics, the word, sequece is used i much the same way as it is i ordiary Eglish. Whe we say
More information1 Correlation and Regression Analysis
1 Correlatio ad Regressio Aalysis I this sectio we will be ivestigatig the relatioship betwee two cotiuous variable, such as height ad weight, the cocetratio of a ijected drug ad heart rate, or the cosumptio
More informationON THE DENSE TRAJECTORY OF LASOTA EQUATION
UNIVERSITATIS IAGELLONICAE ACTA MATHEMATICA, FASCICULUS XLIII 2005 ON THE DENSE TRAJECTORY OF LASOTA EQUATION by Atoi Leo Dawidowicz ad Najemedi Haribash Abstract. I preseted paper the dese trajectory
More informationWeek 3 Conditional probabilities, Bayes formula, WEEK 3 page 1 Expected value of a random variable
Week 3 Coditioal probabilities, Bayes formula, WEEK 3 page 1 Expected value of a radom variable We recall our discussio of 5 card poker hads. Example 13 : a) What is the probability of evet A that a 5
More informationSection 1.6: Proof by Mathematical Induction
Sectio.6 Proof by Iductio Sectio.6: Proof by Mathematical Iductio Purpose of Sectio: To itroduce the Priciple of Mathematical Iductio, both weak ad the strog versios, ad show how certai types of theorems
More informationCS103A Handout 23 Winter 2002 February 22, 2002 Solving Recurrence Relations
CS3A Hadout 3 Witer 00 February, 00 Solvig Recurrece Relatios Itroductio A wide variety of recurrece problems occur i models. Some of these recurrece relatios ca be solved usig iteratio or some other ad
More informationDEFINITION OF INVERSE MATRIX
Lecture. Iverse matrix. To be read to the music of Back To You by Brya dams DEFINITION OF INVERSE TRIX Defiitio. Let is a square matrix. Some matrix B if it exists) is said to be iverse to if B B I where
More information8.1 Arithmetic Sequences
MCR3U Uit 8: Sequeces & Series Page 1 of 1 8.1 Arithmetic Sequeces Defiitio: A sequece is a comma separated list of ordered terms that follow a patter. Examples: 1, 2, 3, 4, 5 : a sequece of the first
More information0.7 0.6 0.2 0 0 96 96.5 97 97.5 98 98.5 99 99.5 100 100.5 96.5 97 97.5 98 98.5 99 99.5 100 100.5
Sectio 13 KolmogorovSmirov test. Suppose that we have a i.i.d. sample X 1,..., X with some ukow distributio P ad we would like to test the hypothesis that P is equal to a particular distributio P 0, i.e.
More informationChapter 4 Multivariate distributions
Chapter 4 Multivariate distributios k Multivariate Distributios All the results derived for the bivariate case ca be geeralized to RV. The joit CDF of,,, k will have the form: P(x, x,, x k ) whe the RVs
More informationTHE ARITHMETIC OF INTEGERS.  multiplication, exponentiation, division, addition, and subtraction
THE ARITHMETIC OF INTEGERS  multiplicatio, expoetiatio, divisio, additio, ad subtractio What to do ad what ot to do. THE INTEGERS Recall that a iteger is oe of the whole umbers, which may be either positive,
More informationChapter 7  Sampling Distributions. 1 Introduction. What is statistics? It consist of three major areas:
Chapter 7  Samplig Distributios 1 Itroductio What is statistics? It cosist of three major areas: Data Collectio: samplig plas ad experimetal desigs Descriptive Statistics: umerical ad graphical summaries
More informationLimits, Continuity and derivatives (Stewart Ch. 2) say: the limit of f(x) equals L
Limits, Cotiuity ad derivatives (Stewart Ch. 2) f(x) = L say: the it of f(x) equals L as x approaches a The values of f(x) ca be as close to L as we like by takig x sufficietly close to a, but x a. If
More informationSECTION 1.5 : SUMMATION NOTATION + WORK WITH SEQUENCES
SECTION 1.5 : SUMMATION NOTATION + WORK WITH SEQUENCES Read Sectio 1.5 (pages 5 9) Overview I Sectio 1.5 we lear to work with summatio otatio ad formulas. We will also itroduce a brief overview of sequeces,
More informationNormal Distribution.
Normal Distributio www.icrf.l Normal distributio I probability theory, the ormal or Gaussia distributio, is a cotiuous probability distributio that is ofte used as a first approimatio to describe realvalued
More informationConvex Bodies of Minimal Volume, Surface Area and Mean Width with Respect to Thin Shells
Caad. J. Math. Vol. 60 (1), 2008 pp. 3 32 Covex Bodies of Miimal Volume, Surface Area ad Mea Width with Respect to Thi Shells Károly Böröczky, Károly J. Böröczky, Carste Schütt, ad Gergely Witsche Abstract.
More informationOverview on SBox Design Principles
Overview o SBox Desig Priciples Debdeep Mukhopadhyay Assistat Professor Departmet of Computer Sciece ad Egieerig Idia Istitute of Techology Kharagpur INDIA 721302 What is a SBox? SBoxes are Boolea
More informationMaximum Likelihood Estimators.
Lecture 2 Maximum Likelihood Estimators. Matlab example. As a motivatio, let us look at oe Matlab example. Let us geerate a radom sample of size 00 from beta distributio Beta(5, 2). We will lear the defiitio
More informationTaking DCOP to the Real World: Efficient Complete Solutions for Distributed MultiEvent Scheduling
Taig DCOP to the Real World: Efficiet Complete Solutios for Distributed MultiEvet Schedulig Rajiv T. Maheswara, Milid Tambe, Emma Bowrig, Joatha P. Pearce, ad Pradeep araatham Uiversity of Souther Califoria
More informationIrreducible polynomials with consecutive zero coefficients
Irreducible polyomials with cosecutive zero coefficiets Theodoulos Garefalakis Departmet of Mathematics, Uiversity of Crete, 71409 Heraklio, Greece Abstract Let q be a prime power. We cosider the problem
More informationCME 302: NUMERICAL LINEAR ALGEBRA FALL 2005/06 LECTURE 8
CME 30: NUMERICAL LINEAR ALGEBRA FALL 005/06 LECTURE 8 GENE H GOLUB 1 Positive Defiite Matrices A matrix A is positive defiite if x Ax > 0 for all ozero x A positive defiite matrix has real ad positive
More informationAQA STATISTICS 1 REVISION NOTES
AQA STATISTICS 1 REVISION NOTES AVERAGES AND MEASURES OF SPREAD www.mathsbox.org.uk Mode : the most commo or most popular data value the oly average that ca be used for qualitative data ot suitable if
More information1 The Gaussian channel
ECE 77 Lecture 0 The Gaussia chael Objective: I this lecture we will lear about commuicatio over a chael of practical iterest, i which the trasmitted sigal is subjected to additive white Gaussia oise.
More information3. Greatest Common Divisor  Least Common Multiple
3 Greatest Commo Divisor  Least Commo Multiple Defiitio 31: The greatest commo divisor of two atural umbers a ad b is the largest atural umber c which divides both a ad b We deote the greatest commo gcd
More informationLecture 3. denote the orthogonal complement of S k. Then. 1 x S k. n. 2 x T Ax = ( ) λ x. with x = 1, we have. i = λ k x 2 = λ k.
18.409 A Algorithmist s Toolkit September 17, 009 Lecture 3 Lecturer: Joatha Keler Scribe: Adre Wibisoo 1 Outlie Today s lecture covers three mai parts: CouratFischer formula ad Rayleigh quotiets The
More informationTHE problem of fitting a circle to a collection of points
IEEE TRANACTION ON INTRUMENTATION AND MEAUREMENT, VOL. XX, NO. Y, MONTH 000 A Few Methods for Fittig Circles to Data Dale Umbach, Kerry N. Joes Abstract Five methods are discussed to fit circles to data.
More information