[PDF] [PDF] 8 Completeness

[PDF] Prove that ) is Cauchy using directly the definition of Cauchy

Assume that (xn)n∈N is a bounded sequence in R and that there exists x ∈ R such that any convergent subsequence (xni )i∈N converges to x Then limn→∞ xn 

[PDF] 14 Cauchy Sequence in R

A sequence xn ∈ R is said to converge to a limit x if • ∀ϵ > 0, ∃N s t n > N ⇒ xn − x < ϵ A sequence xn ∈ R is called Cauchy sequence if • ∀ϵ, ∃N s t n > N m > N ⇒ xn − xm < ϵ Every convergent sequence is a Cauchy sequence Proof

[PDF] Question 1 - Properties of Cauchy sequences Question 2

Show that xn → x as well; i e to prove that a Cauchy sequence is convergent, we only need i e any bounded sequence has a convergent subsequence Show 

[PDF] 1 Cauchy sequences - ntc see result

A sequence {an}is called a Cauchy sequence if for any given ϵ > 0, there exists N ∈ N Proof Since {an}forms a Cauchy sequence, for ϵ = 1 there exists N ∈ N such that (i) lima1/n = 1, if a > 0 (ii) limnαxn = 0, if x < 1 and α ∈ IR Solution: 

ON U-CAUCHY SEQUENCES - Project Euclid

If a sequence of points in X is U-convergent in X then it fulfills U-Cauchy condition Proof Let U − lim n→∞ xn = x and ε > 0 Thus A(ε/2) = {n ∈ N : ρ(xn, x) ≥

[PDF] details - Lecture summary

Definition A sequence (an) is said to be a Cauchy sequence iff for any ϵ > 0 there exists N such prove (over the course of 2 + ϵ lectures) the following theorem: If a subsequence of a Cauchy sequence converges to x, then the sequence

[PDF] be a metric space Let - School of Mathematics and Statistics

Let X = (X, d) be a metric space Let (xn) and (yn) be two sequences in X such that (yn) is a Cauchy sequence and d(xn,yn) → 0 as n → ∞ Prove that (i) (xn) is a 

[PDF] ANALYSIS I 9 The Cauchy Criterion - People Mathematical Institute

Every complex Cauchy sequence is convergent Proof Put zn = x + iy Then xn is Cauchy: xx − xm ⩽ zn − zm (as  

[PDF] Math 431 - Real Analysis I

(c) Show that the sequence xn is bounded below by 1 and above by 2 (d) Use (e) Use (d) in a proof to show that Sn is Cauchy and thus converges In the first case, if k = 0, then we wish to prove that k · f(x) = 0, the zero function, has limit 0

[PDF] 8 Completeness

If (X, d) is a complete metric space and Y is a closed subspace of X, then (Y,d) is complete Proof Let (xn) be a Cauchy sequence of points in Y Then (xn) also 

[PDF] show that x is a discrete random variable

[PDF] show that x is a markov chain

[PDF] show that x is a random variable

[PDF] show that [0

[PDF] show the mechanism of acid hydrolysis of ester

[PDF] show time zone cisco

[PDF] show ∞ n 2 1 n log np converges if and only if p > 1

[PDF] shredded workout plan pdf

[PDF] shredding diet

[PDF] shredding workout plan

[PDF] shrm furlough

[PDF] shuttle paris

[PDF] si clauses french examples

[PDF] si clauses french exercises

[PDF] si clauses french practice

8 CompletenessWe recall the definition of a Cauchy sequence. Let (X,d) be a given metric

space and let (xn) be a sequence of points ofX. Then (xn) (xn) is a Cauchy sequence if for everyε >0 there existsN?Nsuch that d(xn,xm)< εfor alln,m≥N. Properties of Cauchy sequences are summarized in the following propositions Proposition 8.1.(i)If(xn)is a Cauchy sequence, then(xn)is bounded. (ii)If(xn)is convergent, then(xn)is a Cauchy sequence. (iii)If(xn)is Cauchy and it contains a convergent subsequence, then(xn) converges. A Cauchy sequence need not converge. For example, consider the sequence (1/n) in the metric space ((0,1),| · |). Clearly, the sequence is Cauchy in (0,1) but does not converge to any point of the interval. Definition 8.2.A metric space (X,d) is calledcompleteif every Cauchy sequence (xn) inXconverges to some point ofX. A subsetAofXis called completeifAas a metric subspace of (X,d) is complete, that is, if every

Cauchy sequence (xn) inAconverges to a point inA.

By the above example, not every metric space is complete; (0,1) with the standard metric is not complete. Theorem 8.3.The spaceRwith the standard metric is complete. Theorem 8.3 is a consequence of the Bolzano-Weierstrass theorem and

Propositions 8.1

then wherex= (x1,...,xn),y= (y1,...,yn)?X, defines a metric onX. The pair (X,d) is called the product of (Xi,di). Theorem 8.4.If(Xi,di)are complete metric spaces fori= 1,...,m, then the product(X,d)is a complete metric space. 49
Proof.Letxn= (x1n,...,xmn) and (xn) be a Cauchy sequence in (X,d). Then for a givenε >0 there existsksuch thatd(xn,xm)< εfor alln,m≥k. Since d it follows that{xjn}is Cauchy in (Xj,dj) forj= 1,...m. Since (Xj,dj) is complete, forj= 1,...,mthere existsxj?Xjsuch thatxjn→xj. Then settingx= (x1,...,xm), we see, in view of the above definition ofd, that x n→xinX.? Since n? i=1|xi-yi|2? 1/2 nmax{|x1-y1|,...,|xn-yn|}, it follows from Theorem 8.3 and Proposition 8.4 thatRnwith the Euclidean metric is complete. Example 8.5.Denote by?1the set of all real sequences (xk) satisfying?∞k=1|xk|< ∞. Then?1is a vector space if addition and multiplication by a number are defined as follows, (xn) + (yn) = (xn+yn), α(xn) = (αxn).

Ifx= (xk)??1, then

?x?=?(xk)?:=∞?k=1|xk| defines a norm on?1andd(x,y) =?x-y?=?∞k=1|xk-yk|defines a distance on

1. We claim that (?1,d) is a complete metric space.

Indeed, let (Xn)??1be a Cauchy sequence. Then withXn(x(n) k) we have x(n) k-x(m) k??? l=1??? x(n) l-x(m) l??? =?Xn-Xm?.

Hence for everyk≥1, the sequence (x(n)

k) is Cauchy inRand sinceRwith the standard metric is complete, the sequence (x(n) k) converges to somexk. Set X= (xk). We suspect thatXis the limit in?1of the sequence (Xn). To see this we first show thatX??1. Since (Xn) is Cauchy in?1, there isKsuch that ?Xn-Xm?<1 for alln,m≥K. In particular, N k=1??? x(n) k??? k=1??? x(n) k-x(K) k??? +∞?k=1??? x(K) k??? 50
for everyN≥1. FixingNand taking limit asn→ ∞we get N and taking limit asN→ ∞we get So,X??1. Next we show that?Xn-X? →0 asn→ ∞. Givenε >0, there is Ksuch that?Xn-Xm?<1 for alln,m≥K. Consequently, for everyN≥1 we have N k=1??? x(n) k-x(m) k??? Withn > KandNfixed, we letm→ ∞to find that N k=1??? x(n) k-xk???

Since this is true for everyN,

?Xn-X?=∞? k=1??? x(n) k-xk??? forn > K. Henced(Xn,X)→0 and sinceX??1, the space?1is complete. A subspace of a complete metric space may not be complete. Forexample, Rwith the standard metric is complete but (0,1) equipped with the same metric is not complete. Proposition 8.6.If(X,d)is a complete metric space andYis a closed subspace ofX, then(Y,d)is complete. Proof.Let (xn) be a Cauchy sequence of points inY. Then (xn) also satisfies the Cauchy condition inX, and since (X,d) is complete, there existsx?Xsuch that x n→x. ButYis also closed, sox?Yshowing thatYis complete.? Proposition 8.7.If(X,d)is a metric space,Y?Xand(Y,d)is complete, thenYis closed. Proof.Let (xn) be a sequence of points inYsuch thatxn→x. We have to show thatx?Y. Since (xn) converges inX, it satisfies the Cauchy condition inXand so, it also satisfies the Cauchy condition inY. Since (Y,d) is complete, it converges to some point inY, say toy?Y. Since any sequence can have at most one limit, x=y. Sox?YandYis closed.? 51
Let (X,d) and (Y,ρ) be metric spaces. A functionf:X→Yis said to beboundedif the imagef(X) is contained in a bounded subset ofY. Denote byB(X,Y) the set of all functionsf:X→Ywhich are bounded and byCb(X,Y) the space of bounded continuous functionf:X→Y. If Y=R, we simply writeB(X) andCb(X) instead ofB(X,Y) andCb(X,R), respectively. We haveCb(X,Y)?B(X,Y). Forf,g?B(X,Y), we set

D(f,g) := sup{ρ(f(x),g(x))|x?X},

whereρdenotes the metric onY. The metric onCb(X,Y) is defined in the same way. Theorem 8.8.Suppose that(Y,ρ)is a complete metric space. Then the spacesB(X,Y)and(Cb(X,Y),D)are complete. Proof.The verification thatDis a metric is left as an exercise. It suffices to show thatB(X,Y) is a complete and thatCb(X,Y) is a closed subset ofB(X,Y). Let {fn}be a Cauchy sequence inB(X,Y). Then (fn(x)) is a Cauchy sequence in (Y,ρ) for everyx?X. Since, by assumption, (Y,ρ) is complete, the sequence (fn(x)) converges inY. Definef(x) = limn→∞fn(x). Givenε >0, there isNsuch that

ρ(fn(x),fm(x))< εforn,m≥N.

Fixn≥Nand letmtend to∞. Since the functionz→ρ(fn(x),z) is continuous, it follows that which implies that It remains to show thatfis bounded. SincefNis bounded, there arey?Yand r >0 such thatfN(X)?Br(y).From (1) and the triangle inequality, we obtain and thatfis bounded. Hence the space (B(X,Y),D) is complete as claimed. Next we shall show thatCb(X,Y) is a closed subset ofB(X,Y). Take any sequence (fn)?Cb(X,Y) such thatD(fn,f)→0 asn→ ∞. We have to show thatfis continuous. Fixx0?Xand letε >0. Then there isN?Nsuch that

D(fn,f)< ε/3 for alln≥N.

In particular, in view of definition ofD,

ρ(fN(x),f(x))< ε/3 for alln≥N.

SincefNis continuous atx0, there isδ >0 such that ρ(fN(x),fN(x0))< ε/3 for allxsatisfyingd(x0,x)< δ. 52

Hence, ifd(x0,x)< δ, then

This means thatfis continuous atx0and sincex0was an arbitrary point,fis continuous, i.e.,f?Cb(X,Y) as claimed.?

8.0.1 Banach Fixed Point Theorem

Definition 8.9.Let (X,d) be a metric space. A mapf:X→Xis called a contraction if there is constantc?(0,1) such that for allx,y?X. Theorem 8.10(Banach Fixed Point Theorem).Let(X,d)be a com- plete metric space andf:X→Xa contraction. Then there exists exactly one pointu?Xsuch thatf(u) =u. Moreover, for everyx?X, the sequence(fn(x))converges tou.

Proof.Claim: Ifa,b?X, then

1-c[d(a,f(a)) +d(b,f(b))].(2)

Indeed,

from which we conclude (2) Using (2), the mapfcan have at most one fixed point since iff(a) =aandf(b) =b, then the right side of (2) is equal to 0 implying thatd(a,b) = 0. To prove existence of a fixed point, take anyx?X. Sincefis a contraction with the constantc, then-fold compositionfnis a contraction with the constantcn. Using this and (2) witha=fn(x) andb=fn(b), and noticing thatf(a) =fn+1(x) =fn(f(x)),f(b) =fm(f(x)), we conclude

1-c?d(a,f(a)) +d(b),b)]

1

1-c?d(fn(x)fn(f(x))) +d(fm(x),fm(f(x))]

cn+cm

1-cd(x,f(x))

This implies that the sequence (fn(x)) is Cauchy In view of the completeness ofX, there isusuch thatfn(x)→u. In view of the continuity off,f(fn(x)) converges tof(u). On the other handf(fn(x)) =fn+1(x) converges touand so,f(u) =u as required.? 53

8.0.2 Application of Banach Fixed Point TheoremHere is an application of the Banach fixed point theorem to thelocal exis-

tence of solutions of ordinary differential equations. Theorem 8.11(Picard"s Theorem).LetUbe an open subset ofR2and letf:U→Rbe a continuous function which satisfies the Lipschitz condition with respect to the second variable, that is, for all(x,y1),(x,y2)?U, and someα >0. Then for a given(x0,y0)?U there isδ >0so that the differential equation y ?(x) =f(x,y(x)) has a unique solutiony: [x0-δ,x0+δ]→Rsuch thaty(x0) =y0. Proof.Note that it is enough to show that there areδ >0 and a unique function y: [x0-δ,x0+δ]→Rsuch that y(x) =y0+? x x

0f(t,y(t))dt.

Fix (x0,y0)?U. Then we findδ >0 andb >0 such that ifI= [x0-δ,x0+δ] andJ= [y0-b,y0+b], thenI×J?U. Sincefis continuous andI×Jis closed (x,y)?U. Replacingδby a smaller number we may assume thatαδ <1 and αM < b. Denote byXthe set of all continuous functionsg:I→J. The setX with the metricρ(g,h) = sup{|g(x)-h(x)|,x?I}is a complete metric space. For g?X, let (Tg)(x) =y0+? x x

0f(t,g(t))dt.

ThenTg:I→Ris continuous since ifx1,x2?Iandx2> x1, then |(Tg)(x2)-(Tg)(x1)|=????? x2 x

1f(t,g(t))dt????

x2 x |(Tg)(x)-y0|=????? x x

0f(t,g(t))dt????

x x Sincefis Lipschitz with respect to the second variable, we obtain forg,h?Xand 54
x?[x0,x0+δ], |(Tg)(x)-(Th)(x)|=????? x x

0[f(t,g(t))-f(t,h(t))]dt????

x x

0|f(t,g(t))-f(t,h(t))|dt

Sinceαδ <1,Tis a contraction and in view of Banach"s fixed point theorem there exists a unique continuous functiony:I→Jsuch that y(x) = (Ty)(x) =y0+? x x

0f(t,y(t))dt.

55
quotesdbs_dbs19.pdfusesText_25