Homework 2. Show that if h is a bounded sesquilinear form on the Hilbert spaces X and Y, then h has the representation

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1 omework 2 1 Let X ad Y be ilbert spaces over C The a sesquiliear form h o X Y is a mappig h : X Y C such that for all x 1, x 2, x X, y 1, y 2, y Y ad all scalars α, β C we have (a) h(x 1 + x 2, y) h(x 1, y) + h(x 2, y); (b) h(x, y 1 + y 2 ) h(x, y 1 ) + h(x, y 2 ); (c) h(αx, y) αh(x, y); (d) h(x, βy) βh(x, y) A sesquiliear form is bouded if h(x, y) C x y ad the orm of the form h is give by h(x, y) h X Y C x 0,y 0 x y h(x, y) x 1, y 1 Show that if h is a bouded sesquiliear form o the ilbert spaces X ad Y, the h has the represetatio h(x, y) Sx, y Y where S : X Y is a bouded liear operator Moreover, S is uiquely determied by h ad has orm S X Y h X Y C Solutio: The idea is to use the Riesz Represetatio to costruct the liear operator S Fix x ad cosider the fuctioal o the space Y give by L(y) h(x, y) By the properties of h beig a sesquiliear form, we have that L is liear i y, ad if h is a bouded biliear form, we have that L is a bouded liear fuctioal o Y By the Riesz Represetatio Theorem, we have the that there exists a uique elemet z Y such that L(y) y, z Y, or i terms of h that h(x, y) z, y Y While the poit z is uique, it is depedig o the poit x that was fixed We ow defie the map S : X Y by Sx z This map is clearly well-defied sice the poit z is well defied give the x Substitutig i this represetatio we fid, h(x, y) Sx, y Y as desired It remais to show that S is liear, bouded ad uique

2 To see that S is liear, we simply use the liearity of h i the first variable Let α, β C ad x 1, x 2 X, the we have S(αx 1 + βx 2 ), y Y h(αx 1 + βx 2, y) Sice this holds for all y Y we have that ad so S is liear To see that S is bouded, first ote αh(x 1, y) + βh(x 2, y) α Sx 1, y Y + β Sx 2, y Y αsx 1, y Y + βsx 2, y Y αsx 1 + βsx 2, y Y αsx 1 + βsx 2 S(αx 1 + βx 2 ) Sx S X Y Y x 0 x 0,Sx 0 x 0,y 0 x 0,y 0 Sx, Sx Y Sx Y Sx, y Y h(x, y) h X Y C I particular we have S X Y iequality, we have h X Y C, ad so S is bouded To see the other h(x, y) Sx, y Y Sx Y S X Y which gives h X Y C S X Y For uiqueess, pose that there are two liear operators S 1 ad S 2 such that h(x, y) S 1 x, y Y S 2 x, y Y This yields that S 1 x S 2 x for all x X, ad so S 1 S 2 2 Suppose that is ilbert space that cotais a orthoormal sequece {e k } which is total i Show that is separable (it has a coutable dese subset)

3 Solutio: Let A deote the set of all liear combiatio of the form γ kj e kj J where J is a fiite set, k j J, ad γ kj a kj + ib kj, where a kj, b kj Q It is clear that A is coutable We ow will show that A is the coutable dese subset i that we seek We eed to show that for every x ad ɛ > 0 there is a v A such that x v < ɛ Sice the sequece {e k } is total i, there exists a iteger such that Y spa{e 1,, e } cotais a poit y whose distace i x is less tha ɛ I particular, by 2 Parseval, we ca take y x, e k e k ad have x x, e k e k < ɛ 2 Now for each coefficiet x, e k, we ca fid a elemet γ k imagiary parts are ratioal such that ( x, e k γ k ) e k < ɛ 2 C whose real ad The defie v γ ke k A, ad ote that x v x γ k e k x x, e k e k + ( x, e k γ k ) e k < ɛ So we have that A is dese i ad A is coutable, so is separable 3 If p is a subliear fuctioal o a real vector space X, show that there exists a liear fuctioal f o X such that p( x) f(x) p(x) Solutio: Note that by ah-baach we have that for ay liear fuctioal f with f(x) p(x) we have a correspodig extesio f(x) that satisfies f(x) p(x) Now simply ote that if we evaluate this expressio at x the we have f(x) f( x) p( x) ad so we have f(x) p( x) Combiig these iequalities gives the result

4 4 If x i a ormed space X such that f(x) c for all f X of orm at most 1 Show that x c Solutio: We have the followig fact at our disposal: It is easy to show that f X,f 0 f X,f 0 f(x) f X C The result the follows easily, sice f(x) f X C f(x) f X, f X C 1 f(x) c f X, f X C 1 5 Show that for ay sphere cetered at the origi with radius r, S r (0), i a ormed space X ad ay poit y S r (0) there is a hyperplae y y such that B r (0) lies etirely i oe of the two half spaces determied by the hyperplae y Solutio: Let y S r (0), ad let y be the hyperplae cotaiig y As X is a ormed space, it is possible to write X y Ry (here we view thigs as a real vector space, the complex case is ot ay differet) The space Ry is the spa of the vector y Note that the fuctioal f : X R give by f y (h+ry) r has kerel y This algebraic reasoig ca be tured aroud From this decompositio of the space X, we ca the see that a hyperplae is give by the level sets of a liear fuctioal, ie, y fy 1 (0) ker f y Now cosider the bouded liear fuctioal f : Ry R give by f(ry) r The, we ca exted this liear fuctioal to all of X by ah-baach The we have that y f 1 (0), ad it is easy to see that the set B r (0) must lie i oe of the half-spaces It is t too much more work to show that the same coclusio i fact holds whe oe is give a o-empty covex subset C of X The iterested studet should attempt to work this out 6 Of what category is the set of all ratioal umbers Q i R? Solutio: The ratioals Q are meagre i R Note that the ratioals Q are coutable Let q 1, q 2, be a eumeratio of Q, ie, we ca write Q q k The it is clear that each set {q k } is owhere dese Thus, we have writte Q as a coutable uio of owhere dese sets i R

5 7 Show that the complemet M c of a meager subset M of a complete metric space X is o-meager Solutio: Let M be a meager subset of a complete metric space X Suppose, for a cotradictio, that M c is also meager The we ca write M c A k M where each A k ad B k is owhere dese i X But we have that X M M c A k B k C k So we have writte X as a coutable uio of owhere dese sets, ad so X would be of the first category But sice X is complete, it must be of the secod category, ad so we have a cotradictio This the implies that M c is o-meager B k 8 Let X ad Y be Baach spaces Suppose that T B(X, Y ) is such that T + Show that there is a poit x 0 X such that T x 0 Y + The poit x 0 is called a poit of resoace Solutio: Suppose that there does ot exist a poit x 0 X with the desired coclusio The for all x X we have that T x Y < + But, the the operators T satisfy the hypotheses of the Uiform Boudedess Priciple, so we have that the sequece of orms T is fiite, ie, T < owever, this clearly cotradicts the hypotheses of the operators T, ad so we must have that there exists x 0 X such that T x 0 Y + 9 If X ad Y are Baach spaces ad T B(X, Y ), show that the followig are equivalet: (a) { T } is bouded; (b) { T x } is bouded for all x X; (c) { g(t x) } is bouded for all x X ad for all g Y

6 Solutio: It is immediate that (a) implies (b) ad (b) implies (c) So it suffices to prove that (c) implies (a) But, two applicatios of the Uiform Boudedess Priciple gives this result The first applicatio of UBP implies that the sequece {T x} is bouded for all x This is problem 10 below The we ca simply apply the stadard versio of the UBP to get that { T } is bouded 10 If {x } is a sequece i a Baach space is such that for all f X the sequece {f(x )} is bouded, show that { x } is bouded Solutio: Defie a operator T : X C by T (f) f(x ), the we have that T is a liear operator o the space X By the hypothesis, we have that T (f) f(x ) is bouded for all f X The, by the Uiform Boudedess Priciple, we have there there exists some uiversal costat C such that T X C C This implies that for all f X of orm at most 1 that f(x ) T (f) C f X, f X 1 f X, f X 1 But, by the Theorem i class, we have that x X which gives that { x X } is bouded f(x ) C f X, f X 1