数学分析原理部分参考答案.pdf

chapter one 3 Prove that if f is a real function on a measurable space X such that x f x r is measurable for every rational r then f is measurable proof For each real number there exists an descending sequence rn of rational numbers such that lim n rn Moreover we have n 1 rn Hence f 1 n 1 f 1 rn Since sets f 1 rn are measurable for each n the set f 1 is also measurable Then f is measurable 4 Let an and bn be sequences in and prove the following assertions a limsup n an liminf n an b limsup n an bn limsup n an limsup n bn provided none of the sums is of the form c If an bnfor all n then liminf n an liminf n bn Show by an example that strict inequality can hold in b proof a Since sup k n a k inf k n ak n 1 2 Therefore let n it obtains lim n sup k n a k lim n inf k n ak 1 By the defi nations of the upper and the lower limits that is limsup n an liminf n an b Since sup k n a k bk sup k n ak sup k n bk n 1 2 Hence lim n sup k n a k bk lim n sup k n ak sup k n bk lim n sup k n ak lim n sup k n bk By the defi nations of the upper and the lower limits that is limsup n an bn limsup n an limsup n bn example we defi ne an 1 n bn 1 n 1 n 1 2 Then we have an bn 0 n 1 2 But limsup n an limsup n bn 1 c Because an bnfor all n then we have inf k n ak infk n bk n 1 2 By the defi nations of the lower limits it follows liminf n an liminf n bn 5 a Supose f X and g X are measurable Prove that the sets x f x g x f x g x are measurable b Prove that the set of points at which a sequence of measurable real valued functions converges to a fi nite limit is measurable 2 proof a Since for each rational r the set x f x r g x x f x r x r g x is measurable And x f x g x r Q x f x r g x Therefore the set x f x g x is measurable Also beacause f g is measurable function and the set x f x g x n 1 x f x g x 1 n is also measurable b Let fn x be the sequence of measurable real valued functions A be the set of points at which the sequence of measurable real valued functions fn x converges to a fi nite limit Hence A r 1 k 1 n m k x fn x fm x 0 X E 0 then we can prove that the strict inequality in the Fatou s lemma can hold 10 Suppose that X fn is a sequence of bounded complex measurable functions on X and fn f uniformly on X Proved that lim n Z X fnd Z X fd and show that the hypothesis X 0 n 1 2 such that fn x Mnfor every x X n 1 2 Since fn f uniformly on X there exists M 0 such that fn x M f x M on X n 1 2 By the Lebesgue s dominated convergence theorem we have lim n Z X fnd Z X fd counterexample Let X and we defi ne fn 1 n on X n 1 2 12 Suppose f L1 Proved that to each 0 there exists a 0 such that R E f d whenever E 0 there exists a simple measurable function s with 0 s f such that Z X f d Z X sd 2 Suppose that M 0 satisfying 0 s x M on X and let 2M we have Z E f d Z E sd 2 Mm E 2 when m E 4 chapter two 1 Let fn be a sequence of real nonnegative functions on R1 and con sider the following four statements a If f1and f2are upper semicontinuous then f1 f2is upper semicon tinuous b If f1and f2are lower semicontinuous then f1 f2is lower semicon tinuous c If each fn is upper semicontinuous then P 1 fnis upper semicontin uous d If each fn is lower semicontinuous then P 1 fnis lower semicontinu ous Show that tree of these are true and that one is false What happens if the word nonnegative is omitted Is the truth of the statements aff ected if R1is replaced by a general topological space proof a Because for every real number x f1 x f2 x r Q x f1 x r x f2 x r Q x f1 x r x f2 x r and f1and f2are upper semicontinuous therefore f1 f2is upper semicon tinuous c This conclusion is false counterexample for each n N defi ne fn 1 x rn 0 others 1 which rn are the all rationals on R1 Then fn is a sequence of real non negative upper semicontinuous functions on R1 moreover we have f x X 1 fn x Q But f x is not upper semicontinuous d According to the conclusion of b it is easy to obtain If the word nonnegative is omitted a and b are still true but c and d is not The truth of the statements is not aff ected if R1is replaced by a general topological space 2 Let f be an arbitrary complex function on R1 and defi ne x sup f s f t s t x x x inf x 0 Prove that is upper semicontinuous that f is continuous at a point x if and only if x 0 and hence that the set of points of continuity of an arbitrary complex function is a G proof It is obvious that x inf x 1 k k N 0 and x 1 k x 1 k 1 k N For any 0 we will prove that the set x x is open Suppose that x 0 such that x 1 k N Take y U x 1 2k since U y 1 2k U x 1 k and then y y 1 2k x 1 k So we have y x x Therefore U x 1 2k x x 2 Thus it show that x x 0 it exists 0 such that x And so for any y x x we have f x f y x 0 it exists k N satifying f x f y 2 y x 1 k x 1 k and so x 1 k It shows x inf x 1 k k N 0 Finally since x x 0 n 1 x x 1 n then it shows that the set of points of continuity of an arbitrary complex function is a G 3 Let X be a metric space with metric For any nonempty E X defi ne E x inf x y y E Show that Eis a uniformly continuous function on X If A and B are disjoint nonempty closed subsets of X examine the relevance of the function f x A x A x B x to Urysohn s lemma Proof Since for any x1 x2 X and y E we have x1 y x1 x2 x2 y Hence E x1 x1 x2 E x2 3 and so E x1 E x2 x1 x2 Therefore it shows that Eis a uniformly continuous function on X In the following suppose that K V X K is compact subset and V is open subset Defi ne f x Vc x Vc x K x then 0 f 1 and f is continuous on X Moreover K f V 4 Examine the proof of the Riesz theorem and prove the following two statements a If E1 V1and E2 V2 where V1and V2are disjoint open sets then E1 E2 E1 E2 even if E1and E2are not in M b If E MF then E N K1 K2 where Ki is a disjoint countable collection of compact sets and N 0 proof a By the proof of the Riesz theorem there exists a G set G such that E1 E2 G and E1 E2 G Set G1 V1 G G2 V2 G then E1 G1 E2 G2and G1 G2 Therefore E1 E2 G1 G2 G1 G2 G V1 V2 G E1 E2 And since E1 E2 E1 E2 It shows E1 E2 E1 E2 b Since E MF there exist a sequence of disjoint compact subsets Kn such that Kn E n 1 i 0 Ki m E n 1 i 0 Ki m Kn 1 n n 1 2 which we order K0 4 Set K S n 1 Kn Since m E we obtain that m E n i 1 Ki 1 n n 1 2 Whence we have m E K m E n i 1 Ki 1 n n 1 2 Set N E K then it shows that m N 0 and E K N 5 Let E be Cantor s familiar middle thirds set Show that m E 0 even though E and R1have the same cardinality proof According to the construction of Cantor s set it shows that the set E is closed set and m E m 0 1 X n 1 2n 1 3n 0 11 Let be a regular Borel measure on a compact Hausdorff space X assume X 1 prove that there is a compact set K X such that K 1 but H 1 for every proper compact subset H of K Proof Defi ne the set K K X K is compact set with k 1 then X Order K K K Thus K is compact subset Assume that V X is an open set with K V Since X is compact V c is also compact Again since V c K Kc then there exist fi nite sets K 1 K nsuch that V c n k 1 Kc k 5 Because X 1 K Kc k 0 k 1 2 n and so V c 0 We therefore have V 1 Since is regular hence it shows that K 1 By the construction of K it follows that H 1 for any proper compact subset of K If U X is open set with U 0 it follows Uc 1 Ucis compact subset since X is compact and so K Uc by the construction of K Hence U Kc It thus shows that Kcis the largest open set in X whose measure is 0 12 Show that every compact subset of R1is the support of a Borel measure Proof Assume that is the Lebesgue measure on R1and K is a com pact subset of R1 Defi ne f Z K fd f Cc R1 Then is a positive linear functional on Cc R1 By Riesz representation theorem there exists a algebra M in R1which contains all Borel sets in R1 and there exists a unique positive measure on M such that f Z R1 fd f Cc R1 By theorem2 14 it shows K inf f K f K X sup f f X K and so X K K f 0 By the proof of Theorem 2 24 in text we have f x X n 1 tn x x Rk and 2ntnis the characteristic function of a Lebesgue measurable set Tn Rk n N According to the construction of Lebesgue measurable set there exist Borel measurable sets Fn En n N satisfying Fn Tn En m En Fn 0 n N Defi ne g X n 1 2 n Fn h X n 1 2 n En then g h are Borel measurable functions on Rkand g x f x h x Moreover g x h x a e m Secondly it is easy that the conclusion is correct for 0 f 0 such that if n N xn x and yn y 0 8 and for any 0 1 U P0 P X d P0 P P X x x0 and y y0 U P0 P X d P0 P P X x x0 and y y0 Moreover U P0 is compact subset of X Thus X is locally compact Haus dorff space In the following we order the set for f Cc X f x R y R such that f x y 6 0 Let K be the support of f K is compact Then there exists fi nite P1 P2 Pn K for any 0 1 such that K n j 1 U Pj and so f x1 x2 xn Since f n X j 1 Z f xj y dy it is easy that is a positive linear functional on Cc X There thus exist a measure associated with this by Theorem2 14 Let E be the x axis and 0 1 Defi ne V Pi E U Pi where Pi xi 0 it shows V is open and E V For any fi nite elements P1 P2 Pn E the set F n j 1 U Pj 2 is compact with F V By the Urysohn s lemma there exists f Cc X satisfying F f V So we have V f n Hence it shows V as n Therefore E Assume that K E is compact then it is obvious that K is fi nite set Take K P1 P2 Pm 9 For any 0 1 defi ne G m j 1 U Pj then g is open and K G By the Urysohn s lemma there exists g Cc X satisfying K g G Since K g 2 whence K 0 21 If X is compact and f X is upper semicontinuous prove that f attains its maximum at some point of X proof for every t f X defi ne Et x X f x t Assume that f does not attain its maximum at some point of X Then it shows X S t f X Et Since f is continuous Etis open sets Again since X is compact there exists fi nite ti i 1 2 n f X such that X n S i 1 Eti Take t0 max t1 t2 tn then we have f x 0 show that the convexity of ln implies the convexity of but not vice versa Proof Assume that x y a b 0and 1 Then x y x y x y 1 and so is convex on a b Since etis nondecreasing convex on R and eln it shows from the conclusion above that the convexity of ln implies the convexity of for 0 counterexample x x2is convex on 0 but ln x 2lnx is not convex Moreover for 0 it can show that the convexity of logc c 1 implies the convexity of 3 Assume that is a continuous real function on a b such that x y 2 1 2 x 1 2 y for all x and y a b Prove that is convex Proof According to the defi nition of convex function we only need to prove the case 0 1 2 Take E k 2n k 1 2 2n 1 n N For n 1 it is obvious that x y 2 1 2 x 1 2 y for all x and y a b For n 2 suppose that k k 4 k 1 1x 1 1 y 1 2 x y 2 y 1 2 x y 2 1 2 y 1 2 1 2 x 1 2 y 1 2 y 1 4 x 3 4 y 1 x 1 1 y Assume that the case k 2n 1 k 1 2 2n 1 1 is correct For any E with k 2n 1 k 2 n 1 n 2 and for all x and y a b it shows that i x 1 y x y 2 1 2 x 1 2 y x 1 y 2 when k 2n 1 ii for 1 k 2n 1 x 1 y k 2n x 1 k 2n y 1 2 k 2n 1 x 1 k 2n 1 y y Since k 2n 1x 1 k 2n 1 y a b and so x 1 y 1 2 k 2n 1x 1 k 2n 1 y 1 2 y k 2n x 1 2 1 k 2n 1 y 1 2 y x 1 y For any 0 1 there exists a sequence n E with lim n n Thus x 1 y lim n nx 1 n y lim n n x 1 n y x 1 y Therefore it shows that is convex on a b 4 Suppose f is a complex measurable function on X is a positive measure on X and p Z X f pd kfkp p 0 p Let E p p 0 a If r p s r E and s E Prove that p E b Prove that ln is convex in the interior of E and that is continuous on E c By a E is connected Is E necessarily open losed Can E consist of a single point Can E be any connected subset of 0 d If r p s prove that kfkp max kfkr kfks Show that this implies the inclusion Lr Ls Lp e Assume that kfkr for some r and prove that kfkp kfk as p Proof a If r p s there exist 0 1 and 0 1 such that 1 and p r s By H older s inequality it follows that kfkp p p Z X f pd Z X f r sd Z X f rd Z X f sd kfk r r kfk s s 3 As r E and s E it prove that p E b By a it reduces that E is convex Moreover E is connected Since E 0 the interior of E is an interval and denoted by a b For any s t in the interior of E and 0 1 and 0 1 it follows from a that s t s t So it proves ln s t ln s t ln s ln t That is ln is convex in the interior of E Moreover ln is continuous on a b Thus is also continuous on a b Provided that a E For any decreasing sequence sn of a b with lim n sn a it is easy to show that lim n f x sn f x a x X Since f x sn max f x a f x s1 x X and f x a f x s1 L1 by the Lebesgue control convergence theorem it follows that lim n sn a Similarly the other cases can be proved Hence is continuous on E c By a it shows that E is connected and convex E can be any interval on 0 For examples take X 0 and is the Lebesgue measure on X i For E a b 0 a b take f x x 1 b x 0 1 x 1 a x 1 ii For E a b 0 a b take f x x x 0 1 x 1 a 1 x 1 iii For E a b 0 a b take f x x 1 b x 0 1 x 1 a 1 x 1 4 iv For E a b 0 a b take f x x 1 b 1 x 0 1 x 1 a x 1 v For E take f 1 d If r p s by a there exist 0 1 and 0 0 it shows that kfkp 0 for any 0 p 2 0 For any p r 0 2 Z X f pd kfkp p Therefore lim n kfkp Thirdly assume that kfk 1 and kfkr 0 Without generality assume that f x 1 x X Thus For any p r it shows that kfkp p kfk r r That is kfkp kfk r p r And so lim p kfkp 1 kfk The general case can reduce to the case kfk 1 and kfkr 0 This is completed the proof 5 Assume in addition to the hypotheses of Exercise 4 that X 1 a Prove that kfkr kfks if 0 r s b Under what conditions does it happen that 0 r s and kfkr kfks c Prove that Lr Ls if 0 r s Under what conditions do these 5 two spaces contain the same functions d Assume that kfkr 0 and prove that lim p 0 kfkp exp Z X log f d if exp is defi ned to be 0 Proof a When s it is easy that Z X f rd kfkr X and so kfkr kfk If s x x s ris convex on 0 and by the Jensen s Inequality it shows Z X f sd Z X f r s rd Z X f r d s r and whence kfks kfkr b Assume that s and kfkr kfk 0 r s Then for any 1 p it shows by a that kfk1 kfkp kfkq and so f is constant almost everywhere by the Holder s inequality Secondly assume 0 r s 1 then by the Holder s inequality Z X f r 1d Z X f sd r s kfkr s kfkr r Therefore f r constant a e on X So it happen that 0 r s and kfkr kfks if and only if f constant a e on X c If 0 r 0 Firstly assume that there are infi nte pairwise disjoint members in Then it exists a sequence En X of pairwise disjoint measurable sets such that 0 En 1 n 1 2 6 Take 1 s t 1 it has P n 1n t and so f Ls But R X f rd R X P n 1n tr E nd P n 1n tr En P n 1n tr n P n 1n tr which 1 tr 2 Since 0 tr 1 it shows that P n 1n tr Hence f Lr Next we assume that there exist fi nite pairwise disjoint measurable sets at most in Provided that f Lr then there exists an increasing seqnence hn of simple measurable functions such that 0 hn x hn 1 x f x forallx X and hn x f x n x X Put t1 x h1 x tn x hn x hn 1 x n 2 3 then 2ntnis the characteristic function of a measurable set Tn X and Tn are pairwise disjoint Moreover f x X n 1 tn x x X According to the assumption there are fi nite members of Tn with nonzero measure at most Thus f equal to a simple function almost everywhere and so f Ls Therefore Lr Ls 7 d Assume that kfkr 0 then kfkp kfkr kfkp e R X ln f d 0 p 0 Since kfkr 1 F f 1 0 0 such that f x p 1 ln f x p x E f x p 1 ln f x p x F when p Thus kfkp 1 R E ln f pd R F ln f pd 1 p 1 p R E ln f d R F ln f d 1 p Then 0 c lim p 0 kfkp e R Eln f d R F ln f d Let 1 1 then c lim p 0 kfkp e R Eln f d R F ln f d e R X ln f d Hence lim p 0 kfkp e R X ln f d 8 11 Suppose 1 and suppose f and g are positive measurable functions on such that fg 1 Prove that Z fd Z gd 1 Proof Since f and g are positive measurable functions on such that fg 1 f 1 2and g 1 2are positive measurable functions on such that fg 1 2 1 By H older s inequality it follows that 1 Z fg 1 2d 2 Z fd Z gd 12 Suppose 1 and h 0 is measurable If A Z hd prove that 1 A2 Z 1 h2d 1 A Proof It is obvious correct when A or 0 and so we assume 0 A and h 0 Since x 1 x2 is convex on 0 it shows by the Jensen s Inequality that 1 A2 Z 1 h2d Because 1 h2 1 h Z 1 h2d 1 A This complete the proof 14 Suppose 1 p f Lp Lp 0 relative to Lebesgue measure and F x 1 x Z x 0 f t dt 0 x 0 and f L1 prove that F L1 Proof a Firstly assume that f 0 and f Cc 0 Then F x 1 x Z x 0 f t dt 0 x 0 there exists N 0 such that kfn fkp N It shows by the Jensen s Inequality that Fn x F x 1 x Rx 0 fn t f t dt 1 x Rx 0 fn t f t pdt 1 p x 1 q x 1 pkfn fkp and thus Fn x F x n x 0 10 Since kFnkp p p 1kfnkp n 1 2 it shows that Fn Lpand Fn Lpis Cauchy sequence By the completion of Lp it exists G x Lpsatisfying kFn Gkp 0 n and so kFnkp kGkp By the Fatou s Lemma kFkp lim n kFnkp kGkp 0 and f L1 then G R 0 fd 0 such that Z x 0 f t dt G 2 when x N Thus Z 0 F x dx Z N 1 x Z x 0 f t dt G 2 Z N 1 x dx 11 and therefore F L1 18 Let be a positive measure on X A sequence fn of complex measurable functions on X is said to converge in measure to the measurable function f if to every 0 there corresponds an N such that x fn x f x N This notion is of importance in probability theory Assume X 0 we defi ne Ek x X fn x f x k 1 2 Fn k n Ek n 1 2 F n 1 Fn It is easy that F E and so F 0 Since Fn is a sequence of decreasing measurable sets and X 0 suppose that Ek x X fn x f x k 1 2 Then it follows that for 1 p 0 satisfying kfn fkp N Hence it shows En 0 whenever n N It is then proved that fn f in measure c Suppose that fn f in measure For every national k there exists national nksuch that Ek 1 2k Defi ne F n 1 k n Ek then we have F k n Ek X k n Ek 1 2n 1 n 1 2 It is obvious that F 0 For any x X F it follows that there exists national N such that fnk x f x 1 2k for each national k N Therefore we obtain lim k fnk x f x It then shows that fn has a subsequence which converges to f a e 20 Suppose is a real function on R such that Z 1 0 f x dx Z 1 0 f dx for every real bounded measurable f Prove that is then convex 13 Proof Assume that x y R 0 1 Then x 1 y Z 1 0 x 0 t y 1 t dt Thus x 1 y R 1 0 x 0 t y 1 t dt R1 0 x 0 t y 1 t dt x 1 y Therefore is convex 14 chapter four In this set of exercises H always denotes a Hilbert space 1 If M is a closed subspace of H prove that M M Is there a similar true statement for subspaces M which are not necessarily closed Proof It is