Limits and convergence.

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1 Chapter 2 Limits and convergence. 2.1 Limit points of a set of real numbers Limit points of a set. DEFINITION: A point x R is a limit point of a set E R if for all ε > 0 the set (x ε,x + ε) E is infinite. ex Prove that x is a limit point of E if ε > 0, y E,0 < x y < ε; that is: for all ε > 0 the set (x ε,x + ε) contains a point y E, y x Limit points of a sequence DEFINITION: A sequence of real numbers, aka a numerical sequence, is a map n a n R defined for n N, n Z or, more generally, n A Z. The notation is {a n } n A. A sequence is finite if the index set A is finite, otherwise it is infinite. Note that the sequence is finite or infinite depending on the index set and not on the size of the range. Thus {a n } n N is an infinite sequence even if a n = 1 for all n N. DEFINITION: A point x R is a limit point of a numerical sequence {a n } n=1, if for all ε > 0 the set {n : a n x ε} is infinite. Thus in the example above, where a n = 1 for all n, the point x = 1 is a limit point of the sequence {a n } but is not a limit point of the set {1} which is the range of the sequence Convergence of sequences (of real numbers): Convergent sequences; subsequences; 10

2 2.2. Completeness, Bolzano Weierstrass Completeness, Bolzano Weierstrass The least upper bound of a set DEFINITION: The number b is the supremum aka the least upper bound of a set E R, written b = supe, if it is an upper bound for E and no number smaller that b is an upper bound for E: formally a. x E,x b, and b. if c R, c < b, then y E, y > c, (i.e., c is not an upper bound for E). Least upper bound property; consequences; Exercises: quantifiers Move to the appendix: countability of Q vs. uncountability of R Completeness. Cauchy sequences and completeness; Cauchy criterion. ====== Theorem. the completeness axiom for R as stated sometimes, namely: a. Every nonempty subset S of R that is bounded above has a least upper bound in R, (denoted sups). is equivalent to the statement b. Every Cauchy sequence of real numbers has a limit in R. PROOF: We need to show that each statement implies the other a. = b. Assume the completeness axiom. Preparation: Observe that for a bounded monotone non-decreasing sequence y n one has limy n = sup{y n } (The sup exists by the Completeness axiom. By the definition of sup there exist, for any ε > 0, values of N such that y N > sup{y n } ε and if n > N we have y N y n sup{y n } ). Similarly for a bounded monotone non-increasing sequence y n one has limy n = inf{y n }. The completeness axiom now implies that every

3 12 2. Limits and convergence. bounded monotone sequence y n of real numbers has a (real number) limit. This implies the existence, for any bounded sequence {x n }, of the limits limsupx n = lim m sup n>m x n and liminfx n = lim m inf n>m x n. (since y m = sup n>m x n and z m = inf n>m x n are monotone and bounded.) Let {x n } R be a Cauchy sequence; it is clearly bounded. Write S = limsupx n, s = liminfx n then S s and we claim that in fact S = s so that limx n = S and {x n } has a limit in R. We show that S = s by showing that the opposite assumption in not consistent with the Cauchy condition on {x n }. Assume S s > 0 and set ε = S s 3 ; there are arbitrarily large values n, m such that x n < s + ε and x m > S ε so that x m x n > ε, which contradicts the assumption that {x n } is a Cauchy sequence. b. = a. Assume Every Cauchy sequence of real numbers has a limit in R. Let A be a set which is bounded above, and let x 1 be an upper bound. If x 1 = supa we are done. Otherwise let k be the smallest integer such that x 1 1 k is an upper bound (for A) and set x 2 = x 1 1 k. Repeat, defining x n+1 = x n kn 1 where k n is the smallest integer such that x n 1 k n is an upper bound for A; (if at any stage x n = supa we are done ). The sequence {x n } is bounded below (by A) so that k 1 n < Theorem (Bolzano Weierstrass). A bounded infinite set of real numbers has limit points. PROOF: Let E be an infinite set contained in the (finite) interval I = [a,b] R. At least one of the intervals I 0 =[a, a+b 2 ] and I1 =[ a+b 2,b] contains infinitely many elements of E. Write J 1 = I 1 if I 1 E is infinite; otherwise set J 1 = I 0. J 1 E is infinite, and denoting by c 1 is the right end point of J 1, the half line (c,+ ), has finite intersection with E. Let J 2 be the right hand half of J 1 if its intersection with E, J 2 E, is infinite; otherwise set J 2 equal to the left hand half of J 1. The choice guarantees that J 2 E is infinite, while E has only finitely many points

4 2.2. Completeness, Bolzano Weierstrass 13 to the right of J 2. We continue in the sam manner: assuming that we have J m defined so that J m is half of J m 1, J m E is infinite, and E has only finitely many points to the right of J m, we split J m into two equal intervals and set J m+1 equal to the right half of J m if the right half has infinitely many points of E, and equal to the left half of J m otherwise. We denote by c m the right end point of J m and observe that {c m } is a Cauchy sequence since if m j > M, j = 1,2, both c m2 and c m1 are contained in J M so that c m2 c m1 < I 2 M. By the Cauchy criterion, {c m } is convergent. Write c = lim m c m, and verify that [c + ε, ) E is finite for every ε > 0, while [c ε, ) E is infinite. Thus, c is a limit point of E, and in fact the biggest limit point, i.e. c = limsupe Open covers, Heine-Borel. DEFINITIONS: An open cover of a set E in a metric space (X,ρ) is a collection of open sets {O α } α A, such that E α A O α. A subcover of an open cover {O α } α A (of E), is a subcollection, i.e., {O α } α B where B is a subset of the index set A and such that E α B O α. The subcover is a finite subcover if the corresponding index set B is finite. The set E is compact if every open cover of E has a finite subcover. Theorem (Heine-Borel). a. A set E R is compact if and only if it is both bounded and closed. b. A set E R is compact if and only if every infinite subset thereof has limit points in E. PROOF: a. Assume E compact. Let O n =( n,n), then O n covers R, hence E. A finite subcover has a biggest element and it is bounded, so E is bounded. We prove that E is closed by showing that its complement is open. If y E the sets O n = {x : x y > 1 n } are open and their union covers everything but y, hence covers E. If the union of a finite subcollection n N O n = O N covers E, then the neighborhood {x : x y < 1 N } of y is disjoint from E, so that the complement R \ E is open.

5 14 2. Limits and convergence. We prove the converse by contradiction. Let E be bounded and closed, and let A = {O α } α A be an open cover of E. Let I =[a,b] be a finite interval containing E. Assume that there is no finite subcover of E for any finite subset B A the family {O α } α b fails to cover E. As in the proof of the Bolzano-Weiestrass theorem we split I into two equal intervals, I 0 and I 1 and observe that if both I 0 E and I 1 E can be covered by a finite subfamily of A, then the union of the two finite families is a finite cover for E. It follows that at least one of I 0, I 1 has the property that its intersection with E admits no finite subcover. Denote by J 1 either of I 0, I 1 that has this property. By the same argument we have an interval J 2 J 1, which is either the left half or the right half of J 1 that has the property that J 2 E does not admit a finite subcover from A. Continuing, we obtain a sequence {J n } of intervals, each J n is either the left or the right half of J n 1, and for every n the set J n E does not admit a finite subcover from A and in particular, is not empty. Let x n J n E. The sequence {x n } is a Cauchy sequence hence convergent, and since E is closed, x = lim n x n E. Since x E, there exists α A such that x O α, and since O α is open it contains an open interval centered at x and therefore contains J n if n is big enough, contradicting the assumption that J n E is not contained in a finite union of O α s. b. A set E R such that every infinite subset thereof has limit points in E is bounded, otherwise it contains a sequence {x n } such that x n > n that clearly has no limit points; and closed, otherwise it contains a sequence {x n } that converges to point in the complement of E, hence has no limit points in E Total boundedness. DEFINITION: A metric space (X, ρ) is totally bounded if for every ε > 0 there exists a finite set {x j } n j=1 X such that B(x j,ε)=x. ex Prove that a compact metric space is totally bounded and complete. ex Prove that a totally bounded complete metric space is compact.

6 2.3. Series 15 ex Show that the material of this subsection extends verbatim to subsets of R d for any d N. 2.3 Series Series: basic convergence. An infinite series is an expression of the form S = j J a j where J is an ordered infinite index set. We shall focus on the most common case 1 J = N, in which case we write S = j=0 a j. The partial sums of the series are the expressions S N = N j=0 a j. The partial sums of a series are real numbers computed as usual by addition. The sum of the series S can not be computed by (repeated) addition alone, and is commonly defined as the limit, for N, of the partial sums. (2.3.1) S = a j = lim N j=0 N j=0 This definition is consistent with what we expect from a series that has only finitely non-zero elements. DEFINITION: The series S = j=0 a j is convergent if the sequence {S N } of partial sums converges, and the sum of the series is given by (2.3.1). EXAMPLE: Geometric series. These are series of the form 0 an where a < 1. Since S N = N 0 an = 1 an+1 1 a the series converges and its 1 a sum S = lim N+1 N 1 a = 1 1 a Cauchy criterion of convergence. Since the convergence of a series S is defined by means of the convergence of the sequence of its partial sums, conditions that guarantee the latter automatically guarantee the former. For example: Proposition. A necessary and sufficient condition for the convergence of S is that {S N } be a Cauchy sequence. a j. 1 Another common case is J = Z, in which case we write j= a j.

7 16 2. Limits and convergence. By definition, the condition that {S N } be a Cauchy sequence is: for every ε > 0 there exists N = N(ε) such that if m > l > N then S m S l = m l+1 a n < ε. In particular, if a n converges then a n Series with nonnegative terms. When all the entries a n of a series S are nonnegative, the partial sums S N = N 0 a n form a monotone nondecreasing sequence, and hence converge to a (finite) limit if and only if the sequence is bounded. Observe that, being monotone, the sequence is bounded if it has a bounded infinite subsequence Conditional and absolute convergence. DEFINITION: A series S = 0 a n converges absolutely if the series of the absolute values of its summands, 0 a n converges. A convergent series that does not converge absolutely is said to converge conditionally. Proposition. If the series S = 0 a n converges absolutely then it converges. The proof is an immediate consequence of the Cauchy criterion, since (2.3.2) m a n l+1 m l+1 a n. The same observation gives the comparison test: Proposition (Comparison test). Assume b n a n. If a n converges then so does b n. The comparison test can be used in a slightly more general context. Consider the series S = 1 n a, a > 1. The terms are nonnegative and, in order to prove convergence, it suffices to prove the partial sums bounded. This is seen by considering the blocks B m = 2m+1 2 m +1 n a. B m is the sum of 2 m terms, each of which is bounded by 2 ma so that B m 2 (1 a)m, and comparing to the geometric series with ratio 2 1 a < 1 we obtain B m <. Since the partial sums of B m are partial sums of S, the latter are bounded and S converges.

8 2.3. Series 17 ex Prove that S = 1 n a, a 1 does not converge. ex Prove that S = 1 n 1 (logn) a converges if and only if a > 1. Hint: For 2 m < n 2 m+1 we have n 1 (logn) a = O(2 m m a ) Rearrangements. The sum of a finite number of real numbers does not depend on the order in which they are added this is the commutative law. For infinite series, which use addition but also use a limiting process, the order may play a big role. DEFINITION: A permutation of N is a 1 1 map of N onto itself. A rearrangement of a series n=0 a n is a series of the form n=0 a σ(n), where σ is a permutation of N. Theorem. a. If the series a n is absolutely convergent, then every rearangement thereof is absolutely convergent, and to the same sum. b. If the series a n converges conditionally then, given a number c R, there exist rearrangements thereof converging to c. PROOF: a. Denote the sum of the series by a. Let ε > 0 and let N = N(ε) be such that N a n < ε. Observe that if J N, and J [0,N] then a n J a n < ε. If σ is a permutation of N then, given N, there is an M(N) such that (for all M > M(N)) {σ(n) : n < M} [0,N] and we have a M j=0 a σ( j) < ε. b. The conditional convergence guarantees that lima n = 0 and a n =. Split the sequence {a n } into two: the subsequence I + of nonnegative elements and the subsequence I of negative elements. We have (2.3.3) n I + a n =+ and n I a n =. We create a permutation σ by choosing successively elements, in the order they appear, from either I + or I. Assume that c 0. We start by taking (the first) elements from I + stopping when the sum of the chosen elements exceeds c; this is turning point 1. We now take elements from I until the sum goes below c. This is turning point 2. We take the next elements from I + until the (cumulative) sum surpasses c (turning point 3), and then form I until the sum is below c, and by (2.3.3) this procedure can be repeated indefinitely.

9 18 2. Limits and convergence. The partial sums that correspond to turning points are obtained by adding some positive a n to a value smaller that c or by adding some negative a n to a value bigger that c; all other partial sums are between two that correspond to successive turning points. Given ε > 0, once all the terms a n of absolute value exceeding ε are used up, the partial sums of the rearranged series are within ε from c. ex Let a n be conditionally convergent. Prove that the set of limit points of the partial sums of a σ(n), for any permutation of N, is a closed, not necessarily bounded, interval (including a single point). Moreover, any closed interval, as above, is the set of limit points of the partial sums of a σ(n) for some permutations σ of N Summability. It is often very useful to consider infinite series whose partial sums fail to converge. The following method to assign sum to (some) infinite series that fail to converge is the Cesàro (C,1) summability. It extend the standard definition of sum as the limit of the partial sums of the given series to one that assigns a sum to a wider class of series, and does it in a consistent way it gives the same value to convergent series as the standard definition. DEFINITION: A means A series a n is (Cesàro) (C,1)-summable to the sum S 0 + S S n (2.3.4) lim = A n n + 1 Theorem. Let S = a n be a convergent series, S = lim n S n, then S is (C,1)-summable to the same value S. PROOF: The entries a n and the partial sums are bounded. Write σ n = 1 n+1 n j=0 S j. Given ε > 0, there exists N such that for n > N liminfs j ε S n limsups j + ε. It follows that for all m N m infs j + m N m liminfs j ε 1 m m j=1 S j N m sups j + m N m limsups j +ε.

10 2.4. Continuity (R-valued functions ) 19 As m, N m N m 0 and m 1, so that liminfs n ε liminfσ n limsupσ n limsups n + ε. Since ε is arbitrary we obtain (2.3.5) liminfs n liminfσ n limsupσ n limsups n Since we assume the series to converge, all the inequalities above are in fact equalities and limσ n = lims n On the other hand if a n =( 1) n we have S n = 1 if n is even, and S n = 0 for n odd. S n clearly does not have a limit and yet, σ n 1/ Continuity (R-valued functions ) We consider real-valued functions defined on a set E in a metric space (X,ρ). There is no loss of the picture in assuming that (X,ρ) is R with ρ(x,y)= x y, or R d with the Euclidean metric. The metric on E is the metric induced by (X,ρ) Limits of a function at a point. Assume that x is a limit point of a set E in a metric space (X,ρ), and f a real-valued function defined on E. DEFINITION: limsup y x, y E f (y)=lim ε 0 sup{ f (y) : y B(x,ε) E} liminf y x, y E f (y)=lim ε 0 inf{ f (y) : y B(x,ε) E}. Observe that the limits exist since sup{ f (y) : y B(x, ε) E} and inf{ f (y) : y B(x,ε) E} are monotone in ε. The reference to the set E is omitted if E is understood The oscillation o f (x) of a function f at a point x. The oscillation is defined by (2.4.1) o f (x)=limsup y x An equivalent definition is given by f (y) liminf y x f (y) (2.4.2) o f (x)= limsup f (y 1 ) f (y 2 ) y j x, j=1,2

11 20 2. Limits and convergence. in other words, o f (x) is the least upper bound of f (y 1 ) f (y 2 ) with y j arbitrarily close to x. That means that o f (x) > a if for all ε > 0 there exist y 1 and y 2 in B(x,ε) such that f (y 1 ) f (y 2 ) > a. Remark: The definition of o f in terms of limsup and liminf, (2.4.1), uses the order on R. The second definition uses the metric rather than the order and while it is equivalent for real-valued functions, it applies equally, with f (y 1 ) f (y 2 ) replaced by ρ( f (y 1 ), f (y 2 )) to mappings into a general metric space. ex Prove that for any real-valued f and any constant B, the set {c : o c ( f ) B} is closed Continuity at a point. DEFINITION: A real-valued function f defined on a set E is continuous at a point x 0 E if for all ε > 0 there exists δ > 0 such that if y E and y x 0 < δ then f (y) f (x 0 ) < ε. An equivalent definition is: f is continuous at x 0 if o f (x 0 )=0. The number δ depends on the complete configuration: the set E, the function f, the point x 0 and the number ε. We can be explicit by writing δ = δ(e, f,x 0,ε). ex Assume that both f and g are functions defined on a set E and both are continuous at a point x 0 E. Prove that f + g and fg are continuous at x Continuity on a set. DEFINITION: A real-valued function f defined on a set E is continuous on E if it is continuous at every x E. We denote by C R (X) the set of all 2 real-valued functions defined and continuous on X. ex Prove that if f, g C R (X) then f + g and fgare in C R (X). 2 The symbol C(X) commonly denotes the space of all continuous complex valued functions on X.

12 2.4. Continuity (R-valued functions ) Uniform continuity. DEFINITION: A real-valued function f defined on a set E is uniformly continuous on E if for all ε > 0 there exists δ = δ(e, f,ε) > 0 such that if if y,x E, y x < δ then f (y) f (x) < ε. Uniformity means that for every ε > 0 one has a positive δ common to all the points x in E. Theorem. Let f be a real-valued function defined on a compact set E R, and continuous at every x E. Then a. f is bounded on E. b. f is uniformly continuous on E. c. The range of f is closed; in particular f assumes the values sup x E f (x) and inf x E f (x). The continuity of f implies that for every ε > 0, and every x E, there exists δ = δ(x) > 0 such that if y E and y x < 2δ, then f (y) f (x) < ε/2. PROOF: a. The open balls B(x,δ(x)), x E form an open cover of E and since E is compact, there is a finite subcover, that is, a finite set G E such that x G B(x,δ(x)) = E. The boundedness of f follows from the fact that for all y E, f (y) ε + max x G f (x). b. To prove the uniform continuity we have to show that for every ε > 0 there exists δ > 0 such that if y,z E and z y < δ, then f (y) f (z) < ε. Keeping the notation we have above, we claim that δ = min x G δ(x) has the needed property. Let y,z E, z y < δ. Let x G be such that y B(x,δ), and hence z B(x,2δ). Now, f (y) f (x) < ε/2 and f (z) f (x) < ε/2 so that f (y) f (z) < ε. c. Let y be a limit point of the range of f. Let x n E be such that lim n f (x n )=y and let x 0 be a limit point of {x n }, (its existance guaranteed by the Bolzano-Weierstrass theorem) then f (x 0 )=lim n f (x n )=y, and y is in the range of f.

13 22 2. Limits and convergence. The compactness assumption is crucial as can be seen by the following examples: E =(0,1), f (x)=1/x, (E is not closed). E = R, f (x)=x 2, (E is not bounded). ex Prove that the range of a continuous function on a compact set is closed. (Since, by a. above, the range is also bounded this shows that it is in fact compact.) Hint: Repeat the proof of c. above. ex Let f be a continuous map from a compact metric space (X,ρ) into a metric space (Y,ρ ). Assume that f is a bijection (1-1 map) form its domain X onto its range Y = f (X) Y. Prove that the inverse map f 1 is continuous on Y. ex Prove that a real-valued function whose domain is a metric space (X,ρ) is continuous on X if and only if the pre-image f 1 (O)= {x X : f (x) O} of every open set O R is open in X. ex Let f be a continuous function defined on a compact metric space. Use the characterization given in the previous exercise to prove that the range of f is compact. ex The range of a continuous map from a compact metric space into a metric space is compact. ex A uniformly continuous function f on [0, 1] maps Cauchy sequences to Cauchy sequences. ex Let f =(f 1,, f d ) be an R d -valued function on (X,ρ). a. Prove that f is continuous at a point x 0 X if and only if every component function f j, j = 1,...,d, is continuous at x 0 b. Prove that f is uniformly continuous on X if and only if every component function f j, j = 1,...,d, is uniformly continuous Equicontinuity. Let F be a set of continuous real-valued functions defined on a metric space (X,ρ). DEFINITION: F is equicontinuos at a point x X if for every ε > 0 there exist δ = δ(x,ε) such that if y B(x,δ) then for every f F, f (y) f (x) < ε. F is equicontinuous on X if it is equicontinuous at every x X.

14 2.5. Connectedness Connectedness DEFINITION: A metric space (X,ρ) is connected if it is not the union of two disjoint nonempty closed subsets. Since the complement of an open set is closed and the complement of a closed set is open, we can define connectedness by: (X,ρ) is connected if it is not the union of two disjoint nonempty open subsets. Proposition. An interval I R is connected. PROOF: Proof by contradiction. We consider first the case that I = [a,b] is bounded and closed. Assume I = E 1 E 2, with E 1 and E 2 nonempty and closed, numbered so that a E 1. Write c = infe 2. As [a,c) E 1, c is a limit point of E 1 and, as E 1 is closed, we have c E 1. On the other hand, being the infimum of the closed set E 2 we have c E 2, contradicting the assumption that E 1 and E 2 are disjoint. For a general interval J, that may be open, half open, unbounded, we assume J = E 1 E 2 with E j nonempty, closed and disjoint, take a E 1, b E 2, write I =[a,b],e j = E j I, and apply the case where I =[a,b] is bounded and closed.. Theorem. A continuous real-valued function defined on a connected space X (e.g. on [0,1]) has the intermediate value property, that is, if x 0,x 0 X, f (x 0 )=af(x 1 )=b and a < c < b, then there is y X such that f (y)=c. ex A closed subset E R is connected if and only if it is an interval. The interval can be bounded or extend to +, or to or to both. ex An interval is charecterized by the property: if a and b are points in it and a < c < b then c is in it as well. ex The range of a real-valued continuous function on a connected metric space is an interval. ex The range of a continuous map from a connected metric space into a metric space is connected.

15 24 2. Limits and convergence. 2.6 Series of functions Pointwise convergence. We now consider series sequences and series of (real-valued) functions defined on a finite interval, say on [0,1], or, more generally, on a metric space (X,ρ). The convergence of such a sequence or a series at a point x 0 is the convergence of a sequence or of a series of real numbers, which was studied earlier. Here we consider the Pointwise convergence of a sequence (or a series) of continuous functions, and the uniform convergence of such. Since the convergence of a series is defined by the convergence of the sequence of its partial sums (or, in the case of summability, of averages thereof) we focus on sequences. DEFINITION: The sequence { f n } converges pointwise on [0,1] if it converges at every point in the interval. We know that for real-valued functions convergence at a point x 0 is equivalent to the sequence { f n (x 0 )} being a Cauchy sequence, i.e., ε > 0, N = N(ε,x 0 ) such that if n,m > N then f n (x 0 ) f m (x 0 ) < ε. The sequence converges pointwise if the condition is satisfied for every x 0 [0,1] Uniform convergence. In the context and notation of the previous subsection, assume that { f n (x)} converges to f (x) everywhere on [0,1]. DEFINITION: The sequence { f n } converges to f uniformly means: ε > 0 N = N(ε) such that f n (x) f (x) ε for all x [0,1] and n > N. The uniformity is the fact that, for all ε > 0, there is a common N(ε) which accommodates all the points x [0,1]. Theorem. a. The sequence { f n } converges uniformly on on [0,1] if, and only if it is uniformly Cauchy, i.e., ε > 0, N = N(ε) such that for all x [0,1], if n,m > N then f n (x) f m (x) < ε. b. A uniform limit of a sequence { f n } of continuous functions is continuous. ex Assume f continuous on E and g(x) f (x) < ε for all x E. Prove: o g (x) < 2ε for all x E.

16 2.6. Series of functions Equicontinuity A set { f n } of real-valued functions on [0,1] is equicontinuous at a point x 0 if for every ε > 0 the exists δ(ε) > 0 such that if x 0 y < δ then f n (x 0 ) f n (y) < ε for all n. { f n } is uniformly equicontinuous if for every ε > 0 there exists δ(ε) such that if x y < δ then f n (x) f n (y) < ε for all n and for all x. The uniform continuity of a function is the fact that δ(ε) can be chosen so as to accommodate all x for the given function. The equicontinuity is the fact that δ(ε) can be chosen to accommodate all the functions f n (either at a specific point or at every point).

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