Supplementary Notes for W. Rudin: Principles of Mathematical Analysis

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1 Supplementary Notes for W. Rudin: Principles of Mathematical Analysis SIGURDUR HELGASON In 8.00B it is customary to cover Chapters 7 in Rudin s book. Experience shows that this requires careful planning especially since Chapter is quite condensed. These notes include solutions of Exercises 3 6, Chapter because these help in understanding the abstract compactness notion in.3, and makes it more useful in analysis. Since the course is preceded by which deals with the techniques of differential equations without proofs it seems very desirable to go through Exercise 7 in Ch. 5 and Exercise 5, Ch. 7. This gives concrete applications of the general theory in the course, consolidating For reasons of time some omissions seem advisable. This includes the Appendix in Ch., and the section on Rectifiable Curves in Ch. 6. Also taking α(x) = x throughout Chapter 6 seems practical. The notions are then familiar and quite a bit of time is saved. The book has a number of exercises of considerable theoretical interest. However, they are sometimes too difficult to assign to students and as a result they are often omitted altogether. Occasional unproved statements in the text may also baffle the beginner. It is the purpose of these notes to remedy this situation to some extent. Some of this material, like Lusin s Example, related to 3.44, is of course optional. Chapter On p. we define for each x R, x > 0 the set E = {n 0 + n n k 0 k : k = 0,,...}, where for each k, n k is the largest integer such that Proposition. x = supe. n 0 + n n k 0 k x. (Proof not given on p.): We must prove that if y < x then y is not an upper bound of E. Let r k = x [ n 0 + n n ] k 0. k Then r k 0. We shall prove by induction on k that () r k < 0 k. This is clear if k = 0. Assume () for k. If n k+ < 9 then r k+ < 0 k+ (proving ()) because otherwise we could replace n k+ by n k+ +. If n k+ = 9 then r k+ = r k 9 0 k+ < 0 k 9 0 k+ = 0 k+

2 proving () in this case too. Now choose p Q such that 0 < p < x y and then an integer k > 0 such that 0 k < p < x y. Then y < x 0 k < x r k. Since x r k E, y is not an upper bound of E. The following exercise is an analog of Exercise 5. Exercise 5a. Let A be a bounded set of real number x > 0 and A = {x : x = A}. Then () supa = /inf(a ). Proof: The upper bounds of A are just the reciprocals of the lower bounds of A. The least of the former are thus the reciprocals of the largest of the latter. Thus () holds. Exercise 6. Solution (a) If m > 0 then p > 0 and (3) (b m ) q = (b p ) n = b mq. If m < 0 then p < 0 and b m = ( b) m so by (3) (b m ) q = ( ( b) m ) q = ( b) mq = b mq. Similarly (b p ) n = b np. Thus (b m ) n and (b p ) q have the same nq th power so must coincide. (b) By definition if p,q J, α p q = (α p ) q so (4) ( α p q ) q = α p. Let r = m/n, s = p/q and raise the numbers b m n + q = b mq+np nq and b m n b p q to the nq th power. Using (4) the result is the same in both cases. By the uniqueness in Theorem., (5) b r+s = b r b s. (c) Since b > we have b m n > for m,n J because b m n leads to b m. Thus if r Q, t Q,t r we have by (5) (6) b r = b t b r t b t so supb(r) = b r. Hence we define for x R We shall prove b x = supb(x). (7) sup b t = inf t x,t Q r x,r Q br.

3 We have by (6). Let L be the difference between the right and left side in (7). Suppose L > 0. Select n J so large that b x ( b n ) < L (see p. 58). Then select r,t Q such that x < r < n + x, x n < t < x. Then r t < n so b r b t = b t (b r t ) b x (b r t ) < L, which is a contradiction, proving (7). (d) We have supb(x + y) = sup{b t : t Q, t x + y}. The last set contains {b r+s : r x,s y,r,s Q} so by (5) and since r and s are independent this equals Hence supb(x + y) sup{b r b s : r x,s y}, sup{b r : r x}sup{b s : s y}. (8) b x+y b x b y. By Exercise 5a above and (7) (s,t being in Q) (9) b x = supb t = t x If b x b y were < b x+y then by (8) (9) This contradiction proves b x+y = b x b y. Chapter inf t x b t = inf s x b s = b x. b y = b x (b x b y ) < b x (b x+y ) b y. In connection with Theorem. the following result is useful. The proof is quite similar. Theorem.a If E X and E α X for each α, then If f : X Y is a mapping then If F Y, F β Y for each β then E ( α E α ) = α (E E α ), E ( α E α ) = α (E E α ). f( α E α ) = α f(e α ), f( α E α ) α f(e α ). f ( β F β ) = β f (F β ), f ( β F β ) = β f (F β ) f (F c ) = (f (F)) c, f(f (F)) F, E f (f(e)). Remark There are examples such that 3

4 (i) f(a B) f(a) f(b), (ii) f(e c ) (f(e)) c, (iii) f(f (F)) F, (iv) f (f(e)) E. For (i) take a case when A B =. For (ii) take a case when f maps X onto Y. For (iii) take a case when F = Y and f a constant map. For (iv) take a case when E is a single point and f a constant map. Notation If A X, B X, we write A B = {x A : x / B}. Theorem.7 (Improved version.) Let X be a metric space and E X. Then (a) p E if and only if each neighborhood N of p intersects E (i.e., N E ). (b) E is closed. (c) E = E if and only if E is closed. (d) E F for every closed set F X such that E F. (e) Anticipating Definition 3. we have Proof p E if and only if there exists a sequence (p n ) E such that p n p. (a) If p E then either p E or p is a limit point of E. In both cases each neighborhood N of p intersects E. On the other hand suppose p / E. Then p / E. Thus there exists a neighborhood N of p such that N (E p) =. But p / E so N E = proving a). (b) If p / E then by a) there exists a neighborhood N of p such that N E =. We claim that N E = (which would show X E open). Otherwise there is a q E N. But then N is an open set containing q so by a) N E which is a contradiction. (c) If E = E then by b), E is closed. If E is closed then by definition E E so E E. (d) If F is closed and F E, then F F so F E whence F E. (e) Let p E. Using a) we can for each n J pick a point p n N /n (p) E. Then d(p n,p) < n so p n p. On the other hand if (p n ) is a sequence in E such that p n p then p E by a). Theorem.37 (Improved version.) A space X is compact if and only if each infinite subset has a limit point. The only if part is proved in the text. The if part is Exercise 6 which we shall prove along with Exercises 3, 4. Exercise 3(Stronger version.) Let X be a metric space. Then X is separable if and only if X has a countable base. 4

5 Proof: First assume X has a countable dense subset P = {p,p,...}. Consider the countable collection N of neighborhoods N r (p i ), r Q, i J. Let G be open and x G. Then N ρ (x) G for some ρ Q. Since P is dense in X, i.e., P=X, P N ρ/ (x) contains some p i. Then d(p i,x) < ρ/ so x N ρ/ (p i ). If y N ρ/ (p i ) then d(x,y) d(x,p i )+d(p i,y) < ρ so y G. Thus x N ρ/ (p i ) G so G is the union of some members of N. Thus N is a countable base. Conversely, suppose {V i } i J is any countable base for X. Pick x i V i for each i and let F = {x,x,...}. Then F = X because otherwise the open set X F would be the union of some of the V i contradicting x i F. Exercise 4. Let X be a metric space in which each infinite subset has a limit point. Then X is separable. Proof: Let δ > 0 and let x X. Pick x X (if possible) such that d(x,x ) δ. Having chosen x,...,x j choose x j+ (if possible) such that d(x i,x j+ ) δ for i j. This process has to stop because otherwise the set E = (x,x,...) would be an infinite subset of X without limit points. Consequently X is covered by finitely many sets N δ (x i ). Now take δ = n (n =,,...). The construction gives: n =, x,...,x j X = j i= N (x i ) n =, x,...,x j X = j i= N (x i ), etc. Let S consist of these centers x ki k J, i j k. Then S is dense in X (i.e., S = X). In fact let x X and N r (x) any neighborhood of x. Choose k such that k < r and x ki such that x N (x ki ). Then x ki N (x) N r (x) so N r (x) S as claimed. k k Exercise 6. Let X be a metric space in which every infinite subset has a limit point. Then X is compact. Proof: By Exercises 3 and 4, X has a countable base {B i } i J. Let {G α } α A be any open covering of X. If x X then x G α for some α and thus there is a B i such that x B i G α. Let {B i } i I be the sub-collection of {B i } i J consisting of those B i which are contained in some G α. For each B i (i I) let G i be one of the G α containing B i. Then (0) X = i I G i. Thus we have obtained a countable subcovering of the {G α}α A. We shall now show that a suitable finite collection of the G i covers X, thus proving the compactness of X. Let I = {i,i,...}. If no finite sub-collection of {G i } i I covers X then for each n whereas by (0) F n = (G i G in ) c, () F n =. 5

6 Of course we have () F F F n. We select x i F i for each i and let E = {x,x,...}. For each x E let N x denote the set of indices p for which x F p. Then N x = J. x E If E were finite some N x would be infinite and x would belong to F p for infinitely many p whence by () x would belong to F p for all p. This would contradict () so we conclude that E is an infinite subset of X. By assumption it has a limit point z. Let N be any neighborhood of z. Then N (E z) so N E. Thus N intersects some F n so by () N F. Since N was arbitrary we have z F. But F is closed so z F. Now consider the sequence F F 3 F n (with F removed) and E = E x = {x,x 3,...}. The point z is still a limit point of E so by the argument above z F. Continuing we deduce z F i contradicting (). This proves that X is compact. Corollary. Let X be a metric space. The following conditions are equivalent: (i) X is compact. (ii) Each sequence (x n ) in X has a convergent subsequence. (iii) Each infinite subset of X has a limit point. Proof: (i) (ii): This is Theorem 3.6(a). (ii) (iii): Let E X be an infinite subset. Let (x n ) be a sequence of different points in E. By (ii) it has a subsequence (x ni ) converging to a limit x. Then each N r (x) contains infinitely many x ni. Since x ni = x for at most one i this implies N r (x) (E x) so x is a limit point of E. Part (iii) (i) was Exercise 6. Chapter 3 Exercise. Every sequence (x n ) in R has a monotone subsequence. Proof: This is clear if (x n ) is unbounded. If x n is bounded it has a subsequence converging to say x. Considering x n x we may assume x = 0. Considering signs we may assume all terms positive. A positive sequence converging to 0 clearly has a decreasing subsequence. Exercise 3. Solution: We have s =, s n+ = + s n. (s n ) is increasing by induction: If s n+ s n 0 then Also s n < by induction since + <. s n+ s n+ = ( + s n+ ) ( + s n ) 0. 6

7 Exercise 6. Solution: We have α > 0, x > α and ) x n+ = (x n + α n =,3,.... xn Show (x n ) is decreasing and that x n α. For this note that ( x n+ = xn ) α ) xn + α, xn x n+ = (x n αxn. The first relation shows by induction that x n > α and them the second relation shows that (x n ) is decreasing. The definition then shows x n α. Exercise 7. Solution: Here one proceeds similarly using the relations Lusin s Example (Compare Theorem 3.44.) x n+ α = ( α )( α x n ) + x n x n+ x n = (α x + α + x n). n There exists a power series Σa n z n with a n 0 and radius of convergence, diverging everywhere on the circle z =. Proof: Let g m (z) = + z + z m h m (z) = g m (z) + z m g m (e πi/m z) + + z mk g m (e πik/m z) + z m(m ) g m (e πi(m )/m z). Notice that when the individual terms in h m (z) are written out as polynomials in z there is no overlap. Thus h m (z) becomes a polynomial in z (of degree (m+)(m )) in which each coefficient has absolute value. Also if e iϕ The graph of sin x shows Thus Consider now the circular arc g m (e iϕ ) = emiϕ e iϕ = sin mϕ sin ϕ. π x sin x x for 0 x π. g m (e iϕ ) π mϕ ϕ = π m for ϕ π m. A = {e iϕ : ϕ π m }. For each point e iϕ on z = we can find an integer k such that e πki m e iϕ A. In fact the arcs e πki m A (0 k m) cover the circle. It follows that ) (3) Max 0 k m g m ( e πk m i z m for z =. π 7

8 Consider now the series m= m z + + +(m ) h m (z) = n a n z n. Again there is no overlap for different m because the m th term contains powers of z of degree at most (m ) + (m + )(m ) which is less than m which is the lowest degree of z appearing in the (m + ) th term. Now suppose the series did converge at a point z with z =. Then the sequence of terms a n z n must converge to 0. This remains true for each subsequence. The absence of overlaps noted thus implies lim m This would contradict (3). ( ) z + + +(m ) Max 0 k m z mk g m e πki m z = 0. m Corollary. There exists a power series n c nz n with radius of convergence which converges for z = but diverges everywhere else on the circle z =. Proof: With Σa n z n as in Lusin s example put g(z) = a 0 a 0 z + a z a z 3 + a z 4 a z 5 +. Clearly we have convergence for z =. If we had convergence at a point z then the series would also converge and so would the series a 0 ( z) + a z ( z) + a 0 + a z + a z 4 + contradicting the theorem. Exercise 9. Solution (cf. Mattuck: Introduction to Analysis). The Cantor set C is given as follows: C 0 = [0,], C = C 0 ( 3, ) 3, C = C ( 9, ( 9) 7 9, 8 9), etc. and C = n C n. Now represent each x = C 0 by infinite decimal expansion with base 3, i.e., x = α α = α 3 + α 3 +. Here we avoid as much as possible: we replace each ternary decimal ending with by an equivalent ternary decimal ending with an infinite string of. Then 0 if 0 x 3 α = if 3 < x < 3 if 3 x Thus C = {x C 0 : α }. Similarly α is found by taking the interval of length /3 containing x and dividing it into three subintervals of length /9. Then α = 0, or according to which of. 8

9 these intervals contain x just as above. Thus Continuing in this way we see that C = {x C 0 : α,α }. C = {x x = α 3 + α α n 3 n +, each α i }. Exercise 4. (Completion of a metric space.) We prove Part d) before Part c). (a) Easy from the definition. (b) If {p n } {p n} and {q n } {q n} then and exchanging p n and p n,q n and q n d(p m,q n) d(p n,p n ) + d(p n,q n ) + d(q n,q n) d(p n,q n) d(p n,q n ) d(p n,p n ) + d(q n,q n ) so (P,Q) is well defined. Note that if (P,Q) = 0 then {p n } {q n } so P = Q. The inequality (P,R) (P,Q) + (Q,R) is easy. (d) With the notation indicated we have (P p,p q ) = lim d(p,q) = d(p,q), n so the mapping ϕ : p P p from X into X is distance preserving. If {p n } P then by b) (P,P p ) = lim d(p,p n). n In particular, (P pm,p) = lim d(p m,p n ) n so since {p n } is Cauchy sequence, (4) lim m (P p m,p) = 0. Thus given any Cauchy sequence {p n } and ǫ > 0 there is a constant sequence whose equivalence class is within ǫ of the equivalence class of {p n }. This means that ϕ(x) is dense in X. (c) X is complete. Let P,P,... be a Cauchy sequence in X. Given ǫ > 0 choose for each n (by (4)) a p n such that (P pn,p n ) < ǫ. We now prove that P p,p p,... is a Cauchy sequence in X. In fact (P pm,p pr ) (P pm,p m ) + (P m,p r ) + (P r,p pr ) ǫ + (P m,p r ) + ǫ. The left hand side equals d(p m,p r ) so the inequality shows that (p n ) is a Cauchy sequence in X. Let P denote its equivalence class. Since (P,P m ) (P,P pm ) + (P pm,p m ) we deduce from (4) that (P,P m ) 0 proving c). (e) If X was complete to begin with let P X and select {p n } P. Then {p n } is a Cauchy sequence in X so p n p for some p X. Then {p} {p n }, so P p = P, i.e., ϕ(p) = P. 9

10 Chapter 4 Remark 4.3. Given any countable subset E of (a, b) there exists a function f monotonic on (a, b) discontinuous at each point of E and at no other point of (a,b). In fact if n c n is convergent and c n > 0 then the text states that f(x) = c n, x n<x where E consists of the points x,x,..., is such a function. We shall prove this in the following steps: (a) f is increasing. (b) f is continuous at each x (a, b) outside E. (c) f(x) = f(x ) for all x. (d) f(x + n ) f(x n ) = c n. (a) This part is obvious. (b) Here we separate two cases. (i) x / E and x is not a limit point of E. (ii) x / E and x is a limit point of E. In case (i) there is a neighborhood N δ (x) such that N δ (x) E =. Then f(y) = x n<y c n is constant for y (x δ,x + δ), hence continuous at x. In case (ii) let ǫ > 0 and N a number such that N+ c n < ǫ. Let δ > 0 be such that 0 < δ < x x n for n =,...,N. Then if x k (x δ,x + δ) we have k > N. Let y (x δ,x + δ). Then f(y) = c n, f(x) = c n. x n<y Since x k (x δ,x + δ) implies k > N the sums above differ by at most k>n c k which is < ǫ. Thus f(x) f(y) < ǫ if y (x δ,x + δ) so f is continuous at x. (c) Here the relation remains to be proved for x = x n E. Again we consider the two cases (i) x n is an isolated point of E. (ii) x n is a limit point of E. x n<x 0

11 In case (i) choose δ > 0 such that N δ (x n ) E = x n. Then if x n δ < y < x n we have f(x n ) = c k = c k = f(y) x k <x n x k <y so f(x n ) = f(x n ). In case (ii) let ǫ > 0 be given and choose N such that k>n c k < ǫ and then choose δ > 0 such that 0 < δ < x n x m for m =,,...,N, m n. Then if x k E (x n δ,x n ) we have k > N. Thus if y (x n δ,x n ) f(x n ) = c k = c k + x k <x n x k <y y x k <x n c k f(y) + k>n c k f(y) + ǫ. Hence 0 f(x n ) f(y) ǫ so f(x n ) = f(x n ) also in this case. (d) Again we consider two separate cases: (i) x n isolated in E. (ii) x n limit point of E. so In case (i) select δ > 0 such that N δ (x n ) E =. Then if y (x n,x n + δ) f(y) = c k = c n + c k = c n + f(x n ) x k <y x k <x n f(x + n ) = c n + f(x n ) = c n + f(x n ). In the case (ii) let ǫ > 0 be given and select N such that k>n c k < ǫ. Again select δ > 0 such that δ < x n x m for m =,,...,N, m n. If x k E (x n,x n + δ) then k > N. Thus if y (x n,x n + δ) so again f(x + n ) = f(x n) + c n. f(y) = c k = c k + c n + x k <y x k <x n = f(x n ) + c n + η where η < ǫ x n<x k <y c k Exercise 5. A continuous open mapping of R into R is monotonic.

12 Proof: First we show that f is one-to-one. If not suppose x < x and f(x ) = f(x ). Let y and z be points in [x,x ] where f takes its maximum and minimum, respectively. If y (x,x ) then f(y) is an end point of f((x,x )) contradicting the openness of f. Similarly z (x,x ) is impossible. Thus both y and z are end points of [x,x ] so since f(x ) = f(x ), f is constant in [x,x ], contradicting the openness of f. Thus f is one-to-one. Consider any x < x and assume f(x ) < f(x ). We shall prove f increasing on [x,x ]. If not, select x 3 < x 4 in the interval such that f(x 3 ) > f(x 4 ). Put y i = f(x i ) ( i 4). If y 3 y then y 4 < y so by Theorem 4.3 f takes the value y 3 on (x 4,x ). This contradicts the one-to-one property. Thus y 3 > y > y so f takes the value y on (x,x 3 ) contradicting x 3 < x. The argument applies to x = n, x = n and shows f monotonic on ( n,n). Hence f is monotonic on R. Chapter 5 Exercise. (Generalized.) Let X be a complete metric space and f : X X a contractive map, i.e., there is a constant k < such that d(f(x),f(y)) k d(x,y) for all x,y X. Then f has a fixed point. Proof (Richard Palais): By the triangle inequality so d(x,y) d(x,f(x)) + d(f(x),f(y)) + d(f(y),y) d(x, y) [d(x,f(x)) + d(f(y),y)]. k Replace here x by the iterate f (n) (x), y by f (m) (x). Then d(f (n) (x),f (m) (x)) k [d(f(n) (x),f (n+) (x)) + d(f (m) (x),f (m+) (x))] k (kn + k m )d(x,f(x)). Thus for each x X f (n) (x) is a Cauchy sequence and its limit is a fixed point for f. Exercise 5. b) Here one has only x n+ x n (not necessarily x n+ < x n as the text states). In fact the case f(x) = x shows this. In b) one can use the fact that f (x) 0 implies that the curve always lies above its tangent: In fact by the mean value theorem there exists some ξ between x and x and some η between ξ and x such that f(x) [f(x ) + f (x )(x x )] = (f (ξ) f (x ))(x x ) Thus f(x ) 0 and by induction f(x n ) 0. = f (η)(ξ x )(x x ) 0.

13 Chapter 7 Remark on (38) p.54. The sign determination in (38) is made as follows: Consider the intervals I : whose (disjoint) union is the half-open interval 4 m x 4 m δ m t < 4 m x I : 4 m x t < 4 m x + 4 m δ m I : 4 m x 4 m δ m t < 4 m x + 4 m δ m which has length. Exactly one of the intervals I, I contains an integer; if I does we take δ m > 0; if I does we take d m < 0. In the first case the interval and in the second case the interval [4 m x,4 m x + 4 m δ m ) contains no integer [4 m x + 4 m δ m, 4 m x) contains no integer. In both cases ϕ is linear on the interval in question and this results in the crucial fact that γ m = 4 m. Exercise 4. f(x) = (a) If x = 0 series diverges. +n x (b) If x = k for some k J some term is undefined so we have divergence. (c) If k < x < (k+) for some k =,,3 we write Since f(x) = k+ n= + n x + x x +. n k+ (5) (k + ) < x < k we have n + x n (k + ) and k+ converges. By Theorem 7.0 we have uniform convergence on each interval n (k+) (5). (d) If < x < we write (6) f(x) = + x + x Again n +. x n + x n x n so by Theorem 7.0 we have uniform convergence on < x <. 3

14 (e) Writing f in the form (6) it is clear that the series n + x converges uniformly on (0, ) since n + x > n. But what about the series (7) + n x? For this put f n (x) = n k= +k x. If we had uniform convergence of (7) on (0, ) then by Theorem 7. the sequence (A n ) given by (8) A n = lim x 0 f n (x) would converge. However A n = n so this is not the case. Thus (7) converges uniformly on each interval [α, ), α > 0 but not on (0, ). (f) Since the proofs above always involved absolute values we have absolute convergence except for x = k, (k =,,...) and x = 0. The function f is continuous except at these points. (g) Is f(x) bounded? No, if f(x) < K for all x then for x > 0 so by (8) A n K for each n which is false. f n (x) K for each n Then Exercise 4. To see that f(3 k t 0 ) = a k we write t 0 = k i=0 a i 3 i+ + a k 3 k+ + k+ 3 k t 0 = p + a k 3 + z a i 3 i+. where p is an integer and 0 z 3. If a k = 0 then f(3 k t 0 ) = 0 = a k and if a k = then f(3 k t 0 ) = = a k. Referring now to Exercise 9 in Chapter 3 we see that Φ does indeed map the Cantor set onto the unit square. Exercise 5 (Remark on Part b.) In general if F is a family of functions whose derivatives are uniformly bounded then by the mean-value theorem (5.0) the family F is equicontinuous. This does not quite work here since f n is not differentiable at the points x i = i/n. 4

15 However, we can proceed as follows: Let 0 x < y. Select the smallest i for which x x i and the largest k such that x i+k y. (If y x i the relation f n (x) f n (y) M y x is obvious.) Then f n (x) f n (y) = f n (x) f n (x i ) + f n (x i ) f n (x i+ ) + + f n (x i+k ) f n (y). We can use the mean-value theorem on each difference on the right giving an expression which is bounded by M x y. f n (ξ i)(x x i ) + f n (ξ i+)(x i x i+ ) + + f n (ξ i+k+)(x i+k y) 5