Completion. is dense in H. If V is complete, then U(V) = H.

Transcription

1 Competion Theorem 1 (Competion) If ( V V ) is any inner product space then there exists a Hibert space ( H H ) and a map U : V H such that (i) U is 1 1 (ii) U is inear (iii) UxUy H xy V for a xy V (iv) U(V) Ux x V } is dense in H If V is compete then U(V) H Remark 2 (a) H is caed the competion of V (b) U gives a one-to one correspondance between eements of V and eements of U(V) So we can think of U as giving the new name Ux to each x V and we can think of U(V) as being the same as V but with the names of the eements changed Thus we can think of V as being U(V) H Using this point of view the above theorem says that any inner product space can be extended to a compete inner product space Ie can have its hoes fied in Motivation The hard part of the proof is to make a guess as to what H shoud be That s what we do now A good strategy is to work backwards Suppose that somehow we have found a suitabe H with V H If we can find a way to describe each eement of H purey in terms of eements of V then we can turn around and take that as the definition of H Because V is dense in H each eement of H may be written as the imit of a sequence in V and each such sequence is Cauchy Thus specifying an eement of H is equivaent to specifying a Cauchy sequence in V But there is not a one to one correspondance between eements of H and Cauchy sequences in V because many different Cauchy sequences in V converge to the same eement of H For exampe if x n }n IN V converges to x H then x n x n 1 }n IN V and e 1/n x n + ( 1)n n x } 2 V aso converge to x To get a one to one correspondance we can n IN identify each x H with the set of a Cauchy sequences in V that converge to x Outine of Proof First define } V x n } n IN is a Cauchy sequence in V That is V is the set of a Cauchy sequences in V Next we define two Cauchy sequences x n } n IN and y n } n IN in V to be equivaent written x n } n IN y n } n IN if and ony if im x n y n V 0 This definition is rigged so that any two convergent sequences have the same imit if and ony if they are equivaent Next if x n } n IN V we define the equivaence cass of x n } n IN to be the set } y n } n IN V x n } n IN of a Cauchy sequences that are equivaent to x n } n IN We sha shorty prove c Joe Fedman 2011 A rights reserved September Competion 1

2 Lemma 3 is an equivaence reation (1) In particuar if x n } n IN y n } n IN V then either x n } n IN or If you think of a Cauchy sequence as one person and an equivaence cass of Cauchy sequences as a famiy of reated peope then the above Lemma says that the whoe word is divided into a coection of nonoverapping famiies Next we define xn } H } n IN V as the set of a famiies and prove Lemma 4 If x n } n IN y n } n IN V then im x ny n V exists Lemma 5 Define for each x n } n IN H H and α C + xn +y n } n IN α x n } n IN αxn } n IN H im x ny n V Each of these operations is we defined For exampe if x n } n IN x n } n IN (so that x n } n IN ) and y n } n IN then im x ny n V im x n y n V Finay we define U : V H by Ux xxx } The concusions of Theorem 1 are now proven as a series of Lemmata Lemma 6 H with the operations of Lemma 5 is an inner product space Lemma 7 H is compete Lemma 8 U is inear and obeys UxUy H xy V for a xy V Lemma 9 U is one to one Lemma 10 U(V) is dense in H Lemma 11 If V is compete then U(V) H (1) A binary reation on a set S is an equivaence reation if and ony if for a stu S (1) s s (refexivity) (2) if s t then t s (symmetry) and (3) if s t and t u then s u (transitivity) c Joe Fedman 2011 A rights reserved September Competion 2

3 So now we just have to prove a of the Lemmata Lemma 3 is an equivaence reation In particuar if x n } n IN y n } n IN V then either x n } n IN or Proof: The three equivaence reation axioms are triviay verified so we ony prove the ast caim Suppose that x n } n IN but As and y n } n IN intersect there is a Cauchy sequence un } n IN that is in both x n } n IN and Consequenty un } n IN is equivaent to both x n } n IN and y n } n IN so that im u n x n V 0 and im u n y n V 0 Ontheotherhandas x n } n IN and aredifferentthereisacauchysequencevn } n IN that is in one of them (say x n } n IN ) and not in the other (say ) Since vn } n IN is in x n } n IN im v n x n V 0 By the triange inequaity v n y n V v n x n V + x n u n V + u n y n V But this forces im v n y n V 0 which contradicts the assumption that v n } n IN is not in Lemma 4 If x n } n IN y n } n IN V then im x ny n V exists Proof: We first observe that x n y n V x m y m V x n x m y n V + x m y n y m V x n x m V y n V + x m V y n y m V (1) Since x n } n IN and y n } n IN are both Cauchy both x m V } m IN and y n V } n IN are bounded and both x n x m V and y n y m V converge to zero as mn So the same is true for xn y n V x m y m V So xn y n V } n IN is a Cauchy sequence of compex numbers Since C is compete this Cauchy sequence converges Lemma 5 Define for each x n } n IN H H and α C + xn +y n } n IN Each of these operations is we defined α x n } n IN αxn } n IN H im x ny n V c Joe Fedman 2011 A rights reserved September Competion 3

4 Proof: Let x n } n IN x n } n IN and y n } n IN y n } n IN Repace in (1) x m and y m by x n and y n respectivey This gives xn y n V x n y n V xn x n V y n V + x n V y n y n V Since x n } n IN x n} n IN we have that im x n x n V 0 Since y n } n IN y n} n IN we have that im y n y n V 0 Since } y n V and x n IN n } V are bounded n IN im xn y n V x n y n V 0 So the inner product is we defined Simiary (x n +y n ) (x n +y n ) V x n x n V + y n y n V αx n αx n V α x n x n V 0 0 so that addition and scaar mutipication are we defind Lemma 6 H with the operations of Lemma 5 is an inner product space Proof: There are nine vector space axioms and three inner product axioms to be checked The proofs are essentiay trivia and very simiar I just verify the first vector space axiom and first haf of the first inner product axiom Let x n } n IN H and α C + xn +y n } n IN yn +x n } n IN y n } n IN + α H x n } n IN αyn } n IN im x H nαy n V α im x ny n V α x n } n IN H Lemma 7 H is compete Proof: Let X (n) H} n IN be a Cauchy sequence We must prove that it has a imit X H as n Each X (n) is an equivaence cass of Cauchy sequences in V Say X (n) x (n) } IN We sha guess X x n }n IN by choosing each xn to be a x (n) with chosen arger and arger as n increases Here we go: (1) x } IN Cauchy 1 IN st 1 (1) x x (1) V 1 < 1 Pick x 1 x (1) 1 (2) x } IN Cauchy 2 > 1 st 2 (2) x V 2 < 1 2 Pick x 2 2 (3) x } IN Cauchy 3 > 2 st 3 x (3) x (3) V 3 < 1 3 Pick x 3 x (3) 3 x (n) } IN Cauchy n > n 1 st n (n) x n V < 1 n Pick x n x (n) n In the exampe sketched beow the x (n) n s are circed (in the simpest case in which n happens to be n) c Joe Fedman 2011 A rights reserved September Competion 4

5 x (1) 4 x (1) 3 X (1) X (2) X (3) 4 3 x (1) 2 2 x (1) 1 1 Proof that x n } n IN is Cauchy: Let ε > 0 By the triange inequaity xn x V m (n) x n V + (n) x x (m) V + (m) x x (m) V m x (n) n V + x (n) x (m) V X (n) X (m) H for any IN If n the first term is smaer than 1 n By definition X (n) X (m) H im x (n) + X (n) X (m) H + x (m) x (m) m V (2) x (m) V So there is a natura number N nm such that the second term is smaer than ε 4 whenever N nm By hypothesis the sequence X (n) } n IN is Cauchy So there is a natura number Ñ such that the third term is smaer than ε 4 whenever nm Ñ Finay if m the ast term is smaer than 1 m Choose any natura number N max Ñ ε} 4 I caim that xn x m V < ε whenever nm N To see this et nm N Now choose to be any naturanumber bigger than max } N nm n m Then the four terms in (2) are each smaer than ε 4 Proof that X im X(n) : Let ε > 0 By definition By the triange inequaity x (m) X X (n) H im m m m xm x m (n) V im m V (m) x m V n + (n) x (m) x m x m (n) V n x m (n) V (3) Since the sequence x n x (n) n }n IN is Cauchy there is a natura number N such that the first term is smaer than ε 2 whenever nm N By the construction of n the second term is smaer than 1 n whenever m n Choose any natura number N max N 2 ε} I caim that X X (n) H < ε whenever n N To see this et n N Now choose m to be any natura number bigger than max N n } Then the two terms in (3) are each smaer than ε 2 c Joe Fedman 2011 A rights reserved September Competion 5

6 Lemma 8 U is inear and obeys UxUy H xy V for a xy V Proof: UxUy H xxx } yyy } im xy H V xy V The proof of inearity is simiar Lemma 9 U is one to one Proof: If xy V then Ux Uy xxx } yyy } im x y V 0 x y Lemma 10 U(V) is dense in H Proof: Let X x n } n IN H Then I caim that the sequence Uxm }m IN to X To check this it suffices to observe that X Uxm H im x n U(x m ) n V im x n x m V converges in H Since x n } n IN is Cauchy this converges to zero as m Lemma 11 If V is compete then U(V) H Proof: Let X H We must find x V with Ux X By Lemma 10 x n V } st X im n IN Ux n Ux n is Cauchy in H }n IN x n is Cauchy in V by Lemma 8 }n IN x im x n exists since V is compete Ux im Ux n by Lemma 8 Ux X c Joe Fedman 2011 A rights reserved September Competion 6