Walter Rudin 泛函分析答案.pdf

P381.(a)If 3 22,such that 力+2i=y)岔+z2=ythen x+z=力+22,wecan get that x+力+21=x+力+22-Hence=Z2.(b)(0+0)x=Ox+Ox=Ox=Ox=0QO=Q(0+0)=QO+QO=QO=0(c)Vx G 2A,Ba E A,such that x=2a=a+a,so x A-h A,Hence 2A C A+A.Let A be the set of integer,and A A=A D 2A(d)，今If A is convex,then C A,i.e.M+sA C(t+s)A.Furthermore Vx e 4(力+s)x tx sx C tA sA.So(t+s)A C M+sA.Hence(t+s)A=M+sA.今If(t+s)A=M+sA,then we can choose s=1 t,thus we get the conclusion right now.(e)Let F=|J Bi.Vz e A,|a|1,aBi C Then QF=|J C iGA EeABi=F.The situation of intersection can be proved the same way.ieA(f)Let F=口 G,VI e A,G is convex./x、y F,that is,Vi G A,x eCiy e Ci,then 力6+(1 t)y e Ci,/i e t 1.Hence 柢+(1 t)x e n G=F.(g)r=Ci:z G A is a total ordered set with the order of set inclusion relations.Let F=U where Ci is convex.Then Vx,?/G F,3z e A,such EeAthat x e Ci)y e C”Hence VO t 1,to+(1 t)y E Ci C F.Thus F is convex.(h)Since A,B are both convex,Vx+2/eA+B,xeA,2/eB,Vxz+y G A+B,B G A,y G B,and 0 t 1,t(x+9)+(1 力)3+)=tx+(1-t)x+ty+(1-t)y e A-B,0 t 1Hence A+B is convex.(i)Since A,Bare both balanced sets,Vx-hyeA-hB,xeA,1,ax y)=aa ay E A+B.Hence A+B is balanced.(j)The proof is simple and therefore omitted here.2.Let r nco(A)=力1 6 1+tnxn:=1,e A,n e NI i=l1Then Vx,?/co(A),3n N,such that x=y=Eti,wheret,=Eti=i.yo s 0,Vs,Ac tW.V x e co(A),x=七 EZi 行=1,g G 4So nx=2tig t1tW+tntW=t(tW+%+nW),ilNote that W is convex,using 1(d),we have力1W+一+如W=(11+一+tn)W=W.Hence x e tW C tV that is,co(4)C tV.(c)For any neighborhood V containing the origin point,there exists a neighborhood W also containing the origin point,such that W+W C V.Note that A,B are bounded,so 3 si,s2 0,Visi,力2%we have A C B C t2W.This implies thatA+B GhW+hW C(力 1+t2)W+IV)C(力 1+12)上(d)Let Vi covers Then x-B C|JiGr Vx e A.B is compact,thus x+B is also compact.Then x+B G U2i 匕().According to the separating theorem,3 0,such that+x+B C U2 1 匕 Let Vx=Vg x.Thenn匕+8uU 匕，AcJVx.2=1 XEAConcerning A is compact,we have A G U晨 1 匕i，thusm m m n4+8U U%+8=U(%+切0 Uj=l j=l j=l il2We can also get the solution as following.Since both A and B are compact,A x B is compact.AxB A+B(a+b)a+bNote that the addition is continuous,we know it is a continuous map,so A+B is compact.(e)Suppose x 串 A+B.Then(x A)Q B=Note that B is closed and A is compact,i.e.x A is compact.Therefore there exists a neighborhood of origin V,such that(x A+V)Q(B+V)=0.So(rr+V)Q(A+B+V)=0,hence(x+V)Q A+B=0,that implies that x A+B.That is,A+B C A+B C A+B,which completes the proof.(f)Example 1:A=(x,?/)e R:X 0,?/|,B=(劣用)e R：x o,?/一；Then both A and B are closed sets.A+B is the right-half plane,however,it doesnt contain the y axis.Therefore,A+B is open.Example 2:A=n:n e Z+,B=n+:n e%+.nThen A,B are balanced on R.(compatible with the common topology)But n G Z+C A+B,0 g A+B.Then 0 e A-B.Therefore A+B is not closed.4.B=0,22)e C2:|1|z2,/z e B,z=(次,22),I|R.VQ|0,3/;s,such that E g tVi.e.Bxs G xs 串 tV.Sowhens=1,3Xi e Xi 力 1,3Whens=n,3xn E Exn tnV,3tn n,Let F=xi:i e N+C E,which is countable.As the description above,we have,for V,Vs 0,3/：s,such that 力回+i 4 力回+1卜，here we choose t as ts+1.Hence F is unbounded,which leads to contradiction.8.(a)Note that if pi e C,p2 ,then p(x)=max(pi(x),p2()G C.x:px)=x：Pi(x)|c：P2(C):.Therefore V(p,n)=V(pi,n)Q V(p2,九).Hence(*)Vp e p G P,such that V(p,n)C V(pn)Define a topology as Theorem 1.37.Then for,T is the finite intersections of V(p,n)and the union of those intersections,p E C.For P,T is the finite intersections of Vp n)and the union of those intersections,pf e P.Using(*),we have T C rf.Meanwhile,clearly,/pr GG C.So T C T.Accordingly,T=T.Since V(p)n)=八)V(P2,九)，G G,we can obtain that V(p,n)is just the intersection of Vn)%e P.Hence the family of seminorms directly leads to a base rather than a subbase.(b)Note that A is a linear functional.Suppose 3M oc,Vx e X,|Ax|Mp(x).ThenVx e V(p,Mn),p(/)MnHence|Ax|M that is,A is continuous.，今If A is continuous,then 3 U 3 0,U D n),such that Vx e C7,|Ax|1.Clearly,Vx e V(2?,n),Ax 1 also stands.Suppose/p e Cy M Mp(x M 0.That is,1.Here,=匕 In other words,Vp,when x eX,p=心，we have|Ax|1.Picking M n,we can get the contradiction.9.Define/:X/N Y x N AxObviously,Ax=/(lie),i.e.A=/o II.We have known that A is a linear mapping,that is/a1(3 G，y E X,K(ax+13y)=aA力+3Ay.Or+(3y)=&/(b力)+B五y).So/(7T(QX+(3y)=/(a7rx)+f(/37ry)since TT is linear.It is all to know that 7r is a continuous open mapping.If/is a continuous or open mapping,then obviously A is a continuous or open mapping.We first4assume A is continuous,W C Y,V is an open set.TT_1 O=A-1(V)isopen.We have that is open since TV is continuous,so f is continuous.Now we assume A is an open mapping,/U C X/N,U is open,7T-1(C7)is open since TT is continuous,f(U)=A(7r-1(C7)is open,thus f is an open mapping.10.Noting that dim Y=n X/N.Then f is linear,and since Ver,=V+N e r,where/-1 denotes the inverseimage,we have/(V)e r,that is,f is an open mapping.Next define g:X/N Y,where A=g o f.Then g is isomorphic and has the inverse map p_1.Since A=p o/is linear and f is linear,g is linear,so is gT,gi:Y X/N.Considering Y=Cn,gr is continuous.Therefore,g is open,which shows A is also open.(b)If N is closed,we can prove the desired result just in the same way as the proof of Theorem 1.18.Remember to note the range is Cn and that the norm should be the Euclidean norm.11._ _Assume NuLR 0夕 1,dim/N oc.Then N is closed,dim V/N oc.So Lp/N is local convex.Let A:L。Lp/N be a quotient mapping,hence A is a linear continuous mapping.By P37 we have that A=0,in other words,Lp/N=0.Hence N=Lp,and we conclude that N is dense in Lp.12.Without lost of generality,we just consider the local basis B(0,r),z=1,2.For(R,di),B伊,r)=x:c/i(0,x)r=x:x,r)=x:%(0,劣)r=x:r.I 1+JWe just need to consider 0r 1,B2(0,r)=R.Since is increasing on 力#1,81(。/)U-2(。,占1+rrBi r)DB2 t),0t 0,3 0,such that when n N,we have“2(*十)=(1+n)(fn+m)EThat is,劣n旌i is a cauchy sequence.However,obviously,lim一g 力九=oc,which demonstrates that d?is not complete.13.(a)V/e(C,a),d(/)=J Jo1+1/(叫dx 1.So VB G(C,r),E is cr-bounded.(b)Assume there is fn G(C,r)such that%noo.Since V=/:|/(x)|0by Lebesgues dominated convergence theorem.So id:(C,r)(C,a)is sequentially continuous.Following we shall illustrate the mapping is not contin-uous.Choose a member of the local base of(C,r),U=V pX1,ni)nV(Pt2,n2)n-n V(pXrn,nm).Define f E U like this,i 诉，于3=i,、linear,when x=Xi,i=1,m,when 力 e(&+J,&+i J),z=1,m 1,others,where 8=)minxi+i Xi:z=1,m 1.HenceThus id:(C,r)(C,cr)is not continuous.6(c)Assume F is a continuous linear functional on(C,r).We know that 30 E U such that F is bounded on U by Theorem 1.18.So3 V=V(p叫 i)U V(pi22)U-Uand 0 M oc such that F(V)M.Vf,g e V,and/(&)=分=1,m,we conclude that F(f)=F(g).Or else,let h(x)=af +(1 a)g(x)where the value of a is given latter.Then+(1 a)g(xi)=z=1,m.So/z V,furthermore|F(/z)|aF(f)-F(g)|Since F5*F(q),we have|F(/z)|M when a is large enough.This is contradiction.Thus V/,G C0,1 and=g(xi),i=1,m,there is a numbera e R+such that af,ag e V.By F(af)=F(ag we get F(于)=F(p).Thereare/i,72,-Jm W C0,1,力(叼)=际=1,，m,and(supp 九)(supp fj)。=0 JV(UZi=【，1 Defineh=/(6 1)/10)+/(62)/2(劣)4-卜 f(劣,then h(xi)=/(0),z=1,m.SomilWrite Q=F(/i),z=1,-,m,we have F(/)=EZi(d)Assume V#0 is a convex set of(C,cr),then 3r 0,such thatBrl/WI1+l/WIdx rC V.Choose n e N such that-r.V/e C0,1,define于。1(6)=/(x)(2-2nx)、0/(劣)(3+2nx 2z)/S/(x)(3+2nx 2n)q九 0)=(ox e。，3)，x E%，)，others,E r2i-3 2Z-2A 力/君,得),叱皇，Q others,others.x ex z=2,n 1,Then/=乙%=1ngz：=QU=L iJ；昂：?心 -2,pfn-/m)=上 andd（加）=max:T叫;一l 1+Pl Jn Jm)1 P(Jn jm)21+7?l(/n-/m)16So fn has not convergent subsequence,which implies that any set contain-ing fn is not sequentially compactness.Since Vn 2,fn e Vi which is bounded,we have that every closed and bounded set containing fn is not selsequentially compactness,in other words it is not compact set.Hence C(Q)do not have the Heine-Borel property.16.ChooseooKn*0,Kn C Kn+,U Kn=Q,where Kn are compact sets,nlandocEn 子仪 En U 石口+i,|J=Q,where En are compact sets.n=lLet%=/e 0(Q):sup|/(x)：x e Kn ,%=/e C(Q):sup|/(x):x e En ；.According to 1.44 we have known that%,Un are local bases,and induce the topology ri,T?respectively.Hence in order to prove ri=%we should only discuss the local bases.Vn N,Kn C U鼠：i Ei.So exist m e N,such that Kn C U,i E=4n since Kn is compact.This implies that Um C Vn.8V/K,|/(x)|0,力 G N,such that|/(x)|+A since the continuous function reaches maximal or minimal value on compact set.Write I=maxm,力.Vg Ui,that is gx)；for all x in Ei，We haveI/O)+g(/)|/O)|+|g0)|1/0)1+；Y is continuous.Therefore 30 e U,such that d(0,A/)i-.When x e MQC7,pick Vi C UO e V balanced,we have c/(0,Ax)When x e M Q Vi,3 7i 9 仇 such that +帆 U%.三%3U U balanced,d(0,Ax)when x e M Q 匕+%C V stands.-With such procedures,we can get%,。Vn balanced,and Vn+Vn CAx)m.Hence xn xm G M(Since M is a subspace.)xn xm e Vn-Vn C Vn_1.So xn xm e Vn_ QM.This implies that d(0,A(xn xm)21-nHence t/(0,Axn Axm)=A(xn xm)九加)21-n,9so Axn and Ayn converge to the same limit.Consequently,Ax has no relation with the selection of xn.When x e M,we can select xn=x,Vn,andAx=lim Axn=lim Ax=Ax.九一 OO TlTgVx,?/e xy a,(3 e C,separately select xn,yn,xn e(x+Vn)M,yn e(U+KJPI ThenA(QX+/3y)=lim A(axn+/3yn)九一 8=a lim Axn+(3 lim Ayn 71Toe ns=aXx+/3AyThis demonstrates A is linear.To prove the continuity of A,we just need to verify that at the origin.Vx e Vn,pick e(x+Vm)Q M.Since Ax=lim.一g A/g using(*),we have d(Nx,Axn)22-n.Thusd(O,A/)(i(0,Axn)+Axn)21n+22-n Y.20.V/e L2(-7T,7T),oo/(N)=。避次,cn=cn,n=1,2,for allmost everywhere x (7r,TT),andmlim|/-V c.加勺2=0.771-00 zdn=-mf cnein=工 c/.+。_逮一加n=-m n=0 n=lm m=W cneinx+工 cn(/n-nen)Ti=0 n1m m(cn-ncn)en+):cnfn e Xi+X?.71=0 n1Hence+X?is dense in L2(7r,TT).8 1 OOII 工-e-nll2=II 工-I-In=l n=l1n2 oc.10Thus 总i e-n L2(-7r,7r).Howeverlib lib 52 e_n=V Tn-nen)n nn=l n=lm-m=nl n1ooEn=l1e-n nmlim 5mTgn=l1en nm 1 mlim(工九 .).moo z/n z.nl nlBut liniM_g E晨i&n$上2(7r,7r),because the Fourier coefficients do not con-verge to 0.Obviously,-e_n 串 X1+X2.This example also shows X1+X2 is not closed,or X+X?=Xi+X2=L2,which is a contradiction.21.We can easily get such a collection of%/Vn balanced,Vi+V2 C K Kz+i+Kz+i c Vn.Let(mD=1,let A(r)=Vc.If r e Z?,let A(r)=Q(r)%+。2(/)匕+。3(/)4+.Define/(x)=infr:x e A(r),x X).Obviously,/(0)=0,/(x)=l,Vx Vc.Vs 0,3r G Z),0 r e.Vx A(r),we have/(x)r e.Therefore,f is continuous at the origin.Using(8)of P19:f(x+y)/(x)+/(?/),we can get/(%)=f(x-y+y)f(x 9)+f(y)=f(y-x+x)f(y /)+f(x)=f(x +/(x),which implies that|/(x)f(y)0,3 e N,|/n fm N,Thus|/n(0)-/m(0)|+SUp加)：80%11we have that|/n(0)/m(0)|0 0,we conclude that|fn()-于m3-fn(V)+册-)|V 里-W RLet?/=0,we have that|fn(x)加|s(l+6).Hence/九(力)1/(rr),n 7 oo.Then-f-fn(y)+/)|0 e./G Lipa since|/n|0,we have6一/fn)-0,n 0.And limjo 厂加4九)=0,since fn C lipa,we get that厂。5(/)厂0期(%)+厂05(/%)-0,八-oo,8-o.Hence f G lip a,i.e.lip a is closed.23.V/i+/2eUD V(加,ri)n V(坛42)n n V(膈,rn).Then|/i+/2-hi n,/i=l,2,-n.Let 九u；=v偿今？，w=p、/i=i5=v得-短w=Qut /2=1ThenI=Qi+一殳|V72-fy-y+y|=|fl+h 殳|y-In other words,九 e U|,/2 e U,Vz=So 九 e Wn/2 e W2.VpiWi,。2 e W2,gi+92 一九|=51+52-12Hence gi+g Vz=1,n.We obtain pi+2 U,i.e.W1+W2 CU.We get the conclusion that addition is r-continuous.Choose f(x)=x2,x e(0,1),thus|/(x)|0,(1+s)-x2 1 when-7=x 1.So we obtain that(1+s)/(x)串 V(4 1).We get the conclusion that scalar multiplication is not T-continuous.24.In euclidean space R2,we choose V just as the graph 1.V is not convex,obviously,W=U|Q|5,aGRis not convex.In euclidean space R,assume U=(-)U(1,2)U(2,1),then U is notconvex,obviously A=U,and A is not balanced.13P531.Note that X=U9 1 En,where dimEn cx3,Vn and dimX=oc,so every En is a proper closed subspace of X.Since proper subspace does not have interior point of X,En is nowhere dense.Hence X is of the first category.Suppose X is an F-space of infinite dimension with countable Hamel ba-sis ei,e2,en,Vx e=EXi&e.Denote Fn=spanei,62,en.Then X=Fn)and therefore X is of the first cat-egory.While X is an F-space,X should be of the second category.This is a contradiction.Consequently,no infinite-dimensional F-space has countable Hamel basis.2.We construct the set like Cantor set.Write Ao=0,1.i)In the middle of the interval we take an open interval JD；?with measure 卜 the rest interval we call them En and E%respectively.ii)In the middle of E”we take an open interval with measure 晟，i=1,2.We call the rest interval which Bn,E12 have taken out D*,E21,E22,石23,石24iii)Do it like this,we have*:12=1,，2Let Bi=Uti Dij then is an open set.=A0Bi,then Ai is closed,and just as Cantor set Ai is nowhere dense.00 11 3|4|=|4|-0|=i-2-就=1，|3|=nliv)We do step for every D thus we get and nowhere denseset A2 which with measure|xDo it like this we finally get nowhere dense sets An,An=x Let F=4九,then F is of the first category in the unit interval.Furthermore|F|=E14I=L 口4.(a)Denote En=于：J；f(x)dx n.Suppose/m C En,|fm-f 0,/G En.Then fm converges to f in the sense of measure,and therefore there exists subsequence/mJ,such that,frnj oc.f(x)dx=liminf|/m(x)|2cZx liminf fm(x)2dx 0|0./e 0,3m n,such that,F C x e 0,1:f(x)0,and F=Let g=*XF(X).Thenhili=XFdx=.=Jo 8 mhll2=/也=-=mn.f(x)-g(x)2dxJo于3)2dx+2 f(x)g(x)dx+glJO J FIIPII2=mnThat is,/+g En.In other words,f is not an interior point of En in L1.Hence,En has no interior points in L1 and hereby is nowhere dense.Meanwhile L2=En)so L2 is of the first category.(b)Since gn=n%0,n-3(x),when/e L2,n oc,17M C.Obviously,An is linear.Vn,Vs 0,when ll/lli s/n,we have|An(/)|=f(x)dx nfi L1 is obviously linear.Note that J；fxdx|/|2-1=|/|2,we conclude id is continuous.Pick/(x)=x E L1L2,and then id is not an on-to mapping.Since L2 is an F-space,with the help of Theorem 2.11,15we can get that if L2 is of the second category in L1,zcZ(L2)=L1.This is a contradiction.So L2 is of the first category in L1.Through the same proof we can get the conclusion that Lq is of the first category in Lp,where p q.5.1Suppose 0 p q.If x lp,i.e.,xp=x(n)p)N,|x(n)|1.Hence x(n)q|x(n)|p,andOO Q I/F)nN-l/oo q R which is obviously linear.When|x|p 1,we have|x(n)|1,Vn.Then xq (|x|p),using x(n)q x(n)p.So id is con-tinuous.We can come to our conclusion noticing that lp is an F-space and by means of Theorem 2.11.6.V/e L2(T),|/(X)-/He-1|2Denote gn(x)=fn)einx e L2(T).n:n (A/)(0)=f 7n/(n).n=(x)|)is a Banach space with the supremum norm.So(C(T),|)is anF-space.Obviously,T is a linear mapping.Vfm C C(T),fm/,thuslim 加=/(x)e C(T),andOOlim L(/m)=lim V 7n/mH rriTg rriTg z/n=oo oc oc=lim/mH=V 7n/H，z/moo zrn=_g n=-gsince/m uniformly converge.Solim“加)=卬).rngAccording to Theorem 2.15 and its remark,we know that L is continuous.By Riesz representation theorem,there is a complex Borel measure/on T,such that L(/)=/*d匹 that is卬)=I e-*如.ForLef=f H，n=(x)we have7n=yAssume yn=f eznewhere is a complex Borel measure on T.Without loss of generality,we only consider(T)=1.Let7(x)=e-5幽 e i(T).17then 7n=7(n),(n G Z).Write(A1)(n)=6(n)*f),where f e C(T),so A/e C(T).Hence(A/)A(x)=$(/)f(x)=7(x)-f(x),that is8.For all n e N,(A/)A(n)=7n/(n),n e Z.Eg=叁*/nTherefore,every infinite point sequence in 0,6 1,6 2,一 converges to 0.So K is selsequentially compact and hereby compact(Z2 is a metric space).m nHence Vc e K,Amx is bounded.However,Amxm=m is