Problems and solutions - MIT Mathematics

1 CHAPTER 5 Problems and solutions1. Problems Chapter from first principles that ifVis a vector space (overRorC) then for any setXthe space( )F(X;V) ={u:X V}is a linear space over the same field, with pointwise operations . a vector space andS Vis a subset which is closedunder addition and scalar multiplication:( )v1,v2 S, K= v1+v2 Sand v1 SthenSis a vector space as well (called of course a subspace). Vbe a linear subspace of a vector space show that therelation onV( )v1 v2 v1 v2 Sis an equivalence relation and that the set of equivalence classes, denoted usuallyV/S,is a vector space in a natural case you do not know it, go through the basic theory offinite-dimensional vector spaces. Define a vector spaceVto be finite-dimensionalif there is an integerNsuch that anyNelements ofVare linearly dependent ifvi Vfori= 1.

2 N,then there existai K,notallzero, such that( )N i=1aivi= 0 the smallest such integer the dimension ofVand show that a finite dimensionalvector space always has a basis,ei V, i= 1,..,dimVsuch that any element ofVcan be written uniquely as a linear combination( )v=dimV i=1biei, bi the notion of a linear map between vector spaces (dis-cussed above) and show that between two finite dimensional vector spacesVandWover the same field(1) If dimV dimWthen there is an injective linear mapL:V W.(2) If dimV Wthen there is a surjective linear mapL:V W.(3) if dimV= dimWthen there is a linear isomorphismL:V W, aninjective and surjective linear Problems AND that any two norms on a finite dimensional vector spaceare that if two norms on a vector space are equivalent thenthe topologies induced are the same the sets open with respect to the distancefrom one are open with respect to the distance coming from the other.

3 The converseis also true, you can use another result from this section to prove out a proof (you can steal it from one of many places butat least write it out in your own hand) either forp= 2 or for eachpwith 1 p < thatlp={a:N C; j=1|aj|p< , aj=a(j)}is a normed space with the norm a p= j=1|aj|p means writing out the proof that this is a linear space and that the threeconditions required of a norm directly that eachlpas defined in problem is complete, it is a Banach spacel consists of the bounded sequences( )l ={a:N C; supn|an|< }, a = supn|an|.Show that it is a Banach closely related space consists of the sequences con-verging to 0 :( )c0={a:N C; limn an= 0}, a = supn|an|.Check that this is a Banach space and that it is a closed subspace ofl (perhapsin the opposite order).

4 The unit sphere is the set of vectors oflength 1 :S={a lp; a p= 1}.(1) Show thatSis closed.(2) Recall the sequential (so not the open covering definition) characterizationof compactness of a set in a metric space ( by checking in Rudin sbook).(3) Show thatSis not compact by considering the sequence inlpwithkthelement the sequence which is all zeros except for a 1 in thekth slot. Notethat the main problem is not to get yourself confused about sequences ofsequences! that the norm on any normed space is the proof of the completeness of the spaceBconstructedin the second proof of Theorem solutions TO PROBLEMS1392. Hints for some problemsHint1 ( problem ).You need to show that each Cauchy sequence problem here is to find the limit of a given Cauchy sequence.

5 Show that foreachNthe sequence inCNobtained by truncating each of the elements at pointNis Cauchy with respect to the norm in problem that this isthe same as being Cauchy inCNin the usual sense (if you are doingp= 2 it isalready the usual sense) and hence, this cut-off sequence converges. Use this to finda putative limit of the Cauchy sequence and then check that it solutions to ( ).IfVis a vector space (overKwhich isRorC) then forany setXconsider( )F(X;V) ={u:X V}.Addition and scalar multiplication are defined pointwise :( )(u+v)(x) =u(x) +v(x),(cu)(x) =cu(x), u,v F(X;V), c are well-defined functions since addition and multiplication are defined , one needs to check all the axioms of a vector space. Since an equalityof functions is just equality at all points, these all follow from the correspondingidentities ( ).

6 IfS Vis a (non-empty) subset of a vector space andS Vwhich is closed under addition and scalar multiplication:( )v1, v2 S, K= v1+v2 Sand v1 Sthen 0 S,since 0 Kand for anyv S,0v= 0 , ifv Sthen v= ( 1)v all the axioms of a vector space follow from thecorresponding identities Vbe a linear subspace of a vector space consider therelation onV( )v1 v2 v1 v2 say that this is an equivalence relation means that symmetry and transitivityhold. SinceSis a subspace,v Simplies v Ssov1 v2= v1 v2 S= v2 v1 S= v2 , since it is also possible to add and remain inSv1 v2, v2 v3= v1 v2, v2 v3 S= v1 v3 S= v1 this is an equivalence relation and the quotientV/ =V/Sis well-defined where the latter is notation. That is, and element ofV/Sis an equivalence class ofelements ofVwhich can be writtenv+S:( )v+S=w+S v w , we can check the axioms of a linear space once we define addition and scalarmultiplication.

7 Notice that(v+S) + (w+S) = (v+w) +S, (v+S) = v+S1405. Problems AND SOLUTIONSare well-defined elements, independent of the choice of representatives, since addingan lement ofStovorwdoes not change the right , to the axioms. These amount to showing thatSis a zero element foraddition, v+Sis the additive inverse ofv+Sand that the other axioms followdirectly from the fact that the hold as identities ( ).In case you do not know it, go through the basic theory offinite-dimensional vector spaces. Define a vector spaceVto befinite-dimensionalif there is an integerNsuch that anyN+ 1 elements ofVare linearly dependentin the sense that the satisfy a non-trivial dependence relation ifvi Vfori= 1,..N+ 1,then there existai K,notallzero, such that( )N+1 i=1aivi= 0 the smallest such integer the dimension ofV it is also the largest integer suchthat there areNlinearly independent vectors and show that a finite dimensionalvector space always has a basis,ei V, i= 1.

8 ,dimVwhich arenotlinearlydependent and such that any element ofVcan be written as a linear combination( )v=dimV i=1biei, bi ( ).Show that any two norms on a finite dimensional vectorspace are ( ).Show that if two norms on a vector space are equivalentthen the topologies induced are the same the sets open with respect to the distancefrom one are open with respect to the distance coming from the other. The converseis also true, you can use another result from this section to prove ( ).Write out a proof (you can steal it from one of manyplaces but at least write it out in your own hand) either forp= 2 or for eachpwith 1 p < thatlp={a:N C; j=1|aj|p< , aj=a(j)}is a normed space with the norm a p= j=1|aj|p means writing out the proof that this is a linear space and that the threeconditions required of a norm ().

9 The tricky part in problem is the triangle you knew meaning I tell you that for eachN N j=1|aj|p 1pis a norm onCNwould that help?4. Problems CHAPTER ( ).Prove directly that eachlpas defined in problem iscomplete, it is a Banach space. At the risk of offending some, let me say thatthis means showing that each Cauchy sequence converges. The problem here is tofind the limit of a given Cauchy sequence. Show that for eachNthe sequence inCNobtained by truncating each of the elements at pointNis Cauchy with respectto the norm in problem that this is the same as being CauchyinCNin the usual sense (if you are doingp= 2 it is already the usual sense)and hence, this cut-off sequence converges. Use this to find a putative limit of theCauchy sequence and then check that it ( ).

10 The spacel consists of the bounded sequences( )l ={a:N C; supn|an|< }, a = supn|an|.Show that it is a Banach ( ).Another closely related space consists of the sequencesconverging to 0 :( )c0={a:N C; limn an= 0}, a = supn|an|.Check that this is a Banach space and that it is a closed subspace ofl (perhapsin the opposite order). ( ).Consider the unit sphere is the set of vectorsof length 1 :S={a lp; a p= 1}.(1) Show thatSis closed.(2) Recall the sequential (so not the open covering definition) characterizationof compactness of a set in a metric space (e .g . by checking in Rudin).(3) Show thatSis not compact by considering the sequence inlpwithkthelement the sequence which is all zeros except for a 1 in thekth slot. Notethat the main problem is not to get yourself confused about sequences ofsequences!