2021阿里巴巴全球数学竞赛预选赛试题及参考答案.pdf

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1、1n3Cpnn?mmY1,2.3J?-.,z(?vk?A:)g?1,2,.?|,?,35?3y|3*my3b?3*?|?14?S?1 m 1 n m?l dm,nv(m+n)dm,n 1,p?l,:m?l?lo1 JJJKKK(4)e?()SA 3*UNB 8;B 3*UNB?ku 8?;C 3*NB?2 yyyKKK(6)y?1R1 YYY.C S.R2 YYY.).UeS1 1,2,.?.k,?S1 1?n 2,e1 1,2,.,n 1?S,1 n U3=?u 1 m n 1,d dm,n1m+n,?,l1 m?m,!_?r1m+n?l,/?2m+n?lSS1 n?.?l?o2n+1+2n

2、+2+22n 1 2(lnn+1n+lnn+2n+1+ln2n 12n 2)=2ln2n 1n1.5 2ln2,?lUCX?,d1 n oJ?,?1 1,2,.,n 1 m?lvK8.d8B?NB?.)?.?%?:?IX1 1,2,3,4 O3(14,0),(14,0),(0,14),(0,14)?,=?O 0,2,32.d?l?u 2 144=811+2,?d?4 vK8.8ByeK:?k 2,1 1,2,.,2kSu?S?2k/?:?,?pm(3?)lvK8,?1,2,.,2k1?3?.K k=2.e k=c 2k?(.?2kl.1 2k+1,2k+2,.,2k+13?l?:.y35y?9

3、1 2k+1,2k+2,.,2k+1?lvK8.1 2k+1,2k+2 31 2k1?(=1 2k1?22k+1l?);1 2k+3,2k+4 31 2k1 1?;1 2k+2a 1,2k+2a 31 2k1a+1?;1 2k+1 1,2k+131 1?.duc 2k13?,?1?.y3?l(I?“#”?/).?2k+1,zl22k+114=2k+2,u?O m 2k n?,em?kl,Kdm,n2k+121m+n2(m+n)2k+11m+n;em?lTl?n 2k+2a1,2k+2a,K m 2k1a+1,2d(m+n)dm,n2k+2(2k+2a1+2k1a+1)=2k+2(22k1+a)

4、3 2k12k+1=38 1.1 1,2,.,2k+1m?lvK8.d8B,S?.33,4.2019c13Cpnnm?38E?,8NxKI/?m n/d n/eZ?(=3?/?)?m/m/?L R3?C?-N?mk.,.dk/?3?KKK(4)?2021=43 47.o3N,?Ld 43?m 47/?4 KKK(6)?6?)4R3 YYY.R4 YYY.If=.IO?T,?:d5L:T=,:0 ,/,:OAu =2k43.?Dkko“”(u =0),o:(u?43/?:?,u?:?,IP:m?43/?:).PCk,0=(2(k 1)43,0),Ck,1=(2k43,0);Dk,0=(2k+14

5、3,2),Dk,1=(2k+343,2);Ek=(2k+243,2).3 Dk?,?21:,XAk,i=(2(k 1)43+386,i11),i=1,.,21.,?7 z-=243?,21:,P Bk,i,i=1,.,21.(Ck,0Ck,1,Ck,0Ak,1,Ck,1Bk,1,Ak,iAk,i+1,Bk,iBk,i+1,Ak,iBk,i,Ak,iBk,i+1(i=1,.,21),9 Ak,21Dk,0,Bk,21Dk,1,Ak,21Ek,Bk,21Ek,Dk,0Ek EkDk,1.?m 47/.?43?(?E k)m 47/,U?N.:;.?m/?:()N?:(),l?“zg”“2021”?

6、“g”,dd?K?)?.55.?c“F?8c5?/,U“?o“F?o?OuN!?L4?z Rn?,”f?3?oC?8”,?K|J?uS?E?TK|J?od?S!?UJ?uO?x Rn?f(x)?o7L=u54?z f?zgO f?O?3TK?n p(10 m),?J?w?o”kb?f 1w?K4“F?gC?3?f!8I?%/5N5?J?Z3Lvk(/?Jm?Xo“F?o?4?z f LN!?k?:x?zd?f(x)L?,5U5N?zi f?ATkF?Z?xdu 0 k=0,1,2,.,1e1:yk:=argminf(y):y=xk tkei,i=1,.,n#O”2:sk:=1f(xk)f(yk

7、)t2k#e?:sk=1:sk=03:xk+1:=(1 sk)xk+skyk#S“:4:tk+1:=22sk1tk#sk=1:O?sk=0:y3”f:Rn R Xeb?bbb?1.f=?x,y Rn 0,1 kf(1 )x+y)(1 )f(x)+f(y).bbb?2.f 3 Rn f 3 Rn L-Lipschitz Ybbb?3.f?Y8k.=?R8x Rn:f(x)?k.6ubbb?1 bbb?2yhf(x),y xi f(y)f(x)hf(x),y xi+L2kx yk2?x,y Rnbbb?1 bbb?3 K?yf 3 Rn?k?f?5?yyyKKK(20)3bbb?13eu AFBT

8、ylimkf(xk)=f.7R5 yyy.b?f(xk)9 f f(xk)O,?infk0f(xk)f 0Pgk=f(xk),K infk0kgkk 0(?x f(x)=f,K f 5?y hgk,xkxi f(xk)ff(xk)N5 f Y8k.5?y xk x k.?infk0kgkk 0)3 0 kgkk k k 0?k 0,?ik 1,.,nv|hgk,eiki|kgkk ,(1)=1/n?f(yk)minf(xktkeik)f(xk)tk|hgk,eiki|+Lt2k2 f(xk)tk+Lt2k2.(2)tk2L+2,(3)k f(yk)f(xk)t2k,l?sk=1,?k tk+1

9、=2tkddtk t min?t0,L+2?0(4)k k 0?3 k?sk=1(K tk 0);z?k,f(xk)f(xk+1)t2 f ek.g?86.-n?k,P 0k=diag0,.,0|z k k k?-Y=0nAAt0n+1!(2n+1)(2n+1)?,A=(xi,j)1in,1jn+1 n(n+1)?AtP A?=?=(n+1)n?(j,i)?xi,j.(i)yyyKKK(10)E k k?X?A?,XJ3?v=(x1,.,xk)t?Xv=v.y:0 Y?A?Y?A?/X,K AAt?A?(ii)yyyKKK(15)-n=3 a1,a2,a3,a4 4 p?Pa=sX1i4a2i

10、9xi,j=aii,j+aj4,j1a2(a2i+a24)aj(1 i 3,1 j 4),i,j=(1 if i=j0 if i 6=j.y:Y k 7 p?A?9R6 yyy:(i)P In=diag1,.,1|z n n n?.?Cydet(I2n+1 Y)=det(2In AAt).,0 Y?A?Y?A?/X,AAt?KA?.(ii)P u=(a4,a4,a4),v=(a21+a24a,a22+a24a,a23+a24a).O?AAt=diaga21,.,a23+utu vtv.?f(s)=det(sIn AAt)AAt?A?.O?f(a2i)=a2ia2Y1j4,j6=i(a2i a2

11、j)(i 1,2,3,4).-a01,a02,a03,a04 a1,a2,a3,a4-?4S?.d f(x2i)?L?:AAtknp?A?b21,b22,b33,b1,b2,b3va01 b1 a02 b2 a03 b3 a04?.d,d(i)?Y k 7 p?A?.107.u R?Y?E f(x),R?(Sf)(x):(Sf)(x)=Z+e2iuxf(u)du.(i)KKK(10)S(11+x2)S(1(1+x2)2)?wL(ii)KKK(15)?k,P fk(x)=(1+x2)1k.b?k 1,?c1,c2?y=(Sfk)(x)v?xy00+c1y0+c2xy=0.11R7 YYY:(i)

12、S(11+x2)=e2|x|S(1(1+x2)2)=2(1+2|x|)e2|x|.(ii)c1=2k c2=42.)P V R?E!Y!?|?5m.Lemma 0.1.(i)e f(x)V,f0(x)V limxf(x)=0,K(Sf0)(x)=2ix(Sf)(x).(5)(ii)e f(x)V xf(x)V,K(Sf)0=2iS(xf(x).(6)n0.1?y.(i)(Sf0)(x)=Z+e2iuxf0(u)du=e2iuxf(u)|+Z+(e2iux)0f(u)du=2ixZ+e2iuxf(u)du=2ix(Sf)(x)(ii)?a,b R(a 1,i 11+z23 CA.?k.S?4:.

13、d?-A ,?(Sf)(x)=e2x.du f(x),(Sf)(x).?,(Sf)(x)=e2|x|.(ii)P g(x)=e2|x|.?O?(Sg)(x)=Ze2ixue2|u|du=Z0(e2ixu+e2ixu)e2udu=12(e2(1+ix)u1+ix+e2(1ix)u1 ix)|0=11+x2.13Lemma 0.4.(i)u?k 0,Sfk/X(Sfk)x=e2|x|gk(|x|),gk k g.(ii)u?k 0,S(S(fk)=fk S(S(xfk+1(x)=xfk+1(x).Proof.(i)k48fk+1=fk+12(k+1)xf0k(x).dn0.1,?48:Sfk+1=

14、Sfk12(k+1)(x(Sfk)(x)0.dd,d8B?(.(ii)5?f0k(x)=2(k+1)xfk+1(x)(xfk(x)0=(1+2k)fk(x)+2(k+1)fk+1(x).d(i)?(n0.1,?0.2?b?fk(x)xfk+1(x)(k 0).?,d8By S(S(fk)=fk(x)S(S(xfk+1(x)=xfk+1(x)(k 0).?K8?.(i)3n0.3y S(1+x2)1)=e2|x|.d(5)?S(2x(1+x2)2)=22ixe2|x|.2d(6)?S(2x2(1+x2)2)=(1 2|x|)e2|x|.?,S(1+x2)2)=2(1+2|x|)e2|x|.(ii

15、)k,?k 1,xjfk(x)(0 j 2k)?.?,d(6),y=(Sfk)(x)2k gY.dn0.1n0.4?:xy00+c1y0+c2xy=0?du(x2f0k+2xfk)c1xfkc242f0k=0.fk(x)=(1+x2)1k,?c1=2k c2=42.148.?,i#?i?|?%T?r?om?Czr+N?A?N?N n(t,x)Lr?ep t Lm?x LrT?o3 t u 0 x1 0 x kbbb?3.#r?5?k1?i?D mSdO?m?c(t)L2?Pr?DPrgC?!*l?T?r?x kP b(x)b?XJ3,P t=0?n(0,x)=n0(x)?n(t,x)?mzvX

16、e?(tn(t,x)+xn(t,x)+d(x)n(t,x)=0,t 0,x 0,N(t):=n(t,x=0)=c(t)+R0b(y)n(t,y)dy.(7)p N(t)#r?Ob?b,d L+(0,)=b(x)d(x)?(?)k.ekzb?c(t)0=#r?OPr?Dk(i)KKK(10)bbb?1bbb?2/?(7)n(t,x)v?I3?L?.b?Lm?AX2bbb?3)(7)N(t)?(ii)KKK(10)#r?O N(t)b(x)m?Xd?N(t)v?N(t),n0(x),b(x),d(x)?n(t,x)y N(t)vXe?O|N(t)|kbkekbktZ0|n0(x)|dx,(8)p

17、k kL L15(iii)yyyKKK(10)?3?m?n(t,x)koC?duro 0,(0)=R0b(x)(x)dx=1,?Kk?)(x):(0(x)+(0+d(x)(x)=(0)b(x),x 0,(x)0,R0(x)(x)dx=1.,?-?z n(t,x):=n(t,x)e0tyu?H:R+R+v H(0)=0,kddtZ0(x)(x)H?n(t,x)(x)?dx 0,t 0,yZ0(x)n(t,x)dx=e0tZ0(x)n0(x)dx.?zy3b3?.?z?16R8 YYY:(i)?kf1A?dum5OA?x(t)vdx(t)dt=1.?XA?kddtn(t,x(t)=d(x(t)n(

18、t,x(t).?n=?(7)2,?m?t?1b?b?kn(t+t,x+t)=n(t,x)td(x)n(x,t)+o(t).m1Lm?z1?L?r t2-t 0=?du N(t)?IPr?zu?,x?Pr mS0?#r?b(x)n(t,x)?mSkPr0?#rIrk?Pr?z3?LR0b(y)n(t,y)dy(ii)K N(t)?Ikr n(t,x)?N(t)?LI)d5?V-A?)U?ddsn(t+s,x+s)+d(x+s)n(t+s,x+s)=0,KXJ D(x)=Rx0d(y)dyodds?eD(x+s)n(t+s,x+s)?=0.(9)o?s max(t,x)keD(x+s)n(t+s,

19、x+s)=eD(x)n(t,x),x 0,t 0.(10)AO?-x=ys=y?t y n(t,y)=N(t y)eD(y).172-x=ys=t?t y n(t,y)=n0(y t)eD(yt)D(y).?N(t)v?L?N(t)=Z0b(y)n(t,y)dy=Zt0b(y)n(t,y)dy+Ztb(y)n(t,y)dy.A?m1?A?u x=0,t 0?1?A?u x 0,t=0 n(t,y)?LO“?N(t)=Zt0b(y)eD(y)N(t y)dy+Ztb(y)eD(yt)D(y)n0(y t)dy.(11)?n=?N(t)v?N(t)=Zt0b(t x)eD(tx)N(x)dx+Z0

20、b(x+t)eD(x)D(x+t)n0(x)dx.(12)?d(x)0 D 4O?eD(tx),eD(x)D(x+t)u 1u,2|b(x)?k.5 N(t)Xe?O|N(t)|kbkZt0|N(x)|dx+kbkZ0|n0(x)|dx.?|Gronwall n?y?(iii)2?O?Kk(7)U?-?zv?te n(t,x)+xe n(t,x)+(0+d(x)n(t,x)=0.,?/U?n?t n(t,x)(x)+x n(t,x)(x)=0,?ktH?n(t,x)(x)?+xH?n(t,x)(x)?=0.18A?KK?3k(x(x)(x)=(0)b(x)(x),x 0,(x)0,R0(x)(

21、x)dx=1.L?O?t?(x)(x)H?n(t,x)(x)?+x?(x)(x)H?n(t,x)(x)?=(0)b(x)(x)H?n(t,x)(x)?.P d(x)=b(x)(x)dx x 3 R+?,ddtZ0(x)(x)H?n(t,x)(x)?dx=(0)Z0H?n(t,x)(x)?d(x)+(0)H?n(t,0)(0)?.5?d(0)=1 n(t,0)=R0b(x)n(t,x)dxo n(t,0)(0)=n(t,0)=Z0b(x)n(t,x)dx=Z0 n(t,x)(x)d(x),u?ddtZ0(x)(x)H?n(t,x)(x)?dx=(0)?Z0H?n(t,x)(x)?d(x)+H?Z

22、0 n(t,x)(x)d(x)?.2dJensen?ddtZ0(x)(x)H?n(t,x)(x)?dx 0.?-H(u)=u=?y?19Alibaba Global Mathematics Competition-Quanlifying1,2.In a fictional world,each resident(viewed as geometric point)is assigned a number:1,2,.In order to fight against some epidemic,the residents take some vaccine andthey stay at the

23、 vaccination site after taking the shot for observation.Now suppose thatthe shape of the Observation Room is a circle of radius14,and one requires that thedistance dm,nbetween the Resident No.m and the Resident No.n must satisfy(m+n)dm,n 1.Where we consider the distance on the circle,i.e.,the length

24、 of the minor arc betweentwo points.1.Multiple-Choice Question(4 points)Which of the following is correct?A No more than 8 residents can be placed inside the observation room;B The maximal number of residents that can be placed simultaneously is greaterthan 8,but still finite;C Any number of residen

25、ts can be placed inside the observation room.2.Proof Question(6 points)Give a proof of your answer to Question(i).1R1 Answer.The Choice C is correct.R2 Answer.Solution I.We can place the Residents No.1,2,.according to the fol-lowing rule.First,put Resident No.1 arbitrarily.For n 2,if Residents No.1,

26、2,.,n1have already been placed,we consider the positions where Resident No.n cannot beplaced.For 1 m n 1,by dm,n1m+n,we know that the Resident No.n cannot beplaced in the arc that is centered at Resident No.m,and of the length2m+n.The totallength of these arcs is2n+1+2n+2+22n 1 2(lnn+1n+lnn+2n+1+ln2

27、n 12n 2)=2ln2n 1n 1.5 2ln2,sothese arcs would not cover the whole circle,hence it is always possible to find a placefor Resident No.n such that its distances to Residents No.1,2,.,n 1 satisfy therequirement.By induction we conclude that the circle can accomodate any quantity ofresidents.Solution II.

28、We consider the Cartesion coordinate system whose origin is the cen-ter of the circle,and place Residents No.1,2,3 and 4 at(14,0),(14,0),(0,14),(0,14),respectively,or in an equivalent way,we say that(the principle values of)their argu-ments are 0,2and32.Now the distance between any two residents is

29、no less than2 144=811+2,so the placement of these 4 residents satisfies the requirement.We prove the following assertion by induction:for any integer k 2,we can placeResidents No.1,2,.,2kat the vertices of a regular 2k-gon inscribed to the circle,suchthat their mutual distances fulfill the requireme

30、nt,and no two residents among the first2k1occupy adjacent vertices.The assertion holds for k=2.If it is valid for k,so that the first 2kresidentsare placed.They divide the circle into 2kequal arcs.We need to put Residents No.2k+1,2k+2,.,2k+1at the midpoints of these arcs.So we just need to prove the

31、distances related to Residents No.2k+1,2k+2,.,2k+1satisfy the requirement.We put Residents No.2k+1,2k+2 in positions next to Resident No.2k1(i.e.thecorresponding arguments differ to that of Resident No.2k1by22k+1);put Residents No.2k+3,2k+4 next to Resident No.2k11;put Residents No.2k+2a1,2k+2anext

32、to Resident No.2k1a+1;put Residents No.2k+11,2k+1next to ResidentNo.1.As the first 2k1residents do not occupy any consecutive positions,the above place-ment would not cause any problem.Now we consider distance between two residents(only the cases where at least one resident in the pair is“new”need t

33、o be considered).2Since the circle is now divided into 2k+1arcs,each has length22k+114=2k+2.ForResidents No.m(2k)and n,if they are separated by at least two pieces of arcs,thendm,n2k+121m+n2(m+n)2k+11m+n;If they are just separated by one piece of arc,then for n 2k+2a 1,2k+2a,wehave m 2k1 a+1,hence(m

34、+n)dm,n2k+2(2k+2a1+2k1a+1)=2k+2(22k1+a)3 2k12k+1=38 1.Therefore,the distances between each pair of Residents No.1,2,.,2k+1satisfies therequirement.Then by induction,we conclude that any number of residents can beaccomodated in that way.33,4.Two years ago,the winners of 2018 Alibaba Global Mathematic

35、s Competitionmade a paper polyhedron as a present to the organizers.As shown in the photoes,thepolyhedron has 60 equal triangular faces,and its surface can also be divided into 60congruent non-planar quadrilaterals.A non-planar n-gon is a non-planar figure obtained from a planar n-gon by folding ita

36、long some diagonals(i.e.,to form certain dihedral angle at each of the chosen diagonals).Two non-planar figures are congruent if and only if each can be obtained from the otherone by a certain isometry of R3.A polyhedron is a bounded region in R3whose boundaryis the union of a finite collection of p

37、lanar polygons along common edges.3.True-False Question(4 points)We know that 2021=43 47.Is there apolyhedron whose surface can be formed by gluing together 43 equal non-planar47-gons?4.Question and Answer(6 points)Please justify your answer to Question(i)with a rigorous argument.4R3,Answer.The answ

38、er is YES.R4 Answer.All we need to do is to construct an example.Lets consider a standardtorus T,whose points can be represented by two parameters:T=,:0 ,0.For k=0,1,2,.,do the following.1:yk:=argmin?f(y):y=xk tkei,i=1,.,n?#Evaluate loss function.2:sk:=1?f(xk)f(yk)t2k?#Sufficient decrease?Yes:sk=1;N

39、o:sk=0.3:xk+1:=(1 sk)xk+skyk#Update iterate.4:tk+1:=22sk1tk#Update step size.sk=1:double;sk=0:halve.Now we make the following assumptions on the loss function f:Rn R.6Assumption 1.f is convex.This means thatf(1 )x+y)(1 )f(x)+f(y)for all x,y Rnand 0,1.Assumption 2.f is differentiable on Rnand f is L-

40、Lipschitz on Rn.Assumption 3.f is level-bounded,meaning that x Rn|f(x)is a boundedset for any given R.Based on Assumptions 1 and 2,we can prove thathf(x),y xi f(y)f(x)hf(x),y xi+L2kx yk2for all x,y Rn;Assumptions 1 and 3 ensure that f has a finite minimum fon Rn.For more properties of convex functio

41、ns,see any textbook on convex analysis.Proof Question(20 points)Under Assumptions 13,prove for AFBT thatlimkf(xk)=f.7R5 Proof.Assume that f(xk)9 f.Then infk0f(xk)f 0 since f(xk)is non-increasing,Denote gk=f(xk).Then infk0kgkk 0(Pick an xwith f(x)=f.Then f(xk)f hgk,xkxi by the convexity of f;meanwhil

42、e,xk is bounded dueto the monotonicity of f(xk)and the level-boundedness of f.Thus infk0kgkk 0).In other words,there exists an 0 such that kgkk for all k 0.Given any k 0,we can pick an ik 1,.,n satisfying|hgk,eiki|kgkk (1)with =1/n.Hencef(yk)minf(xktkeik)f(xk)tk|hgk,eiki|+Lt2k2 f(xk)tk+Lt2k2.(2)Ther

43、efore,whenevertk2L+2,(3)we will have f(yk)f(xk)t2k,which renders sk=1 and hence tk+1=2tk.It is theneasy to see that tk has a positive lower bound,namelytk t min?t0,L+2?0for eachk 0.(4)Thus sk=1 for infinitely many k(otherwise,tk 0),and f(xk)f(xk+1)t2for eachof such k,contradicting the lower-boundedn

44、ess of f.The proof is complete.86.Let n be a positive integer.For any positive integer k,write 0k=diag0,.,0|z k forthe k k zero matrix.LetY=0nAAt0n+1!be a(2n+1)(2n+1)matrix,where A=(xi,j)1in,1jn+1is an n(n+1)real matrixand Atdenotes the transpose of A,that is,the(n+1)n matrix whose(j,i)-entry isxi,j

45、.(i)Proof Question(10 points)A complex number is called an eigenvalue of ak k matrix X if Xv=v for some nonzero column vector v=(x1,.,xk)t.Show that:0 is an eigenvalue of Y and every other eigenvalue of Y is of the form where is a non-negative real eigenvalue of AAt.(ii)Proof Question(15 points)Let

46、n=3 and a1,a2,a3,a4be four distinct positivereal numbers.Puta=sX1i4a2iandxi,j=aii,j+aj4,j1a2(a2i+a24)aj(1 i 3,1 j 4),where i,j=(1 if i=j0 if i 6=j.Show that:Y has 7 distincteigenvalues.9R6 Proof.(i)Write In=diag1,.,1|z n for the nn identity matrix.By an elementaryreduction the characteristic polynom

47、ialdet(I2n+1 Y)=det(2In AAt).Then,0 is an eigenvalue of Y and every other eigenvalue of Y is of the form where is a non-negative real eigenvalue of AAt.(ii)Put u=(a4,a4,a4)and v=(a21+a24a,a22+a24a,a23+a24a).By calculation we haveAAt=diaga21,.,a23+utu vtv.Let f(s)=det(sInAAt)be the characteristic pol

48、ynomial of AAt.Then,by calculationone shows thatf(a2i)=a2ia2Y1j4,j6=i(a2i a2j)for each i 1,2,3,4.Let a01,a02,a03,a04be the descending re-ordering of a1,a2,a3,a4.Then,we get:AAthas three distinct eigenvalues b21,b22,b33where b1,b2,b3are positivereal numbers such thata01 b1 a02 b2 a03 b3 a04.Then,by(i

49、)Y has 7 distinct real eigenvalues.107.For a continuous and absolutely integrable complex-valued function f(x)on R,definea function(Sf)(x)on R by(Sf)(x)=Z+e2iuxf(u)du.(i)Question and Answer(10 points)Find explicit forms of S(11+x2)and S(1(1+x2)2).(ii)Question and Answer(15 points)For any integer k,w

50、rite fk(x)=(1+x2)1k.When k 1,find constants c1,c2such that the function y=(Sfk)(x)solves asecond order differential equationxy00+c1y0+c2xy=0.11R7 Answer.Write V for the space of complex-valued,continuous and absolutelyintegrable functions on R.Lemma 0.1.(i)If f(x)V,f0(x)V and limxf(x)=0,then(Sf0)(x)