Weekly Notices 3. September 26, October 2 J. Hoffmann-Jørgensen September 26, 2005

Transcription

1 Weekly Notices 3 September 26, October 2 J. Hoffmann-Jørgensen September 26, 2005 Last lectures: Lectures: Last time I introduce the complex numbers and the n-dimensional space R n. I introduced the notion of a limits of sequences of real or complex numbers or vectors and I introduced notions of limits and continuity of vector functions. Friday September 21 (13:15 15:00): Metric spaces: limits and Cauchy sequences (Rudin Chap.3 p.45 55) 1. Cauchy sequence Augustin Louis Cauchy ( ) was a French mathematician teaching mathematical analysis at the University of Paris. In his famous book Cours d analyse from 1820 he observes that every convergent sequence (an) of real numbers have the following property (called the Cauchy property): the distance between an and ak becomes arbitraly small when n and k are sufficiently large and then he writes le contraire et évidente ; that is, every sequence with the Cauchy property is convergent to some real number c 2 R. Cauchy did not prove that statement actually he was not able to tdo it, due to the lack of a precise definition of limits and real numbers. Weierstrass rephrased the Cauchy property as follows. If (an) is a sequence of real numbers, then we say that (an) is a Cauchy sequence if and only if (*) 8 ">0 9 n" 1 so that jak 0 anj <" 8 n; k n" If (an) is a sequence of complex numbers or a sequence of vectors in R k, we say that (an) is a Cauchy sequence if and only if (**) holds. With this rephrasing, Cauchy s stament: (an) is convergent to some c 2 R if and only if (an) is Cauchy sequence becomes true. However, to proved the statement we the following notions: The limes superior and the limes inferior: Let a 1 ;a 2 ;... 2 R be a given sequence of real mumbers. Then we define the limes superior, and the limes inferior of (an) as follows: lim sup n!1 a n := inf sup an ; lim inf k1 nk n!1 a n := sup k1 inf nk a n Suppose that an denote the hight of landscape at n for every n 1. If you stand at k, then sup nk a n is the height of the heighest mountain ahead of you, and inf nk an is the depth of the deepest valley ahead of you. Hence, we see that lim sup an is the 1

2 height of the heighest mountain infinitely far way, and that lim inf a n the deepest valley infinitely far way. In particular, we have is the depth of (1) inf a n nk lim inf n!1 a n lim sup n!1 a n sup a n 8 k 1 nk NB: Rudin (p.56) gives an different definition of lim sup and lim inf. However, the two definitions are equivalent and the definition chosen here is more convenient for our purpose. Theorem 1.1: (The Cauchy-Weierstrass theorem) Let a 1 ;a 2 ;... 2 R be a given sequence of real numbers and let c 2 R be a real number. Then we have (i) c = lim n!1 a n, c = lim sup n!1 a n = lim inf n!1 a n (ii) (a n ) is convergent to some number b 2 R if and only if (a n ) is a Cauchy sequence Remark: Suppose that we want to show that (a n ) is convevergent. If use the definition (*), we have to specify the limit c. The importance of (ii) lies in the fact that we can prove convergence by verifying (**) without specifying the limit c. The fact that Cauchy observed (ii) without to be able to proof it, shows his ingenious insight in mathematics. Let me illustrate (ii) by an example: Euler s constant: Let R x 1 log x := dy denote the natural logarithm of x for 1 y all x 1 and let us define a n := log n 8 n =1; 2; 3;... n It can be shown that (a n ) is a Cauchy sequence (see Rudin Exc.9 p.197) and so by (ii) we have that (a n ) converges to some number wich is called Euler s constant and is denoted. Euler s constant owes it existence to Cauchy property and it is still today unknown if is rational or irrational (but it is generraly believed that is irrational). Euler s constant can be computed to any given precision; for instance, we have = 0: Proof: (i): Let us define a 3 := lim sup n!1 a n and a3 := lim inf n!1 a n. Suppose that c = lim n!1 a n and let ">0 be given. Then there exists n " 1 such that jc 0 a n j < " for all n n". Let n n" be given integers. Since jc 0 anj <",we have 0" <c0 an and an 0 c<" and since an = c +(an 0 c) =c 0 (c 0 an), we have c 0 "<an <c+ ". So by (1), we have c 0 " inf nn " an a3 a 3 sup nn " an c + " Since this holds for every ">0, we see that a3 = a 3 = c. Conversely, suppose that a3 = a 3 = c and let ">0 be given. By the definition of lim sup, there exists 2

3 n 1 1 such that sup kn 1 ak <a3 + " = c + " and by the definition of lim inf, there exists n 2 1 such that sup kn 2 ak >a3 0 " = c 0 ". Let us define n" := n 1 + n 2 and let n n" be given. Since n n 1 and n n 2, we have c 0 "< sup kn 2 ak an sup kn 1 ak <c0 " 8 n n" Hence, we see that an 2 N"(c) for all n n" and so by Thm.4.1.(ii) in Weekly Notices 2, we conclude that c = lim n!1 an which completes the proof of (i). (ii): Suppose that (an) converges to some number b 2 R and let " >0 be given. Applying (3) with " replaced by ", there exists and integer 2 n " 1 such that an 2 N " (c) for all n n ". Let n; k n" be given integers. Then we 2 have an 2 N " (c) and a k 2 N " (c) and since N " (c) =(c0 " ;c+ " ) is an open 2 interval of lenght ", we have jak 0 anj < ". Hence, se that (an) is a Cauchy sequence. Conversely, suppose that (an) is a Cauchy sequence and let us define a 3 := lim sup n!1 a n and a3 := lim inf n!1 an. Let ">0 be given. By (*), there exists an integer n" such that jak 0 anj <" for all n; k n". Let n n" be given. Since jan " 0 anj <", we have 0" <an " 0 an and an 0 an " <" and since an = an " +(an 0 an " )=an " 0 (an " 0 an), we have an " 0 " 2 <a n <an " + ". Since inf nk a n a3 a 3 sup nk for all k 1, we have an " 0 " inf nn " an a3 a 3 sup nn " an an " + " In particular, we see that a3 and 3 a are finite and that 0 3 a 0 a3 < 2 ". Since this holds for all ">0, we see that a3 3 = a and that b := a3 2 R. So by (i) we see that b = lim n!1 an which completes the proof of (ii). 2. Metric spaces Observe that the all the definitons and theorems on limits and continuity from Weekly Notices 2 only depend on the following properties of the distance (a): jx 0 xj =0 for all x ; (b): 0 < jx 0 yj = jy 0 xj for all x 6= y ; (c): jx 0 yj jx 0 zj + jz 0 yj for all x; y; z. This observation suggests the following generalization: Let X be an arbitrary set. Then a metric (also call a distance function) on X is a function d : X 2 X! [0; 1) satisfying (A) d(x; x) =0 8 x 2 X and 0 <d(x; y) =d(y; x) 8 x 6= y ; x; y 2 X (B) d(x; y) d(x; z) +d(z; y) 8 x; y; z 2 X Let a 1 ;a 2 ;... 2 X be a given sequence and let a 2 X be a given element. Then we say that (a n ) is convergent with limit a if and only if (*) 8 ">0 9 n " 1 so that d(a; a n ) <" 8 n n " 3

4 and we say that (a n ) is a Cauchy sequence if and only if (**) 8 ">0 9 n " 1 so that d(a k ;a n ) <" 8 n; k n " If a 2 X and r 0, we let N r (a) := fx 2 X j d(x; a) < rg and N r [a] := fx 2 X j d(x; a) rg denote the open and closed d-balls with center a and radius r, and we let N (a) :=N r r(a) nfag and N [a] :=N r r[a] nfag denote the open and closed punctuated d-balls with center a and radius r. Let A X be a given set and let a 2 X be a given elements. Then we introduce the following definitions: (1) We say that a is an interior point of A if and only if there exists a positive number r>0 satisfying N r (a) A. The set of all interior points of A is denoted A and is called the interior of A. Note that A A (2) We say that A is open if and only if all points in A are interior poins; that is, if and only if A A and if and only if A = A (3) We say that a is a limit point of A if and only if A \ N (a) 6= ; for all r>0. r The set of all limits of A is denoted A 0 and the set A := A [ A 0 is called the closure of A. Note that A A (4) We say that A is closed if and only if every limit point of A belongs to A ; that is, if and only if A 0 A and if and only if A A and if and only if A = A (5) We say that a is an isolated point of A if and only if a 2 A and a is not a limit point of A ; that is, if and only if a 2 A n A 0 and if and only if there exist a positive number r>0 satisfying A \ N r (a) =fag (6) A is perfect if every element in A is a limit point of A ; that is, if and only if A A 0 and if and only if A has no isolated points (7) We say that A is dense in X if and only if A = X (8) We say that A is bounded if and only if there exist q 2 X and r > 0 satisfying A N r (q) (9) If r 0 is a given non-negative number, we say that A is r-dicrete if and only if d(x; y) >r for all x; y 2 A satisfying x 6= y. Note that every set A X is 0-discrete The standard metric on R : Let x; y 2 R be real numbers. Then jx 0 yj is the lenght of the interval connecting x and y and d(x; y) :=jx 0 yj is a metric on R called the standard metric on R. Let :[0; 1)! R be a concave increasing function satisfying (0) = 0 and (x) > 0 for all x > 0. By concavity, we obtain the following inequality (u + v) (u) +(v) for all u; v 0, and if we define 4

5 p d (x; y) :=(jx 0 yj) for all x; y 2 R, then d is metric on R. For instance, we have that jx 0 yj and log (1 + jx 0 yj) are metrics on R. The Euclidian metric on R 2 : Let A =(x; y) and B =(u; v) be two points in the plane R 2. By Phytagoras theorem we have that ja0bj := p(u 0 x) 2 +(v 0 y) 2 is the distance between the points A and B and by Euclid s famous triangle theorem we have that ja 0 Bj is a metric on R 2 which is called the Euclidian metric; see the figure below: B=(u,v) A-B v-y A=(x,y) u-x The taxi metric on R 2 : The Euclidian metric, is the shortest rute between the points A and B provided that you actually are able to move along the straight line between A and B. However, if you are a taxidriver in New York city you have to follow the streets and if so, then the shortest route becomes ja0bj 1 := ju0xj+jv0yj. It is easily verified that jja 0 Bjj 1 is a metric on R 2 which is called the taxi metric. Metric on R n : Let n 1 be a given integer and let A =(x 1 ;...;x n ) and B =(u 1 ;...;u n ) be two vectors in R 2. In analogy with the previous examples we define the Euclidian metric and the taxi metric as follows ja 0 Bj := q (x 1 0 u 1 ) (x n 0 u n ) 2 ja 0 Bj 1 := jx 1 0 u 1 j jx n 0 u n j and we define the maximum metric as follows: ja 0 Bj1 := maxfjx 1 0 u 1 j ;...; jx n 0 u n jg The maximum metric is used in connection with election methods. Let p>0 be a given number, the we define the `p-metric as follows ja 0 Bj p := jx 1 0 u 1 j p jx n 0 u n j p if 0 <p 1 ja 0 Bj p := 8jx 1 0 u 1 j p jx n 0 u n j p 9 1 p if p>1 Then ja 0 Bj p is a metric on R n for every fixed number 0 <p1, and we have ja 0 Bj = ja 0 Bj 2. Metrics on function spaces: Let I = [a; b] be a closed interval and let f; g : I! R be two continuous functions. In analogy with the example above 5

6 we define the L p -metric as follows: kf 0 gk p := kf 0 gk p := kf 0 gk1 := Z b a Z b a jf (x) 0 g(x)j p dx if 0 <p 1 sup axb jf (x) 0 g(x)j p jf (x) 0 g(x)j p 1 p dx if 1 <p<1 Then jjf 0 gjj p is a metric on the set of all continuous functions on I for every fixed number 0 <p1. 3. Covering and density numbers If M is a given set, we let # M denote the number of elements in M if M is finite and define # M := 1 if M is infinite. Let (X; d) be a metric space and let A X be a given set and let r 0 be a non-negative number. Then we define the covering number C r (A) and the density number D r (A) as follows: (1) C r (A) :=0 if A = ; and if A 6= ;, we let C r (A) denote the smallest number of closed balls of radius r needed to cover the set A, with the convention C r (A) =1 if we cannot cover the set A with finitely many balls (2) D r (A) :=supf # D j D A and D is r-discrete g Note that C 0 (A) =D 0 (A) =#A and that C r (;) =D r (;) for all r 0. We say that A is totally bounded if C r (A) is finite for every positive number r>0. Example: Let us consider the real line R with its standard metric jx 0 yj. If A = [0; 1], we have C r (A) =1 if r 1 2 ; C r(a) =2 if 1 4 r<1 2 ; C r(a) =3 if 1 6 r<1 4 C r (A) =4 if 1 8 r< 1 6 etc.: C r(a) =n if n 2 and 1 2n r< 1 2(n01) In particular, we see that A is totally bounded. Theorem 3.1: Let (X; d) be a metric space and let A X is a given set, then we have (i) Every open ball is open and every closed ball is closed (ii) If A is bounded, then for every p 2 X there exists a number > 0 such that A N [p] (iii) If there exists a number r 0 such that C r (A) < 1, then A is bounded 6

7 (iv) If A is totally bounded, then A is bounded (v) C r (A) D r (A) C r 2 (A) 8 r 0 (vi) A is totally bounded if and only if D r (A) < 1 for all r >0 Proof: (i): Let p 2 X and r 0 be given. If r = 0, we have N0(p) =; which is open by convention. Suppose that r > 0 and let y 2 N r (p) be given. Then d(y; p) < r and := r 0 d(y; p) is strictly positive. Let x 2 N (y) be given. Then we have d(x; y) < and by the triangle inequality we have d(x; p) d(x; y) + d(y; p) < r+ d(y; q) =. Since that holds for all x 2 N (y), we have N (y) N r (p). Hence, we see that N r (p) is open. To show that N r [p] is closed we shall consider the complement N r [p] c := fx 2 X j d(x; p) > rg. Let y 2 N r [p] c be given. Then d(y; p) >r and := d(y; p) 0 r is strictly positive. Let x 2 N (y) be given. Then we have d(x; y) < and by the triangle inequality we have + r = d(y; p) d(y; p) +d(p; x) <+ d(x; p). Hence, we have d(x; p) >r and consequently, we have x 2 N r [p] c. Since that holds for all x 2 N (y), we have N (y) N r [p] c. Hence, we see that N r [p] c is open and so by Thm.2.23 in Rudin p.34, we conclude that N r [p] is closed. (ii): Let p 2 X be given. Since A is bounded, there exists q 2 X and r > 0 such that A N r [q]. Let us define := r + d(p; q) and let x 2 A be given. Since x 2 A N r [q], we have d(x; q) r and by the triangle inequality we have d(x; p) d(x; q) +d(q; p) r + d(p; q) =. Since this holds for all x 2 A, we see that A N [p] which completes the proof of (ii). (iii): Suppose that n := C r (A) < 1. Then there P exist p1;...;p n 2 X such that n A N r [p1] [111[ N r [p n ]. Let us define := r + i=1 d(p 1;p i ) and let x 2 A be given. Since A N r [p1] [111[N r [p n ], there exists and integer 1 i n such that x 2 N r [p i ]. Hence, we have d(x; p i ) r and by the triangle inequality we have d(x; p1) d(x; p i )+d(p i ;p1) r + d(p i ;p1) =. Since this holds for all x 2 A, we see that A N [p1] which completes the proof of (iii). (iv): Immediate consequence of (iii). (v): Let me first show that C r (A) D r (A). If D r (A) =1, this is evident. So suppose that d := D r (A) < 1. Then there exists an r-discrete set D A with exactly d elements. Then I claim that A [ u2d N r [u]. So let x 2 A be given. If x 2 D, we have x 2 N r [x] [ u2d N r [u]. If x 2 A n D, we see that D [fxg has d +1 elements and since d +1 > D r (A), we see that D [fxg is not r-discret. Since D is r-discrete, there exists v 2 D such that d(v; x) r or equivalently x 2 N r [v] [ u2d N r [u]. Hence, we have A [ u2d N r [u] and since D has d = D r (A) elements, we see that C r (A) D r (A). Suppose that D r (A) >C r=2 (A). Then we have n := C r=2 (A) < 1 and there exist closed balls N r=2 (u 1 );...;N r=2 (u n ) satisfying A N r=2 (u 1 )[111[N r=2 (u n ). Since D r (A) >n, we have D r (A) n +1 ans so there exists an r-discrete set D A with n +1 elements. Since D A N r=2 (u 1 )[111[N r=2 (u n ), there exist an integer 1 n and elements x; y 2 D such that x 6= y and x; y 2 N r=2 (u ). Hence, we have 7

8 d(x; y) >r and d(u ;x) r and d(u 2 ;y) r 2 have r<d(x; y) d(x; u )+d(u ;y) r that D r (A) N r=2 which completes the proof of (v). (vi): Immediate consequence of (v).. So by the triangle inequality we which is impossible. Thus, we conclude 8