实分析与复分析部分答案.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, f1(,+) = n=1 f1(rn,+). Since sets f1(rn,+) are measurable for each n, the set f1(,+) 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 kn(a k) = inf kn ak,n = 1,2,. Therefore, let n , it obtains lim n sup kn(a k) = lim n inf kn ak. 1 By the defi nations of the upper and the lower limits, that is limsup n (an) = liminf n an. (b) Since sup kn(a k+ bk) sup kn ak+ sup kn bk,n = 1,2,. Hence lim n sup kn(a k+ bk) lim nsup kn ak+ sup kn bk = lim n sup kn ak+ lim n sup kn 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 kn(ak) infkn 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) 0, then we can prove that the strict inequality in the Fatous lemma can hold. 10.Suppose that (X) 0 such that |fn(x)| M,|f(x)| M on X,n = 1,2,. By the Lebesgues 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 0, satisfying 0 s(x) M on X, and let = 2M, we have Z E |f|d Z E sd + 2 Mm(E) + 2 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) 0 such that (x,) N,xn= x;and|yn y| 0. 8 and for any 0 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 0 for any 0 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 0 such that kfn fkp N. It shows by the Jensens 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 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= kn 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 kn Ek then we have (F) ( kn Ek) X k=n (Ek) = 1 2n1 , 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 0 such that kfn fmk N and so |fn(x) fm(x)| kfn fmk kxk N, x X. Thus fn (x) C is also a cauchy sequence and so we can defi ne f(x) = lim n fn(x);x X. It is easy that f is a linear functional on X. Since kfnk is a cauchy sequence, ther is M 0 satisfying kfnk M, n = 1,2,. Hence, for ?,N as above, it shows for n N the following |f(x)| |fn(x)| + kxk? kxk(M + ?);x X. It forces |f(x)| Mkxk;x X and so kfk M. That is, f X. Therefore, Xis a Banach space. 3 (b) For each x X, defi ne (f) = f(x);f X Then for any f,g Xand , C, (f + g) = (f + g)(x) = f(x) + g(x) = (f) + (g) and we see that is a linear functional on X. Since |(f)| = |f(x)| kfk kxk;f X whence kk kxk. From the Hahn-Banach Theorem, there is g Xsuch that g(x) = kxk and kgk = 1. It implies that kk = kxk. (c) Suppose that xn X such that f(xn) is bounded for every f X. By (a),(b) and the Banach-Steinhaus Theorem, it is easy to prove that kxnk is bounded. 9. Let c0,l1,lbe the Banach spaces consisting of all complex sequences x = i , i = 1,2, defi ne as follows: x l1if and only ifkxk1= X |i| 0, there is N 0 such that |fn(a) fm(a)| 0 such that kxk xk0k N. and so kyk yk0k = lim n kn(xk xk0)k lim n knk kxk xk0k M ? M = ? It hence implies that yk Y is a cauchy sequence. Since Y is a Banach space, there is y Y with kyk yk 0. Therefore, knx yk= knx nxk+ nxk yk+ yk yk knk kxk xk + knxk ykk + kyk yk M kxk xk + knxk ykk + kyk yk This implies that lim n nx = y. 10 ? chapter six 2.Prove that the example given at the end of Sec.6.10 has stated properties. Proof: 3. Prove that the vector space M(X) of all complex regular Borel mea- sures on a locally compact hausdorff space X is a Banach space if kk = |(X). Proof: Firstly, assume , M(X) and C. Then M(X) and |,| are regular Borel measures. For any measurable set E in X and ? 0, it exist compact sets K1,K2 E and open sets G1,G2 E satisfying: |(E) 0 for any x X with x Enand so |g(x)| = |gn(x)| kgnk M. it shows that kgk M and g L(). In the case 1 q , because |gn(x)| |gn+1(x)|,n = 1,2,;|gn(x)| |g(x)|,x X By the Lebesgues monotone convergence Theorem, it shows Z x |gn|qd Z X |g|qd,n and so ( Z X |g|qd) 1 q M. Therefore, kgkq M and g Lq(). 6. Sppose 1 p and prove that Lq() is the dual space of Lp() even if is not -fi nite. Proof: Sppose 1 p 0. 3 Put fn= En|g|q1 where g = |g|; | = 1. Then fn Lp() and kfnkp= ( Z En |g|qd) 1 p; n = 1,2,. Moreover, |fn| converges increasingly to |g|q1. Therefore, Tg(fn) = Z X fn gd = Z En |g|qd;n = 1,2,. and so Z En |g|qd kTgk kfnkp= kTgk ( Z En |g|qd) 1 p It shows that Z En |g|qd kTgkq;n = 1,2,. By the Lebesgues monotone convergence Theorem, it shows that Z En |g|qd = Z X |fn| |g|d Z X |g|qd;n . Hence, R X |g|qd kTgkqand so kgkq= kTgk. (b) Assume (Lp() and kk 6= 0. At fi rst, for each f Lp(), defi ne Df= x X : f(x) 6= 0;En= x X : |f(x)| 1 n. Since Df= S n=1En, and (En)( 1 n )p Z En |f|pd Z X |f|pd + it thus follows that (En) +, n = 1,2,; and so Df is -fi nite measur- able set. Secondly, given E X is -fi nite measurable set, and put M = f Lp() : Df E then M Lp() is a Banach subspace. In fact, provided that fn M is a cauchy sequence, there is f Lp() such that kfn fkp 0,n . 4 So it exists a subsequence of fn M which converges to f almost every- where. It implies that f M. Therefore, |Mis a bounded linear functional on M. Since it also that M Lp(E,), by the Hahn-Banach Theorem, it can extend |Mto a bounded linear functional |Eof Lp(E,) and so there is unique hE Lq(E,) such that (f) = Z E f hEd,f M;khEkq= k |Ek = k |Mk kk. Take gE= hEon